meromorphic functions in the whole p-adic field $K$ or “unbounded”
meromorphic function inside an “open” disk of $K$. We also consider
the problem for rational functions on the field $\mathbb{E}$.

On the other hand, in each main claim, instead of assuming that
the equality $h(f(x)) = h(g(x))$ holds in the whole set of definition,
thanks to properties of analytic sets, we check that it is sufficient to
have the equality on a bounded sequence having no cluster point at
the poles of $f$ and $g$. For that, we use properties of analytic elements
on infraconnected subsets of $K$.

**Notation.** In the following, we denote by $h$ a meromorphic function either in $K$ or inside a disk $d(a, R^{-})$ and we denote by $c_1, \dots, c_n, \dots$ the finite or infinite sequence of zeros of $h'$.

**Lemma 55.1.** Let $h \in \mathcal{M}(K)\setminus K$ and let $f, g \in \mathcal{M}(K)$ (respectively $f, g \in \mathcal{M}(d(0, r^{-}))$) satisfy $h \circ f(x) = h \circ g(x) \quad \forall x \in K\setminus S(f, h) \cup S(g, h)$ (respectively $h \circ f(x) = h \circ g(x) \quad \forall x \in d(0, R^{-})\setminus S(f, h) \cup S(g, h)$). Moreover, if $f, g$ do not belong to $\mathcal{A}(K)$ (respectively if $f, g$ do not belong to $\mathcal{A}(d(0, r^{-}))$, we assume that $h \notin K(x)\setminus K[x]$). Then $f, g$ satisfy $\mathcal{P}(f) = \mathcal{P}(g), \mathcal{S}(f, h) = \mathcal{S}(g, h)$.

For each $j = 1, \dots, k$, let $q_j = \omega_{c_j}(h')$. We assume that $h(c_j) \neq h(c_n) \quad \forall j = 1, \dots, k, \forall n \neq j$. Then $f, g$ satisfy

$$
\begin{align*}
\overline{N}(r, f) + \sum_{j=1}^{k} \overline{Z}(r, f - c_j) &\le \overline{Z}\left(r, \frac{1}{f} - \frac{1}{g}\right) + \sum_{j=1}^{k} \frac{1}{q_j} Z(r, g' \mid f(x) = c_j, \\
&\qquad g(x) \notin C(h')).
\end{align*}
$$

Furthermore, if $f, g \in \mathcal{A}(K)$ (respectively if $f, g \in \mathcal{A}(d(0, r^{-}))$), then

$$
\begin{align*}
\sum_{j=1}^{k} \overline{Z}(r, f - c_j) &\le \overline{Z}(r, f - g) + \sum_{j=1}^{k} \frac{1}{q_j} Z(r, g' \mid f(x) = c_j, \\
&\qquad g(x) \notin C(h')).
\end{align*}
$$

**Proof.** First, we show that $\mathcal{P}(f) = \mathcal{P}(g)$. Suppose $\alpha$ is a pole of $f$. If $h \in \mathcal{A}(K)$ (respectively if $h \in \mathcal{A}(d(0, R^{-})))$, then by Lemma 54.9, $\alpha$ is a singular point for $h \circ f$, hence for $h \circ g$, hence $\alpha$ is a singular point for $g$, hence it is a pole for $g$. Now suppose that $h \notin \mathcal{A}(K)$, hence $h \in \mathcal{M}(K)\setminus K(x)$ (respectively suppose $h \notin \mathcal{A}(d(0, R^{-})))$, hence
$h \in \mathcal{M}(d(0, R^-))\mathbb{K}(x)$). Since $h$ does not lie in $\mathbb{K}(x)$, by Lemma 54.9, $\alpha$ is a point of high singularity for $h \circ f$, hence for $h \circ g$. But if $\alpha$ is not a pole for $g$, it is a regular point for $g$, hence it is either a regular point or a pole for $h \circ g$, a contradiction. Consequently, $\alpha$ is a pole for $g$ and therefore, since $f$ and $g$ play the same role, we have $\mathcal{P}(f) = \mathcal{P}(g)$.

Now, suppose $\alpha \in \mathcal{T}(f, h)$, hence $h \circ f$ has a pole at $\alpha$ and so does $h \circ g$, thereby $\alpha \in \mathcal{T}(g, h)$. Consequently, $\mathcal{T}(f, h) \subset \mathcal{T}(g, h)$, hence $\mathcal{T}(f, h) = \mathcal{T}(g, h)$, and therefore, $S(f, h) = S(g, h)$.

We now assume that $h(c_j) \neq h(c_n) \forall j = 1, \dots, k, \forall n \neq j$. Without loss of generality, we may assume that $0 \notin \mathcal{C}(h')$. Indeed, if $0 \in \mathcal{C}(h')$, we can find $b \in \mathbb{K}$ such that $c_n + b \neq 0 \forall n \in \mathbb{N}^*$. Setting $\bar{f}(x) = f(x) - b$, $\bar{g}(x) = g(x) - b$ and $\bar{h}(z) = h(z + b)$, we have $\bar{h}(\bar{f}(x)) = \bar{h}(\bar{g}(x) \forall x \in \mathbb{K}\setminus S(\bar{f}, \bar{h})$ (respectively $\bar{h}(\bar{f}(x)) = \bar{h}(\bar{g}(x)) \forall x \in d(0, R^-)\setminus S(\bar{f}, \bar{h})$), thereby we may process in the same way with $\bar{f}, \bar{g}, \bar{h}$. Moreover, we note that if $f, g$ lie in $\mathcal{A}(\mathbb{K})$ (respectively in $\mathcal{A}(d(0, r^-))$), then so do $\bar{f}, \bar{g}$. Consequently, in order to simplify a deduction, we assume that $c_n \neq 0 \forall n \in \mathbb{N}^*$.

Let $\phi = \frac{1}{f} - \frac{1}{g}$. Since $\mathcal{P}(f) = \mathcal{P}(g)$, for each pole $\alpha$ of $f$, we have $\phi(\alpha) = 0$, therefore

$$ (1) \quad \bar{N}(r, f) \le \bar{Z}(r, \phi | x \in \mathcal{P}(f)). $$

Let us fix $j \in \{1, \dots, k\}$ and let $\alpha \in \mathbb{K}$ (respectively $\alpha \in d(0, R^-)$) satisfy $f(\alpha) = c_j$. Suppose first that $g(\alpha)$ lies in $\mathcal{C}(h')$. Thanks to the hypothesis $h(c_n) \neq h(c_j) \forall n \neq j$, if $g(\alpha) \neq c_j$, then $h(g(\alpha)) \neq h(c_j)$, a contradiction to $h(g(\alpha)) = h(f(\alpha))$. So we have $g(\alpha) = f(\alpha) = c_j$ and since $c_j \neq 0$, then $\phi(\alpha) = 0$. Consequently,

$$ (2) \quad \bar{Z}(r, f - c_j | g(x) \in \mathcal{C}(h')) \le \bar{Z}(r, \phi | f(x) = c_j), $$

and similarly if $f, g \in \mathcal{A}(\mathbb{K})$ or if $f, g \in \mathcal{A}_u(d(0, R^-))$,

$$ (3) \quad \bar{Z}(r, f - c_j | g(x) \in \mathcal{C}(h')) \le \bar{Z}(r, f - g | f(x) = c_j). $$

Consequently, since $\mathcal{P}(f) \cap \mathcal{C}(h') = \emptyset$, in the general case, by (1) and (2), we can derive

$$ (4) \quad \bar{N}(r, f) + \sum_{j=1}^{k} \bar{Z}(r, f - c_j | g(x) \in \mathcal{C}(h')) \le \bar{Z}(r, \phi). $$
Similarly, if $f, g \in \mathcal{A}(\mathbb{K})$ (respectively if $f, g \in \mathcal{A}(d(0, r^{-}))$), then
by (3), we have

$$ (5) \quad \sum_{j=1}^{k} \overline{Z}(r, f - c_j \mid g(x) \in \mathcal{C}(h')) \le \overline{Z}(r, f - g). $$

In order to complete the proof, we will show

$$ (6) \quad \overline{Z}(r, f - c_j \mid g(x) \notin \mathcal{C}(h')) \le \frac{1}{q_j} Z(r, g' \mid f(x) = c_j, g(x) \notin \mathcal{C}(h')). $$

Indeed, consider $\alpha$ such that $g(\alpha) \notin \mathcal{C}(h')$. Since $h'(f(\alpha)) = h'(c_j) = 0$, we note that $f'(\alpha)h'(f(\alpha)) = g'(\alpha)h'(g(\alpha)) = 0$. But since $g(\alpha) \notin \mathcal{C}(h')$, we have $h'(g(\alpha)) \ne 0$, hence $g'(\alpha) = 0$. Consequently, we obtain

$$ (7) \quad \overline{Z}(r, f - c_j \mid g(x) \notin \mathcal{C}(h')) \le \overline{Z}(r, g' \mid f(x) = c_j, g(x) \notin \mathcal{C}(h')). $$

On the other hand, since $f(\alpha) = c_j$, we see that $\omega_{\alpha}(f'(x)h'(f(x))) \ge q_j$, hence

$$ (8) \quad \omega_{\alpha}(g'(x)h'(g(x))) \ge q_j. $$

But since $g(\alpha) \notin \mathcal{C}(h')$, we have $h'(g(\alpha)) \ne 0$, hence by (8), $\omega_{\alpha}(g') \ge q_j$, and consequently,

$$ (9) \quad \overline{Z}(r, g' \mid f(x) = c_j, g(x) \notin \mathcal{C}(h')) \le \frac{1}{q_j} Z(r, g' \mid f(x) = c_j, g(x) \notin \mathcal{C}(h')) $$

Thus, by (7) and (9), we obtain (6) which by (4) proves

$$ \begin{aligned} \overline{N}(r, f) + \sum_{j=1}^{k} \overline{Z}(r, f - c_j) &\le \overline{Z}(r, \frac{1}{f} - \frac{1}{g}) + \sum_{j=1}^{k} \frac{1}{q_j} Z(r, g' \mid f(x) = c_j, g(x) \notin \mathcal{C}(h')). \\ &\qquad g(x) \notin \mathcal{C}(h'). \end{aligned} $$

Similarly, by (5) and (6), we have

$$ \begin{aligned} \sum_{j=1}^{k} \overline{Z}(r, f - c_j) &\le \overline{Z}(r, f - g) + \sum_{j=1}^{k} \frac{1}{q_j} Z(r, g' \mid f(x) = c_j, g(x) \notin \mathcal{C}(h')). \\ &\qquad g(x) \notin \mathcal{C}(h'). \end{aligned} \quad \square $$
**Theorem 55.2.** Let $h \in \mathcal{M}(\mathbb{K})\backslash\mathbb{K}$, let $f, g \in \mathcal{A}(\mathbb{K})\backslash\mathbb{K}$, and let $(a_m)_{m \in \mathbb{N}}$ be a bounded sequence of $\mathbb{K}\backslash(\mathcal{T}(f,h) \cup \mathcal{T}(g,h))$ satisfying $h \circ f(a_m) = h \circ g(a_m) \quad \forall m \in \mathbb{N}$. Then, $\mathcal{T}(f,h) = \mathcal{T}(g,h)$ and $h \circ f(x) = h \circ g(x) \quad \forall x \in \mathbb{K}\backslash\mathcal{T}(f,h)$. Moreover, if $\Upsilon(h) \ge 2$, then $f = g$.

**Remark.** Throughout Corollaries 55.3, 55.4, and 55.11, we apply to polynomials (respectively rational functions) with coefficients in $\mathbb{E}$ results proven for analytic (respectively meromorphic) functions in $\mathbb{K}$. Indeed, as it was often previously done, since $\mathbb{E}$ has characteristic 0, there exists a finite extension $\mathbb{M}$ of $\mathbb{Q}$ containing all coefficients, zeros, and poles of all functions involved and consequently we can consider $\mathbb{M}$ as a subfield of $\mathbb{C}_p$.

**Corollary 55.3.** Let $h \in \mathcal{M}(\mathbb{K})$ (respectively $h \in \mathbb{E}(x)$) satisfy $\Upsilon(h) \ge 2$. Then $h$ is a function of uniqueness for $\mathcal{A}(\mathbb{K})$ (respectively for $\mathbb{E}[x]$).

**Notation.** Given a meromorphic function $f \in \mathcal{M}(\mathbb{K})$ or $f \in \mathcal{M}(d(0, R^{-}))$, we denote by $(\mathcal{F})$ the hypothesis: the restriction of $f$ to the set of zeros of $f'$ is injective [60], [3], [49], [50].

Thus, particularly, if a function $f$ satisfies Hypothesis (F) and if $f'$ has $q$ zeros, then $\Upsilon(f) = q$.

**Examples.** (1) Let $h(x) = \frac{x(x-1)}{x-2}$. Hence, $h'(x) = \frac{x^2-4x+2}{(x-2)^2}$. Let $\sqrt{2}$ denote a square root of 2 in the field $\mathbb{K}$. The zeros of $h'$ are $c_1 = 2 + \sqrt{2}$, $c_2 = 2 - \sqrt{2}$. Thus, $h(c_1) = 3 - 2\sqrt{2}$, $h(c_2) = 3 + \sqrt{2}$, hence $h$ satisfies the hypothesis of Corollary 55.3.

(2) Let $b \in \mathbb{E}^*$ be a zero of the polynomial $Q(x) = \frac{x^5}{20} - \frac{x^3}{6} + \frac{x}{4} + \frac{2}{15}$ and let $h(x) = \frac{x^5}{5} - \frac{bx^4}{4} - \frac{x^3}{3} + \frac{bx^2}{2}$. Then, $h'(x) = x(x-1)(x+1)(x-b)$. Now, we note that $h(0) = 0$, $h(1) = \frac{2}{15} + \frac{b}{4}$, $h(-1) = -\frac{2}{15} + \frac{b}{4}$, $h(b) = -\frac{b^5}{20} + \frac{b^3}{6}$. Since $Q(b) = 0$, we have $h(1) = h(b)$ and clearly $h(1) \ne h(0)$, $h(-1) \ne h(1)$, $h(0) \ne h(b)$, $h(-1) \ne h(b)$. Consequently, $h'$ has four zeros $c_1 = 0$, $c_2 = -1$, $c_3 = 1$, $c_4 = b$ satisfying $h(c_j) \ne h(c_l) \quad \forall j = 1, 2, l \ne j$, $1 \le l \le 4$. Therefore, $h$ satisfies the hypothesis of Corollary 55.3. However, $h$ does not satisfy Hypothesis (F) because $h(c_3) = h(c_4)$.

**Corollary 55.4.** Let $P(x) \in \mathbb{K}[x]$ be of degree $n \ge 3$ and assume that $P'$ has exactly two distinct zeros. Then $P$ is a polynomial of uniqueness for $\mathcal{A}(\mathbb{K})$.
**Proof.** Let $a_1, a_2$ be the two zeros of $P'$, of order $s_1, s_2$ respectively. Suppose $\Upsilon(P) < 2$. Then, $P(a_1) = P(a_2)$. Let $b = P(a_1)$. Then, $P - b$ admits each $a_j$ as a zero of order $s_j + 1$, $j = 1, 2$. Consequently, $\deg(P) = \deg(P - b) \ge s_1 + s_2 + 2$, and hence, $\deg(P') = s_1 + s_2 + 1$, a contradiction since $a_1, a_2$ are the only zeros of $P'$. Consequently, $\Upsilon(P) = 2$. $\square$

**Corollary 55.5.** Let $P(x) \in \mathbb{E}[x]$ be a of degree 3. Then $P$ is a polynomial of uniqueness for $\mathcal{A}(\mathbb{K})$ if and only if it is not of the form $c(x-a)^3 + b, a, b, c \in \mathbb{K}$.

**Proof.** Indeed, $P'$ has one or two distinct zeros. On one hand, if $P$ is of the form $c(x-a)^3 + b, a, b, c \in \mathbb{K}$, obviously it is not a polynomial of uniqueness for $\mathcal{A}(\mathbb{K})$. On the other hand, if $P$ is not of the form $c(x-a)^3 + b, a, b, c \in \mathbb{K}$, then $P'$ has two distinct zeros, hence by Corollary 55.4, $P$ is a polynomial of uniqueness for $\mathcal{A}(\mathbb{K})$. $\square$

**Corollary 55.6.** Let $h, f, g \in \mathcal{A}(\mathbb{C})$ have all coefficients in $\mathbb{Q}$ and also lie in $\mathcal{A}(\mathbb{C}_p)$ for some prime $p$. Let $(a_m)_{m \in \mathbb{N}}$ be a bounded sequence of $\mathbb{C}$ satisfying $h \circ f(a_m) = h \circ g(a_m) \forall m \in \mathbb{N}$. Then, $h \circ f(x) = h \circ g(x) \forall x \in \mathbb{C}$. Moreover, if $\Upsilon(h) \ge 2$, then $f = g$.

**Proof.** We know that the identity $h \circ f = h \circ g$ is obvious in $\mathcal{A}(\mathbb{C})$, which means the coefficients of the two functions are the same, hence the identity also holds in $\mathbb{C}_p$. Therefore, if $\Upsilon(h) \ge 2$, by Theorem 55.2, we have $f = g$ in $\mathcal{A}(\mathbb{C}_p)$ i.e. the coefficients of $f, g$ are the same, hence this identity obviously holds in $\mathcal{A}(\mathbb{C})$. $\square$

**Remarks.** Conversely, a polynomial of degree 2 is never a function of uniqueness for any family of functions because through a suitable translation of the variable, it is possible to put it in the form of an even polynomial.

The condition $\Upsilon(h) \ge 2$ is not a necessary condition to assure that $h$ is a function of uniqueness for entire or meromorphic functions: for instance, a linear fractional function has a derivative which has no zero but obviously is a function of uniqueness for meromorphic functions in $\mathbb{K}$ or in $d(a, R^-)$.

The proofs of Theorems 55.7, 55.10, and 55.12 are gathered in the following:

**Theorem 55.7.** Let $h \in \mathcal{M}(\mathbb{K})\backslash\mathbb{K}$, let $f, g \in \mathcal{A}_u(d(a, R^-))$, and let $(a_m)_{m \in \mathbb{N}}$ be a sequence of $d(a, R^-)$ such that $\sup_{n \in \mathbb{N}} |a_n - a| < R$,
satisfying $h \circ f(a_m) = h \circ g(a_m) \quad \forall m \in \mathbb{N}$. Then, $S(f, h) = S(g, h)$ and $h \circ f(x) = h \circ g(x) \quad \forall x \in d(a, R^-) \setminus S(f, h)$. Moreover, if $\Upsilon(h) \ge 3$, then $f = g$.

**Example.** Let $h(x) = \frac{x^3-x^2+x-2}{x-2}$. Hence, $h'(x) = \frac{2x^3-7x^2+4x}{(x-2)^2}$. Let $\sqrt{17}$ denote a square root of 17 in the field $\mathbb{K}$. The zeros of $h'$ are $c_1 = 0$, $c_2 = \frac{7-\sqrt{17}}{4}$, $c_3 = \frac{7+\sqrt{17}}{4}$. Thus, $h(c_1) = 1$, $h(c_2) = \frac{73-17\sqrt{17}}{2}$, $h(c_3) = \frac{73+17\sqrt{17}}{2}$, hence $h$ satisfies the hypothesis of Theorem 55.7.

**Corollary 55.8.** Let $h \in \mathcal{A}(\mathbb{K})$ satisfy $\Upsilon(h) \ge 3$. Then $h$ is a function of uniqueness for $\mathcal{A}_u(d(a, R^-))$.

Similar to Corollary 55.6, we can state Corollary 55.9:

**Corollary 55.9.** Let $h, f, g \in \mathcal{A}(\mathbb{C})$ have all coefficients in $\mathbb{Q}$ and assume that $h$ also lies in $\mathcal{A}(\mathbb{C}_p)$ for some prime $p$ and $f, g$ lie in $\mathcal{A}_u(d(a, R^-))$ (with respect to the field $\mathbb{C}_p$). Let $(a_m)_{m \in \mathbb{N}}$ be a bounded sequence of $\mathbb{C}$ satisfying $h \circ f(a_m) = h \circ g(a_m) \quad \forall m \in \mathbb{N}$. Then, $h \circ f(x) = h \circ g(x) \quad \forall x \in \mathbb{C}$. Moreover, if $\Upsilon(h) \ge 3$, then $f = g$.

**Remark.** In this chapter, we see that if $P$ is of the form $x^n-bx^{n-1}+t$ ($t \in \mathbb{K}$), then it is not a function of uniqueness for $\mathcal{M}(\mathbb{K})$ and an immediate generalization shows that the same holds when $P'$ has exactly two distinct zeros, one of them being of order 1.

Assuming again that the set of zeros $S$ of a polynomial $P$ is affinely rigid and $P$ satisfies Hypothesis (F), it is shown in [3] (Theorem 1) that if $P'$ has exactly two distinct zeros $c_j$ of order $m_j$ ($j = 1, 2$), then it is a function of uniqueness for $\mathcal{M}(\mathbb{K})$ if and only if $\min(m_1, m_2) \ge 2$.

On the other hand, Theorem 47.6 shows that a polynomial such that $\Upsilon(P) = 2$ may be a polynomial of uniqueness for both $\mathcal{M}(\mathbb{K})$ and $\mathcal{M}(d(0, R^-))$.

**Theorem 55.10.** Let $h \in \mathcal{M}(\mathbb{K}) \setminus (\mathbb{K} \cup (\mathbb{K}(x) \setminus \mathbb{K}[x]))$, let $f, g \in \mathcal{M}(\mathbb{K}) \setminus \mathbb{K}$, and let $(a_m)_{m \in \mathbb{N}}$ be a bounded sequence of $\mathbb{K} \setminus (S(f, h) \cup S(g, h))$ of diameter $< R$ admitting no cluster point in $P(f) \cup P(g)$, satisfying further $h \circ f(a_m) = h \circ g(a_m) \quad \forall m \in \mathbb{N}$. Then, $S(f, h) = S(g, h)$ and $h \circ f(x) = h \circ g(x) \quad \forall x \in \mathbb{K} \setminus S(f, h)$. Moreover, if $\Upsilon(h) \ge 3$, then $f = g$.
**Corollary 55.11.** Let $h \in \mathcal{A}(\mathbb{K})$ (respectively $h \in \mathbb{E}[x]$) satisfy $\Upsilon(h) \ge 3$. Then $h$ is a function of uniqueness for $\mathcal{M}(\mathbb{K})$ (respectively for $\mathbb{E}(x)$).

**Remark.** In [3], it is shown that a polynomial $P \in \mathbb{E}[x]$ satisfying (F), whose set of zeros is affinely rigid, is a function of uniqueness for $\mathcal{M}(\mathbb{K})$ if and only if either $P'$ has at least three distinct zeros, or $P'$ has just two distinct zeros, both of order $\ge 2$. By Theorems 55.10, we can find other polynomials of uniqueness for $\mathcal{M}(\mathbb{K})$ having a set of zeros which is not affinely rigid.

**Example.** Suppose $p \ne 3$. Let $P(x) = x^4 - 4x$, let $j$ be a cubic root of 1 different from 1, and let $a \in \mathbb{K}$ be a cubic root of 4. Then $P$ has four distinct zeros $\{0, a, ja, j^2a\}$. Thus, the set of zeros of $P$ is not affinely rigid (but is centered). Next, the set of zeros of $P'$ is $\{1, j, j^2\}$ and we can check that $P$ satisfies (F), hence $\Upsilon(P) = 3$, therefore $P$ is a function of uniqueness for $\mathcal{M}(\mathbb{K})$.

**Remark.** By Theorem 1.10, given a polynomial $P(x) \in \mathbb{E}[x]$ of degree 4, we can check that $P$ satisfies the hypotheses of Theorem 55.10 and Corollaries 55.3 and 55.11, if and only if it is not of the form $[(x - a + l)(x - a - l)]^2 + A$ with $A \in E$, $l \in E^*$ (which means $P$ is not an even function, up to some affine change of variable).

**Example.** $P(x) = (x^2 - a^2)^2 + x$ satisfies the hypotheses of Corollaries 55.8 and 55.11.

Now, let $P$ be a polynomial of degree 5 such that $P'$ admits four distinct zeros $c_1, c_2, c_3, c_4$. If $\Upsilon(P) > 0$, then $\Upsilon(P) \ge 2$. Indeed, suppose $\Upsilon(P) = 1$. We may assume that $P(c_1) = P(c_2) = P(c_3)$ and $P(c_1) \ne P(c_4)$. But then, $P - P(c_1)$ admits three zeros of order 2, a contradiction with $\deg(P) = 5$.

Similarly, if $\Upsilon(P) = 0$, then up to a good indexation we have $P(c_1) = P(c_2)$ and $P(c_3) = P(c_4)$.

**Theorem 55.12.** Let $h \in \mathcal{M}(\mathbb{K})\setminus(\mathbb{K} \cup (\mathbb{K}(x)\setminus\mathbb{K}[x]))$, let $f, g \in \mathcal{M}_u(d(a, R^{-}))$, and let $(a_m)_{m \in \mathbb{N}}$ be a bounded sequence of $d(a, R^{-})\setminus(S(f, h) \cup S(g, h))$ admitting no cluster point in $\mathcal{P}(f) \cup \mathcal{P}(g)$, satisfying further $h \circ f(a_m) = h \circ g(a_m) \quad \forall m \in \mathbb{N}$. Then, $S(f, h) = S(g, h)$ and $h \circ f(x) = h \circ g(x) \quad \forall x \in d(a, R^{-})\setminus S(f, h)$. Moreover, if $\Upsilon(h) \ge 4$, then $f = g$.
**Corollary 55.13.** Let $h \in \mathcal{A}(\mathbb{K})$ satisfy $\Upsilon(h) \ge 4$. Then $h$ is a function of uniqueness for $M_u(d(a, R^-))$.

**Proof.** of Theorems 55.2, 55.7, 55.10, and 55.12. By Proposition 54.10, we have $h \circ f(x) = h \circ g(x) \forall x \in \mathbb{K} \setminus (S(f, h) \cup S(g, h))$ in Theorems 55.2 and 55.7 and $h \circ f(x) = h \circ g(x) \forall x \in d(a, R^-) \setminus (S(f, h) \cup S(g, h))$ in Theorems 55.10 and 55.12. Suppose that $f$ and $g$ are not identical.

Then by Lemma 55.1, we have $\mathcal{P}(f) = \mathcal{P}(g)$ and $S(f, h) = S(g, h)$. Without loss of generality, we can obviously assume that $a = 0$ in Theorems 55.10 and 55.12. Since $c_1, \dots, c_k$ lie in $\mathcal{C}(h')$, clearly by applying Theorem 45.7, we obtain in Theorems 55.2 and 55.10

$$ (k-1)T(r, f) \le \sum_{j=1}^{k} \overline{Z}(r, f - c_j) + \overline{N}(r, f) - Z(r, f' | f(x) \notin \mathcal{C}(h')) - \log r + O(1) \quad (r > 0), $$

$$ (k-1)T(r,g) \le \sum_{j=1}^{k} \overline{Z}(r,g - c_j) + \overline{N}(r,g) - Z(r,g' | g(x) \notin \mathcal{C}(h')) - \log r + O(1) \quad (r > 0). $$

In Theorems 55.7 and 55.12, we have

$$ (1\text{bis}) \qquad (k-1)T(r,f) \le \sum_{j=1}^{k} \overline{Z}(r,f - c_j) + \overline{N}(r,f) - Z(r,f' | f(x) \notin \mathcal{C}(h')) + O(1) \quad (r \in]0, R[), $$

$$ (2\text{bis}) \qquad (k-1)T(r,g) \le \sum_{j=1}^{k} \overline{Z}(r,g - c_j) + \overline{N}(r,g) - Z(r,g' | g(x) \notin \mathcal{C}(h')) + O(1) \quad (r \in]0, R[). $$
Now, let $\phi = \frac{1}{f} - \frac{1}{g}$, and for each $j = 1, \dots, k$, let $q_j = \omega_{c_j}(h')$.
By Lemma 55.1, in Theorems 55.2 and 55.10, we obtain

$$
(3) \quad (k-1)T(r, f) \le \bar{Z}(r, \phi) + \sum_{j=1}^{k} \frac{1}{q_j} Z(r, g' \mid f(x) = c_j, \\
g(x) \notin \mathcal{C}(h')) - Z(r, f' \mid f(x) \notin \mathcal{C}(h')) - \log r + O(1),
$$

and similarly,

$$
(k-1)T(r,g) \le \bar{Z}(r, \phi) + \sum_{j=1}^{k} \frac{1}{q_j} Z(r, f' \mid g(x) = c_j, \\
f(x) \notin \mathcal{C}(h')) - Z(r, g' \mid f(x) \notin \mathcal{C}(h')) - \log r + O(1).
$$

In Theorems 55.7 and 55.12, we obtain

$$
(3\text{bis}) \quad (k-1)T(r,f) \le \bar{Z}(r,\phi) + \sum_{j=1}^{k} \frac{1}{q_j} Z(r,g' \mid f(x) = c_j, \\
g(x) \notin \mathcal{C}(h')) - Z(r,f' \mid f(x) \notin \mathcal{C}(h')) + O(1),
$$

$$
(4\text{bis}) \quad (k-1)T(r,g) \le \bar{Z}(r,\phi) + \sum_{j=1}^{k} \frac{1}{q_j} Z(r,f' \mid g(x) = c_j, \\
f(x) \notin \mathcal{C}(h')) - Z(r,g' \mid g(x) \notin \mathcal{C}(h')) + O(1).
$$

By adding in each case the two inequalities we have respectively
obtained, in Theorems 55.2 and 55.10, by (3) and (4), we obtain

$$
(5) \quad (k-1)(T(r,f) + T(r,g)) \le 2\bar{Z}(r,\phi) + \sum_{j=1}^{k} \frac{1}{q_j} [Z(r,f' \mid g(x) = c_j, \\
f(x) \notin \mathcal{C}(h')) + Z(r,g' \mid f(x) = c_j, g(x) \notin \mathcal{C}(h'))] \\
\qquad - Z(r,f' \mid f(x) \notin \mathcal{C}(h')) \\
\qquad - Z(r,g' \mid g(x) \notin \mathcal{C}(h')) \\
\qquad - 2\log r + O(1),
$$
and in Theorems 55.7 and 55.12, by (3bis) and (4bis), we have

$$
\begin{align*}
(5\text{bis}) \qquad &(k-1)(T(r,f) + T(r,g)) \le 2\bar{Z}(r,\phi) + \sum_{j=1}^{k} \frac{1}{q_j} [Z(r,f' \mid g(x) = c_j, f(x) \notin C(h')) \\
&\phantom{(5\text{bis}) \qquad (k-1)(T(r,f) + T(r,g)) \le 2\bar{Z}(r,\phi) + \sum_{j=1}^{k} \frac{1}{q_j} [} -Z(r,f' \mid f(x) \notin C(h')) \\
&\phantom{(5\text{bis}) \qquad (k-1)(T(r,f) + T(r,g)) \le 2\bar{Z}(r,\phi) + \sum_{j=1}^{k} \frac{1}{q_j} [} -Z(r,g' \mid g(x) \notin C(h')) + O(1).
\end{align*}
$$

Now, in each inequality, we note that in the left-side member we
have the term $\sum_{j=1}^k \frac{1}{q_j} [Z(r, f' \mid g(x) = c_j, f(x) \notin C(h'))] - Z(r, f' \mid f(x) \notin C(h'))$ which is clearly inferior or equal to zero and
similarly $\sum_{j=1}^k \frac{1}{q_j} [Z(r, g' \mid f(x) = c_j, g(x) \notin C(h')) - Z(r, g' \mid g(x) \notin C(h'))] \le 0$. Consequently, in Theorems 55.2 and 55.10, we obtain

$$
(6) \quad (k-1)(T(r,f) + T(r,g)) \le 2\bar{Z}(r,\phi) - 2\log r + O(1), \text{ and}
$$

in Theorems 55.7 and 55.12, we have

$$
(6\text{bis}) \quad (k-1)(T(r,f) + T(r,g)) \le 2\bar{Z}(r,\phi) + O(1).
$$

Now, by Theorem 43.3, in Theorems 55.10 and 55.12, we have
$Z(r, \phi) \le T(r, f) + T(r, g) + O(1)$. Consequently, in Theorem 55.10,
we have $k \le 2$, and in Theorem 55.12, we have $k \le 3$.

Now, assume the hypotheses of Theorems 55.2 and 55.7. By
Lemma 55.1, we can replace $Z(r, \phi)$ by $Z(r, f - g)$. Next, by
Lemma 54.8, $T(r, f) - T(r, g)$ is bounded, and hence, by Theorem
43.8, we derive $T(r, f - g) \le T(r, f) + O(1) = T(r, g) + O(1)$. Con-
sequently, in place of (6), in Theorem 55.2, we can obtain

$$
\begin{align*}
(7) \quad (k-1)(T(r,f) + T(r,g)) &\le 2\bar{Z}(r,f-g) \\
&\phantom{\le} -2\log r + O(1) \le T(r,f) + T(r,g) - 2\log r + O(1),
\end{align*}
$$

and in place of (6bis), in Theorem 55.7, we have

$$
\begin{align*}
(7\text{bis}) \quad (k-1)(T(r,f) + T(r,g)) &\le 2\bar{Z}(r,f-g) \\
&\phantom{{}\le{}} -2\log r + O(1) \le T(r,f) + T(r,g) + O(1).
\end{align*}
$$
Thus, we can conclude that $k \le 1$ in Theorem 55.2 and $k \le 2$ in Theorem 55.7. □

**Example.** Let $\Omega$ be an algebraic closure of $\mathbb{Q}$. Let $R \in \mathbb{Q}(x)$, let $C(R') = \{c_1, \dots, c_q\}$, and assume that $R(c_j) \neq R(c_n) \forall j = 1, \dots, k$, $\forall n \le q$. Let $h(x) = e^{R(x)}$. Then, $C(h') = \{c_1, \dots, c_q\}$ and we check that $h(c_j) \neq h(c_n) \forall j = 1, \dots, k$, $\forall n \le q$.

Indeed, suppose that $h(c_j) = h(c_l)$ with $j \neq l$ and $j \le k$. Then, $R(c_j) - R(c_l)$ is of the form $2im\pi$ with $m \in \mathbb{Z}$, which is impossible because $R(c_j) - R(c_l)$ lies in $\Omega$.

For instance, let $\alpha$ be a zero of the polynomial $D(x) = \frac{x^4}{42} - \frac{x^2}{4} + \frac{2}{3}$ and let $P(x) = x^7 - 7x^5 + \frac{28}{3}x^3 - \frac{7\alpha}{6}x^6 + \frac{35\alpha}{4}x^4 - 14\alpha x^2 + A$ (with $A \in \Omega$).

We check that $\alpha \notin \mathbb{Q}$ (because it is a square root of a zero of $\frac{u^2}{42} - \frac{u}{4} + \frac{2}{3}$). Then, $P'(x) = 7x^6 - 35x^4 + 28x^2 + 7\alpha x^5 + 35\alpha x^3 - 28\alpha x$ admits six distinct zeros: $c_1 = 1$, $c_2 = -1$, $c_3 = 2$, $c_4 = -2$, $c_5 = 0$, $c_6 = \alpha$. We note that $P(c_5) = P(c_6) = A$. Next, for all $j, l$, $(1 \le j < l \le 6)$, $P(c_j) - P(c_l)$ is of the form $s + t\alpha$ with $s, t \in \mathbb{Q}$ and $t \neq 0$, except if $j = 5$ and $l = 6$. Consequently, $P(c_j) \neq P(c_l)$ for all $j, l, 1 \le j < l \le 6$ such that $j < 5$. Therefore, $P$ (playing the role of $h$) satisfies the hypothesis of Theorem 55.12. However, $P$ does not satisfy Hypothesis (F).

**Theorem 55.14.** Let $P \in \mathbb{K}[x]$ (respectively $P \in \mathbb{E}[x]$) be of degree 4. Then $P$ is not a polynomial of uniqueness for $M(\mathbb{K})$ (respectively for $\mathcal{A}(\mathbb{K})$, respectively $\mathbb{E}(x)$, respectively for $\mathbb{E}[x]$) if and only if the set of zeros of $P'$ is of the form $\{a-l, a, a+l\}$.

**Proof.** Let $S$ be the set of zeros of $P'$. Suppose first that $P$ is not a polynomial of uniqueness for $M(\mathbb{K})$ (respectively for $\mathbb{E}(x)$). Then, by Corollary 55.11, $P$ we have $\Upsilon(P) < 3$, hence $P'$ admits two zeros $a, b$ such that $P(a) = P(b)$. Through an affine change of variable, we can assume that $a = 1$, $b = -1$ and we can also assume that $P$ is monic and satisfies $P(-1) = P(1) = 0$. Consequently, $P$ admits 1 and $-1$ as zeros of order 2 and hence $P(x) = (x^2 - 1)^2$. Then the zeros of $P'$ are $-1, 0, 1$. Thus, due to the affine change of variable we made, in general, $S$ is of the form $\{a-l, a, a+l\}$. Conversely, if $S$ is of the form $\{a-l, a, a+l\}$, then through a similar change of
variable we can assume $S = \{-1, 0, 1\}$, so $P(x) = (x^2 - 1)^2 + C$
with $C \in \mathbb{K}$ and hence $P(f) = P(-f)) \quad \forall f \in \mathcal{M}(\mathbb{K})$ (respectively
$\forall f \in E(x)$). $\quad \square$

Lemma 55.15 is useful for the following theorem. It is an immedi-
ate consequence of properties of zeros of power series (see Corollaries
22.3 and 22.34).

**Lemma 55.15.** Let $h(x) = \sum_{n=0}^{\infty} a_n x^n \in \mathcal{A}(\mathbb{K})$ satisfy
$|\frac{a_n}{a_{n+1}}| < |\frac{a_{n+1}}{a_{n+2}}| \quad \forall n \ge t$ and $|a_t| |\frac{a_t}{a_{t+1}}|^t \ge |a_n| |\frac{a_t}{a_{t+1}}|^n \quad \forall n < t$. Then
$h$ admits $t$ zeros in $d(0, |\frac{a_t}{a_{t+1}}|)$ (taking multiplicities into account),
admits a unique zero of order 1 in each circle $C(0, |\frac{a_m}{a_{m+1}}|)$ for each
$m > t$, and admits no other zero in $\mathbb{K}$.

**Theorem 55.16.** (1) Let $(b_n)_{n \in \mathbb{N}}$ be a sequence of $\mathbb{K}$ satisfying $b_0 = b_1 = 0$, $b_2 = 1$, $|2b_4| < |9(b_3)^2|$, $|3b_3b_5| < |4(b_4)^2|$, $|4b_4| \ge |5b_5||\frac{b_4}{b_5}|$, $|4b_4| > |nb_n||\frac{b_4}{b_5}|^{n-4} \quad \forall n > 5$ and such that the sequence $|\frac{b_n}{b_{n+1}}|_{n \ge 2}$ is strictly increasing of limit $+\infty$. Let $h(x) = \sum_{n=0}^{\infty} b_n x^n$ and let $(c_n)_{n \in \mathbb{N}^*}$ be the sequence of zeros of $h'$ ordered in such a way that $|c_n| \le |c_{n+1}|$. Then $h$ belongs to $\mathcal{A}(\mathbb{K})$ and satisfies $h(c_i) \ne h(c_n) \quad \forall i = 1, 2, 3 \quad \forall n \ne i$.

(2) Let $(b_n)_{n \in \mathbb{N}}$ be a sequence of $\mathbb{K}$ satisfying $b_0 = b_1 = 0$, $b_2 = 1$, $|2b_4| < |9(b_3)^2|$, $|3b_3b_5| < |4(b_4)^2|$, $|4b_4| \ge |5b_5||\frac{b_4}{b_5}|$, $|4b_4b_6| < |5(b_5)^2|$, $|5b_5| \ge |6b_6||\frac{b_5}{b_6}|$, $|5b_5| > |nb_n||\frac{b_5}{b_6}|^{n-5} \quad \forall n > 6$ and be such that the sequence $|\frac{b_n}{b_{n+1}}|_{n \ge 2}$ is strictly increasing of limit $+\infty$. Let $h(x) = \sum_{n=0}^{\infty} b_n x^n$ and let $(c_n)_{n \in \mathbb{N}^*}$ be the sequence of zeros of $h'$ ordered in such a way that $|c_n| \le |c_{n+1}|$. Then $h$ belongs to $\mathcal{A}(d(0, R^-))$ and satisfies $h(c_i) \ne h(c_n) \quad \forall i = 1, 2, 3, \forall n \ne i$.

**Proof.** By construction, we can see that $h$ has a zero of order 2 at 0. For each $n \ge 2$, we set $r_n = \left|\frac{b_n}{b_{n+1}}\right|$. By hypothesis, the sequence $\left(\left|\frac{b_n}{b_{n+1}}\right|\right)_{n \ge 2}$ is strictly increasing. Hence, by Lemma 55.15, $h$ has a unique zero $b_n$ in the circle $C(0, r_n)$ and this zero is of order 1. This is true for each $n \ge 1$ and $h$ does not admit any other zero in $\mathbb{K}$.

Now, let $\lambda \in \mathbb{K}$ and $r > 0$ be such that $|\lambda| < |h|(r)$. We know that
$\Psi(h-\lambda, \mu) = \Psi(h, \mu) \quad \forall \mu \ge \log r$ and $\nu^+(h-\lambda, \mu) = \nu^+(h, \mu)$, $\nu^-(h-\lambda, \mu) = \nu^-(h, \mu) \quad \forall \mu \ge \log r$, hence $h - \lambda$ admits a unique zero in
$C(0, r_n)$ for each $n$ such that $r_n \ge r$, this zero being of order 1. And then $h - \lambda$ does not admit any other zero in $K\setminus d(0, r^{-})$.

Next, since $|2b_4| < |9(b_3)^2|$, we see that $\left|\frac{2}{3b_3}\right| < \left|\frac{b_3}{4b_4}\right|r_2$, hence
$|c_2| < r_3$. Similarly, since $|3b_3b_5| < |4(b_4)^2|$, we see that
$\left|\frac{3b_3}{4b_4}\right| < \left|\frac{4b_4}{5b_5}\right|r_4$, hence $|c_3| < r_4$.

Now, suppose that there exist $m \in \mathbb{N}$ and $i \in \{1, 2, 3\}$ such that $h(c_m) = h(c_i)$ with $m \neq i$. We first note that $h(c_m) \neq 0$ $\forall m \neq 1$ because if $h(c_m) = 0$, then $c_m$ is a zero of order 2 of $h$, hence $c_m = 0$. Thus, we have $m \ge 3$. Suppose $c_m \ge r_3$. Since $|c_2| < r_3$, we have $|h(c_2)| < |h|_{r_3}$, hence as it was seen, $h - h(c_2)$ admits a unique zero of order 1 in each circle $C(0, r_n)$ for each $n \ge 3$ and has no other zero in $K\setminus d(0, r_3^{-})$, as does $h$. But if $|c_m| \ge r_3$, then it is a zero of order 2 for $h - h(c_m)$, a contradiction showing that $|c_m| < r_3$. Let $\rho = \max(|c_2|, |c_m|)$. So, $\rho < r_3$ and $h - h(c_2)$ must admit at least two multiple zeros in $d(0, \rho)$. But since $|b_2| = r_3$, we know that $\nu^+(h, \mu) = \nu^-(h, \mu) = 3 \forall \mu \in ]-\log(r_3), -\log(r_2)]$. On the other hand, when $\mu < -\log(\rho)$, we have seen that $\nu^+(h, \mu) = \nu^+(h - h(c_3), \mu)$, $\nu^-(h, \mu) = \nu^-(h - h(c_3), \mu)$, hence $\nu^+(h - h(c_3), \mu) = 3 \forall \mu \in ]-\log(r_3), -\log(\rho)]$. Consequently, $h - h(c_3)$ admits at most three zeros in $d(0, \rho)$, taking multiplicities into account. This is a contradiction to the assumption $c_m \in d(0, r_3^{-})$ and finishes showing that $h(c_m) \neq h(c_2) \forall \neq 2$. Thus, we have shown that $h(c_m) \neq h(c_j)$ for $j = 1, 2$ and $m \neq j$. We now suppose that there exists $m \neq 3$ such that $h(c_m) = h(c_3)$. Clearly, as for $h(c_2)$, we have $h(c_3) \neq 0$, and by the above, $h(c_3) \neq h(c_2)$, hence $m > 3$. Since $|c_3| < r_4$, we have $|h(c_3)| < |h|_{r_4}$, hence $\nu^+(h - h(c_2), \mu) - \nu^-(h - h(c_2), \mu) \le 1 \forall \mu \le -\log r_4$, which shows that $c_m \in d(0, r_4^{-})$. Thus, in $d(0, r_4^{-})$, $h'$ admits at least 4 zeros: $c_1, c_2, c_3, c_m$. But by the hypothesis $|4b_4| \ge |5b_5|\frac{b_4}{b_5}|$ and $|4b_4| > |nb_n|\frac{b_4}{b_5}|^{n-4} \forall n > 5$, we can see that $|4b_4|r_4^3 \ge |5b_5|r_4$ and $|4b_4|r_4^3 > |nb_n|r_4^{n-1} \forall n > 5$. Therefore, we have $\nu^+(h', -\log r_4) \le 4$ and $\nu^+(h', -\log r) \le 3 \forall r < r_4$. Consequently, $h'$ admits at most three zeros in $d(0, r_4^{-})$, a contradiction to the existence of a $c_m \in d(0, r_4^{-})$ such that $h(c_m) = h(c_3)$ with $m \ge 4$. Thus, the first conclusion is now established.

We now assume further that $|5b_5| \ge |6b_6|\left|\frac{b_4}{b_5}\right|$, $|4b_4b_6| < |5(b_5)^2|$ and that $|5b_5| > |nb_n|\left|\frac{b_4}{b_5}\right|^{n-5} \forall n > 6$. Suppose that there exists $m \neq 4$ such that $h(c_m) = h(c_4)$. By what precedes, we have $m > 4$.
Thanks to the hypothesis $|4b_4b_6| < |5(b_5)^2|$, we see that $|c_4| < r_5$. Consequently, $|h(c_m)| < |h|(r_5)$, hence $c_m$ lies in $d(0, r_5^-)$. Therefore, $d(0, r_5^-)$ contains 5 zeros of $h': c_1, c_2, c_3, c_4, c_m$. But thanks to the hypotheses $|5b_5| \ge |6b_6||\frac{b_5}{b_6}|, |5b_5| > |nb_n||\frac{b_5}{b_6}|^{n-5} \forall n > 6$, we can see that $\nu^+(h - h(c_4), -\log(r)) - \nu^-(h, -\log(r)) \le 4 \forall r < r_5$, hence $h - h(c_4)$ admits at most 4 zeros in $d(0, r_5^-)$, a contradiction to the assumption $h(c_4) = h(c_m)$ for some $m \ne 4$. This ends the proof of Theorem 55.16. □
# Chapter 56

## Urscm and Ursim

Throughout this chapter, $\mathbb{K}$ is of characteristic 0.

We introduce urscm and ursim for $p$-adic meromorphic functions. Many studies were made in the eighties and the nineties concerning functions in $\mathbb{C}$, [7], [59], [77], [78]. Studies were also made in the non-Archimedean context since the late nineties [3], [4], [21], [22], [27].

Here, we only consider the situation in an ultrametric field.

**Definition and notation.** Throughout this chapter, $\mathbb{E}$ is an algebraically closed field of characteristic 0 without any assumption on the existence of an absolute value and we denote by $\mathbb{P}^1(\mathbb{E})$ the projective space of dimension 1 over $\mathbb{E}$ and $\mathbb{K}$ is also of characteristic 0.

Given a family of functions $\mathcal{F}$ defined in $\mathbb{K}$ or in a subset of $\mathbb{K}$ (respectively in $\mathbb{E}$ or in a subset of $\mathbb{E}$), with values in $\mathbb{P}^1(\mathbb{K})$ (respectively in $\mathbb{P}^1(\mathbb{E}))$, a subset $S$ of $\mathbb{K}$ (respectively of $\mathbb{E}$) is called an *ursim for $\mathcal{F}$* if for any two non-constant functions $f, g \in \mathcal{F}$ satisfying $f^{-1}(S) = g^{-1}(S)$, these functions are equal. This definition particularly applies to $\mathcal{A}(\mathbb{K})$, $\mathcal{M}(\mathbb{K})$, $\mathcal{A}_u(d(a, R^{-}))$, $\mathcal{M}_u(d(a, R^{-}))$, $\mathbb{K}[x]$, $\mathbb{K}(x)$, $\mathbb{E}[x]$, $\mathbb{E}(x)$.

We now define URSCM. Given a subset $S$ of $\mathbb{E}$ and $f \in \mathbb{E}(x)$, we denote by $\mathcal{E}(f, S)$ the set in $\mathbb{E} \times \mathbb{N}^*$: $\bigcup_{a \in S} \{(z, q) \in \mathbb{E} \times \mathbb{N}^* | z$ a zero of order $q$ of $f(x) - a\}$.

And given a subset $S$ of $\mathbb{P}^1(\mathbb{E})$ containing $\{\infty\}$ and $f \in \mathbb{E}(x)$, we denote by $\mathcal{E}(f, S)$ the subset of $\mathbb{E} \times \mathbb{N}^*$: $\mathcal{E}(f, S \cap \mathbb{E}) \cup \{(z, q) \mid z$ a pole of order $q$ of $f\}$.
Consider now meromorphic functions in the field $K$. For a subset $S$ of $K$ and $f \in \mathcal{M}(K)$ (respectively $f \in \mathcal{M}(d(a, R^{-})))$, we denote by $\mathcal{E}(f, S)$ the set in $K \times N^*$: $\bigcup_{a \in S} \{(z, q) \in K \times N^* | \text{za zero of order } q \text{ of } f(x) - a\}$.

And given a subset of $\mathbb{P}^1(K)$ containing $\{\infty\}$ and $f \in \mathcal{M}(K)$ (respectively $f \in \mathcal{M}(d(a, R^{-})))$, we denote by $\mathcal{E}(f, S)$ the subset of $K \times N^*$: $\mathcal{E}(f, S \cap K) \cup \{(z, q) \mid z \text{ a pole of order } q \text{ of } f\}$.

Similarly, given a subset of $\mathbb{P}^1(\mathbb{E})$ containing $\{\infty\}$ and $f \in \mathbb{E}(x)$, we denote by $\mathcal{E}(f, S)$ the subset of $\mathbb{E} \times N^*$: $\mathcal{E}(f, S \cap \mathbb{E}) \cup \{(z, q) \mid z \text{ a pole of order } q \text{ of } f\}$.

Let $\mathcal{F}$ be a non-empty subset of $\mathcal{M}(K)$ (respectively of $\mathcal{M}(d(a, R^{-})))$. We say that two non-constant functions $f, g \in \mathcal{F}$ share $S$, counting multiplicity if $\mathcal{E}(f, S) = \mathcal{E}(g, S)$. And the set $S$ is called a unique range set counting multiplicity (an *urscm* in brief) for $\mathcal{F}$ if for any two non-constant $f, g \in \mathcal{F}$ sharing $S$ counting multiplicity, one has $f = g$.

Similarly, let $\mathcal{F}$ be a non-empty subset of $\mathbb{E}(x)$. We say that two non-constant functions $f, g \in \mathcal{F}$ share $S$, counting multiplicity if $\mathcal{E}(f, S) = \mathcal{E}(g, S)$. And the set $S$ is called a unique range set counting multiplicity (an *urscm* in brief) for $\mathcal{F}$ if for any two non-constants $f, g \in \mathcal{F}$ sharing $S$ counting multiplicity, one has $f = g$.

In other words, if we consider a set $S = \{a_1, \dots, a_t\} \subset K$ with $a_i \neq a_j \forall i \neq j$, we can set $P(X) = \prod_{j=1}^t (X - a_j)$ and then the set $S = \{a_1, \dots, a_t\}$ is an *urscm* for $\mathcal{F}$ if for any two functions $f, g \in \mathcal{F}$ such that $P \circ f$ and $P \circ g$ have the same zeros with the same multiplicity, then $f = g$.

An immediate lemma appears:

**Lemma 56.1.** If a subset of $K$ (respectively of $\mathbb{E}$) is an *ursim* for a family $\mathcal{F}$ of functions, it is an *urscm* for $\mathcal{F}$.

**Proof.** Indeed, suppose that a subset $S$ of $K$ (respectively of $\mathbb{E}$) is an *ursim* for a family $\mathcal{F}$ of functions and consider two functions $f, g \in \mathcal{F}$ such that $\mathcal{E}(f, S) = \mathcal{E}(g, S)$. So much the more, we have $f^{-1}(S) = g^{-1}(S)$, hence $f = g$. $\square$

Similar definitions were given for meromorphic functions on $\mathbb{C}$ before these questions were examined on the field $K$. *Urscm* of only 11 points for complex meromorphic functions in the whole field $\mathbb{C}$ where found in [59] and the same method showed the existence of
urscm of only 7 points for complex entire functions. So far, they are
the smallest known in $\mathbb{C}$.

In the field $\mathbb{K}$, the same method lets us find urscm of 11 points
for $M_u(d(a, R^{-}))$ and urscm of 10 points for $M(\mathbb{K})$.

Concerning analytic functions, urscm of 3 points were found for
entire functions and polynomials [22] and urscm of 7 points were
found for unbounded analytic functions in a disk $d(a, R^{-})$.

Actually, urscm for polynomials in a the field $\mathbb{E}$ where characterized as the affinely rigid sets of $\mathbb{E}$. And urscm of 3 points for entire functions in $\mathbb{K}$ where also characterized as the affinely rigid sets of 3 points in $\mathbb{K}$. That led to the conjecture: *URSCM for entire functions are the affinely rigid sets in $\mathbb{K}$ [22]*. The proof is given at the end of this chapter.

A first basic proposition is useful to understand the role of
URSCM:

**Proposition 56.2.** Let $S = \{a_1, \dots, a_n\} \subset \mathbb{K}$ (respectively $S = \{a_1, \dots, a_n\} \subset \mathbb{E}$), let $a \in \mathbb{K}$, let $R \in \mathbb{R}_+^*$, and let $P(x) = \prod_{i=1}^n (x - a_i)$. Then given any two functions $f, g \in \mathcal{A}(\mathbb{K})$ (respectively $f, g \in \mathcal{A}(d(a, R^{-}))$, respectively in $\mathbb{E}[x]$), then $\mathcal{E}(f, S) = \mathcal{E}(g, S)$ if and only if $\frac{P(f)}{P(g)}$ is a constant in $\mathbb{K}^*$ (respectively is an invertible function in $\mathcal{A}(d(a, R^{-}))$, respectively is a constant in $\mathbb{E}^*$).

**Proof.** Indeed, given $f, g \in \mathcal{A}(\mathbb{K})$ (respectively $f, g \in \mathcal{A}(d(a, R^{-}))$, respectively $f, g \in \mathbb{E}[x]$), for each $\alpha \in \mathbb{K}$ (respectively $\alpha \in d(a, R^{-})$, respectively $\alpha \in \mathbb{E}$), $f$ and $g$ share $S$ if and only if we have $\omega_{\alpha}(f - f(\alpha)) = \omega_{\alpha}(g - g(\alpha))$, which means that $\frac{P(f)}{P(g)}$ has no zero and no pole. Consequently, $\frac{P(f)}{P(g)}$ is a constant in $\mathcal{A}(\mathbb{K})$ (respectively is an invertible element in $\mathcal{A}(d(a, R^{-}))$, respectively is a constant in $\mathbb{E}$). $\square$

**Corollary 56.3.** Let $S = \{a_1, \dots, a_n\} \subset \mathbb{K}$ (respectively let $S = \{a_1, \dots, a_n\} \subset \mathbb{E}$) with $a_i \neq a_j \forall i \neq j$ and let $P(x) = \prod_{i=1}^n (x - a_i)$. Then $P$ is a polynomial of strong uniqueness for $\mathcal{A}(\mathbb{K})$ (respectively for $\mathbb{E}[x]$) if and only if $S = \{a_1, \dots, a_n\}$ is an URSCM for $\mathcal{A}(\mathbb{K})$ (respectively for $\mathbb{E}[x]$).

**Remark.** Let $P(x) = x^4 - 4x^3$ and let $j$ be a primitive third root of 1. Clearly, $C(P') = \{1, j, j^2\}$ and $\Upsilon(P) = 3$, hence by Corollary 55.3,
$P$ is a polynomial of uniqueness for $\mathcal{A}(\mathbb{K})$. But $P(jf) = jP(f) \quad \forall f \in \mathcal{A}(\mathbb{K})$, hence $P$ is not a polynomial of strong uniqueness for $\mathcal{A}(\mathbb{K})$ (or for $\mathbb{E}[x]$).

The equivalence for a subset $S = \{a_1, \dots, a_t\}$ to be an urscm and to define a polynomial of strong uniqueness $P$ for $\mathcal{A}(\mathbb{K})$ does not expand to $\mathcal{M}(\mathbb{K})$ because given $f, g \in \mathcal{M}(\mathbb{K})$ such that $P(f)$ and $P(g)$ have the same zeros, that does not imply that they have the same poles. However, we can note the following lemma:

**Lemma 56.4.** Let $P(x) = \prod_{j=1}^t (x - a_j)$ with $a_i \neq a_j \quad \forall i \neq j$. If $\{a_1, \dots, a_t\}$ is an urscm for $\mathcal{M}(\mathbb{K})$, then $P$ is a polynomial of strong uniqueness for $\mathcal{M}(\mathbb{K})$.

**Proof.** Indeed, let $f, g \in \mathcal{M}(\mathbb{K})$ be such that $P \circ f = \lambda P \circ g$ with $\lambda \neq 0$. Then $P \circ f$ and $P \circ g$ obviously have the same zeros. But since $\{a_1, \dots, a_t\}$ is an urscm for $\mathcal{M}(\mathbb{K})$, we have $f = g$. $\square$

In the proof of the following theorems, we use Lemma 56.5 that comes from [59]. It holds in $\mathcal{M}(\mathbb{K})$ as well as in $\mathcal{M}(d(a, R^-))$.

**Lemma 56.5.** Let $F, G \in \mathcal{M}(d(a, R^-))$ (respectively $F, G \in \mathcal{M}(\mathbb{K})$) have the same poles, ignoring multiplicity, and let $H = \frac{F''}{F'} - \frac{G''}{G'}$. Every pole of $H$ has multiplicity order 1. Let $\alpha$ be a pole of $F$ and $G$. If $\alpha$ has same multiplicity for $F$ and $G$, then $H$ has no pole at $\alpha$. Moreover, if $\alpha$ has a multiplicity order 1 for both $F$ and $G$, then $\alpha$ is a zero of $H$.

**Proof.** Without loss of generality, we can assume $\alpha = 0$. Since both $\frac{F''}{F'}$, $\frac{G''}{G'}$ are logarithmic derivative, all their poles are of order one. Therefore, all poles of $H$ are of order one.

Now, suppose $\alpha$ is a pole of order $q$ for both $\frac{F''}{F'}$ and $\frac{G''}{G'}$. Consider the Laurent series of $F$ and $G$ at $0$: $F(x) = \frac{a-q}{x^q} + \dots + \frac{a-1}{x} + a_0 + \sum_{n=1}^{\infty} a_n x^n$ and $G(x) = \frac{b-q}{x^q} + \dots + \frac{b-1}{x} + b_0 + \sum_{n=1}^{\infty} b_n x^n$.

We can check that both $\frac{F''}{F'}$, $\frac{G''}{G'}$ have a residue at $0$ equal to $\frac{-1}{q+1}$. Consequently, the residue of $\frac{F''}{F'} - \frac{G''}{G'}$ is null and hence $\frac{F''}{F'} - \frac{G''}{G'}$ has no pole at $0$.
Suppose now that $q = 1$. Then $\frac{F''}{F'}$ is of the form

$$ \left(\frac{1}{x}\right) \left( \frac{2a_{-1} + 2a_2x^2 + x^3\phi(x)}{-a_{-1} + x\theta(x)} \right) $$

with $\phi$, $\theta$ meromorphic functions having no pole at 0. Similarly, $\frac{G''}{G'}$ is of the form

$$ \left(\frac{1}{x}\right) \left( \frac{2b_{-1} + 2b_2x^2 + x^3\psi(x)}{-b_{-1} + x\eta(x)} \right) $$

with $\psi$, $\eta$ meromorphic functions having no pole at 0. Consequently, we can check that

$$ H(x) = \left(\frac{1}{x}\right) \left( \frac{2a_2x^2 + x^3\phi(x)}{-a_{-1} + x\theta(x)} - \frac{2b_2x^2 + x^3\psi(x)}{-b_{-1} + x\eta(x)} \right), $$

and hence $H(0) = 0$. □

In the proof of the following theorems, we also need the following lemma:

**Lemma 56.6.** Let $f, g \in \mathcal{M}(\mathbb{K})$ (respectively $f, g \in \mathcal{M}(d(a, R^{-}))$) be two different non-constant functions satisfying $\frac{f'}{f} = \frac{g''}{g}$. Then $f$ and $g$ are linked by a relation of the form $f = ag + b$.

**Proof.** By Theorem 50.4, the space of solutions of the equation $y'g'' - yg' - yg$ is of dimension at most 1. Consequently, $f, g$ satisfy a relation of the form $f' = ag'$ with $a \in \mathbb{K}$, and therefore, $f = ag + b$ with $b \in \mathbb{K}$. □

**Notation.** Let $n \in \mathbb{N}$ and let $c \in \mathbb{E}$ (respectively $c \in \mathbb{K}$). We denote by $S(n, c)$ the set of zeros of the polynomial $P_{n,c}(x) = (n-1)(n-2)x^n - 2n(n-2)x^{n-1} + n(n-1)x^{n-3} + c$ and let $S(n, c)$ be the set of zeros of $P_{n,c}$ in $\mathbb{E}$ (respectively in $\mathbb{K}$).

**Theorem 56.7.** Let $f, g \in \mathcal{M}(\mathbb{K})$ be two different non-constant functions satisfying $f^{-1}(S(n, c)) = g^{-1}(S(n, c))$. Then, $n \le 15$. Moreover, if $f, g \in \mathcal{A}(\mathbb{K})$, then $n \le 8$.

**Corollary 56.8.** Let $n \ge 16$. Then $S(n, c)$ is an ursim for $\mathcal{M}(\mathbb{K})$.

**Corollary 56.9.** Let $n \ge 9$. Then $S(n, c)$ is an ursim for $\mathcal{A}(\mathbb{K})$.
**Theorem 56.10.** Let $f \in \mathcal{M}_u(d(a, R^-))$ and let $g \in \mathcal{M}(d(a, R^-))$ be two different non-constant functions satisfying $f^{-1}(S(n, c)) = g^{-1}(S(n, c))$. Then, $n \le 16$. Moreover, if $f \in \mathcal{A}_u(d(a, R^-))$ and if $g \in \mathcal{A}(d(a, R^-))$, then $n \le 9$.

**Corollary 56.11.** Let $n \ge 17$. Then $S(n, c)$ is an ursim for $\mathcal{M}_u(d(a, R^-))$.

**Corollary 56.12.** Let $n \ge 10$. Then $S(n, c)$ is an ursim for $\mathcal{A}_u(d(a, R^-))$.

**Theorem 56.13.** Let $f, g \in \mathcal{M}(\mathbb{K})$ be two distinct non-constant functions satisfying $\mathcal{E}(f, S(n, c)) = \mathcal{E}(g, S(n, c))$. Then $n \le 9$. Moreover, if $f, g \in \mathcal{A}(\mathbb{K})$, then $n \le 5$.

**Corollary 56.14.** For every $n \ge 10$, $S(n, c)$ is an urscm for $\mathcal{M}(\mathbb{K})$, and for every $n \ge 6$, $S(n, c)$ is an urscm for $\mathcal{A}(\mathbb{K})$.

**Remark.** In Theorem 47.6, we saw that $P_{n,0}$ is a polynomial of uniqueness for $\mathcal{M}(\mathbb{K})$ for every $n \ge 2$ and for $\mathcal{M}(d(a, R^-))$ for every $n \ge 3$. Here we can add some precisions:

**Corollary 56.15.** For every $n \ge 10$, $P_{n,c}$ is a polynomial of strong uniqueness for $\mathcal{M}(\mathbb{K})$. And for every $n \ge 6$, $P_{n,c}$ is a polynomial of strong uniqueness for $\mathcal{A}(\mathbb{K})$.

**Theorem 56.16.** Let $f, g \in \mathcal{M}_u(d(a, R^-))$ be two different non-constant functions satisfying $\mathcal{E}(f, S(n, c)) = \mathcal{E}(g, S(n, c))$. Then, $n \le 10$. Moreover, if $f, g \in \mathcal{A}_u(d(a, R^-))$, then $n \le 6$.

**Corollary 56.17.** For every $n \ge 11$, $S(n, c)$ is an urscm for $\mathcal{M}_u(d(a, R^-))$. And for every $n \ge 7$, $S(n, c)$ is an urscm for $\mathcal{A}_u(d(a, R^-))$.

**Corollary 56.18.** For every $n \ge 10$, $P_{n,c}$ is a polynomial of strong uniqueness for $\mathcal{M}_u(d(a, R^-))$. And for every $n \ge 7$, $P_{n,c}$ is a polynomial of strong uniqueness for $\mathcal{A}_u(d(a, R^-))$.

**Proof (of Theorems 56.7, 56.10, 56.13, and 56.16).** In order to simplify the notation, we just denote by $P$ the polynomial $P_{n,c}$ and by $S$ the set $S(n, c)$. Without loss of generality, we may assume $a = 0$ in Theorems 56.10 and 56.16.
We then consider two distinct non-constant meromorphic functions $f, g \in \mathcal{M}(\mathbb{K})$ in Theorems 56.7 and 56.16 and $f, g \in \mathcal{M}(d(0, R^{-}))$ in Theorems 56.10 and 56.16. And we assume $f^{-1}(S(n, c)) = g^{-1}(S(n, c))$ in Theorems 56.7 and 56.10 and $\mathcal{E}(f, S) = \mathcal{E}(g, S)$ in Theorems 56.13 and 56.16. Without loss of generality, we may assume that all functions we consider have no zero and no pole at 0.

Let $F = \frac{1}{P(f)}, G = \frac{1}{P(g)}$, and $H = \frac{F''}{F'} - \frac{G''}{G'}$.

We first suppose that $H$ is identically 0. By Lemma 56.6, it is immediate to derive that there exist $A \in \mathbb{K}^*$ and $B \in \mathbb{K}$ such that $P(f) = \frac{P(g)}{BP(g)+A}$. Therefore, by Theorems 43.8, we have

$$ (1) \quad T(r, f) = T(r, g) + O(1) \quad (r \in ]0, R[). $$

We can check that $P(X) + c$ is of the form $bX^{n-2}(X - e_1)(X - e_2) + c$, with $b, e_1, e_2 \in \mathbb{E}$ (respectively $b, e_1, e_2 \in \mathbb{K}$) and $e_1e_2 \neq 0$, $e_1 \neq e_2$.

We have to distinct three cases: (i) $B \neq 0$, (ii) $B = 0$, $A = 1$, (iii) $B = 0$, $A \neq 1$.

(i) Since $AB \neq 0$, every zero of $P(g) + \frac{A}{B}$ is a pole of $P(f)$ and therefore is a zero of order at least $n$ of $P(g) + \frac{A}{B}$. On the other hand, we check that whenever $A, B \in \mathbb{K}^*$, the polynomial $P(X) + \frac{A}{B}$ admits at least two distinct zeros $b_1$ and $b_2$ of order 1 and therefore it admits another zero $l$ of order at most $n-2$. Consequently, every zero of $g-b_1$ or $g-b_2$ has order at least $n$ and every zero of $g-l$ has order at least 2. By a change of variable, we can obviously assume that $g, g-l, g-b_1, g-b_2$ have neither any zeros nor any poles at 0. Then, by applying Theorem 45.7, we obtain

$$ \begin{aligned} 2T(r, g) - \bar{N}(r, g) &\le \bar{Z}(r, g-l) + \bar{Z}(r, g-b_1) + \bar{Z}(r, g-b_2) + O(1) \\ &\le \frac{1}{2}Z(r, g-l) + \frac{1}{n}(Z(r, g-b_1) + Z(r, g-b_2)) \\ &\quad + O(1) \quad (r \in J), \end{aligned} $$

which leads to $n \le 4$.

(ii) In this case, we have $P(f) = P(g)$. But by Theorem 47.6, $P$ is a polynomial of uniqueness, for $M_u(d(0, R^{-}))$, hence $f = g$.

(iii) Let $\lambda = \frac{1}{A}$. We check that at least one of the two polynomials $Q_1(X) = (n-1)(n-2)X^n - 2n(n-2)X^{n-1} + n(n-1)X^{n-2} + c(\lambda-1)$
and $Q_2(X) = (n-1)(n-2)X^n - 2n(n-2)X^{n-1} + n(n-1)X^{n-2} + c(\frac{1}{\lambda} - 1)$ admits $n$ distinct zeros. Indeed, suppose this not true. We note that $Q'_1(X) = Q'_2(X) = P'(X) = n(n-1)(n-2)X^{n-3}(X-1)^2$. Consequently, if both $Q_1$ and $Q_2$ admit at least one zero of order $>1$, then this must be 1 because it cannot be 0. But then, we have $c(\lambda - 1) = c(\frac{1}{\lambda} - 1) = -2$. Since $c \neq 0$, and since $\lambda \neq 1$, this implies $\lambda = -1$ and $c = 1$, which is excluded by hypothesis.

Now, since $\lambda$ and $\frac{1}{\lambda}$ play symmetric roles, without loss of generality, we can assume that $Q_1$ admits $n$ distinct zeros $b_1, \dots, b_n$. Thus, putting $\gamma(X) = \lambda X^{n-2}(n-1)((n-2)X^2 - 2n(n-2)X + n(n-1))$, we have

$$ (2) \quad Q_1(f) = \gamma(g). $$

So, applying Theorem 45.7, we obtain

$$ (n-1)T(r, f) \le \sum_{j=1}^{n} \overline{Z}(r, f - b_j) + \overline{N}(r, f) + O(1) \quad (r \in J). $$

On the other hand, we have

$$ \begin{aligned} \sum_{j=1}^{n} \overline{Z}(r, f - b_j) &\le \overline{Z}(r, Q_1(f)) = \overline{Z}(r, g) + \overline{Z}(r, g - e_1) + \overline{Z}(r, g - e_2) \\ &\le 3T(r, g). \end{aligned} $$

But by (2), it is seen that $T(r, g) \le T(r, f) + O(1) \quad (r \in J)$, hence finally

$$ (3) \quad (n-1)T(r, f) \le 3T(r, f) + \overline{N}(r, f) + O(1) \quad (r \in J). $$

Consequently, we have $n \le 5$. Moreover, if $f, g \in A(d(0, R^-))$, then $N(r, f) = 0$, and then (3) leads to $n \le 4$. This finishes proving the claims of the theorems when $H = 0$.

Henceforth, we suppose that $H$ is not identically 0 and apply Fujimoto's method [60]. Let $\alpha$ be a zero of $f - a_j$, for some $j$. Then both $P(f)$ and $P(g)$ vanish at $\alpha$. So, by Lemma 43.4, we have $\sum_{j=1}^n \overline{Z}(r, f - a_j) = \overline{Z}(r, P(f)) = \overline{Z}(r, P(g))$. According to Lemma 56.5, if both $P(f), P(g)$ have a zero of order one, then $H$ also has a zero. Else, at least one of the two functions $P(f)$ and $P(g)$ admits $\alpha$
as a zero of order strictly greater than 1, and then (since all zeros of $P$ are simple), at least one of the two functions $f', g'$ has a zero at $\alpha$, while $\alpha$ is a zero $f - a_i$ for some $i$ and a zero of $g - a_j$ for some $j$. Consequently, we obtain

$$ (3A) \quad \sum_{j=1}^{n} \overline{Z}(r, f - a_j) \le \overline{Z}(r, H) + \overline{Z}(r, f') + \overline{Z}(r, g') - \overline{Z}_0^S(r, f') - \overline{Z}_0^S(r, g'). $$

Moreover, in the hypothesis of Theorems 56.13 and 56.16, since any zero $\alpha$ of $f - a_i$ is a zero of certain $g - a_j$ with the same multiplicity order, we have $\omega_\alpha(f') = \omega_\alpha(g')$. Consequently, we obtain this improvement of (3A):

$$ (3B) \quad \sum_{j=1}^{n} \overline{Z}(r, f - a_j) \le \overline{Z}(r, H) + \frac{1}{2} \left[ \overline{Z}(r, f') + \overline{Z}(r, g') - \overline{Z}_0^S(r, f') - \overline{Z}_0^S(r, g') \right]. $$

By Corollary 43.6, we know that $\overline{Z}(r, H) \le N(r, H) + O(1) (r \in J)$, hence

$$ \sum_{j=1}^{n} \overline{Z}(r, f - a_j) \le N(r, H) + \overline{Z}(r, f') + \overline{Z}(r, g') - \overline{Z}_0^S(r, f') - \overline{Z}_0^S(r, g') + O(1) (r \in I) $$

(respectively $r \in J$) and similarly

$$ \sum_{j=1}^{n} \overline{Z}(r, g - a_j) \le N(r, H) + \overline{Z}(r, f') + \overline{Z}(r, g') - \overline{Z}_0^S(r, f') - \overline{Z}_0^S(r, g') + O(1) (r \in I) $$

(respectively $r \in J$), hence

$$ (4A) \quad \sum_{j=1}^{n} \overline{Z}(r, f - a_j) + \overline{Z}(r, g - a_j) \le 2N(r, H) + 2[\overline{Z}(r, f') + \overline{Z}(r, g') - \overline{Z}_0^S(r, f') - \overline{Z}_0^S(r, g')]. $$
And in the hypothesis of Theorems 56.13 and 56.16, by (3B), we obtain

$$ (4B) \quad \sum_{j=1}^{n} \overline{Z}(r, f - a_j) + \overline{Z}(r, g - a_j) \\ \le 2N(r, H) + [\overline{Z}(r, f') + \overline{Z}(r, g') - \overline{Z}_0^S(r, f') - \overline{Z}_0^S(r, g')]. $$

Consider now $N(r, H)$, and let $\eta$ be a pole of $H$. Either $\eta$ is a zero of $F'G'$ or it is a pole of $FG$. Let $\varphi$ be the counting function of the poles of $H$ occurring when $FG$ has a pole. So, we have

$$ (5) \qquad N(r, H) = \overline{Z}(r, F'G') + \varphi(r). $$

Suppose first $\eta$ is a zero of $F'$ but is not a pole of $FG$. Either $\eta$ is a zero of $f'$, or it is a zero of $P'(f)$, or it is a pole of $f$. Consequently, $\eta$ is not a zero of $P(f)$ and we have $\overline{Z}(r, F') \le \overline{Z}_0^S(r, f') + \overline{Z}(r, P'(f)) + \overline{N}(r, f)$. Since

$$ P'(X) = n(n-1)(n-2)(X-1)^2 X^{n-3}, $$

we have $\overline{Z}(r, P'(f)) \le 2T(r, f)$, $\overline{Z}(r, P'(g)) \le 2T(r, g)$, and therefore,

$$ (6) \qquad \overline{Z}(r, F') \le \overline{Z}_0^S(r, f') + 2T(r, f) + \overline{N}(r, f). $$

Similarly, if $\eta$ is a zero of $G'$ but is not a pole of $FG$, we have

$$ (7) \qquad \overline{Z}(r, G') \le \overline{Z}_0^S(r, g') + 2T(r, g) + \overline{N}(r, g). $$

Suppose now $\eta$ is a pole of $FG$. Then $\eta$ is a zero of $P(f)$ and $P(g)$ (we note that when $\eta$ is a zero of $P(f)$, it is a zero of $P(g)$, and vice-versa), and then by Lemma 56.5, it may not be a pole of $H$ when it is a zero of the same order of $P(f)$ and $P(g)$. Consequently, in the hypothesis of Theorems 56.7 and 56.10, $\varphi(r)$ satisfies

$$ \varphi(r) \le \overline{Z}(r, f') - \overline{Z}_0^S(r, f') + \overline{Z}(r, g') - \overline{Z}_0^S(r, g'), $$

and therefore, by (5), (6), and (7), we obtain

$$ N(r, H) \le \overline{Z}_0^S(r, f') + \overline{Z}_0^S(r, g') + 2T(r, f) + 2T(r, g) + \overline{N}(r, f) \\ + \overline{N}(r, g) + \overline{Z}(r, f') - \overline{Z}_0^S(r, f') + \overline{Z}(r, g') - \overline{Z}_0^S(r, g'), $$
hence

$$N(r, H) \leq 2T(r, f) + 2T(r, g) + \overline{N}(r, f) + \overline{N}(r, g) \\ + \overline{Z}(r, f') + \overline{Z}(r, g').$$

Thus, by (4A), in the hypothesis of Theorems 56.7 and 56.10, we obtain

$$(8A) \quad \sum_{j=1}^{n} \overline{Z}(r, f - a_j) + \overline{Z}(r, g - a_j) \\ \leq 4(T(r, f) + T(r, g)) + 2(\overline{N}(r, f) + 2\overline{N}(r, g)) \\ + 4(\overline{Z}(r, f') + \overline{Z}(r, g')) + O(1).$$

Now, in the hypotheses of Theorems 56.13 and 56.16, since the order of a zero is the same for $P(f)$ and $P(g)$, the counting function $\varphi$ is identically 0, so by, (5) (6), and (7), we have

$$N(r, H) \leq 2T(r, f) + 2T(r, g) + \overline{N}(r, f) + \overline{N}(r, g) \\ + \overline{Z}_0^S(r, f') + \overline{Z}_0^S(r, g'),$$

and therefore, by (4B), we obtain

$$(8B) \quad \sum_{j=1}^{n} \overline{Z}(r, f - a_j) + \overline{Z}(r, g - a_j) \\ \leq 4(T(r, f) + T(r, g)) + 2(\overline{N}(r, f) + 2\overline{N}(r, g)) \\ + (\overline{Z}(r, f') + \overline{Z}(r, g')) + (\overline{Z}_0^S(r, f') + \overline{Z}_0^S(r, g')) + O(1).$$

Now, applying Theorem 45.7 to $f$ and $g$ at the points $a_j$ ($1 \leq j \leq n$), we have

$$(n-1)T(r, f) \leq \overline{N}(r, f) + \sum_{j=1}^{n} \overline{Z}(r, f - a_j) - Z_0^S(r, f') \\ + O(1) - \log r \ (r \in I)$$
(respectively $r \in J$)

$$ (n-1)T(r,g) \le \bar{N}(r,g) + \sum_{j=1}^{n} \bar{Z}(r, g - a_j) - Z_0^S(r, g') - \log r + O(1), \quad (r \in I) $$

(respectively $r \in J$), hence

$$ (n-1)(T(r,f) + T(r,g)) \le \bar{N}(r,f) + \bar{N}(r,g) + \sum_{j=1}^{n} \bar{Z}(r, f - a_j) + \bar{Z}(r, g - a_j) - Z_0^S(r, f') - Z_0^S(r, g') - 2\log r + O(1) \quad (r \in I) $$

(respectively $r \in J$), hence by (8A), in the hypotheses of Theorems 56.7 and 56.10, we obtain

$$ (9A) \quad (n-1)(T(r,f) + T(r,g)) \le 3(\bar{N}(r,f) + \bar{N}(r,g)) + 4(T(r,f) + T(r,g)) + 4(\bar{Z}(r,f') + \bar{Z}(r,g')) - 2\log r + O(1) \quad (r \in I) $$

(respectively $r \in J$).

By Corollary 15, we can derive

$$ (n-1)(T(r,f) + T(r,g)) \le 3(\bar{N}(r,f) + \bar{N}(r,g)) + 4(T(r,f) + T(r,g)) + 4(Z(r,f) + Z(r,g)) + 4(\bar{N}(r,f) + \bar{N}(r,g)) - 10\log r + O(1) \quad (r \in I) $$

(respectively $r \in J$).

Consequently, in Theorem 56.7, we have $n \le 15$, and in Theorem 56.10, we have $n \le 16$. Moreover, if $f, g$ belong to $\mathcal{A}(\mathbb{K})$, then in Theorem 56.7, we have $n \le 8$, and if $f, g$ belong to $\mathcal{A}_u(d(0, R^{-}))$, in Theorem 56.10, we obtain $n \le 9$.
By (8B), in the hypotheses of Theorems 56.13 and 56.16, we obtain

$$ (n-1)(T(r, f) + T(r, g)) \\ \leq 3(\overline{N}(r, f) + \overline{N}(r, g)) + 4(T(r, f) + T(r, g)) \\ + \overline{Z}(r, f') + \overline{Z}(r, g') - 2 \log r + O(1) \quad (r \in I) $$

(respectively $r \in J$).

Hence, by Corollary 43.15, we obtain $(n-1)(T(r, f) + T(r, g)) \leq 3(\overline{N}(r, f) + \overline{N}(r, g)) + 4(T(r, f) + T(r, g)) + (Z(r, f) + Z(r, g)) + \overline{N}(r, f) + \overline{N}(r, g) - 4 \log r + O(1) \quad (r \in I) \text{ (resp. } r \in J\text{)}. Consequently, $n \leq 9$ in Theorem 56.13 and $n \leq 10$ in Theorem 56.16. Moreover, if $f, g$ belong to $\mathcal{A}(\mathbb{K})$ or to $\mathcal{A}_u(d(0, R^{-}))$, then we obtain $(n-1)(T(r, f) + T(r, g)) \leq 4(T(r, f) + T(r, g)) + (Z(r, f) + Z(r, g)) - 4 \log r + O(1) \quad (r \in J)$. So, if $f, g$ belong to $\mathcal{A}(\mathbb{K})$, we have $n \leq 5$, and if $f, g \in \mathcal{A}_u(d(0, R^{-}))$, we have $n \leq 6$. $\square$

We now characterize all URSCM for $\mathcal{A}(\mathbb{K})$. First, we have to recall Picard-Berkovich Theorem [9].

**Theorem 56.19 (Berkovich) [9].** Let $F(x, y) = 0$ be the equation of an algebraic curve of genus $\geq 1$. There exist no functions $f, g \in \mathcal{M}(\mathbb{K})$ such that $F(f(t), g(t)) = 0$.

We also recall a classical theorem in algebraic geometry:

**Theorem 56.20 (Picard) [81].** Let $F(x, y) \in \mathbb{E}[x, y]$ be irreducible. If the algebraic curve of equation $F(x, y) = 0$ has genus 0, there exist rational functions $\phi, \theta \in \mathbb{K}(t)$ and $R(x, y) \in \mathbb{K}(x, y)$ such that, setting $t = R(x, y)$, we have $F(\phi(t), \theta(t)) = 0 \quad \forall t \in \mathbb{E}$.

We can now explain the method to characterize URSCM for $\mathcal{A}(\mathbb{K})$ in the general case.

**Proposition 56.21.** Let $F(x, y) \in \mathbb{K}[x, y]$ and let $f, g \in \mathcal{A}(\mathbb{K})$ be non-constant and satisfy $F(f(x), g(x)) = 0$. There exists $h \in \mathcal{A}(\mathbb{K})$ and $A, B \in \mathbb{K}[x]$ such that $f = A(h), g = B(h)$.
**Proof.** There obviously exists an irreducible factor $F_0(x, y)$ of $F$ such that $F_0(f, g) = 0$. Since $f, g$ are not constant, by Theorem 56.19, the curve $\Lambda$ defined as $F_0(x, y) = 0$ has genus zero. Therefore, by Theorem 56.20, there exist rational functions $\phi, \theta \in \mathbb{K}(t)$ and $R(x, y) \in \mathbb{K}(x, y)$ such that, setting $t = R(x, y)$, we have $F_0(\phi(t), \theta(t)) = 0 \quad \forall t$. Now, let $h = R(f, g)$. Hence by definition, $h$ belongs to $\mathcal{M}(\mathbb{K})$. Then, $f = \phi(h)$, $g = \theta(h)$.

Now, since $f, g$ belong to $\mathcal{A}(\mathbb{K})$, $h$ must omit $\phi^{-1}(\infty)$ and $\theta^{-1}(\infty)$. But by Theorem 29.23, $h$ can't have two exceptional values in $\mathbb{K} \cup \{\infty\}$. Consequently, $\phi^{-1}(\infty) = \theta^{-1}(\infty)$ and this is reduced to a single point. Therefore, up to a projective linear change in coordinates, we can assume that $\phi^{-1}(\infty) = \theta^{-1}(\infty) = \{\infty\}$ and that $h, \phi, \theta$ omit $\infty$. But then $h$ belongs to $\mathcal{A}(\mathbb{K})$ and both $\phi, \theta$ are polynomials. $\square$

**Lemma 56.22.** Let $h \in \mathcal{A}(\mathbb{K})$ be non-constant and let $\phi, \theta, P \in \mathbb{K}[x]$, with $\deg(P) = n \ge 1$, $\deg(\phi) = t \ge \deg(\theta)$ and $t \ge 1$. Suppose there exists $C \in \mathbb{K}^*$ such that $P(\phi(h)) = CP(\theta(h))$. Then $\phi(h)$ is of the form $A\theta(h) + B$, with $A^n = C$.

**Proof.** We can write $\phi(x) = \sum_{j=0}^{t} a_j x^j$, $\theta(x) = \sum_{j=0}^{t} b_j x^j$. Let $P(x) = \sum_{k=0}^{n} l_k x^k$ and let $M = \mathbb{Q}[a_0, \dots, a_t, b_0, \dots, b_t, l_0, \dots, l_n]$. Then, $\mathbb{K}$ contains elements $\nu$ transcendental over $M$. Since $h$ is not constant, we can take $z \in \mathbb{K}$ such that $h(z) = \nu$. Then, $\phi(h(z)) = \sum_{j=0}^{t} a_j \nu^j$, $\theta(x) = \sum_{j=0}^{t} b_j \nu^j$. Now, we note that all powers of $\nu$ strictly greater than $t(n-1)$ in $\sum_{j=0}^{t} a_j \nu^j$ involve $l_n$. First, we see that $\deg(\theta) = t$ and that $(\frac{at}{bt})^n = C$. Let $A = \frac{at}{bt}$. By comparing all powers strictly greater than $t(n-1)$ in $\sum_{j=0}^{t} a_j \nu^j$ and in $\sum_{j=0}^{t} b_j \nu^j$, we can see that $a_j = Ab_j \quad \forall j = 1, \dots, t$, which ends the proof. $\square$

**Lemma 56.23.** Let $S$ be an affinely rigid subset of $\mathbb{K}$. Let $g \in \mathcal{A}(\mathbb{K})$ be non-constant and let $f = ag + b$, with $a \in \mathbb{K}^*$ and $b \in \mathbb{K}$. If $f^{-1}(S) = g^{-1}(S)$, then $f = g$.

**Proof.** Since $f$ has no exceptional value, we have $f(f^{-1}(S)) = S$. Suppose $f^{-1}(S) = g^{-1}(S)$. Then,

$$ (1) \quad f(g^{-1}(S)) = S. $$

Now, $aS + b = ag(g^{-1}(S) + b = f(g^{-1}(S))$, and therefore by (1), $aS + b = S$. Thus, the similarity $\phi$ defined by $\phi(x) = ax + b$ preserves $S$. But since $S$ is affinely rigid, $\phi$ is the identity, hence $f = g$. $\square$

<figure><img src="image_23.png" /></figure>
We can now present the conclusion of this method due to W. Cherry and C.C. Yang:

**Theorem 56.24 (W. Cherry and C.C. Yang) [31].** A finite subset $S$ of $\mathbb{K}$ is an URSCM for $\mathcal{A}(\mathbb{K})$ if and only if it is affinely rigid.

**Proof.** On one hand, it is obvious that a finite subset $S$ of $\mathbb{K}$ which is not affinely rigid is not an URSCM for $\mathbb{K}[x]$ and therefore for $\mathcal{A}(\mathbb{K})$. Conversely, let $S = \{s_1, \dots, s_n\}$ be a finite affinely rigid subset $S$ of $\mathbb{K}$ and let $f, g \in \mathcal{A}(\mathbb{K})$ satisfy $\mathcal{E}(f, S) = \mathcal{E}(g, S)$. Let $P(x) = \prod_{j=1}^n (x - s_j)$. Since $\mathcal{E}(f, S) = \mathcal{E}(g, S)$, $P(f)$ and $P(g)$ have the same zeros, counting multiplicity. Consequently, $\frac{P(f)}{P(g)}$ belongs to $\mathcal{M}(\mathbb{K})$ but has no zero and no pole. Hence, it is a constant $C$. Now, setting $F(x, y) = P(x) - CP(y)$, we have $F(f, g) = 0$. Consequently, by Proposition 56.21, there exists $h \in \mathcal{A}(\mathbb{K})$ and $\phi, \theta \in \mathbb{K}[x]$ such that $f = \phi(h)$, $g = \theta(h)$. Now, by Lemma 56.22, $f$ is of the form $ag + b$ with $a \in \mathbb{K}^*$ and $b \in \mathbb{K}$, satisfying further $a^n = C$. Then, since $S$ is affinely rigid, by Lemma 56.23, we have $f = g$, which completes the proof. $\square$

Similarly, we can derive Theorem 56.25:

**Theorem 56.25.** A finite subset $S = \{a_1, \dots, a_n\}$ of $\mathbb{E}$ is affinely rigid if and only if it is an urscm for $\mathbb{E}[x]$.

**Proof.** If $S$ is an urscm, of course, it is affinely rigid. Now suppose $S$ is affinely rigid, let $P, Q \in \mathbb{E}[x]$ be such that $\mathcal{E}(P, S) = \mathcal{E}(Q, S)$ and let $F$ be an algebraic extension of $\mathbb{Q}$ containing $S$ and all coefficients of $P$ and $Q$. There exists a $\mathbb{Q}$-isomorphism $\phi$ from $F$ into $\mathbb{C}_p$ and then we can check that $\phi(S)$ is affinely rigid and hence is an urscm in $\mathbb{C}_p$ for $\mathcal{A}(\mathbb{C}_p)$ and hence for $\mathbb{C}_p[x]$. Consequently, $\phi(P) = \phi(Q)$ and hence $P = Q$. $\square$

**Corollary 56.26.** A finite subset $S = \{a_1, \dots, a_n\}$ of $\mathbb{K}$ (respectively of $\mathbb{E}$) is affinely rigid if and only if the polynomial $P(x) = \prod_{i=1}^n (x - a_i)$ is a polynomial of strong uniqueness for $\mathcal{A}(\mathbb{K})$ (respectively for $\mathbb{E}[x]$).
**Remark.** The characterization made in characteristic 0 doesn't hold in characteristic $p \neq 0$. The following example is given in [23]. Let $F$ be an algebraically closed field of characteristic $p \geq 3$ and let $P(X) = X^p + (X - 1)^{p-1}$. Let $f(x) = x^{p-1}$ and $g(x) = (x - 1)^{p-1}$. Then we can check that $P(f(x)) = P(g(x))$, hence the set $S$ is not an URSCM for $\mathcal{A}(\mathbb{K})$.

However, $S$ is a set of $p$ distinct points which are the $p$ roots of 1 of order $p+1$, other than 1, because $P(X) = \frac{X^{p+1}-1}{X-1}$. Consequently, $\sum_{s \in S} s = -1$. We check that $S$ is affinely rigid. Suppose there exists a similarity $\phi$, say $\phi(x) = ax + b$, satisfying $\phi(S) = S$. Then, $\sum_{s \in S} \phi(s) = \sum_{s \in S} s = -1$, hence $\sum_{s \in S} as + b = -1$. But $\sum_{s \in S} as + b = a \sum_{s \in S} s + pb = a \sum_{s \in S} s = -a$, hence $a = 1$. Thus, $\phi(s) = s + b$, hence $(s + b)^{p+1} = 1$, therefore $s^{p+1} + sb^p + sb^{p+1} = 1$, and hence

$$ (1) \qquad b^p + sb^{p-1} + s^p = 0. $$

This is true for all $s \in S$, so we can add these $p$ relations and obtain

$$ 0 = pb^p + b^{p-1} \sum_{s \in S} s + \sum_{s \in S} s^p = -b^{p-1} + \left( \sum_{s \in S} s \right)^p = -b^{p-1} + (-1)^p. $$

Thus, $b^{p-1} = -1$. Consequently, by (1), we have $b = s^p - s$.

Here we note that $-1$ belongs to $S$ because $p \geq 3$. Let $S^* = S \setminus \{-1\}$. Then,

$$ (q-1)b = \sum_{s \in S^*} s^p - \sum_{s \in S^*} s = 0, $$

hence $b = 0$, which ends the proof (another argument lets us conclude when $p = 2$).
# Chapter 57

## Other Urscm, Usim, and Non-Urscm

Throughout this chapter, $\mathbb{K}$ is of characteristic 0.

We denote by $\mathbb{E}$ an algebraically closed field of characteristic 0.
The field $\mathbb{K}$ is supposed to have characteristic 0.

In Chapter 56, we have constructed sets of $n$ points which are ursims for $p$-adic entire functions whenever $n \ge 9$ and are ursims for $p$-adic meromorphic functions whenever $n \ge 16$. Here, we construct new sets which are ursims of $n$ points for $\mathbb{E}[x]$ and for $\mathcal{A}(\mathbb{K})$ for every $n \ge 9$ but then, we show that such sets are never urscm and of course are not ursim for $\mathcal{M}(\mathbb{K})$ and for $\mathbb{E}(x)$, although they are not preserved by any Moebius functions, a contradiction to a natural expectation.

Next, we also construct urscm for unbounded analytic functions inside a disk $d(a, R^{-})$, with the use of the Nevanlinna theory on three small functions.

**Lemma 57.1.** Let $Q(x) = x^n - x^m + k \in \mathbb{E}[x]$ with $m < n$, g.c.d.($m, n$) = 1, $k \in \mathbb{K}^*$. If $Q$ admits a multiple zero $c$, then $m^m(n-m)^{(n-m)} = n^m k^{n-m}$.

**Proof.** Suppose $Q$ admits a multiple zero $c$. Then $c \neq 0$ and we have

$$ (1) \quad m = nc^{n-m}. $$
Since g.c.d.($m, n$) = 1, $p$ cannot divide both $m$ and $n$. Hence, by (1), $p$ cannot divide $m$ or $n$. Now, we can derive $P(c) = c^n - \frac{n}{m}c^n + k = 0$, and hence

$$ (2) \quad (n-m)c^m = kn. $$

Therefore, by (1) and (2), we have

$$ c^{n(n-m)} = \frac{k^{n-m}m^{n-m}}{(n-m)^{n-m}} = \frac{m^n}{n^n}, $$

hence

$$ k^{n-m} = \frac{m^m(n-m)^{n-m}}{n^n}. \quad \square $$

**Corollary 57.2.** Let $P(x) = x^n - x^m + k \in \mathbb{E}[x]$ with $m < n$, g.c.d.($m, n$) = 1, $k \neq 0$, and $m^m(n-m)^{(n-m)} \neq k^{n-m}n^n$. Then $P$ admits $n$ distinct zeros.

**Lemma 57.3.** Let $P(x) = x^n - x^m + k \in \mathbb{E}[x]$ with $m < n$, g.c.d.($m, n$) = 1, $k \neq 0$ and let $c \in \mathbb{E}\setminus\{0, 1\}$. Moreover, we assume that

$$ m^m(n-m)^{(n-m)} \neq n^n(1-u)^{n-m}k^{n-m} $$

for every $u \in \mathbb{E}$ such that $u^{n-m} = (-1)^{n-m}$. Then one of the polynomials $P - ck$ and $P - \frac{k}{c}$ admits $n$ distinct zeros.

**Proof.** Suppose that both $P - kc$ and $P - kc^{-1}$ have a zero of multiple order. By Corollary 57.2, we have

$$ n^nk^{n-m}(1-c)^{n-m} = n^nk^{n-m}(1-c^{-1})^{n-m} = m^m(n-m)^{(n-m)}, $$

hence $(\frac{1-c}{1-c^{-1}})$ is an $(n-m)$th root $t$ of 1. Resolving the equation shows that either $c = 1$ or $c = -t$. But $c = 1$ contradicts the relation $n^nk^{n-m}(1-c)^{n-m} = m^m(n-m)^{(n-m)}$, hence $c = -t$. But then, $1-c = 1+t$. Putting $u = -t$, we have $(u)^{n-m} = (-1)^{n-m}$ whereas $n^nk^{n-m}(1-u)^{n-m} = m^m(n-m)^{(n-m)}$, which contradicts the hypothesis. So, one of the two polynomials $P-kc$ and $P-\frac{k}{c}$ has two distinct zeros. $\square$
**Proposition 57.4.** Let $P(x) = x^n - x^m + k \in \mathbb{K}[x]$ with $m < n$ be such that $m^m(n - m)^{(n-m)} \neq k^{n-m}n^n(1 - u)^{(n-m)}$ for every $u \in \mathbb{K}$ such that $u^{n-m} = (-1)^{n-m}$. Let $f, g \in \mathcal{A}(\mathbb{K})\backslash\mathbb{K}$ (respectively let $\mathcal{A}_u(d(\alpha, R^{-}))$ satisfy $P(f) = \lambda P(g)$ with $\lambda \in \mathbb{K}^*$. In the following two cases, we have $\lambda = 1$:

(i) $f, g \in \mathcal{A}(\mathbb{K})\backslash\mathbb{K}, m \ge 2,$

(ii) $f, g \in \mathcal{A}_u(d(\alpha, R^{-}))$, $m \ge 3$.

**Proof.** Suppose $\lambda \neq 1$. By Theorem 43.3, we first note that $T(r, f) = T(r, g) + O(1)$. If $f, g \in \mathcal{A}_u(d(\alpha, R^{-}))$, we can obviously assume $\alpha = 0$.

By Lemma 57.3, one of the two polynomials $x^n - x^m + k(1 - \frac{1}{\lambda})$ and $x^n - x^m + k(1 - \lambda)$ admits $n$ distinct zeros. Without loss of generality, we can suppose that the polynomial $Q(x) = x^n - x^m + k(1 - \frac{1}{\lambda})$ has $n$ distinct zeros. And by hypothesis, we have $\lambda Q(g) = f^n - f^m$.

Now, $Q(x)$ is of the form $\prod_{i=1}^n (x - e_i)$ with $e_i \neq e_j \forall i \neq j$. Then we have $f^n - f^m = \lambda \prod_{i=1}^n (g - e_i)$. Consequently,

$$Z(r, Q(g)) - \bar{Z}(r, Q(g)) = \sum_{j=1}^{n} Z(r, g - e_j) - \bar{Z}(r, g - e_j) \\ = Z(r, f^n - f^m)) - \bar{Z}(r, f^n - f^m),$$

and of course $Z(r, f^n - f^m)) - \bar{Z}(r, f^n - f^m) \ge (m-1)Z(r, f)$, hence
(1) $Z(r, Q(g)) - \bar{Z}(r, Q(g)) \ge (m-1)Z(r, f)$.

But since $f \in \mathcal{A}(\mathbb{K})$ or $f \in \mathcal{A}_u(d(0, R^{-}))$, we have $Z(r, f) = T(r, f) + O(1)$, therefore, by (1), we obtain

(2) $Z(r, Q(g)) - \bar{Z}(r, Q(g)) \ge (m-1)T(r, f) + O(1)$.

On the other hand, by Lemma 43.14, we have $Z(r, Q(g)) - \bar{Z}(r, Q(g)) \le T(r, g) - \log r + O(1)$. Therefore, by (2), $(m-1)T(r, f) \le T(r, g) - \log r + O(1)$. And since $T(r, f) = T(r, g) + O(1)$, we can derive $(m-1)T(r, f) \le T(r, f) - \log r + O(1)$. Hence, if $f, g$ belong to $\mathcal{A}(\mathbb{K})$, we have $m \le 1$, and if $f, g$ belong to $\mathcal{A}_u(d(0, R^{-}))$, we have $m \le 2$, a contradiction in both cases. Consequently, if $f, g$ belong to $\mathcal{A}(\mathbb{K})\backslash\mathbb{K}$ and if $m \ge 2$, we can conclude $\lambda = 1$. And similarly, if $f, g$ belong to $\mathcal{A}_u(d(0, R^{-}))$ and if $m \ge 3$, we have $\lambda = 1$. □
Now, we use a slightly different method to consider the problem $P(f) = \lambda P(g)$ when $f, g$ are meromorphic.

**Theorem 57.5.** Let $P(x) = x^n - x^m + k$ with $m < n$ and let $f, g \in \mathcal{M}(\mathbb{K})\backslash\mathbb{K}$ (respectively let $f, g \in \mathcal{M}_u(d(0, R^{-}))$ be such that $P(f) = \lambda P(g)$, with $\lambda \in \mathbb{K}^*$. If $f, g \in \mathcal{M}(\mathbb{K})\backslash\mathbb{K}$ and $2m \ge n + 3$, or if $f, g \in \mathcal{M}_u(d(0, R^{-}))$ and $2m \ge n + 4$, then $\lambda = 1$. Moreover, if $f, g \in \mathcal{A}(\mathbb{K})\backslash\mathbb{K}$ and $2m \ge n + 2$ or if $f, g \in \mathcal{A}_u(d(0, R^{-}))$ and $2m \ge n + 3$, then $\lambda = 1$.

**Proof.** Suppose $\lambda \ne 1$. By Theorem 43.3, we first note that $T(r, f) = T(r, g) + O(1)$. If $f, g \in \mathcal{M}_u(d(\alpha, R^{-}))$, we can obviously assume $\alpha = 0$. Let $F(x) = f(x)^n - f(x)^m$. Applying Theorem 45.7 to $F$, we have $T(r, F) \le \bar{Z}(r, F) + \bar{Z}(r, F - (\lambda - 1)) + \bar{N}(r, F) - \log r + O(1)$, hence $nT(r, f) \le \bar{Z}(r, f) + \bar{Z}(r, f^{n-m} - 1) + \bar{Z}(r, g) + \bar{Z}(r, g^{n-m} - 1) + N(r, f) - \log r + O(1)$. Consequently, $nT(r, f) \le (2(n - m + 1) + 1)T(r, f) - \log r + O(1)$. Finally, if $f, g \in \mathcal{M}(\mathbb{K})\backslash\mathbb{K}$, we have $2m \le n + 2$, and if $f, g \in \mathcal{M}_u(d(0, R^{-}))$, we have $2m \le n + 3$. Consequently, we have proven that $\lambda = 1$ in the following two cases:

(i) $f, g \in \mathcal{M}(\mathbb{K})\backslash\mathbb{K}$ with $2m \ge n + 3$,

(ii) $f, g \in \mathcal{M}_u(d(0, R^{-}))$ with $2m \ge n + 4$.

Particularly, if $f, g \in \mathcal{A}(\mathbb{K})\backslash\mathbb{K}$, we have $2m \le n + 1$, and if $f, g \in \mathcal{A}(d(0, R^{-}))$, we have $2m \le n + 2$. Consequently, we have proven that $\lambda = 1$ in the following two cases:

(i) $f, g \in \mathcal{A}(\mathbb{K})\backslash\mathbb{K}$ with $2m \ge n + 2$,

(ii) $f, g \in \mathcal{A}_u(d(0, R^{-}))$ with $2m \ge n + 3$. $\square$

<figure><img src="image_24.png" /></figure>

**Remark.** If $f, g \in \mathcal{A}(\mathbb{K})\backslash\mathbb{K}$ and if $2m \le n + 2$, by Proposition 57.5, we have $\lambda = 1$, even if the hypothesis $m^m(n - m)^{(n-m)} \ne k^{n-m}n^n (1 - u)^{(n-m)}$ for every $u \in \mathbb{K}$ such that $u^{n-m} = (-1)^{n-m}$ is not satisfied. Similarly, when $f, g \in \mathcal{A}_u(d(0, R^{-}))$ with $2m \ge n + 3$, when $m^m(n - m)^{(n-m)} \ne k^{n-m}n^n(1 - u)^{(n-m)}$ for every $u \in \mathbb{K}$ such that $u^{n-m} = (-1)^{n-m}$ is not satisfied.

**Lemma 57.6.** Let $P(x) = x^n - x^m + k$ with $k \ne 0$, $m < n$, $n \ge 3$, g.c.d.($m, n$) = 1. Let $f, g \in \mathcal{A}(\mathbb{K})\backslash\mathbb{K}$ (respectively $f, g \in \mathcal{A}_u(d(\alpha, R^{-}))$) satisfy $P(f) = P(g)$. Then, $f = g$.
**Proof.** If $f, g$ belong to $\mathcal{A}_u(d(\alpha, R^{-}))$, we can obviously suppose $\alpha = 0$. Let $h = \frac{f}{g}$ and suppose $h \neq 1$. Then, $g^{n-m} = \frac{h^m-1}{h^{n-1}}$. We note that if $f, g$ belong to $\mathcal{A}_u(d(0, R^{-}))$, then $h$ belongs to $\mathcal{M}_u(d(0, R^{-}))$ because $\mathcal{M}_b(d(0, R^{-}))$ is a field.

Let $G$ be the group of $n$th roots of 1 and let $G'$ be the group of $m$th roots of 1. Since g.c.d.($n, m$) = 1, $G \setminus G'$ contains no element different from 1. So, $G$ has $n$ distinct elements and $G \setminus (G \cap G')$ has $n-1$ distinct elements. Now, $h(x)$ cannot take a value in $G \setminus G'$ because this would be a zero of $h^n - 1$ but not a zero of $h^m - 1$ and therefore it would be a pole of $g^{n-m}$ that lies in $\mathcal{A}(\mathbb{K})$ (resp. in $\mathcal{A}_u(d(0, R^{-}))$). Since $G \setminus G \cap G'$ contains at least two different elements $u_1$ and $u_2$, $h$ must take at least one of these two values, a contradiction as we just saw. Consequently, $h$ is identically 1. $\square$

**Notation.** Given $k \in \mathbb{K}^*$ and $m, n \in \mathbb{N}^*$, we set $Q_{n,m,k}(x) = x^n - x^m + k \in \mathbb{K}[x]$ and $Q_{n,k}(x) = x^n - x^{n-1} + k \in \mathbb{K}[x]$, and we denote by $Y(n, k)$ the set of zeros of $Q_{n,k}$. Moreover, we put $u_n = \frac{(n-1)^{n-1}}{n^n}$.

**Theorem 57.7.** If $2 \le m$ (respectively $3 \le m$), with g.c.d.($m, n$) = 1 and $m^m(n-m)^{(n-m)} \ne k^{n-m}n^n(1-u)^{(n-m)}$ for every $u \in \mathbb{K}$ such that $u^{n-m} = (-1)^{n-m}$, then $Q_{n,m,k}$ is a polynomial of strong uniqueness for $\mathcal{A}(\mathbb{K})$ (respectively for $\mathcal{A}_u(d(\alpha, R^{-}))$).

**Proof.** By Proposition 57.4, we have $\lambda = 1$, therefore, $f^n - f^m = g^n - g^m$, and then, by Lemma 57.6, $f = g$. That ends the proof. $\square$

**Theorem 57.8.** Let $k \in \mathbb{K} \setminus \{0, \frac{u_n}{2}\}$. Then $Y_{n,k}$ is an ursim for $\mathcal{A}(\mathbb{K})$ for every $n \ge 9$. Moreover, $Y_{n,k}$ is an ursim for $\mathcal{A}(\mathbb{K})$ for every $n \ge 3$, provided that $k \ne u_n$ whenever $3 \le n \le 8$.

**Proof.** Suppose first that $n \ge 9$, and suppose that $f, g \in \mathcal{A}(\mathbb{K})$ are not constant and satisfy $f^{-1}(Y(n,k)) = g^{-1}(Y(n,k))$. We put $F = \frac{1}{Q_{n,k}(f)}$, $G = \frac{1}{Q_{n,k}(g)}$, $H = \frac{F''}{F'} - \frac{G''}{G'}$. Then $F, G$ have the same poles, ignoring multiplicity. Without loss of generality, we may obviously assume that $\omega_0(F) = \omega_0(G) = \omega_0(F') = \omega_0(G') = \omega_0(f) = \omega_0(g) = 0$, and $F(0) \ne c_j$, $G(0) \ne c_j \forall j = 1, 2, 3$. Following the same method as in Theorem 56.7, we can prove that $H$ is identically 0.

So, we may assume $H = 0$. Then there exist $a, b \in \mathbb{K}$ such that $Q_{n,k}(f) = \frac{\dot{Q}_{n,k}(g)}{aQ_{n,k}(g)+b}$, with $b \ne 0$. But, since $f, g$ lie in $\mathcal{A}(\mathbb{K})$, $a$ must be equal to zero. Thus, we have $Q_{n,k}(f) = \frac{1}{b}Q_{n,k}(g)$. Now, let $c \in \mathbb{K}$
satisfy $c^n = \frac{1}{k}$. Then we have $Q_{n,k}(x) = k((cx)^n - c(cx)^{n-1} + 1)$. Setting $P(y) = y^n - cy^{n-1} + 1$, we have $P(cf) = \frac{1}{b}P(cg)$. But as $n \ge 9$, by Proposition 57.4, we check that $b = 1$, hence $cf^n - f^{n-1} = cg^n - g^{n-1}$.

Let $h = \frac{f}{g}$. From this equality, we can easily obtain $cg = -\frac{h^{n-1}-1}{h^{n-1}}$. Clearly, if $h$ is not constant, $h^n - 1$ admits zeros which are not zeros of $h^{n-1} - 1$, a contradiction with the hypothesis $g \in \mathcal{A}(\mathbb{K})$. Hence, $h$ is constant and so are $f, g$, a contradiction with the hypothesis.

Suppose now that $f, g \in \mathcal{A}(\mathbb{K})$ satisfy $\mathcal{E}(f, Y(n, k)) = \mathcal{E}(g, Y(n, k))$ with $n \ge 3$ and that $k \ne u_n$ whenever $3 \le n \le 8$. We first note that if $n \ge 9$, $Y_{n,k}$ is an urscm for $\mathcal{A}(\mathbb{K})$, hence by Lemma 56.1, it is an ursim for $\mathcal{A}(\mathbb{K})$. Suppose now $3 \le n \le 8$. Then, by Proposition 56.2, there exists $\lambda \in \mathbb{K}^*$ such that $Q_{n,k}(f) = \lambda Q_{n,k}(g)$. But by Proposition 57.4, $\lambda = 1$. Consequently, $f^n - f^{n-1} = g^n - g^{n-1}$, and hence, by Lemma 57.6, we have $f = g$. Therefore, $Q_{n,k}$ is a polynomial of strong uniqueness for $\mathcal{A}(\mathbb{K})$. Moreover, by Corollary 57.2, $Q_{n,k}$ has no multiple zero since $k \ne u_n$. Consequently, by Corollary 56.3, $Y(n, k)$ is an urscm for every $n \ge 3$, provided $k \ne u_n$ whenever $3 \le n \le 8$. That finishes the proof of Theorem 57.8. $\square$

**Corollary 57.9.** Let $k \in \mathbb{E}\{0, \frac{u_n}{2}\}$. $Y(n, k)$ is an ursim for $\mathbb{E}[x]$ for every $n \ge 9$ and $Y(n, k)$ is an urscm for $\mathbb{E}[x]$ for every $n \ge 3$ provided that $k \ne u_n$ whenever $3 \le n \le 8$.

**Proof.** Indeed, there exists a finite extension of $\mathbb{Q}$ containing $Y(n, k)$ and all coefficients of $f$ and $g$. As we can embed this extension into $\mathbb{C}_p$ so we can transfer results from $\mathbb{C}_p[x]$ to $\mathbb{E}[x]$. $\square$

**Remarks.** Concerning urscm of 3 points for $\mathcal{A}(\mathbb{K})$ or for polynomials, the last statement of Theorem 57.8 provides us with another proof that every affinely rigid subset of 3 elements in $\mathbb{K}$ is urscm for $\mathcal{A}(\mathbb{K})$ (and therefore every affinely rigid subset of 3 elements in $\mathbb{E}$ is urscm for $\mathbb{E}[x]$). Indeed, given a subset of 3 points that is not affinely rigid, by Corollary 1.7, through an affine transformation, either they are the zeros of a polynomial of the form $x^3 - k$ or they are the zeros of a polynomial of the form $x^3 - x$ which can also be transformed to $3\sqrt{3}(u^3 - u^2 + \frac{2}{27})$ and then the case $k = \frac{2}{27}$ is excluded by hypothesis on $k$. Consequently, if a subset of $\mathbb{K}$ of 3 elements is affinely rigid, it is an urscm for $\mathcal{A}(\mathbb{K})$, and of course, if a subset of $\mathbb{E}$ of 3 elements is affinely rigid, it is an urscm for $\mathbb{E}[x]$. Recall that more generally,
in [22], it was proven just by classical techniques that all affinely rigid subsets of $\mathbb{E}$ are urscm for polynomials.

As we saw in Theorem 56.24, urscm for $\mathbb{E}[x]$ and for $\mathcal{A}(\mathbb{K})$ are characterized as the affinely rigid sets, similarly, it would be natural to think that the urscm for $\mathbb{E}(x)$ and for $\mathcal{M}(\mathbb{K})$ are the sets which are not preserved by any Moebius function other than the identity. This is definitely false, as we are going to see [43]. See also [96].

**Lemma 57.10.** Let $f, g \in \mathcal{M}(\mathbb{K})$ such that $f^n(f - a) = g^n(g - a)$ and let $h = \frac{f}{g}$. Then, $f = \frac{a(n+2)}{n+1} \left( \frac{h^{n+1}-1}{h^{n+2}-1} \right) h$ and $g = \frac{a(n+2)}{n+1} \left( \frac{h^{n+1}-1}{h^{n+2}-1} \right)$.

**Proof.** We have $g^n h^n - g^{n-1} = g^n - g^{n-1}$, hence $g^n(h^n - 1) = g^{n-1}(h^{n-1} - 1)$, hence $g = \frac{h^{n-1}-1}{h^{n-1}}$ and then $f = hg$. $\square$

**Corollary 57.11.** Let $h \in \mathbb{E}(x)$ be non-constant, and let $n \in \mathbb{N}$ be such that $n \ge 3$. Let $g = \frac{h^{n-1}-1}{h^{n-1}}$, and let $f = gh$. Then, $f^n(f-1) = g^n(g-1)$, $\mathcal{E}(f,Y(n,k)) = \mathcal{E}(g,Y(n,k))$. Moreover, $f$ and $g$ have the same poles, counting multiplicity.

**Corollary 57.12.** Let $k \in \mathbb{E}\setminus\{0, u_n\}$. Then $Y(n,k)$ is not an urscm for $\mathbb{E}(x)$ whenever $n \in \mathbb{N}^*$.

**Remark.** Obviously, any set which is not an urscm for $\mathbb{E}(x)$ cannot be an urscm for $\mathcal{M}(\mathbb{K})$, cannot be an ursim for $\mathbb{E}(x)$, and cannot be an ursim for $\mathcal{M}(\mathbb{K})$. In fact, we do not know any example of an urscm for a family of functions $\mathcal{F}$ which is not also an ursim for $\mathcal{F}$. For example, as noted in [22], sets of the form $\{a, a+h, a+3h\}$, with $a \in W$, $h \in E^*$, are urscm for $\mathbb{E}[x]$ but are they also ursims for $\mathbb{E}[x]$?

**Theorem 57.13.** Let $k \in \mathbb{E}\setminus\{0, u_n\}$. For every $n \ge 5$, there exists no Moebius function $h$ preserving $Y(n,k)$, except the identity.

**Proof.** We suppose that a Moebius function $h$ defined as $h(x) = \frac{ax+b}{cx+d}$ (with $ad-bc \ne 0$) satisfies $h(Y(n,k)) = Y(n,k)$ for some $n \ge 5$. Let $R(x) = Q_{n,k}(h(x))$. Then $R$ is a rational function of the form $\sum_{j=0}^{n} q_j x^j$. Since $h(Y(n,k)) = Y(n,k)$, we note that the set of zeros of $R$ is $Y(n,k)$ and therefore $Y(n,k)$ is the set of zeros of $\sum_{j=0}^{n} q_j x^j$. Consequently, this polynomial is of the form $\lambda Q_{n,k}$, with $\lambda \in W^*$. In particular, we have

$$ (1) \quad q_1 = q_2 = q_3 = 0. $$
On the other hand, we have $\sum_{j=0}^{n} q_j x^j = (ax + b)^n - (cx + d)(ax + b)^{n-1} + k(cx + d)^n$, so we obtain

$$q_1 = nab^{n-1} - (n-1)ab^{n-2}d - b^{n-1}c + kncd^{n-1},$$

$$q_2 = \binom{n}{2} a^2 b^{n-2} - \binom{n-1}{2} a^2 b^{n-3} d - (n-1)ab^{n-2}c + k \binom{n}{2} c^2 d^{n-2},$$

$$q_3 = \binom{n}{3} a^3 b^{n-3} - \binom{n-1}{3} a^3 b^{n-4} d - \binom{n-1}{2} a^2 b^{n-3} c + k \binom{n}{3} c^3 d^{n-3}.$$

We now compute $A = \frac{a^2q_1}{n} - \frac{4abq_2}{n(n-1)} + \frac{6b^2q_3}{n(n-1)(n-2)}$ and obtain $A = cd^{n-3}k(ad - bc)^2$. But of course, by (1), we have $A = 0$. Since $k(ad - bc) \neq 0$, we must have $cd = 0$.

First, suppose $d = 0$. Then, we have

(2) $bc \neq 0$.

But, as $q_1 = 0$, this implies $nab^{n-1} = b^{n-1}c$, i.e.

(3) $na = c$.

And as $q_2 = 0$, we have $\frac{n(n-1)}{2}a^2b^{n-2} = (n-1)ab^{n-2}c$, i.e.

(4) $na^2 = 2ac$.

By (3) and (4), we check that $ac = 0$, hence by (2), $a = 0$, which contradicts (2) and (3).

Thus, we are led to the hypothesis $d \neq 0$, $c = 0$. Then, $ad \neq 0$. And then by (1) we obtain $nab^{n-1} = (n-1)ab^{n-2}d$, i.e.

(5) $nb^{n-1} = (n-1)b^{n-2}d$.

And $na^2b^{n-2} = (n-2)a^2b^{n-3}d$, i.e.

(6) $nb^{n-2} = (n-2)b^{n-3}d$.
hence by (5) and (6), $(n-2)b^{n-2}d = (n-1)b^{n-2}d$, and finally, $b=0$. So, $h(x) = \frac{ax}{d}$. Taking $d=1$, we obtain $\sum_{j=0}^{n} q_j x^j = a^n x^n - a^{n-1} x^{n-1} + k$, which must be proportional to $Q_{n,k}(x)$. In particular, $x^n - \frac{x^{n-1}}{a} + \frac{k}{a^n} = x^n - x^{n-1} + k$ and therefore $a=1$. This finishes showing that the identity is the only Moebius function preserving $Y(n,k)$. □

**Remark.** In Theorem 57.13, the hypothesis $n \ge 5$ is necessary as shown in Theorem 57.14:

**Theorem 57.14.** Every urscm for $\mathbb{E}(x)$ or for $\mathcal{M}(\mathbb{K})$ has at least five points.

**Proof.** Indeed, given any four distinct points $\alpha, \beta, \gamma, \delta \in E$, there does exist a Moebius function $h$, different from the identity, satisfying $h(\alpha) = \beta, h(\beta) = \alpha, h(\gamma) = \delta, h(\delta) = \gamma$. Indeed, $h$ is of the form

$$h(x) = \frac{x(\alpha\beta - \gamma\delta) - \alpha\beta(\gamma + \delta) + \gamma\delta(\alpha + \beta)}{x(\alpha + \beta - \gamma - \delta) - \alpha\beta + \gamma\delta}.$$

And of course, every urscm for $\mathcal{M}(\mathbb{K})$ has at least five points. □

**Conclusion.** By Corollary 57.12, the sets $Y(n,k)$ in $\mathbb{E}$ are not urscm for $\mathbb{E}(x)$ though, by Theorem 57.13, there exist no Moebius functions $h$ (different from the identity) preserving $Y(n,k)$ as soon as $n \ge 5$. Thus, the set of finite urscm for $\mathbb{E}(x)$ is strictly smaller than the set of finite sets $S$ such that there exist no Moebius functions $h$ preserving $S$. So, it is not easy to imagine a characterization of urscm for $\mathbb{E}(x)$. And a characterization of urscm for $\mathcal{M}(\mathbb{K})$ appears even more difficult.

Next, by Theorem 2 of [83], we know that given a compact $A$ in $\mathbb{C}$ which is not preserved by any non-trivial similarity of the form $\varphi(x) = a + (x-a)e^{i\alpha}$ and given two monic polynomials $F, G$ of same degree such that $F^{-1}(A) = G^{-1}(A)$, we have $F = G$. Now, we note that, by choosing for $h$ a rational function such as $\frac{x}{x+1}$, we can obtain two rational functions $f$ and $g$ both equal to a quotient of monic polynomials, satisfying $f^{-1}(Y(n,k)) = g^{-1}(Y(n,k))$, though the sets $Y(n,k)$ are not preserved by similarities, at least when $n \ge 4$. Thus, one doesn't see any clear generalization of Theorem 2 of [83] to rational functions.
**Example of sets Y(5, k).** Let $Q(x) = x^5 - x^4 + \frac{3}{32}$. Then $Q$ factorizes in $\mathbb{Q}[x]$ in the form $(x + \frac{1}{2})(x^4 - \frac{3}{2}x^3 + \frac{3}{22}x^2 - \frac{3}{23}x + \frac{3}{24})$.

In $\Omega_p$, we can define

$$a = \frac{\sqrt{2}}{4}(\sqrt{3}\sqrt{7} - 1)\sqrt{\sqrt{3}\sqrt{7} + 3}, \quad b = \sqrt{2}\sqrt{3}\sqrt{\sqrt{3}\sqrt{7} - 3},$$

$$\alpha = \frac{1}{8}(3 + a + i\sqrt{2}\sqrt{\sqrt{3}\sqrt{7} - 3a + 2b}),$$

$$\beta = \frac{1}{8}(3 - a + i\sqrt{2}\sqrt{\sqrt{3}\sqrt{7} + 3a - 2b}).$$

Then we have $T(5, (\frac{3}{32})) = \{\frac{1}{2}, \alpha, \bar{\alpha}, \beta, \bar{\beta}\}$.

We can check that the derivative of $(\frac{h^m-1}{h^n-1}) = g^{n-m}$ is not identically zero and therefore $h'$ is not identically zero.

Now, since g.c.d.($m, n$) = 1, $G'$ has $m$ elements and $G'\backslash G$ has $m-1$ elements $u_1, \dots, u_{m-1}$. Then, for each zero $\beta$ of $h - u_i$ ($1 \le i \le m-1$), $\beta$ is a zero of $h^m - 1$ and is not a zero of $h^n - 1$. Hence, it is a zero of $g^{n-m}$ and consequently it is a zero of order at least $n-m$ of $h - u_i$. Thus, for each $i = 1, \dots, m-1$, we have

$$(1) \quad \bar{Z}(r, h - u_i) \le \frac{1}{n-m} Z(r, h - u_i).$$

Now, let us apply Theorem 45.7 to $h$. Since $h'$ is not identically zero, we have

$$(m-2)T(r, h) \le \sum_{i=1}^{m-1} \bar{Z}(r, h - u_i) + N(r, h) - \log r + O(1),$$

hence by (1), we have

$$(2) \quad (m-2)T(r, h) \le \left( \frac{m-1}{n-m} + 1 \right) T(r, h) - \log r + O(1).$$

Concerning ursc for $\mathcal{A}(d(0, R^{-}))$, we can improve the results obtained in Chapter 54 thanks to the Nevanlinna theorem on small functions.
**Notation.** Given $k \in \mathbb{K}^*$ and $n, m \in \mathbb{N}^*$ with $m < n$, we set by $Q_{n,m,k}(x) = x^n - x^m + k$ and we denote by $Y(n, m, k)$ the set of zeros of $Q_{n,m,k}$.

**Theorem 57.15.** If $2m \ge n + 3$, then $Y(n, m, k)$ is an urscm for $\mathcal{A}_u(d(a, R^-))$ ($a \in \mathbb{K}, k \in \mathbb{K}^*, R > 0$).

**Proof.** We can obviously assume that $a = 0$. Suppose $Y(n, m, k)$ is not an urscm for $\mathcal{A}(d(0, R^-))$ and let $f, g \in \mathcal{A}_u(d(0, R^-))$ be such that

$$ \mathcal{E}(f, Y(n, m, k)) = \mathcal{E}(f, Y(n, m, k)). $$

By Proposition 56.2, $\frac{Q_{n,k}(f)}{Q_{n,k}(g)}$ is an invertible function $\phi \in \mathcal{A}_b(d(0, R^-))$. Consequently, we have $f(x)^n - f(x)^m + k = \phi(x)(g(x)^n - g(x)^m + k)$. Set $F(x) = f(x)^n - f(x)^m$. By Corollary 50.13, we can obtain

$$ T(r, F) \le \bar{Z}(r, F) + \bar{Z}(r, F - k(\phi - 1)) + o(T(r, f)). $$

Now, clearly, $\bar{Z}(r, F) \le \bar{Z}(r, f) + \bar{Z}(r, f^{n-m} - 1) + O(1)$ and $\bar{Z}(r, f^{n-m} - 1) \le (n - m)T(r, f)$, hence

$$ (1) \quad \bar{Z}(r, F) \le (n - m + 1)T(r, f) + o(T(r, f)). $$

And similarly,

$$ (2) \quad \bar{Z}(r, F - k(\phi - 1)) = \bar{Z}(r, g^n - g^m) \le \bar{Z}(r, f) + \bar{Z}(r, f^{nm} - 1). $$

Of course, since $\phi$ is bounded, we have $T(r, f) = T(r, g) + O(1)$, hence, by (1) and (2), we obtain $T(r, F) \le 2(n - m + 1)T(r, f^{n-m} - 1) + o(T(r, f))$.

On the other hand, we have $T(r, F) = nT(r, f) + O(1)$, hence $nT(r, f) \le 2(n - m + 1)T(r, f) + o(T(r, f))$. That yields $2m \le n + 2$, a contradiction to the hypothesis. $\square$

**Corollary 57.16.** $Y(5, k)$ is an urscm for $\mathcal{A}_u(d(a, R^-))$ ($a \in \mathbb{K}, k \in \mathbb{K}^*, R > 0$).
**Open question.** Do URSCM of 3 or 4 points for $\mathcal{A}_u(d(a, R^{-}))$ exist?

For instance, consider an affinely rigid set of 3 or 4 points, such as $\{0, 1, 3\}$, and let $P(X) = X(X-1)(X-3)$. Do functions $f, g \in \mathcal{A}_u(d(0, R^{-}))$ exist such that $P(f) = P(g)$?
# Chapter 58

## Nevanlinna Theory in
Characteristic *p*

Results in characteristic *p* were published in [25], [26], and [27].

**Notation.** In this chapter, we denote by *p* the characteristic of $\mathbb{K}$ and by *q* its characteristic exponent, i.e. $q = p$ if $p \neq 0$, and $q = 1$ if $p = 0$.

As usual, given $a \in \mathbb{K}$ and $n \in \mathbb{N}$, we denote by $a^n\sqrt[p]{a}$ the unique $b \in \mathbb{K}$ such that $b(p^n) = a$.

Given $m, n \in \mathbb{N}$, we set $m < n$ if $m$ divides $n$ and $m \not< n$ if $m$ does not divide $n$.

When $p \neq 0$, we denote by $S$ the $\mathbb{F}_p$-automorphism of $\mathbb{K}$ defined by $S(x) = \sqrt[p]{x}$. More generally, this mapping has continuation to a $\mathbb{K}$-algebra automorphism of $\mathbb{K}[X]$ as $S(c \prod_{j=1}^n (X - a_j)) = S(c) \prod_{j=1}^n (X - S(a_j))$, $c \in \mathbb{K}$.

Lemma 58.1 is easily verified.

**Lemma 58.1.** Let $f \in \mathcal{M}(d(0, R^{-}))$ be such that $f'$ is not identically zero and let $\alpha \in d(0, R^{-})$. We have $\omega_{\alpha}(f') = \omega_{\alpha}(f) - 1$ if $p \neq \omega_{\alpha}(f)$ and $\omega_{\alpha}(f') \ge \omega_{\alpha}(f)$ if $p < \omega_{\alpha}(f)$.

**Proposition 58.2.** Suppose $p \neq 0$. Let $r > 0$ and let $f \in \mathcal{M}(d(a, r^{-}))$. Then $\sqrt[p]{f}$ belongs to $\mathcal{M}(d(a, r^{-}))$ if and only if $f' = 0$. Moreover, there exists a unique $t \in \mathbb{N}$ such that $t\sqrt[p]{f} \in \mathcal{M}(d(a, r^{-}))$ and $(\sqrt[p]{f})' \neq 0$.
**Proof.** If $f$ is of the form $l^p$ with $l \in \mathcal{M}(d(a, r^{-}))$, then of course we have $f' = 0$. Now, suppose that $f' = 0$. Suppose first $f \in \mathcal{A}(d(a, r^{-}))$. Since $f' = 0$, $f(x)$ is of the form $\sum_{n=0}^{\infty} (a_n)^p x^n p$. Put $l(x) = \sum_{n=0}^{\infty} (a_n)x^n$. The radius of convergence of $l$ is equal to that of $f$. Hence, $l \in \mathcal{A}(d(a, r^{-}))$, and of course, we have $f = l^p$. We now consider the general case when $f \in \mathcal{M}(d(a, r^{-}))$. Let $(b_n, t_n)_{n \in \mathbb{N}}$ be the sequence of poles of $f$ inside $d(a, r^{-})$ where $t_n$ is the multiplicity order of $b_n$. By Theorem 27.14, we can find $h \in \mathcal{A}(d(a, r^{-}))$ such that $\omega_{b_n}(h) \ge t_n \quad \forall n \in \mathbb{N}$. Clearly, $fh^p$ belongs to $\mathcal{A}(d(a, r^{-}))$ and satisfies $(fh^p)' = 0$. Consequently, $fh^p$ is of the form $g^p$, with $g \in \mathcal{A}(d(a, r^{-}))$, therefore $f = \left(\frac{g}{h}\right)^p$. On the other hand, by Theorem 29.11, the set of integers $s$ such that $\sqrt[s]{f}$ belongs to $\mathcal{M}(d(a, r^{-}))$ is obviously bounded and therefore admits a biggest element, which ends the proof. $\square$

**Definition and notation.** Suppose $p \neq 0$. Given, $f \in \mathcal{M}(d(a, r^{-}))$, we call ramification index of $f$ the integer $t$ such that $\sqrt[s]{f} \in \mathcal{M}(d(a, r^{-}))$ and $(\sqrt[s]{f})' \neq 0$.

In the same way, given an algebraically closed field $B$ of characteristic $p \neq 0$ and $P(x) \in B[x]$, we call ramification index of $P$ the unique integer $t$ such that $\sqrt[s]{P} \in B[x]$ and $(\sqrt[s]{P})' \neq 0$. This ramification index is denoted by $\text{ram}(f)$ for any $f \in \mathcal{M}(d(a, r^{-}))$ or $f \in \mathcal{M}(\mathbb{K})$ and similarly it is denoted by $\text{ram}(P)$ for any $P \in B[x]$.

**Remark.** Suppose $p \neq 0$ and let $f \in \mathcal{M}(d(a, r^{-}))$ have ramification index $t$. For every $r' \in ]0, r[$, $f$ has the same ramification index as an element of $\mathcal{M}(d(a, r'^{-}))$ because of course, on one hand, $\sqrt[s]{f} \in \mathcal{M}(d(a, r'^{-}))$, and on the other hand, by properties of analytic functions, $(\sqrt[s]{f})'$ is not identically zero inside $d(a, r')$.

**Corollary 58.3.** Suppose $p \neq 0$ and let $f, g \in \mathcal{M}(d(a, R^{-})) \setminus \mathbb{K}$ satisfy an equation of the form $g(x) = G(f(x))$ and $G \in \mathbb{K}(Y)$. Then, $\text{ram}(g) \ge \text{ram}(f)$ and, putting $t = \text{ram}(f)$, $g_t = \sqrt[s]{g}$, $f_t = \sqrt[s]{f}$, $G_t = S^t(G)$, they satisfy $g_t = G_t \circ f_t$.

**Remark.** This property does not hold for meromorphic functions inside a disk $d(a, r^{-})$.

**Corollary 58.4.** Let $p \neq 0$. Let $f(x) \in \mathcal{M}(\mathbb{K})$, let $S$ be the set of zeros and poles of $f$ in $\mathbb{K}$, and let $d$ be the g.c.d. of $\{\omega_\alpha(f) \mid \alpha \in S\}$. Then, $\text{ram}(f) = -\log_p(|d|_p)$.
**Notation.** In the sequel, *I* denotes an interval of the form $[\rho, +\infty[$, with $\rho > 0$, and *J* denotes an interval of the form $[\rho, R[$. Let $f \in \mathcal{M}(d(0, R^{-}))$. The counting functions of zeros and poles of $f$: $Z(r, f)$ and $N(r, f)$ are defined as usual and we put again $T(r, f) = \max(Z(r, f), N(r, f))$.

Next, denoting by $E(r, f)$ the set $\{a \in d(0, r) \mid \omega_a(f) > 0, p^{\text{ram}(f)+1} \nleq \omega_a(f)\}$, if $0 \notin E(r, f)$, we set $\tilde{Z}(r, f) = \sum_{\alpha \in E(r, f)} \log \frac{r}{|\alpha|}$, and if $0 \in E(r, f)$, we set $\tilde{Z}(r, f) = \log r + \sum_{\alpha \in E(r, f), \alpha \neq 0} \log \frac{r}{|\alpha|}$.

Similarly, we define $\tilde{N}(r, f) = \tilde{Z}(r, \frac{1}{f})$.

Now, assume that $f'$ is not identically 0.

Let $V(r, f) = \{a \in d(0, r) \mid \omega_a(f) < 0, p^{\text{ram}(f)+1} < \omega_a(f)\}$. We put

$$N_0(r, f') = \sum_{\alpha \in V(r, f)} [\omega_{\alpha}(f') - \omega_{\alpha}(f)] \log \frac{r}{|\alpha|}.$$

Given a finite subset $S$ of $\mathbb{K}$, we put $\Lambda'(r, f, S) = \{a \in d(0, r) \mid f'(a) = 0, f(a) \notin S\}$ and $\Lambda''(r, f, S) = \{a \in d(0, r) \mid p^{\text{ram}(f)+1} < \omega_a(f - f(a)), f(a) \in S\}$. Then we can define

$$Z_0^S(r, f') = \sum_{\alpha \in \Lambda'(r, f, S)} \omega_\alpha(f') \log \frac{r}{|\alpha|} + \sum_{\alpha \in \Lambda''(r, f, S)} [\omega_\alpha(f') - \omega_\alpha(f - f(\alpha))] \log \frac{r}{|\alpha|}.$$

**Remarks.** (1) It is easily verified that all the above functions are positive.

(2) If $p = 0$, we have $\bar{Z}(r, f) = \tilde{Z}(r, f)$ and $\bar{N}(r, f) = \tilde{N}(r, f)$.

**Lemma 58.5.** Let $f \in \mathcal{M}(d(0, R^{-}))$, let $t = \text{ram}(f)$, and let $g = \sqrt[t]{f}$. Then, $\tilde{Z}(r, f) = \tilde{Z}(r, g)$ and $\tilde{N}(r, f) = \tilde{N}(r, g)$.

**Proof.** Let $a$ be a zero of $f$ and let $s = \omega_a(f)$. Then $s$ is of the form $nt$ with $n \in \mathbb{N}^*$. If $n = 1$, then $a$ belongs to both $E(r, f)$ and $E(r, g)$. And if $n > 1$, then $a \notin E(r, f)$. But then $a$ is a zero of order $n$ of $g$ and hence $a$ does not belong to $E(r, g)$. $\square$
Similar to Lemma 43.4, we have Lemma 58.6:

**Lemma 58.6.** Let $\alpha_1, \dots, \alpha_n \in \mathbb{K}$ be pairwise distinct, let $P(u) = \prod_{i=1}^n (u - \alpha_i)$, and let $f \in \mathcal{M}(d(0, R^{-}))$. Then, $Z(r, P(f)) = \sum_{i=1}^n Z(r, f - \alpha_i)$ and $\tilde{Z}(r, P(f)) = \sum_{i=1}^n \tilde{Z}(r, f - \alpha_i)$.

Lemmas 58.7 and 58.8 are consequences of Lemma 58.1.

**Lemma 58.7.** Let $f \in \mathcal{M}(d(0, R^{-}))$ be such that $f' \neq 0$ and let $S$ be a finite subset of $\mathbb{K}$. Then,

$$ \sum_{b \in S} (Z(r, f-b) - \tilde{Z}(r, f-b)) = Z(r, f') - Z_0^S(r, f'). $$

**Lemma 58.8.** Let $f \in \mathcal{M}(d(0, R^{-}))$ be such that $f' \neq 0$. Let $0 < r < R$. Then,

$$ N(r, f') = N(r, f) + \tilde{N}(r, f) - N_0(r, f'). $$

**Lemma 58.9.** Let $f \in \mathcal{M}(d(0, R^{-}))$ be such that $f' \neq 0$. Let $0 < r < R$. Then,

$$ Z(r, f') \leq Z(r, f) + \tilde{N}(r, f) - N_0(r, f') - \log r + O(1), (r \in J). $$

**Proof.** Without loss of generality, up to a change of variable, we can assume that both $f, f'$ have no zero and no pole at 0. By Theorem 43.2, we have $Z(r, f) - N(r, f) = \Psi(f, \log r) - \Psi(f(0))$, and $Z(r, f') - N(r, f') = \Psi(f', \log r) - \Psi(f'(0))$. But, by Corollary 36.11, $\Psi(f', \log r) \leq \Psi(f, \log r) - \log r$, hence we obtain $Z(r, f') \leq N(r, f') - N(r, f) + Z(r, f) - \log r + O(1)$. And by Lemma 58.7, we have $N(r, f') - N(r, f) = \tilde{N}(r, f) - N_0(r, f')$, which completes the proof. □

The previous results enable us to prove the ultrametric Nevankinna main theorem in a basic form:

**Theorem 58.10.** Let $\alpha_1, \dots, \alpha_n \in \mathbb{K}$, with $n \geq 2$, and let $f \in \mathcal{M}(d(0, R^{-}))$ (respectively $f \in \mathcal{M}(\mathbb{K})$) of ramification index $t$. Let $S = \{\sqrt[q]{\alpha_1}, \dots, \sqrt[q]{\alpha_n}\}$. Then we have

$$ \frac{(n-1)T(r, f)}{q^t} \leq \sum_{i=1}^{n} \tilde{Z}(r, f - \alpha_i) + Z(r, (\sqrt[q]{f})') - Z_0^S(r, (\sqrt[q]{f})') + O(1) \quad \forall r \in J $$

(respectively $\forall r \in I$).
Moreover, if $f$ belongs to $\mathcal{A}(d(0, R^{-}))$ (respectively $f \in \mathcal{A}(\mathbb{K}))$, then

$$ \frac{nT(r, f)}{q^t} \le \sum_{i=1}^{n} \tilde{Z}(r, f - \alpha_i) + Z(r, (q^t \sqrt{f})') - Z_0^S(r, (q^t \sqrt{f})') + O(1) \quad \forall r \in J $$

(respectively $\forall r \in I$).

Now, following the same method as in Theorem 45.7, we can obtain that classical form of the Nevanlinna inequality where $\bar{Z}$ and $\bar{N}$ are replaced by $\tilde{Z}$ and $\tilde{N}$.

**Theorem 58.11.** Let $\alpha_1, \dots, \alpha_n \in \mathbb{K}$, with $n \ge 2$, and let $f \in \mathcal{M}(d(0, R^{-}))$ (respectively $f \in \mathcal{M}(\mathbb{K}))$ of ramification index $t$. Let $S = \{q^t \sqrt{\alpha_1}, \dots, q^t \sqrt{\alpha_n}\}$. Then we have

$$ \frac{(n-1)T(r, f)}{q^t} \le \sum_{i=1}^{n} \tilde{Z}(r, f - \alpha_i) + \tilde{N}(r, f) - Z_0^S(r, (q^t \sqrt{f})') - N_0(r, (q^t \sqrt{f})') - \log r + O(1) \quad \forall r \in J $$

(respectively $\forall r \in I$).

**Corollary 58.12.** Let $f \in \mathcal{M}(d(0, R^{-}))$ (respectively $f \in \mathcal{M}(\mathbb{K}))$ be such that $f' \ne 0$ and let $\alpha_1, \dots, \alpha_n \in \mathbb{K}$, with $n \ge 2$. Then we have

$$ \sum_{i=1}^{n} Z(r, f - \alpha_i) - \tilde{Z}(r, f - \alpha_i) \le T(r, f) + \tilde{N}(r, f) - Z_0^S(r, f') - N_0(r, f') - \log r + O(1) \quad \forall r \in J $$

(respectively $\forall r \in I$).

**Remark.** In Theorems 45.2 and 45.4 as in Corollaries 45.3 and 45.5, the field $\mathbb{K}$ is not supposed to be of characteristic 0. Consequently, all those statements apply to fields of characteristic $p$.

**Proof of Theorems 58.10 and 58.11.** For convenience, we put $g = q^t \sqrt{f}$, and $\beta_i = q^t \sqrt{\alpha_i}$ for every $i = 1, \dots, n$. So, $S = \{\beta_1, \dots, \beta_n\}$.

By Theorem 29.14, there exist $\phi, \theta \in \mathcal{A}(d(0, R^{-}))$ such that $g = \frac{\phi}{\theta}$, and
$$ (1) \quad Z(r, \phi) \leq Z(r, g) + 1, $$

$$ (2) \quad Z(r, \theta) \leq N(r, g) + 1. $$

By Theorems 45.2 and 45.4, there exists $A \in \mathbb{R}$, and for any $r \in J$ (respectively $r \in I$), there exists $l(r) \in \{1, \dots, n\}$ such that $Z(r, \phi - \beta_j\theta) \geq \max(Z(r, \phi), Z(r, \theta)) + A \quad \forall j \neq l(r)$, therefore there exists $B \in \mathbb{R}$ such that

$$ (3) \quad Z(r, \phi - \beta_i\theta) \geq T(r, g) + B \quad \forall i \neq l(r), \quad \forall r \in J $$

(respectively $\forall r \in I$).

We check that $D(\phi) - D(\frac{\phi}{g}) = D(\theta) - D(\frac{\theta}{\phi})$, therefore

$$ \begin{aligned} D(\phi - \beta_i\theta) &= D(g - \beta_i) + D(\theta) - D\left(\frac{1}{g - \beta_i}\right) \\ &= D(g - \beta_i) + D(\theta) - D\left(\frac{1}{g}\right). \end{aligned} $$

Then, applying counting functions, we have $Z(r, \phi - \beta_i\theta) = Z(r, g - \beta_i) + Z(r, \theta) - N(r, g)$, and therefore, by (2), we obtain

$$ (4) \quad Z(r, \phi - \beta_i\theta) \leq Z(r, g - \beta_i) + 1. $$

Then, by (3) and (4), we obtain $(n-1)(T(r,g) + B) \leq \sum_{1 \leq i \leq n, i \neq l(r)} Z(r, \phi - \beta_i\theta) \leq \sum_{1 \leq i \leq n, i \neq l(r)} Z(r, g - \beta_i) + n - 1 \quad \forall r \in J$ (respectively $\forall r \in I$).

Putting $M = (n-1)(1-B)$, we obtain

$$ (5) \quad (n-1)T(r,g) \leq \sum_{i=1}^{n} Z(r,g - \beta_i) + M - Z(r,g - \beta_{l(r)}) \quad \forall r \in J $$

(respectively $\forall r \in I$).

By Lemma 58.7, we have

$$ \sum_{i=1}^{n} Z(r, g - \beta_i) = \sum_{i=1}^{n} \tilde{Z}(r, g - \beta_i) + Z(r, g') - Z_{0}^{S}(r, g'), $$
hence by (5), we obtain

$$ (6) \quad (n-1)T(r,g) \le \sum_{i=1}^{n} \tilde{Z}(r, g - \beta_i) + Z(r, g') - Z_0^S(r, g') \\ \qquad - Z(r, g - \beta_{l(r)}) + O(1) \quad \forall r \in J $$

(respectively $\forall r \in I$).

Now, since $T(r,g) = \frac{T(r,f)}{q^t}$ and since $\tilde{Z}(r,g - \beta_i) = \tilde{Z}(r, f - \alpha_j)$
$\forall j = i, \dots, n$, we obtain

$$ \frac{(n-1)T(r,f)}{q^t} \le \sum_{i=1}^{n} \tilde{Z}(r, f - \alpha_i) + Z(r, (\sqrt[q]{f})') \\ - Z_0^S(r, (\sqrt[q]{f})') + O(1) \quad \forall r \in J $$

(respectively $\forall r \in I$).

Suppose now that $f$ belongs to $\mathcal{A}(d(a, R^-))$ or to $\mathcal{A}(\mathbb{K})$. Then so does $g$. By Theorem 43.7, we have $Z(r, g - \beta_{l(r)}) = T(r, g) + O(1)$ $\forall r \in J$ (respectively $\forall r \in I$) so, by (6), we obtain

$$ nT(r,g) \le \sum_{i=1}^{n} \tilde{Z}(r,g - \beta_i) + Z(r,g') \\ - Z_0^S(r,g') + O(1) \quad \forall r \in J $$

(respectively $\forall r \in I$), and consequently,

$$ \frac{nT(r,f)}{q^t} \le \sum_{i=1}^{n} \tilde{Z}(r, f - \alpha_i) + Z(r, (\sqrt[q]{f})') \\ - Z_0^S(r, (\sqrt[q]{f})') + O(1) \quad \forall r \in J $$

(respectively $\forall r \in I$).

Now, returning to the general case, we have $g' = (g - \beta_{l(r)})'$ and
$\tilde{N}(r,g) = \tilde{N}(r,g - \beta_{l(r)})$. So, by Lemma 58.9, we have

$$ (7) \quad Z(r, g') - Z(r, g - \beta_{l(r)}) \le \tilde{N}(r, g) - N_0(r, g') - \log r + O(1). $$
Finally, by (6) and (7), we obtain

$$ \frac{(n-1)T(r, f)}{q^t} \le \sum_{i=1}^{n} \tilde{Z}(r, f - \alpha_i) + \tilde{N}(r, f) - Z_0^S(r, (\sqrt[q]{f})') \\ - N_0(r, (\sqrt[q]{f})') - \log r \quad \forall r \in J $$

(respectively $\forall r \in I$).

That completes the proof.

□

**Theorem 58.13.** Let $f \in \mathcal{M}(\mathbb{K})$ (respectively $f \in \mathcal{M}_u(d(0, R^{-})))$ and let $u_1, u_2, u_3 \in \mathcal{M}_f(\mathbb{K})$ (respectively $u_1, u_2, u_3 \in \mathcal{M}_f(d(0, R^{-})))$ be pairwise distinct. Let $\phi(x) = \frac{(f(x)-u_1(x))(u_2(x)-u_3(x))}{(f(x)-u_3(x))(u_2(x)-u_1(x))}$ and let $t$ be the ramification index of $\phi$.

Then, $\frac{T(r,f)}{q^t} \le \sum_{j=1}^3 \tilde{Z}(r, f - u_j) + o(T(r, f))$.

**Proof.** By Theorem 58.11, we have

$$ (1) \quad \frac{T(r, \phi)}{q^t} \le \tilde{Z}(r, \phi) + \tilde{Z}(r, \phi - 1) + \tilde{N}(r, \phi) + O(1). $$

Next, we follow the same way as in Theorem 47.9. We have $T(r, f) \le T(r, \frac{u_3-u_1}{f-u_3}) + o(T(r, f))$, hence

$$ \begin{aligned} T(r, f) &\le T\left(r, \frac{u_3 - u_1}{f - u_3} + 1\right) + o(T(r, f)) \\ &= T\left(r, \frac{f - u_1}{f - u_3}\right) + o(T(r, f)). \end{aligned} $$

Now, $T(r, \frac{u_2-u_1}{u_2-u_3}) = o(T(r, f))$. Consequently, by writing $\frac{f-u_1}{f-u_3} = \phi(\frac{u_2-u_1}{u_2-u_3})$, we have $T(r, \frac{f-u_1}{f-u_3}) \le T(r, \phi) + T(r, \frac{u_2-u_1}{u_2-u_3}) \le T(r, \phi) + o(T(r, f))$ and finally $T(r, f) \le T(r, \phi) + o(T(r, f))$. Thus, by (1), we obtain

$$ (2) \quad \frac{T(r, f)}{q^t} \le \tilde{Z}(r, \phi) + \tilde{Z}(r, \phi - 1) + \tilde{N}(r, \phi) + o(T(r, f)). $$
Now, we can check that

$$ 
\begin{aligned}
\tilde{Z}(r, \phi) + \tilde{Z}(r, \phi - 1) + \tilde{N}(r, \phi) & \\
& \leq \sum_{j=1}^{3} \tilde{Z}(r, f - u_j) + \sum_{1 \leq j < k \leq 3} \tilde{Z}(r, u_k - u_j) \\
& \leq \sum_{j=1}^{3} \tilde{Z}(r, f - u_j) + o(T(r, f))
\end{aligned}
$$

which, by (2), completes the proof. □

Similar to Theorem 47.10, we can derive Theorem 58.14.

**Theorem 58.14.** Let $f \in \mathcal{M}(\mathbb{K})$ (respectively $f \in \mathcal{M}_u(d(0, R^-))$), let $u_1, u_2 \in \mathcal{M}_f(\mathbb{K})$ (respectively $u_1, u_2 \in \mathcal{M}_f(d(0, R^-))$) be distinct, and let $t$ be the ramification index of $\frac{f(x)-u_1(x)}{f(x)-u_2(x)}$. Then,

$$ \frac{T(r, f)}{q^t} \leq \tilde{Z}(r, f - u_1) + \tilde{Z}(r, f - u_2) + \tilde{N}(r, f) + o(T(r, f)). $$

**Proof.** Let $g = \frac{1}{f}$, $w_j = \frac{1}{u_j}$, $j = 1, 2$, $w_3 = 0$. Clearly,

$$ T(r, g) = T(r, f) + O(1), \quad T(r, w_j) = T(r, u_j), \quad j = 1, 2, $$

so we can apply Theorem 58.13 to $g$, $w_1$, $w_2$, $w_3$. On the other hand, we note that

$$ \frac{(g(x) - w_1(x))w_2(x)}{(g(x) - w_2(x))w_1(x)} = \frac{f - u_1}{f - u_2}. $$

Thus, we have $\frac{T(r,g)}{q^t} \leq \tilde{Z}(r,g - w_1) + \tilde{Z}(r,g - w_2) + \tilde{Z}(r,g) + o(T(r,g))$.

But $\tilde{Z}(r,g - w_j) = \tilde{Z}(r, f - u_j)$ for $j = 1, 2$ and $\tilde{Z}(r,g) = \tilde{N}(r, f)$. Moreover, we know that $o(T(r,g)) = o(T(r,f))$. Consequently, the claim is proved when $u_1 u_2$ is not identically zero. Now, we can generalize exactly like in Theorem 47.10. □
Next, by setting $g = f - u_1$ and $u = u_2 - u_1$, we can write Corollary 58.15:

**Corollary 58.15.** Let $g \in \mathcal{M}(\mathbb{K})$ (respectively $g \in \mathcal{M}_u(d(0, R^-))$), let $u \in \mathcal{M}_g(\mathbb{K})$ (respectively $u \in \mathcal{M}_g(d(0, R^-))$), and let $t$ be the ramification index of $\frac{g-u}{g}$.

Then, $\frac{T(r,g)}{q^t} \le \tilde{Z}(r,g) + \tilde{Z}(r,g-u) + \tilde{N}(r,g) + o(T(r,g))$.

**Corollary 58.16.** Let $f \in \mathcal{A}(\mathbb{K})$ (respectively $f \in \mathcal{A}_u(d(0, R^-)$) and let $u_1, u_2 \in \mathcal{A}_f(\mathbb{K})$ (respectively $u_1, u_2 \in \mathcal{A}_f(d(0, R^-)$) be distinct and let $t$ be the ramification index of $\frac{f-u_1}{f-u_2}$. Then, $\frac{T(r,f)}{q^t} \le \bar{Z}(r, f - u_1) + \bar{Z}(r, f - u_2) + o(T(r, f))$.

**Corollary 58.17.** Let $f \in \mathcal{A}(\mathbb{K})$ (respectively $f \in \mathcal{A}_u(d(0, R^-)$) and let $u \in \mathcal{A}_f(\mathbb{K})$ (respectively $u \in \mathcal{A}_f(d(0, R^-)$) be non-identically zero and let $t$ be the ramification index of $\frac{f-u}{f}$. Then, $\frac{T(r,f)}{q^t} \le \bar{Z}(r, f) + \bar{Z}(r, f-u) + o(T(r, f))$.
Chapter 59

Strong Uniqueness and URSCM
in Characteristic *p*

**Notation.** Throughout Chapter 54, $\mathbb{K}$ is a field of characteristic $p \neq 0$, $\alpha$ belongs to $\mathbb{K}$, and $R$ belongs to $]0, +\infty[$. We try to generalize the results of Chapters 52 and 53 when the characteristic is $p$. Many results come from [22] and [23].

**Lemma 59.1.** Let $P(x) = x^n - x^m + k$ with $m < n$ and $k \neq 0$. Let $f, g \in A(\mathbb{K})\backslash\mathbb{K}$ (respectively $f, g \in A_u(d(\alpha, R^{-}))$) satisfy $P(f) = \lambda P(g)$ with $\lambda \in \mathbb{K}^*$. Then $f$ and $g$ have the same ramification index.

**Proof.** Let $s$ be the ramification index of $f$ and let $t$ be this of $g$. We can obviously suppose $s \le t$. Let $f_s = {}^p\sqrt[f]{f}$, $g_s = {}^p\sqrt[g]{g}$, let $k_s = {}^p\sqrt[k]{k}$, let $P_s(x) = x^n - x^m + k_s$, and let $\lambda_s = {}^p\sqrt[\lambda]{\lambda}$. Then $f_s, g_s$ satisfy $P_s(f_s) = \lambda_s P_s(g_s)$, hence $f'_s(n(f_s)^{n-1} - m(f_s)^{m-1}) = g'_s(n(g_s)^{n-1} - m(g_s)^{m-1})$. Since both $f_s, g_s$ belong to $A(\mathbb{K})\backslash\mathbb{K}$ (respectively to $A_u(d(\alpha, R^{-}))$, $n(f_s)^{n-1} - m(f_s)^{m-1}$, $n(g_s)^{n-1} - m(g_s)^{m-1}$ are not identically zero. Now, by definition of the ramification index, $f'_s$ is not identically zero, hence neither is $g'_s$ and therefore $t = s$. $\square$

Lemma 59.12 is a generalization of Lemma 57.6 when $p \neq 0$:

**Lemma 59.2.** Let $P(x) = x^n - x^m + k$ with $k \neq 0$, $m < n$, $n \ge 3$, g.c.d.($m, n$) = 1 and further, $m, n$ satisfy one of the following two conditions:

(A) $n$ is not a power of $p$.

(B) $(m-2)(n-m) \ge n-1$ (respectively $(m-2)(n-m) \ge n$).
Let $f, g \in \mathcal{A}(\mathbb{K})\mathbb{K}$ (respectively $f, g \in \mathcal{A}_u(d(\alpha, R^{-}))$) satisfy
$P(f) = P(g)$. Then, $f = g$.

**Proof.** If $f, g$ belong to $\mathcal{A}_u(d(\alpha, R^{-}))$, we can obviously suppose
$\alpha = 0$. Now, set $n = p^t q$ with g.c.d.($q, p$) = 1. Let $h = \frac{f}{g}$ and
suppose $h \neq 1$. Then, $g^{n-m} = \frac{h^{m-1}}{h^n-1}$. As in Lemma 57.6, we note
that if $f, g$ belong to $\mathcal{A}_u(d(0, R^{-}))$, then $h$ belongs to $M_b(d(0, R^{-}))$
because $M_b(d(0, R^{-}))$ is a field.

Let $G$ be the group of $n$th roots of 1 and let $G'$ be the group of
$m$th roots of 1. Since $\text{g.c.d.}(n,m) = 1$, $G\backslash G'$ contains no element
different from 1.

Suppose that Condition (A) is satisfied. Let $n = qp^t$ with $\text{g.c.d.}(q,p) = 1$ and $q > 1$. Suppose first $q > 2$. Then $G\backslash G'$ contains at least two different elements $u_1$ and $u_2$, hence $h$ must take at least one of these two values, a contradiction as we just saw. Consequently, we must have $q = 2$ and hence $p \neq 2$. Since $n \ge 3$, $n$ is of the form $2p^t$ and $G$ has two elements: $1$ and $-1$. Therefore, each zero of $h^n - 1 = (h^2 - 1)p^t$ is of order $p^t$. Since $-1 \notin G'$, $h(x)$ cannot take the value $-1$. Hence, $h(x)$ must take the value $1$ at a point $\beta \in \mathbb{K}$ (respectively $\beta \in d(0, R^{-})$) and then $\beta$ is a zero of order $p^t$ for $h^n - 1$. But since $\text{g.c.d.}(n,m) = 1$, it is a zero of order $1$ for $h^m - 1$, and hence $g^{n-m}$ admits $\beta$ as a pole of order $p^t - 1$, a contradiction again. Thus, Condition (A) implies that $h$ is identically $1$.

Now, suppose that Condition (B) is satisfied. Since we don't need Condition (A), we can assume that $n = p^t$. If $p$ divides $n-m$, it divides neither $m$ nor $n$ because $\text{g.c.d.}(n,m) = 1$. Consequently, we can assume that $p$ divides neither $m$ nor $n-m$. Let us first assume that $f'g'$ is not identically zero. Since $n-m$ is prime to $p$, we can check that the derivative of $\frac{h^{m-1}}{h^n-1} = g^{n-m}$ is not identically zero and therefore $h'$ is not identically zero.

Now, since $\text{g.c.d.}(m,p) = \text{g.c.d.}(m,n) = 1$, $G'$ has $m$ elements and $G'\backslash G$ has $m-1$ elements $u_1, \dots, u_{m-1}$. Then, for each zero $\beta$ of $h-u_i$ ($1 \le i \le m-1$), $\beta$ is a zero of $h^m-1$ and is not a zero of $h^n-1$. Hence, it is a zero of $g^{n-m}$, and consequently, it is a zero of order at least $n-m$ of $h-u_i$. Thus, for each $i=1, \dots, m-1$, we have

$$
(1) \quad \tilde{Z}(r, h - u_i) \le \frac{1}{n-m} Z(r, h - u_i).
$$
Now, let us apply Theorem 58.11 to $h$. Since $h'$ is not identically zero, we have

$$ (m-2)T(r, h) \leq \sum_{i=1}^{m-1} \tilde{Z}(r, h - u_i) + N(r, h) - \log r + O(1), $$

hence by (1), we have

$$ (2) \quad (m-2)T(r, h) \leq \left( \frac{m-1}{n-m} + 1 \right) T(r, h) - \log r + O(1). $$

Therefore, if $f, g$ belong to $\mathcal{A}(\mathbb{K})$, by (2), that yields $m-2 < \frac{n-1}{n-m}$, and if $f, g$ belong to $\mathcal{A}_u(d(0, R^{-}))$, we have $m-2 \leq \frac{n-1}{n-m}$, both conditions excluded by hypotheses. Consequently, $h=1$ and therefore $f=g$.

Suppose now that the ramification index of $f$ is $s$. By Lemma 59.11, the ramification index of $g$ is $s$ too. Let $k_s = \sqrt[s]{k}$, and let $P_s(x) = x^n - x^m + k_s$, $f_s = \sqrt[s]{f}$, $g_s = \sqrt[s]{g}$. Then, $P_s(f_s) = P_s(g_s)$, and we are taken back to the same problem with $f_s$ and $g_s$ that satisfy $(f_s)'(g_s)'$ not identically zero. So, we have $f_s = g_s$ and therefore $f = g$. $\square$

**Proposition 59.3.** Let $P(x) = x^n - x^m + k$ with $2 \leq m < n$ (respectively $3 \leq m < n$), g.c.d.($m, n$) = 1 and $a \neq 0$. Assume that $m^m(n-m)^{(n-m)} \neq n^n(1-u)^{(n-m)}(\sqrt[s]{k})^{n-m}$ for every $s \in \mathbb{N}$ and for every $u \in \mathbb{K}$ such that $u^{n-m} = (-1)^{n-m}$. Next, we assume that one of the following two conditions (a) and (b) is satisfied:

(a) $n$ is not a power of $p$,

(b) $(m-2)(n-m) \geq n-1$ (respectively $(m-2)(n-m) \geq n$).

Let $f, g \in \mathcal{A}(\mathbb{K})$ (respectively for $f, g \in \mathcal{A}_u(d(\alpha, R^{-}))$) be such that $P(f) = \lambda P(g)$. Then, $\lambda = 1$.

**Proof.** If $f, g \in \mathcal{A}_u(d(\alpha, R^{-}))$, we can assume $\alpha = 0$. By Lemma 57.3, $P$ has $n$ distinct zeros, hence its set of zeros $S$ has $n$ elements. Suppose there exists $\lambda \in \mathbb{K}^*$ such that $P(f) = \lambda P(g)$.

If $f'g'$ is not identically zero, we can make a proof that is very similar to that of in Proposition 57.4.
Suppose first that $f'g'$ is not identically zero. We prove that $\lambda = 1$. Indeed, suppose $\lambda \neq 1$. Let $Q(x) = x^n - x^m + k - \frac{1}{\lambda}$. Then,

$$ (1) \qquad Q(g) = P(g) - \frac{1}{\lambda} = \frac{1}{\lambda}(\lambda P(g) - 1) = \frac{1}{\lambda}(P(f) - 1) = \frac{1}{\lambda}(f^n - f^m). $$

Similarly, let $R(x) = P(x) - \lambda$. Then, $R(f) = P(f) - \lambda = \lambda(\lambda^{-1}P(f) - 1) = \lambda(P(g) - 1) = \lambda(g^n - g^m)$. Now, by Lemma 57.3, at least one of the two polynomials $Q, R$ has all its zeros distinct. Without loss of generality, we may assume that $Q$ has all its zeros distinct. So, $Q(x)$ is of the form $\prod_{i=1}^n (x - e_i)$ with $e_i \neq e_j \forall i \neq j$. Then by (1), we have $f^n - f^m = \lambda \prod_{i=1}^n (g - e_i)$. Consequently, by Lemma 58.6, we have $Z(r, Q(g)) - \tilde{Z}(r, Q(g)) = \sum_{j=1}^n Z(r, g - e_j) - \tilde{Z}(r, g - e_j) = Z(r, f^n - f^m) - \tilde{Z}(r, f^n - f^m)$ and of course $Z(r, f^n - f^m) - \tilde{Z}(r, f^n - f^m) \geq (m-1)Z(r, f)$, hence

$$ (2) \quad Z(r, Q(g)) - \tilde{Z}(r, Q(g)) \geq (m-1)Z(r, f). $$

On the other hand, since $f'g'$ is not identically zero, by Corollary 58.12, we have $Z(r, f^n - f^m) - \tilde{Z}(r, f^n - f^m) \leq T(r, f) - \log(r) + O(1)$, hence $Z(r, Q(g)) - \tilde{Z}(r, Q(g)) \leq T(r, g) - \log r + O(1)$, and hence by (2), we obtain

$$ (3) \quad (m-1)Z(r, f) \leq T(r, g) - \log r + O(1). $$

But by Theorem 43.8, we have $T(r, P(f)) = nT(r, f) + O(1)$, and by (1), we have $T(r, P(g)) = +O(1) = nT(r, g) + O(1)$, hence $T(r, f) = T(r, g) + O(1)$, and hence by (3), we have $(m-1)Z(r, f) \leq T(r, f) - \log r + O(1)$. By Theorem 43.3, this yields $(m-1)T(r, f) \leq T(r, f) - \log r + O(1)$ and therefore $m < 2$ when $f, g$ belong to $\mathcal{A}(\mathbb{K})$. And if $f, g$ belong to $\mathcal{A}(d(0, \mathbb{R}^-))$, then $m \leq 2$. Consequently, since $m \geq 2$ (respectively $m \geq 3$), we have proven in both cases that $\lambda = 1$.

Suppose now that $f' = 0$ and let $s$ be the ramification index of $f$. Then by Lemma 59.1, this of $g$ is also $s$. Putting $k_s = \sqrt[s]{k}$, $P_s(x) = x^n - x^m + k_s$, $f_s = \sqrt[s]{f}$, $g_s = \sqrt[s]{g}$, and $\lambda_s = \sqrt[s]{\lambda}$, now we have $P(f_s) = \lambda_s P(g_s)$ and we are taken back to the same situation with $P_s, f_s, g_s$. On one hand, we note that the zeros of $P_s$ are the $s$th roots of those of $P$. Hence, $P_s$ also admits $n$ distinct zeros. Next, the condition $m^m(n - m)^{(n-m)}(\sqrt[s]{k})^{n-m} \neq n^n(1 - u)^{(n-m)}$ for every $s \in \mathbb{N}$ and for every $u \in \mathbb{K}$ such that $u^{n-m} = (-1)^{n-m}$ also applies in
the same way to $P_s$. Consequently, we can generalize the reasoning
to the case when the ramification index of $f$, $g$ is $s$. That ends the
proof. $\square$

By Lemma 59.2 and Proposition 59.3, we have Corollary 59.4:

**Corollary 59.4.** Let $P(x) = x^n - x^m + k$ with $m < n$ and $k \neq 0$, $2 \le m < n$ (respectively $3 \le m < n$), g.c.d.($m, n$) = 1, $m^m(n - m)^{(n-m)} \neq n^n(1-u)^{(n-m)}(\sqrt[k]{u})^{n-m}$ for every $s \in \mathbb{N}$ and for every $u \in \mathbb{K}$ such that $u^{n-m} = (-1)^{n-m}$. Assume that one of the following two conditions (a) and (b) is satisfied:

(a) $n$ is not a power of $p$.

(b) $(m-2)(n-m) \ge n-1$ (respectively $(m-2)(n-m) \ge n$).

Then $P$ is a polynomial of strong uniqueness for $\mathcal{A}(\mathbb{K})$ (respectively for $\mathcal{A}_u(d(\alpha, R^{-})))$. The set of zeros of $P$ is an urscm for $\mathcal{A}(\mathbb{K})$.

**Proposition 59.5.** Let $P(x) = x^n - x^m + k$ with $m < n$, $k \neq 0$ and let $f, g \in \mathcal{M}(K)\K$ (respectively let $f, g \in \mathcal{M}_u(d(0, R^{-})))$ be such that $P(f) = \lambda P(g)$, with $\lambda \in \K^*$. If $f, g \in \mathcal{M}(K)\K$ and $2m \ge n+3$, or if $f, g \in \mathcal{M}_u(d(0, R^{-}))$ and $2m \ge n+4$, then $\lambda = 1$. Moreover, if $f, g \in \mathcal{A}(K)\K$ and $2m \ge n+2$ or if $f, g \in \mathcal{A}_u(d(0, R^{-}))$ and $2m \ge n+3$, then $\lambda = 1$.

**Proof.** The proof is the same as that of Theorem 57.5, just by
replacing $\mathbb{Z}$ and $\mathbb{N}$ by $\tilde{\mathbb{Z}}$ and $\tilde{\mathbb{N}}$, respectively. $\square$

**Theorem 59.6.** Let $\mathbb{K}$ have characteristic $p \neq 2$ and let $P(x) = x^n - x^{n-2} + k$ with $k \neq 0$, $n = p^t \ge 7$ (respectively $n = p^t \ge 8$) with $t \in \mathbb{N}^*$ such that $(n-2)^{n-2} \neq n^k k^2$.

Then $P$ is a polynomial of strong uniqueness for $\mathcal{A}(\mathbb{K})$ (respectively for $\mathcal{A}_u(d(\alpha, R^{-})))$. Moreover, the set of zeros of $P$ has $n$ elements and is an urscm for $\mathcal{A}(\mathbb{K})$.

**Proof.** In order to show that we can apply Corollary 58.4, we show first that $m^m(n-m)^{(n-m)} \neq n^n(1-u)^{(n-m)}(\sqrt[k]{u})^{n-m}$, which yields

(1) $(n-2)^{n-2} \neq n^n(1-u)^2(\sqrt[k]{u})^{n-m}$ for every $s \in \mathbb{N}^*$ and for every $u \in \mathbb{K}$ such that $u^2 = (-1)^2$ i.e. for $u=1$ or $u=-1$.
Next, we need to have $(m-2)(n-m) \ge n-1$ (respectively $(m-2)(n-m) \ge n$) i.e.

(2) $2(n-4) \ge n-1$, which is trivial when $u=1$ and becomes

(3) $4(n-2)^{n-2} \ne 4n^n(\sqrt[n]{k})^2$ when $u=-1$.

On the other hand, since $b^p = b \forall b \in \mathbb{N}$, we can see that (3) is equivalent to

$$ (4) \quad (n-2)^{n-2} \ne n^k k^2. $$

Now, we note that by hypothesis, either $n = p = 7$ with $t = 1$ or $n \ge 8$. Finally, (2) is equivalent to $2(n-4) \ge n-1$ (respectively $2(n-4) \ge n$). Thus, (1) and (2) are satisfied and therefore Corollary 59.4 shows the conclusion. $\square$

**Theorem 59.7.** Let $\mathbb{K}$ have characteristic 2, let $n = 2^s$, $s \ge 3$, and let $P(x) = x^n - x^{n-3} + k$ with $k \ne 0$.

Then $P$ is a polynomial of strong uniqueness for $\mathcal{A}(\mathbb{K})$ and for $\mathcal{A}_u(d(\alpha, R^-))$. Moreover, the set of zeros of $P$ has $n$ elements and is an urscm for $\mathcal{A}(\mathbb{K})$.

**Proof.** Indeed, in characteristic 2, all hypotheses of Theorem 59.7 involving $n^n$ are trivially satisfied since in $\mathbb{K}$, $n = 2^s = 0$. Next, the hypotheses $n = 2^s$, $m = n-3$, and $s \ge 3$ imply $(m-2)(n-m) \ge n$. $\square$

Theorem 59.8 shows that in characteristic $p \ne 0$, an affinely rigid set is not necessarily an urscm for $\mathcal{A}(\mathbb{K})$ and even for polynomials.

**Theorem 59.8.** Let $\mathbb{E}$ be an algebraically closed field of characteristic 3. There exists no subset of three elements that is an urscm for $\mathbb{E}[x]$.

**Proof.** Let $S$ be a subset of $\mathbb{E}$ of 3 elements $a_1, a_2, a_3$. We can find an affine change of variable $h$ such that the images $e_i = h(a_i)$ are the zeros of a monic polynomial $P(Y)$ of the form $Y^3 + bY + c$. Then $h(x)$ is of the form $x+m$ and therefore $(h^{-1})(x) = x-m$. Let $T = \{e_1, e_2, e_3\}$. Now, let $\alpha \in \mathbb{E}$ satisfy $\alpha^2 = -b$ and consider $f(x) = x$ and $g(x) = x+\alpha$. Then we can check that $P(f) = P(g)$.
Strong Uniqueness and URSCM in Characteristic p

601

Indeed, $P(g(x)) = (x + \alpha)^3 + b(x + \alpha) + c = x^3 + \alpha^3 + bx + c\alpha + c = P(x) + \alpha^3 + b\alpha = P(f(x))$ since $\alpha^3 + b\alpha = 0$. Now, let $\phi = (\frac{1}{h}) \circ f$ and $\theta = (\frac{1}{h}) \circ g$. Then,

$$ 
\begin{aligned}
P(f(x)) &= \prod_{j=1}^{3} (f(x) - e_j) = \prod_{j=1}^{3} (h(\phi(x) - h(a_j))) \\
&= \prod_{j=1}^{3} (\phi(x) + m - (a_j + m)) = \prod_{j=1}^{3} (\phi(x) - a_j).
\end{aligned}
$$

Let $Q(Y) = \prod_{j=1}^{3}(Y - a_j)$. Then, $P(f(x)) = Q(\phi(x))$ and similarly $P(g(x)) = Q(\theta(x))$. Consequently, $Q(\phi) = Q(\theta)$. Thus, $Q$ is not a polynomial of uniqueness for $\mathbb{E}[x]$, and hence, by Corollary 56.3, $S$ is not an urscm for $\mathbb{E}[x]$. $\square$
# Chapter 60

## The Functional Equation

$$P(f) = Q(g)$$

Throughout this chapter, the field $K$ is supposed to have characteristic $p \ge 0$ and characteristic exponent $q$. $R$ is a strictly positive number.

Most of results come from [46] and [64].

In the sequel, we use the following basic lemmas:

**Lemma 60.1.** Let $\mathbb{E}$ be a field, and let $P \in \mathbb{E}[x]$ satisfy $\deg(P) = 3$ and be such that $P'$ has two distinct zeros $c_1, c_2$. Then, $P(c_1) \neq P(c_2)$.

**Proof.** Suppose $P(c_1) = P(c_2)$. Without loss of generality, we can suppose $P(c_1) = P(c_2) = 0$. Then $P$ admits two distinct zeros of order $\ge 2$, a contradiction since $\deg(P) = 3$. $\square$

<figure><img src="image_25.png" /></figure>

Henceforth, we look for general results linked to the property $P(a) \neq Q(b)$, where $a$ and $b$ are zeros of $P'$ and $Q'$, respectively. In order to state and prove our main results (Theorems 60.4 and 60.12), we need to recall the notion of the so-called Nevanlinna characteristic function and its associated functions. However, here, for the sake of convenience, we define counting functions of zeros and poles without multiplicities which, in characteristic $p \neq 0$, are the same as in characteristic 0, contrary to some other results that require a more accurate form. Moreover, we use the ramification index for analytic functions, defined in Chapter 58.
**Lemma 60.2.** Let $f, g$ be in $\mathcal{M}(d(0, R^{-}))$, let $P, Q$ be in $\mathbb{K}[x]$, and assume that $f$ and $g$ satisfy $P(f) = Q(g)$. Then $f$ belongs to $\mathcal{M}_b(d(0, R^{-}))$ if and only if so does $g$.

**Proof.** By Theorem 43.10, a function $h \in \mathcal{M}(d(0, R^{-}))$ belongs to $\mathcal{M}_b(d(0, R^{-}))$ if and only if $T(r, h)$ is bounded in $]0, R[$. And by Corollary 43.12, $P(h)$ belongs to $\mathcal{M}_b(d(0, r^{-}))$ if and only if so does $h$. Consequently, the claim is immediate. $\square$

**Lemma 60.3.** Let $P, Q \in \mathbb{K}[x]$ satisfy $P'Q' \neq 0$. Let $f, g \in \mathcal{M}(\mathbb{K})$ satisfy $P(f) = Q(g)$. Then, $\text{ram}(f) = \text{ram}(g)$ and $P_t(\sqrt[q]{f}) = Q_t(\sqrt[q]{g})$.

**Proof.** Clearly, $f'P'(f) = g'Q'(g)$. Since $P'Q' \neq 0$, either $f' = g' = 0$ or $f'g' \neq 0$. Suppose $f' = g' = 0$. By Proposition 58.2, there exist $f_1, g_1 \in \mathcal{M}(\mathbb{K})$ such that $(f_1)^q = f, (g_1)^q = g$. Then, we have $P_1(f_1) = Q_1(g_1)$ and therefore we are led to the same problem with $f_1$ and $g_1$. Thus, by induction, after $t$ similar operations, we finally obtain $P_t(\sqrt[q]{f}) = Q_t(\sqrt[q]{g})$ with $\sqrt[q]{f}, \sqrt[q]{g} \in \mathcal{M}(\mathbb{K})$ and $(\sqrt[q]{f})'(\sqrt[q]{g})' \neq 0$ and of course $t = \text{ram}(f) = \text{ram}(g)$. $\square$

**Theorem 60.4.** Let $P, Q$ be in $\mathbb{K}[x]$ with $P'Q'$ not identically zero and let $m = \text{deg}(P)$, $n = \text{deg}(Q)$ satisfy $2 \le \min(m, n)$. Assume that there exist $k$ distinct zeros $c_1, \dots, c_k$ of $P'$ such that $P(c_i) \neq P(c_j) \ \forall i \neq j$ and $P(c_i) \neq Q(\ell)$ for every zero $\ell$ of $Q'$ ($i = 1, \dots, k$). Assume that there exist two functions $f, g \in \mathcal{M}_u(d(a, R^{-}))$ such that $P(f) = Q(g)$ and let $t = \text{ram}(f)$. Then, $\tilde{N}(r, f) \ge \frac{T(r, f)(kn-m)}{q^t n} + O(1)$. Moreover, suppose $\frac{2m}{3} < n$. Then, $k \le 2$. Further, if $m \neq n$, then $k = 1$, $c_1$ is a simple zero of $P'$ and either $n < m$ or $\text{g.c.d.}(m, n) = n - m$.

**Remark.** In Corollaries 60.5–60.10, when we assume that one of the functions $f$ and $g$ belongs to $\mathcal{M}_u(d(a, R^{-}))$, then by Corollary 59.2, both $f$ and $g$ belong to $\mathcal{M}_u(d(a, R^{-}))$.

**Corollary 60.5.** Let $P, Q$ be in $\mathbb{K}[x]$ with $P'Q'$ not identically 0 and such that $\text{deg}(P) = \text{deg}(Q)$. Assume that there exist two distinct zeros $c_1, c_2$ of $P'$ such that $P(c_1) \neq P(c_2)$ and $P(c_i) \neq Q(\ell)$ for every zero $\ell$ of $Q'$. If two functions $f, g \in \mathcal{A}(d(a, R^{-}))$ satisfy $P(f) = Q(g)$, then $f, g \in \mathcal{A}_b(d(a, R^{-}))$.
The Functional Equation $P(f) = Q(g)$

605

**Proof.** Without loss of generality, we can assume $a = 0$. Let $\deg(P) = \deg(Q)$ and assume that the conclusion is wrong, hence by Lemma 60.2, $f$ and $g$ are not bounded, hence by Theorem 43.10, $\lim_{r \to +\infty} T(r, f) = +\infty$. Now, by Theorem 60.4, we have $\lim_{r \to +\infty} \tilde{N}(r, f) = +\infty$. But since $f \in \mathcal{A}(d(0, R^{-}))$, this is absurd. $\square$

**Corollary 60.6.** Let $P, Q$ be in $\mathbb{K}[x]$ where $P'Q'$ is not identically zero, and let $m = \deg(P)$, $n = \deg(Q)$ with $3 \le m$ and $n \ge \min(2, \frac{m}{2})$. Assume that there exist two distinct zeros $c_1, c_2$ of $P'$ such that $P(c_i) \ne Q(\ell)$ for every zero $\ell$ of $Q'$ ($i = 1, 2$). Assume further that, if $m \ge 4$, then $P(c_1) \ne P(c_2)$. If two functions $f, g \in \mathcal{A}(d(a, R^{-}))$ satisfy $P(f) = Q(g)$, then $f, g \in \mathcal{A}_b(d(a, R^{-}))$.

**Proof.** Without loss of generality, we can assume $a = 0$. If $m = 3$, by Lemma 60.1, we have $P(c_1) \ne P(c_2)$, hence that inequality is satisfied anyway. So, we can apply Theorem 60.4 with $k = 2$ which shows that $\tilde{N}(r, f)$ is unbounded (because $2\deg(Q) - m > 0$), a contradiction since $f \in \mathcal{A}(d(0, R^{-}))$. $\square$

**Corollary 60.7.** Let $P, Q$ be in $\mathbb{K}[x]$ with $P'Q'$ not identically zero and $\deg(P) < \deg(Q)$. Assume that there exists a zero $c$ of $P'$ such that $P(c) \ne Q(\ell)$ for every zero $\ell$ of $Q'$. If two functions $f, g \in \mathcal{A}(d(a, R^{-}))$ satisfy $P(f) = Q(g)$, then $f, g \in \mathcal{A}_b(d(a, R^{-}))$.

**Proof.** Without loss of generality, we can assume $a = 0$. Let $m = \deg(P)$, $n = \deg(Q)$ and assume that $f$ or $g$ is not bounded. By Lemma 60.2, both $f$ and $g$ are not bounded. We can apply Theorem 59.4 with $k = 1$, $m < n$, which shows that $\tilde{N}(r, f)$ is unbounded, a contradiction since $f \in \mathcal{A}(d(0, R^{-}))$. $\square$

We now apply Theorem 59.4 to meromorphic functions in $d(a, R^{-})$.

**Corollary 60.8.** Let $P, Q$ be in $\mathbb{K}[x]$ with $P'Q'$ not identically zero and $m = \deg(P)$, $n = \deg(Q)$ satisfying $m \ne n$ and $2m < 3n$. Assume that there exist two distinct zeros $c_1$ and $c_2$ of $P'$ such that $P(c_1) \ne P(c_2)$ and $P(c_i) \ne Q(\ell)$ for every zero $\ell$ of $Q'$ ($i = 1, 2$). If $f, g \in \mathcal{M}(d(a, R^{-}))$ satisfy $P(f) = Q(g)$, then $f, g \in \mathcal{M}_b(d(a, R^{-}))$.

**Proof.** Let $k$ be the number of zeros $c_i$ of $P$ such that $P(c_i) \ne P(c_j) \forall i \ne j$ and $P(c_i) \ne Q(\ell)$ for every zero $\ell$ of $Q'$ ($i = 1, 2$). By Lemma 60.2, if one of the functions $f$ and $g$ belongs to
$\mathcal{M}_u(d(a, R^{-}))$, then both belong to $\mathcal{M}_u(d(a, R^{-}))$. So, if $f$ and $g$ belong to $\mathcal{M}_u(d(a, R^{-}))$, then by Theorem 60.4 and the assumption that $2m < 3n$, we have $k = 1$, a contradiction. $\square$

**Corollary 60.9.** Let $P, Q$ be in $\mathbb{K}[x]$ with $P'Q'$ not identically zero and $3 \le \deg(P) \le 4$, $\deg(Q) \ge 3$, $\deg(P) \ne \deg(Q)$. Assume that there exist two distinct zeros $c_1$ and $c_2$ of $P'$ such that $P(c_i) \ne Q(\ell)$ for every zero $\ell$ of $Q'$ ($i = 1, 2$). Moreover, if $\deg(P) = 4$, we assume that $P(c_1) \ne P(c_2)$. If $f, g \in \mathcal{M}(d(a, R^{-}))$ satisfy $P(f) = Q(g)$, then $f, g \in \mathcal{M}_b(d(a, R^{-}))$.

**Proof.** Let $m = \deg(P)$, $n = \deg(Q)$. We note that the inequality $P(c_1) \ne P(c_2)$ is assured by Lemma 59.1 when $m = 3$ and is assumed when $m = 4$ and hence holds in all cases. Assume that one of the two functions $f, g$ belongs to $\mathcal{M}_u(d(a, R^{-}))$, hence by Lemma 60.2, both $f$ and $g$ belong to $\mathcal{M}_u(d(a, R^{-}))$. By Theorem 60.4, we have $k = 1$, a contradiction to the hypothesis. $\square$

**Corollary 60.10.** Let $P, Q$ be in $\mathbb{K}[x]$ with $P'Q'$ not identically 0 and $m = \deg(P)$, $n = \deg(Q)$, satisfying $m < n$ and $n - m \ne$ g.c.d.($m, n$). Assume that there exists a zero $c$ of $P'$ such that $P(c) \ne Q(\ell)$ for every zero $\ell$ of $Q'$. If $f, g \in \mathcal{M}(d(a, R^{-}))$ satisfy $P(f) = Q(g)$, then $f, g \in \mathcal{M}_b(d(a, R^{-}))$.

**Proof.** Assume that one of the two functions $f, g$ belongs to $\mathcal{M}_u(d(a, R^{-}))$, hence by Lemma 60.2, both $f$ and $g$ belong to $\mathcal{M}_u(d(a, R^{-}))$. Since $m \ne n$, by Theorem 60.4, we have that either $n < m$ or $\text{g.c.d.}(m, n) = n - m$, both cases that are excluded from the hypotheses of the corollary. Hence, both $f$ and $g$ belong to $\mathcal{M}_b(d(a, R^{-}))$. $\square$

**Corollary 60.11.** Let $P, Q$ be in $\mathbb{K}[x]$ with $P'Q'$ not identically zero and $\deg(P) = \deg(Q) \ge 4$. Assume that there exist three distinct zeros $c_1, c_2, c_3$ of $P'$ such that $P(c_i) \ne P(c_j) \forall i \ne j$ and $P(c_i) \ne Q(\ell)$ for every zero $\ell$ of $Q'$ ($i = 1, 2, 3$). If $f, g \in \mathcal{M}(d(a, R^{-}))$ satisfy $f' \ne 0$ and $P(f) = Q(g)$, then $f, g \in \mathcal{M}_b(d(a, R^{-}))$.

**Proof.** Without loss of generality, we can assume $a = 0$. Let $n = \deg(P) = \deg(Q)$ and assume that $f$ or $g$ belongs to $\mathcal{M}_u(d(a, R^{-}))$, hence by Lemma 60.2, both $f$ and $g$ belong to $\mathcal{M}_u(d(a, R^{-}))$.
The Functional Equation $P(f) = Q(g)$

607

Since $f' \neq 0$, we have $\mathrm{ram}(f) = 0$, hence by Theorem 60.4, $\tilde{N}(r, f) \ge T(r, f)(\frac{3n-n}{n}) + O(1)$, therefore $\lim_{r \to +\infty} \tilde{N}(r, f) - T(r, f) = +\infty$, which is absurd. □

**Theorem 60.12.** Let $P, Q \in K[x]$ with $P'Q'$ not identically zero and let $m = \deg(P)$, $n = \deg(Q)$ with $2 \le \min(m, n)$. Assume that there exist $k$ distinct zeros $c_1, \dots, c_k$ of $P'$ such that $P(c_i) \ne P(c_j) \forall i \ne j$ and $P(c_i) \ne Q(\ell)$ for every zero $\ell$ of $Q'$ ($i = 1, \dots, k$). Assume that there exist two non-constant functions $f, g \in \mathcal{M}(K)$ such that $P(f) = Q(g)$ and let $t = \mathrm{ram}(f)$. Then, $q \le p$ and $f$ satisfies $\tilde{N}(r, f) \ge \frac{T(r, f)(kq-p)}{q^t q} + \log r + O(1)$. Moreover, if $\frac{p}{2} < q$, then $k=1$ and $c_1$ is a simple zero of $P'$.

**Corollary 60.13.** Let $P, Q$ be in $K[x]$ with $P'Q'$ not identically zero and $\deg(P) \le \deg(Q)$. Assume that there exists a zero $c$ of $P'$ such that $P(c) \ne Q(\ell)$ for every zero $\ell$ of $Q'$. If there exist $f, g \in \mathcal{A}(K)$ satisfying $P(f) = Q(g)$, then $f$ and $g$ are constants.

**Proof.** Let $m = \deg(P)$, $n = \deg(Q)$. Assume that $f$ and $g$ are not constant. Then by Theorem 60.12, we have $\tilde{N}(r, f) \ge T(r, f)(\frac{n-m}{q^t n}) + \log r + O(1) \ge \log r + O(1)$, hence $\lim_{r \to +\infty} \tilde{N}(r, f) = +\infty$, which is absurd. □

**Corollary 60.14.** Let $P, Q \in K[x]$ with $P'Q'$ not identically zero and let $m = \deg(P)$, $n = \deg(Q)$ with $2 \le \min(m, n)$ and $\frac{m}{2} < n$. Assume that there exist 2 distinct zeros $c_1, c_2$ of $P'$ such that $P(c_1) \ne P(c_2)$ and $P(c_i) \ne Q(\ell)$ for every zero $\ell$ of $Q'$ ($i = 1, 2$). Assume that there exist two functions $f, g \in \mathcal{M}(K)$ such that $P(f) = Q(g)$. Then $f$ and $g$ are constants.

**Proof (Theorems 60.4 and 60.12).** In Theorem 60.4, without loss of generality, we can assume $a=0$. We first assume that both $f, g$ have a ramification index equal to 0. Let $b = \mathrm{g.c.d.}(m, n)$ and $m = b\hat{m}, n = b\hat{n}$. We can clearly assume that $a=0$ and that none of the meromorphic functions to be considered attains 0 or $\infty$ at 0. We denote by $I$ (respectively $J$) an interval of the form $[u, +\infty[$ (respectively $[u, \log r]$).

Clearly, $mT(r, f) = nT(r, g) + O(1)$, hence

$$ (1) \quad \hat{m}T(r, f) = \hat{n}T(r, g) + O(1). $$
Let $\alpha$ be a pole of $f$ of order $k$. Then $\alpha$ is a pole of $g$ of order $l$ such that $k\hat{m} = l\hat{n}$, hence $\hat{n}$ divides $k$, therefore

$$ (2) \quad N(r, f) \geq \hat{n}\tilde{N}(r, f). $$

Since $c_i$ is a zero of $P'$, $P - P(c_i)$ has a factorization of the form $(x - c_i)^{s_i}R_i(x)$, with $R_i \in \mathbb{K}[x]$, $R_i(c_i) \neq 0$. Consequently,

$$ (3) \quad (f - c_i)^{s_i} R_i(f) = Q(g) - P(c_i) \quad (i = 1, \dots, k). $$

We put $S = \sum_{i=1}^t (s_i - 1)$. Since by definition we have $s_i \geq 2 \ \forall i = 1, \dots, t$, then $S \geq t$.

Since, by hypothesis, $Q - P(c_i)$ does not vanish at any zero of $Q'$; this implies that it has no multiple zero and hence it has a factorization in the form $\prod_{j=1}^q (x - b_{i,j})$, where the $b_{i,j}$ are all distinct for each fixed $i$. It follows that

$$ (4) \quad \tilde{Z}(r, Q - P(c_i)) = \sum_{j=1}^{w} \tilde{Z}(x - b_{i,j}) \quad (i = 1, \dots, t). $$

On the other hand, we note that $b_{i,j} \neq b_{k,l}$ whenever $(i, j) \neq (k, l)$. Indeed, suppose that $b_{i,j} = b_{k,l}$ for some $(i, j) \neq (k, l)$. Then $i \neq k$, hence $P(c_i) \neq P(c_k)$, therefore $(Q - P(c_i)) - (Q - P(c_k))$ is a constant different from 0. But since $b_{i,j} = b_{k,l}$, the point $b_{i,j}$ is a zero of $(Q - P(c_i)) - (Q - P(c_k))$, a contradiction. Thus, all the points $b_{i,j}$ are distinct $(i = 1, \dots, t; j = 1, \dots, w)$.

Now, by applying Theorem 58.11 to $g$ at the points $b_{i,j}$, for all $i = 1, \dots, t$ and $j = 1, \dots, w$, we have

$$ (5) \quad (nt - 1)T(r, g) \leq \sum_{i=1}^{t} \sum_{j=1}^{w} \tilde{Z}(r, g - b_{i,j}) + \tilde{N}(r, g) - \log r + O(1). $$

Hence, by (4), we have

$$ (6) \quad (tn - 1)T(r, g) \leq \sum_{i=1}^{t} \tilde{Z}(r, Q - P(c_i)). $$

But by (3) and (6), we obtain

$$ \begin{aligned} (7) \quad & \tilde{Z}(r, Q(g) - P(c_i)) \leq \tilde{Z}(r, f - c_i) + \tilde{Z}(r, R_i(f)) + O(1) \\ & \leq (m - s_i + 1)T(r, f) + O(1) \\ & \leq (m - 1)T(r, f) + O(1), \quad (i = 1, \dots, t). \end{aligned} $$
The Functional Equation $P(f) = Q(g)$

609

On the other hand, it is obvious that $\tilde{N}(r, g) = \tilde{N}(r, f)$. Hence, it follows from (2) that

$$ \tilde{N}(r, g) = \tilde{N}(r, f) \le \frac{1}{\tilde{n}}N(r, f) \le \frac{1}{\tilde{n}}T(r, f). $$

Next, by (5) and (7), we have $\tilde{N}(r, f) \ge (tn - 1)T(r, g) - \sum_{i=1}^{t} \tilde{Z}(r, f - c_i) + \tilde{Z}(r, R_i(f)) + \log r + O(1) \ge (tn - 1)T(r, g) - T(r, f) \sum_{i=1}^{t} (1 + m - s_i) + \log r + O(1) = (tn - 1)T(r, g) - T(r, f) (tm - S) + \log r + O(1)$, hence

$$ (8) \quad \tilde{N}(r, f) \ge \left( \frac{Sn - m}{n} \right) T(r, f) + \log r + O(1). $$

Since $t \le S$, the inequality $\tilde{N}(r, f) \ge \frac{T(r, f)(tn - m)}{n} + O(1)$ is then proven.

Now by (1), we can write $(tn - 1)T(r, g) = (\frac{tn - 1}{n})mT(r, f) + O(1)$, hence by (2), (5), (6), and (7), we obtain

$$ (9) \quad \left( \frac{tn - 1}{n} \right) mT(r, f) \le \left( \sum_{i=1}^{t} (m - s_i + 1) + \frac{1}{\tilde{n}} \right) T(r, f) - \log r + O(1). $$

Therefore, since $f, g$ belong to $M_u(d(0, R^-))$ in Theorem 60.4 and in $M(\mathbb{K})$ in Theorem 60.12, then by Theorem 43.10, (9) implies

$$ (10) \quad (tn - 1)m \le ntm - n \left( \sum_{i=1}^{t} s_i \right) + nt + b. $$

We can then deduce $(nt - 1)\hat{m} \le \hat{ntm} - \hat{n}(\sum_{i=1}^{t} s_i) + \hat{nt} + 1$. But since $n\hat{m} = \hat{nm}$, we have $\sum_{i=1}^{t}(s_i - 1)\hat{n} \le \hat{m} + 1$. It follows that

$$ (11) \quad S\hat{n} \le \hat{m} + 1. $$

Thus, in Theorem 60.4, by (11), we see that either $n \le m$ or $\hat{n} = \hat{m} + 1$, hence g.c.d.($m, n$) = $n - m$ and then $S = 1 = t$.

In Theorem 60.12, both $f$ and $g$ belong to $M(\mathbb{K})$. By (8), we have already shown $\tilde{N}(r, f) \ge (\frac{nt - m}{n})T(r, f) + \log r + O(1)$.
Moreover, since $\log r$ tends to $\infty$ with $r$, by (9), we obtain $(\frac{tn-1}{n})m < \sum_{i=1}^{t}(m-s_i+1)+\frac{1}{n}$, hence $S\hat{n} < \hat{m}+1$, i.e.

$$ (12) \quad S\hat{n} \le \hat{m}. $$

Thus, in Theorem 60.12, by (12), we have $n \le m$.

Now assume the hypotheses of Theorem 60.4 hold and suppose that $m < \frac{3n}{2}$. So we have $\hat{m} < \frac{3\hat{n}}{2}$ in Theorem 60.4. Therefore, by (11), we obtain

$$ (13) \quad \frac{2}{3}S\hat{m} < \hat{m}+1, $$

hence $S \le 2$.

Suppose $S=2$. By (13), we have $\hat{m}=1$ or 2. If $\hat{m}=2$, since $S\hat{m} \le \hat{m}+1$, and since g.c.d.($\hat{m}, \hat{n}$) = 1, we are led to $\hat{n}=1$, a contradiction to the hypothesis $m \le \frac{3n}{2}$. Consequently, $\hat{m}=1$, and then $m=n=1$. Thus, if $m \ne n$, we have $S=1$, hence $t=1$, $s_1=2$, therefore $c_1$ is a simple zero of $P'$. This also completes the proof of Theorem 60.4.

Now, in Theorem 60.12, assuming $\frac{m}{2} < n$, by (12), we see that the inequality $S \ge 2$ is impossible, hence $S=t=1$, therefore $c_1$ is a simple zero of $P'$.

We must now consider the general case when the ramification index of one of the two functions $f, g$ is not 0. By Lemma 60.3, $f$ and $g$ have the same ramification index $h$ and satisfy $P_h(\sqrt[h]{f}) = Q_h(\sqrt[h]{g})$. Moreover, one can easily verify that all hypotheses on the degrees satisfied by $P$ and $Q$ are also satisfied by $P_h$ and $Q_h$. And of course, $\sqrt[h]{f}, \sqrt[h]{g}$ lie in $M_u(d(0, R^-))$ in Theorem 60.4 and in $M(K)\backslash K$ in Theorem 60.12. Consequently, putting $f_{(h)} = \sqrt[h]{f}, g_{(h)} = \sqrt[h]{g}$, we have

$$ \tilde{N}(r, f_{(h)}) \ge \frac{T(r, f_{(h)})(tn-m)}{n} + O(1). $$

Actually, $\tilde{N}(r, f_{(h)}) = \tilde{N}(r, f)$ and $T(r, f_{(h)}) = \frac{T(r, f)}{q^h}$, thereby the inequality $\tilde{N}(r, f_{(h)}) \ge \frac{T(r, f_{(h)})(tn-m)}{q^h n} + O(1)$ is established. Now, suppose $\frac{2m}{3} < n$. Then we can check that all statements on $t, m, n$, true for $P_h, Q_h$, hold for $P, Q$. Similarly, we can conclude in Theorem 60.12, when $\frac{m}{2} < n$. $\square$
The Functional Equation $P(f) = Q(g)$

611

Henceforth, we suppose that the characteristic of $K$ is null and we consider functional equations of the form $P(x)(g(x))^{n} = Q(x)(f(x))^{m} + S(x)$, where $S$ is a small function with respect to $f$ and $g$ and we show that the equations have no solutions $f, g$ when $f, g$ are transcendental or are unbounded inside a disk.

**Theorem 60.15.** Let $P, Q, S \in K(x)$ and suppose that two functions $f, g \in M(K)$ have finitely many poles and satisfy $P(x)(g(x))^{n} = Q(x)(f(x))^{m} + S(x) \forall x \in K$ with $\min(m, n) \ge 2$. Then, $f$ and $g$ belong to $K(x)$.

**Proof.** Let $F(x) = Q(x)(f(x))^{m} + S(x)$. Then $F$ belongs to $M(K)$ and has finitely many poles. Since $P(x)(g(x))^{n} = F(x)$, $0$ is a perfectly branched value for $F$. On the other hand, $F(x) - S(x) = Q(x)(f(x))^{m}$, hence $Q(x)$ is a perfectly branched polynomial for $F$. But then by Theorem 49.9, $F$ is not transcendental and hence belongs to $K(x)$ and then so does $(f(x))^{m}$. But since the entire function $f$ is algebraic over $K(x)$, actually by Corollary 51.7, it belongs to $K(x)$. But then, so does $g^{n}$ and similarly, so does $g$. $\square$

**Corollary 60.16.** Let $P, Q, S \in K[x]$ and suppose that two entire functions $f, g \in A(K)$ satisfy $P(x)(g(x))^{n} = Q(x)(f(x))^{m} + S(x) \forall x \in K$ with $\min(m, n) \ge 2$. Then, $f$ and $g$ belong to $K[x]$.

In Theorem 29.23, we saw that the function $\sqrt[q]{\cdot}$ is defined in $d(1, 1^{-})$ when $q$ is prime to $p$ and that $\sqrt[q]{\cdot}$ is defined in $d(1, (\frac{1}{2})^{-})$ if $p = 2$.

**Theorem 60.17.** Let $h \in A(K)$ have all its zeros of order multiple of $q \ge 2$ and let $w \in K[x]$ be non-identically 0. Then the functional equation

$$ (\mathcal{E}) \quad (g(x))^{q} = h(x)(f(x))^{q} + w(x) $$

has no solution $f, g \in A(K) \setminus K[x]$.

**Proof.** Let $f, g \in A(K)$ satisfy $(\mathcal{E})$. If $h \in K[x]$, by Corollary 60.16, both $f$ and $g$ are polynomials. Now suppose that $h \notin K[x]$. By Theorem 29.7, there exists $\phi \in A(K)$ such that $\phi^{q} = h$. Consequently, since $w$ is not 0, we have $w(x) = (g(x))^{q} - (\phi(x)f(x))^{q}$, and hence, by Theorem 50.4, $(g(x))^{q} - (\phi(x)f(x))^{q}$ does not belong to $A_{f}(K)$, a contradiction to the hypothesis $(g(x))^{q} - h(x)(f(x))^{q} = w(x)$. $\square$
We can now consider analytic functions inside a disk $d(a, R^{-})$.

**Theorem 60.18.** Let $q \in \mathbb{N}^{*}$ be prime to $p$ and let $R \in ]0, +\infty[$. Let $h \in \mathcal{A}(d(a, R^{-}))$ be such that that all zeros of $h$ are of order multiple of $q$ and let $w \in \mathcal{A}_{b}(d(a, R^{-}))$. Then the functional equation

$$ (\mathcal{E}) \quad (g(x))^{q} = h(x)(f(x))^{q} + w(x) $$

has no solution in $\mathcal{A}_{u}(d(a, R^{-}))$.

Similarly, we can prove Theorem 60.19:

**Theorem 60.19.** Suppose $\mathbb{K}$ has residue characteristic 2. Let $a \in \mathbb{K}$ and $R \in ]0, +\infty[$ and let $h, w \in \mathcal{A}_{b}(d(a, R^{-}))$ be such that that all zeros of $h$ are of even order. Then the equation $(\mathcal{E})$ $g^{2} = hf^{2} + w$ has no solution $f, g \in \mathcal{A}_{u}(d(a, R^{-}))$.

**Proof of Theorems 60.18 and 60.19.** We can obviously suppose $a = 0$. Suppose that $(\mathcal{E})$ has solutions $f, g \in \mathcal{A}_{u}(d(0, R^{-}))$. By Theorem 29.23, we can apply the root function $\sqrt[q]{\cdot}$ to the function $h$ in both Theorems 60.18 and 60.19. Therefore, there exists $\phi \in \mathcal{A}_{b}(d(0, R^{-}))$ such that $\phi^{q} = h$. Consequently, we have $g^{q} - (\phi f)^{q} = w$ (with $q = p = 2$ in Theorem 60.19). But by Theorem 50.4, $g^{q} - (\phi f)^{q}$ is not bounded, which ends the proof. □

By Theorem 60.18, we can also derive the following corollary:

**Corollary 60.20.** Suppose $\mathbb{K}$ has residue characteristic $p \neq 2$. Let $R \in ]0, +\infty[$, and let $h, w \in \mathcal{A}_{b}(d(a, R^{-}))$ be such that $h(x) \neq 0$ $\forall x \in d(a, R^{-})$. Then the equation $(\mathcal{E})$ $g^{2} = hf^{2} + w$ has no solution $f, g \in \mathcal{A}_{u}(d(a, R^{-}))$.
# Chapter 61

## Yoshida's Equation in the Field $K$

Throughout this chapter, the field $K$ has characteristic $p \ge 0$.

We call Yoshida's equation a differential equation of the form $(\mathcal{E})$ $(y')^m = F(x, y)$ (with $F(x, y) \in K(x, y)\K$). Several results were obtained in characteristic 0 and in characteristic $p \ne 0$, for meromorphic functions in the whole field $K$ or inside a disk $d(a, R^{-})$. In [24], it was shown that if $(\mathcal{E})$ admits solutions in $M(\K)\K(x)$, then $F \in \K(x)[y]$ and $\deg_y(F) \le 2m$. Moreover, it was shown that if $F \in \K(y)$, then any solution of the equation lying in $M(\K)$ is a rational function with a very specific form. That was generalized in characteristic $p \ne 0$ in [27]. Theorems 61.1 and 61.2 were proven when the ground field is $\mathbb{C}_p$. They have an immediate generalization in any algebraically closed field complete for an ultrametric absolute value, such as $K$.

**Theorem 61.1.** Let $F, G \in \K(x, Y)$ be irreducible. If the differential equation $F(x, y^{(n)}) = G(x, y)$ admits a solution $f(x) \in M(\K)\K(x)$, then $\deg(G) \le (n+1)\deg(F)$.

**Proof.** Suppose the given equation admits a solution $f(x) \in M(\K)\K(x)$. Let $w = \deg(F)$, $\ell = \deg(G)$.

By Theorem 51.5, we have $T(r, G(x, f(x))) = \ell T(r, f) + o(T(r, f))$ and $T(F(x, f^{(n)}(x))) = w T(r, f^{(n)}) + o(T(r, f))$. Now by Corollary 43.15, we have $T(r, f^{(n)}) \le (n+1)T(r, f) + O(1)$. Consequently, $\ell T(r, f) \le w(n+1)T(r, f) + o(T(r, f))$. Thus, since $T(r, f)$ is unbounded, $\ell \le (n+1)w$. $\square$
**Notation.** Let $F(x, y) \in \mathbb{K}(x, y)$. We denote by $\deg_x(F)$ (respectively $\deg_y(F)$) the degree of $F$ as a rational function that belongs to $\mathbb{K}(y)(x)$ (respectively to $\mathbb{K}(x)(y)$).

Theorem 61.2 comes from [24] in characteristic 0 and from [27] in characteristic $p \neq 0$.

**Theorem 61.2 (Yoshida).** Let $F(x, y) \in \mathbb{K}(x, y)\setminus\{0\}$ and suppose that there exists a non-constant solution $f \in \mathcal{M}(\mathbb{K})\setminus\mathbb{K}(x)$ of the differential equation

$$ (\mathcal{E}) \quad (y')^m = F(x, y). $$

Then, $F$ belongs to $\mathbb{K}(x)[y]$ and $\deg_y(F) \leq 2m$.

**Proof.** Let us write $G(x, y)$ in an irreducible form $\frac{P(x, y)}{Q(x, y)}$ with $P, Q \in \mathbb{K}[x, y]$, $\deg(P) = s$, $\deg(Q) = t$. Let $\alpha \in \mathbb{K}$ be such that $P(x, \alpha)$ is not identically zero and set $z = \frac{1}{y-\alpha}$. Equation (E) is equivalent to

$$ (\mathcal{E}') \quad (z')^n = R(x, z), $$

where $R(x, z)$ is an irreducible rational function that we may write in the form $\frac{\sum_{j=0}^{s} a_j(x)z^{2n-j}}{\sum_{j=0}^{t} b_j(x)z^{-j}}$, with $a_j \in \mathbb{K}[x]$, $(0 \le j \le s)$, $b_j \in \mathbb{K}[x]$, $(0 \le j \le t)$. We note that $a_0(x) = P(\alpha) \neq 0$.

Suppose first $s - 2n \geq t$. Then,

$$ G(x, z) = (-1)^n \frac{\sum_{j=0}^{s} a_j(x) z^{s-j}}{\sum_{j=0}^{t} b_j(x) z^{s-2n-j}}. $$

So, $\deg_z(R(x, z)) = s$. By Theorem 61.1, we have $s \leq 2n$, hence $t = 0$ and therefore $G$ belongs to $\mathbb{K}[x, y]$.

Suppose now $s - 2n < t$. We may write $R(x, z)$ in the form

$$ (-1)^n \frac{\sum_{j=0}^{s} a_j(x) z^{2n+t-j}}{\sum_{j=0}^{t} b_j(x) z^{t-j}}. $$

Thus, we can see that $\deg_z(R) = 2n + t$, hence by Theorem 61.1, we have $2n \geq 2n + t$, thereby $t = 0$ again. $\square$
**Theorem 61.3.** Let $F(Y) \in \mathbb{K}(Y)\backslash\mathbb{K}$ and let $(\mathcal{E})$ be the equation $(y')^m = F(y)$. Then there exists non-constant solutions $f \in \mathcal{M}(\mathbb{K})$ of $(\mathcal{E})$ if and only if $F$ is of the form $B(Y-c)^\ell$ $(0 \le \ell \le 2m)$, whereas $m - \ell$ divides $m$. If those hypotheses are satisfied, then the solutions are the functions of the form $f(x) = c + (ax + \phi(x^p))^{\frac{m}{m-\ell}}$, where a satisfies

$$ (\frac{am}{m-\ell})^m = B \text{ when } m > \ell, $$

$$ (\frac{am}{m-\ell})^m = (-1)^m B \text{ when } m < \ell $$

and

$\phi$ is a constant when $p=0$,

$\phi(u) \in \mathcal{M}(\mathbb{K})$ when $p \neq 0$.

**Proof.** Suppose first that a solution $f$ is such that $f' = 0$. Then we have $F(f(x)) = 0$. Therefore, $f$ must be equal to one of the zeros of $F$ for every $x \in K$, and finally, $f$ is a constant. Consequently, a solution $f$ is such that $f'$ is not identically 0. Therefore, $\text{ram}(f) = 0$.

We first verify that $F$ is a polynomial of degree $\ell \le 2m$. Let $f \in \mathcal{M}(\mathbb{K})\backslash\mathbb{K}$ be a solution of the equation $(\mathcal{E})$. Let $b$ be a pole of $F(X)$ and $\beta$ be a zero of $f-b$. Then $\beta$ is a pole of $F \circ f$, hence by equation $(\mathcal{E})$, $\beta$ is a pole of $f'$, a contradiction since $f$ has no pole at $\beta$. Consequently, $f$ must avoid any pole of $F(X)$. Therefore, by Corollary 36.23, $F(X)$ admits at most one pole $b$ and hence is of the form $\frac{Q(X)}{(X-b)^s}$, with $Q(X) \in \mathbb{K}[X]$, $Q(b) \neq 0$ and $s \in \mathbb{N}$. Suppose $s \ge 1$. Since $f$ must avoid any pole of $F$, $f$ is of the form $f = b + \frac{1}{h}$ with $h \in \mathcal{A}(\mathbb{K})\backslash\mathbb{K}$ and hence $h' \neq 0$. From $(\mathcal{E})$, we derive that $h$ satisfies $(h')^m = (-1)^m h^{2m+s} Q(b + \frac{1}{h})$. Since $h \in \mathcal{A}(\mathbb{K})$ and is not a constant, $h$ admits at least one zero $\gamma$ and then we can find $M \in \mathbb{R}$ such that $M \ge \Psi(\gamma)$ and $\Psi(h, \mu) > -\Psi(b) \forall \mu \ge M$. Then, $\Psi(b + \frac{1}{h}, \mu) = \Psi(b) \forall \mu \ge M$, hence $\Psi(Q(b + \frac{1}{h})) = \Psi(Q(b)) \forall \mu \ge M$. Since $h' \neq 0$, we obtain $m\Psi(h', \mu) = (2m + s)\Psi(h, \mu) + \Psi(Q(b)) \forall \mu \ge M$. But by Theorem 37.10, we have $\Psi(h', \mu) \le \Psi(h, \mu) - \mu \forall \mu \in \mathbb{R}$, hence $m(\Psi(h, \mu) - \mu) \ge (2m + s)\Psi(h, \mu) + \Psi(Q(b))$ and therefore

$$ (1) \quad (m + s)\Psi(h, \mu) + m\mu + \Psi(Q(b)) \le 0. $$

Now, since $M \ge \Psi(\gamma)$ and since $h$ belongs to $\mathcal{A}(\mathbb{K})$, the function $m(\Psi(h, \mu) + \mu) + \Psi(Q(b))$ is strictly increasing and tends to $+\infty$
with $\mu$, a contradiction to Relation (1). Consequently, $F(X)$ has no pole and hence is a polynomial $a_0 + a_1 X + \cdots + a_\ell X^\ell$. By Theorem 43.8, we have $T(r, F \circ f) = \ell T(r, f) + O(1)$, ($r \in J$), so, then, using the inequality $T(r, f') \le 2T(r, f)$, we can derive from $(\mathcal{E})$ that $\ell \le 2m$.

Now, let $c$ be a zero of $F(X)$ of order $n$ and let $\alpha$ be a zero of $f-c$ of order $w$ such that $p \not\| w$. We check that

$$ (2) \quad w = \frac{m}{m-n} \quad \text{and} \quad m > n. $$

So, all zeros of $f-c$ are multiple, and, if their order is not a multiple of $p$, they have the same multiplicity order. Furthermore, we have $\frac{m}{m-n} \ge 2$.

If we consider a pole of $f$ of order $t$ such that $p \not\| t$, we have

$$ (3) \quad t = \frac{m}{\ell - m} \quad \text{and} \quad m < \ell. $$

So, all poles of $f$ are multiple and the poles whose order is not multiple of $p$ have the same multiplicity order.

Let $c_1, \dots, c_k$ be the distinct zeros of $F$ and for each $j = 1, \dots, k$, and let $n_j$ be the order of $c_j$ as a zero of $F$. Let us apply Theorem 58.11 to $f$. Since $\text{ram}(f) = 0$, we have

$$ (k-1)T(r, f) \le \sum_{j=1}^{k} \tilde{Z}(r, f - c_j) + \tilde{N}(r, f) - \log r + O(1), $$

and then, by (2) and (3), we can derive

$$ (k-1)T(r, f) \le (Tr, f) \left( \sum_{j=1}^{k} \left( \frac{m-n_j}{m} \right) + \frac{\ell-m}{m} \right) - \log r + O(1), $$

hence

$$ (k-1)T(r, f) \le \frac{\left( \sum_{j=1}^{k} (m-n_j) \right) + \ell - m}{m} T(r, f) - \log r + O(1). $$

But $\sum_{j=1}^{k} n_j = \ell$, so we obtain

$$ (k-1)T(r, f) \le (k-1)T(r, f) - \log r + O(1), $$

which is impossible whenever $k \ge 2$. That shows that $k = 1$. Therefore, $F(X)$ is of the form $B(X-c)^\ell$. Now, by (2) and (3), we can see
that $f-c$ can't have simultaneously a zero and a pole whose order is not multiple of $p$.

So, by (2) and (3), either $m > \ell$ and then $f$ belongs to $\mathcal{A}(K)$ and admits at least a zero, or $\ell > m$ and then $f$ has no zero but admits at least a pole and is of the form $c + \frac{1}{h}$ with $h \in \mathcal{A}(K)$. And since $f'$ is not identically zero, we note that $f$ must have at least one zero or one pole, therefore $f$ is not a constant.

(A) Suppose $m > \ell$ and hence $f$ belongs to $\mathcal{A}(K)$. Let $\gamma$ be a zero of $f-c$. Then $m\omega_\gamma(f') = \ell\omega_\gamma(f-c)$. Let $\rho = \text{g.c.d.}(m, \ell)$ and set $m = \sigma\rho$, $\ell = \tau\rho$. Then $\sigma\omega_\gamma(f') = \tau\omega_\gamma(f-c)$. Now, since $\sigma, \tau$ are relatively prime, $\sigma$ divides $\omega_\gamma(f-c)$ for every zero $\gamma$ of $f-c$, and therefore, by Theorem 27.7, there exists $g \in \mathcal{A}(K)$ such that $g^\ell = f-c$. On the other hand, from the given equation $(f')^{\sigma\rho} = C_2(f-c)^{\tau\rho}$, the function $\theta(x) = \frac{(f')^\sigma}{C_2(f-c)^\tau}$ is a meromorphic function such that $\theta^\rho = 1$. Then by Lemma 36.1, $\theta$ is a constant. Consequently, $(f')^\sigma$ is of the form $C_3(f-c)^\tau$, with $(C_3)^\rho = C_2$, hence $(f')^\ell = C_3g^{\tau\ell}$, therefore $f' = C_4g^\tau$ with $C_4^\sigma = C_3$. But $f' = \sigma g'g^{\sigma-1}$, hence $\sigma g'g^{\sigma-1} = C_4g^\tau$ and finally $g'g^{\sigma-\tau-1} = \frac{C_4}{\sigma}$. Now, since we have noted that $\ell < m$, we have $\sigma - \tau - 1 \ge 0$. Consequently, since $g \in \mathcal{A}(K)$, the only solution is that both $g'$ and $g^{\sigma-\tau-1}$ are constant i.e. $g(x)$ is of the form $ax + v(x)$ with $\sigma = \tau + 1$, with $a \in K^*$ and $v \in \mathcal{M}(K)$ has a null derivative i.e. if $p = 0$, then $v$ is just a constant $b \in K$, and if $p \ne 0$, then $v$ is of the form $\phi(x^p)$, whereas $\phi(u)$ belongs to $\mathcal{M}(K)$.

More precisely, we note that $\sigma = \frac{m}{m-\ell}$ because $\sigma = \tau + 1$. Consequently, $f(x) = c + (ax + v)^{\frac{m}{m-\ell}}$ and therefore $f'(x) = \frac{am}{m-\ell}(ax + v)^{\frac{\ell}{m-\ell}}$. Now, since $f$ is solution of $(\mathcal{E})$ and since $v'(x)$ is identically zero, we can derive

$$ \left( \frac{am}{m-\ell} \right)^m (ax + v(x))^{\frac{m\ell}{m-\ell}} = B(ax + v(x))^{\frac{\ell}{m-\ell}}, $$

therefore $(\frac{am}{m-\ell})^m = B$.

(B) Suppose $\ell > m$ and hence $f$ is of the form $c + \frac{1}{h}$ with $h \in \mathcal{A}(K)$. Then $h$ satisfies $h'^m = (-1)^m Bh^{2m-\ell}$. Thus, we are lead to the same problem with $2m - \ell$ instead of $\ell$ and a polynomial $G(Y)$ of the form $(-1)^m BY^{2m-\ell}$ instead of $F$. Consequently, $h$ is of the form $(ax + v(x))^{\frac{m}{\ell-m}}$ with $v$ of the form $\phi(x^p)$, whereas $\phi(u)$ belongs to
$\mathcal{M}(\mathbb{K})$ and with $(\frac{am}{\ell-m})^m = (-1)^m B$. So, $f(x) = c + (ax + v(x))^{\frac{m}{m-\ell}}$ with $(\frac{am}{\ell-m})^m = (-1)^m B$.

Conversely, suppose that $m-\ell$ divides $m$ and suppose that $F(Y)$ is of the form $B(Y-c)^\ell$. Let $f(x) = c + (ax + \phi(x^p))^{\frac{m}{m-\ell}}$, where $a$ satisfies the following: $(\frac{am}{m-\ell})^m = B$ when $m > \ell$, $(\frac{am}{m-\ell})^m = (-1)^m B$ when $m < \ell$ and satisfies further the following: $\phi$ is a constant when $p=0$, $\phi(u) \in \mathcal{M}(\mathbb{K})$ when $p \neq 0$. Then we can check that $f$ is a solution of $(\mathcal{E})$. $\square$
## Chapter 62

# Yoshida's Equation Inside a Disk

Here we want to consider Yoshida's equation for meromorphic functions inside a disk $d(a, R^{-})$ [23]. Several methods look like parts of the proof of Theorem 61.3. However, we do not obtain a result as general as in Chapter 61.

Throughout this chapter, $\mathbb{K}$ has characteristic 0.

**Proposition 62.1.** Let $F(Y), G(Y) \in \mathcal{M}_b(d(a, R^{-}))(Y)$ be irreducible as rational functions. If the differential equation $F(y^{(n)}) = G(y)$ admits a solution $f \in \mathcal{M}_u(d(a, R^{-}))$, then $\deg(G) \le (n+1)\deg(F)$.

**Proof.** Suppose there exists $f \in \mathcal{M}_u(d(a, R^{-}))$ such that $F(f^{(n)}) = G(f(x))$.

On one hand, by Corollary 43.15, we have $T(r, (f^{(n)})) \le (n+1)T(r, f) + O(1)$. Hence, by Theorem 51.5 and Corollary 43.15, we have $T(r, F(f^{(n)})) \le (n+1)\deg(F)T(r, f) + o(T(r, f))$. On the other hand, by Theorem 51.5, we have $T(r, G(f)) = \deg(G)T(r, f) + o(T(r, f))$. Now, since $T(r, f)$ is unbounded, the conclusion is clear. $\square$

**Theorem 62.2.** Let $m \in \mathbb{N}^*$ and let $F \in \mathcal{M}_b(d(a, R^{-}))(Y)$. If the differential equation $(\mathcal{E})$ $(y')^m = F(y)$ admits a solution $f \in \mathcal{M}_u(d(a, R^{-}))$, then $F$ belongs to $\mathcal{M}_b(d(a, R^{-}))[Y]$ and $\deg(F) \le 2m$.

**Proof.** Let us write $f(Y) = \frac{\sum_{i=0}^{k} c_i(x)Y^i}{\sum_{i=0}^{l} b_i(x)Y^i}$ and suppose that $(\mathcal{E})$ admits a solution $f \in \mathcal{M}_u(d(a, R^{-}))$. Let us take $\alpha \in \mathbb{K}$ such
that $\sum_{i=0}^{k} c_i(x)\alpha^i$ is not identically zero and let $g = \frac{1}{f-\alpha}$. Let $G \in \mathcal{M}_b(d(a, R^{-}))(Y)$ be defined by $F(Y) = G(\frac{1}{Y-\alpha})$. Then $g$ satisfies an equation

$$ (g')^m = G(g) \qquad (1) $$

with $G(Y)$ of the form $(-1)^m \frac{\sum_{i=0}^{k} c_i(x)Y^{2m-i}}{\sum_{i=0}^{l} e_i(x)Y^{-i}}$ and $c_0 = \sum_{i=0}^{k} c_i(x)\alpha^i$.

Suppose first $k - 2m \ge l$. Then, $G(Y) = (-1)^m \frac{\sum_{i=0}^{k} c_i(x)Y^{k-i}}{\sum_{i=0}^{l} e_i(x)Y^{k-2m-i}}$.

So, the degree of $G$ is $k$. Then by Proposition 62.1, we have $2m \ge k$, hence $l = 0$ and hence $F$ is a polynomial of degree $k \le 2m$. Suppose now $k - 2m < l$. We have $G(Y) = (-1)^m \frac{\sum_{i=0}^{k} c_i(x)Y^{2m+l-i}}{\sum_{i=0}^{l} e_i(x)Y^{l-i}}$. So, $G$ is of degree $2m + l$ and then, by applying Proposition 62.1 to (1), we obtain $l = 0$ and hence $k < 2m$ i.e. $F$ is a polynomial of degree $k < 2m$. $\square$

**Notation.** In proofs of Theorems 62.3 and 62.4, we have to consider a spherically complete extension $\mathbb{K}$ of $\mathbb{K}$ and the disk $\hat{d}(a, R^{-})$.

Theorem 62.3 looks like Theorem 61.3 when we assume that $F$ is a polynomial with a unique zero.

**Theorem 62.3.** Suppose $\mathbb{K}$ is of characteristic 0. Let $F(Y) \in \mathbb{K}[Y]$ be of the form $B(Y - c)^l$. The solutions $f \in \mathcal{M}(d(a, R^{-}))$ of $(\mathcal{E})$ are the functions of the form $c + (ax + b)^{\frac{m}{m-l}}$ such that

$$ \text{either } m > l \text{ and then } B = \left( \frac{am}{m-l} \right)^m $$

$$ \text{or } m < l \text{ and then } (-1)^m B = \left( \frac{am}{m-l} \right)^m. $$

**Proof.** Suppose $(\mathcal{E})$ admits a solution $f \in \mathcal{M}_u(d(a, R^{-}))$. Let $\alpha$ be a zero of $f - c$ of order $s$. So, we have

$$ (1) \quad s = \frac{m}{m-l} \quad \text{and} \quad m > l. $$

Consequently, all zeros of $f - c$ are multiple and of same order. Now, let $\beta$ be a pole of $f$ of order $t$. Then we have

$$ (2) \quad t = \frac{m}{l-m} \quad \text{with } m < l. $$
So, all poles are of same order. Moreover, we note that $f$ cannot have both zeros and poles.

Consequently, either $m > \ell$ and $f$ belongs to $A_u(d(a, R^-))$ or $m < \ell$ and $f$ is of the form $c + \frac{1}{h}$ with $h \in A_u(d(a, R^-))$.

Suppose first that $m > \ell$. Then $f$ belongs to $A_u(d(a, R^-))$. By (1), every zero $\alpha$ of $f - c$ is of order $s = \frac{m}{m-\ell}$, hence $\alpha$ is a zero of $f'$ of order $\frac{\ell}{m-\ell}$. By Theorem 36.12, there exists $g \in A(d(a, R^-))$ such that $g^s = f - c$. Then, by Equation (E), we have $(sg')^m(g)^{m(s-1)} = Bg^{s\ell}$ which, by (1), yields $(sg')^m = B$. Consequently, since $K$ is of characteristic 0, by Lemma 36.1, $g'$ is a constant and hence $g(x)$ is of the form $(ax + b)$. So, $f - c = (ax + b)^s$ and (similar to Theorem 61.3) we obtain $s = \frac{m}{m-\ell}$, $B = (\frac{am}{m-\ell})^m$, and $f(x) = c + (ax + b)^{\frac{m}{m-\ell}}$.

Suppose now that $m < \ell$ and then $f$ is of the form $c + \frac{1}{h}$ with $h \in A_u(d(a, R^-))$. Then $h$ satisfies

$$ (3) \quad (h')^m = (-1)^m B h^{2m-\ell} $$

and by (2) every zero $\beta$ of $h$ is of order $t = \frac{m}{\ell-m}$. Consequently, $\beta$ is a zero of $h'$ of order $t - 1 = \frac{2m-d}{\ell-m}$. Consider again a function $g \in A(\hat{d}(a, R^-))$ such that $g^t = h$. In the same way as in the previous case, we can check that $g'$ is a constant and therefore, since $K$ is of characteristic 0, $f$ is of the form $c + \frac{1}{(ax+b)^t}$ and the conclusion follows as in Theorem 61.3. $\square$

**Proposition 62.4.** Suppose $K$ is of characteristic 0. Let $F(Y) \in K[Y]$ have at least two distinct zeros. Then the differential equation $(\mathcal{E})$ $(y')^m = F(y)$ does not admit solutions $f \in A_u(d(a, R^-))$ or of the form $c + \frac{1}{h}$ with $h \in A_u(d(a, R^-))$.

**Proof.** Let $F(Y) = B \prod_{j=1}^k (Y - c_j)^{\ell_j}$ with $k \ge 2$, $c_i \ne c_j \forall i \ne j$ and let $\ell = \deg(F)$. Suppose first that a solution $f$ of $(\mathcal{E})$ belongs to $A_u(d(a, R^-))$. Since, by Corollary 18.3, $|f'|(r) \le \frac{|f|}{r} \forall r \in ]0, R[$, we note that $m \ge \ell$, therefore $\ell_j < m \forall j = 1, \dots, k$.

For each $j = 1, \dots, k$, let $t_j = \text{g.c.d.}(m, \ell_j)$ and set $m = t_j q_j$, $\ell_j = t_j w_j$. So, $q_j$ and $w_j$ are relatively prime. For each pair $(i, j)$, $i \ne j$, since $c_i \ne c_j$, $f - c_i$ and $f - c_j$ have no common zero. So, if $\alpha$ is a zero of $f - c_j$, we have $m\omega_\alpha(f') = \ell_i\omega_\alpha(f - c_i)$ and therefore $q_i\omega_\alpha(f') = w_i\omega_\alpha(f - c_i)$. Thus, $q_i$ divides $\omega_\alpha(f - c_i)$, $(1 \le i \le k)$.
Now, by Theorem 36.12, there exists $g_i \in \mathcal{A}_u(\hat{d}(a, R^{-}))$ such that $f - c_i = (g_i)^{q_i}$.

We note that if $q_i = 1$, then $m$ must divide $\ell_i$, which is excluded because $\ell_j < m$. Consequently, $q_i \ge 2 \ \forall i = 1, \dots, k$. Next, for any pair $(i, j)$ of two distinct integers among $1, \dots, k$, we have $(g_i)^{q_i} - (g_j)^{q_j} = c_i - c_j \ne 0$. But since $\min(q_i, q_j) \ge 2$, by Theorem 47.11, both $g_i, g_j$ are bounded a contradiction since $f$ is unbounded.

Suppose now that $f = c + \frac{1}{h}$ with $h \in \mathcal{A}_u(d(a, R^{-}))$. At most one of the $c_i$ is equal to $c$. So, we can assume $c_i \ne c \ \forall i = 1, \dots, k-1$. Set $a_j = c - c_j \ \forall j = 1, \dots, k$ and $\ell = \sum_{j=1}^{k-1} \ell_j$. Then $f$ must satisfy

$$ (1) \quad (h')^m = (-1)^m Bh^{2m-\ell} \prod_{i=1}^k \left(c - c_i + \frac{1}{h}\right)^{\ell_i} \\ = (-1)^m Bh^{2m-\ell} \left(a_k + \frac{1}{h}\right)^{\ell_k} \prod_{i=1}^{k-1} \left(a_i + \frac{1}{h}\right)^{\ell_i}. $$

Now, since $h$ is unbounded, when $r$ is close enough to $R$, we have

$$ (|h'|(r))^m = |B|(|h|(r))^{2m} \left( \left|a_k + \frac{1}{h}\right|(r) \right)^{\ell_k} \prod_{i=1}^{k-1} \left( \left|a_i + \frac{1}{h}\right|(r) \right)^{\ell_i} \\ = |B|(|h|(r))^{2m} \left( \left|a_k + \frac{1}{h}\right|(r) \right)^{\ell_k} \prod_{j=1}^{k-1} |a_j|^{\ell_j}. $$

Set $C = |B| \prod_{i=1}^{k-1} (|a_i|)^{\ell_i}$. By Corollary 18.3, when $r$ is close enough to $R$, we obtain

$$ C(|h|(r))^{2m} \left( \left|a_k + \frac{1}{h}\right|(r) \right)^{\ell_k} \le C \left( \frac{|h|(r)}{r} \right)^m, $$

therefore $Cr^m(|h|(r))^{m} (|a_k + \frac{1}{h}|(r))^{l_k} \le 1$ and hence $Cr^m(|h|(r))^{m-l_k} \le 1$ when $r$ tends to $R$. But since $m < \ell_k$, that inequality is impossible when $r$ tends to $R$. $\square$

By Theorems 62.3 and 62.4, we have the following corollary:

**Corollary 62.5.** Suppose $\mathbb{K}$ is of characteristic 0. Let $F(Y) \in \mathbb{K}[Y]$ and let $m \in \mathbb{N}^*$. Then the differential equation $(\mathcal{E})$ $(y')^m = F(y)$ does not admit any solution $f \in \mathcal{A}_u(d(a, R^{-}))$ or of the form $c + \frac{1}{h}$ with $h \in \mathcal{A}_u(d(a, R^{-}))$.
**Definition.** In Theorem 62.6, we have to present Equation (E) when it is in the form $(y')^{mq} = (F(y))^q$, where $q \in \mathbb{N}^*$. We say that (E) is a qth power of the equation $(y')^m = (F(y))$ and an equation (E) is said to be irreducible if it is not a qth power of another equation of the form (E), for some $q > 1$.

**Remark.** According to Theorems 62.3 and 62.4, we have characterized all solutions of (E) that either belong to $A(d(a, R^{-}))$ or admit an exceptional value. In Theorem 62.6, we examine the form of Equation (E) when a solution is not of the form as shown in Theorem 62.3.

**Theorem 62.6.** Suppose $\mathbb{K}$ is of characteristic 0. Let $F(Y) \in \mathbb{K}[Y]$ and let $m \in \mathbb{N}^*$. Suppose the differential equation $(\mathcal{E})$ $(y')^m = F(y)$ admits a solution $f \in \mathcal{M}_u(d(a, R^{-}))\setminus A(d(a, R^{-}))$ that has no exceptional value in $\mathbb{K}$. Then $(\mathcal{E})$ is a power of one of the following forms:

$$ (\mathcal{E}_1) (y')^2 = A(y - c_1)(y - c_2)(y - c_3), $$

$$ (\mathcal{E}_2) (y')^2 = A(y - c_1)(y - c_2)(y - c_3)(y - c_4), $$

$$ (\mathcal{E}_3) (y')^3 = A(y - c_1)^2(y - c_2)^2, $$

$$ (\mathcal{E}_4) (y')^3 = A(y - c_1)^2(y - c_2)^2(y - c_3)^2, $$

$$ (\mathcal{E}_5) (y')^6 = A(y - c_1)^3(y - c_2)^4(y - c_3)^5, $$

$$ (\mathcal{E}_6) (y')^4 = A(y - c_1)^2(y - c_2)^3(y - c_3)^3, $$

$$ (\mathcal{E}_7) (y')^6 = A(y - c_1)^4(y - c_2)^5, $$

$$ (\mathcal{E}_8) (y')^4 = A(y - c_1)^3(y - c_2)^3, $$

$$ (\mathcal{E}_9) (y')^6 = A(y - c_1)^3(y - c_2)^5, $$

$$ (\mathcal{E}_{10}) (y')^4 = A(y - c_1)^2(y - c_2)^3, $$

$$ (\mathcal{E}_{11}) (y')^6 = A(y - c_1)^3(y - c_2)^4. $$

**Proof.** Let $F(Y) = A \prod_{i=1}^{k}(Y - c_i)^{\ell_i}$ with $c_i \neq a_j \forall i \neq j$ and $\ell = \sum_{i=1}^{k} \ell_i$. We can note that each zero of $f - c_i$ is a zero of $f'$ and hence is a zero of order $\frac{m}{m-\ell_i} \geq 2$. Particularly, we have

$$ (1) \quad \frac{m}{2} \leq \ell_i \leq m - 1 $$

and hence $\frac{km}{2} \leq \sum_{i=1}^{k} \ell_i \leq k(m-1)$ i.e. $\frac{km}{2} \leq \ell \leq k(m-1)$. In the same way, for each pole $\beta$ of $f$, we have $\omega_{\beta}(f) = \frac{m}{\ell-m} \geq 1$,
624 Value Distribution in Ultrametric Analysis and Application

hence $m + 1 \le \ell \le 2m$, therefore

$$ (2) \quad \max \left[ m + 1, \frac{km}{2} \right] \le \ell \le \min[2m, k(m-1)], $$

and hence

$$ (3) \quad \frac{m+1}{m-1} \le k \le \frac{2\ell}{m} \le 4. $$

Let $u = \frac{m}{\ell - m}$ and $u_i = \frac{m}{m - \ell_i}$, $i = 1, \dots, k$.

Then,

$$ (4) \quad \frac{1}{m} \le \frac{1}{u_i} \le \frac{1}{2}. $$

By (2), we have

$$ (5) \quad \max \left[ \frac{1}{m}, \frac{k-2}{2} \right] \le \frac{1}{u} \le \min \left[ 1, \frac{k(m-1)}{m} - 1 \right]. $$

On the other hand, by definition of $u$ and $u_i$, we also note that

$$ (6) \quad \frac{1}{u} + \sum_{i=1}^{k} \frac{1}{u_i} = k - 1. $$

(A) First, suppose $m = 2$. By (1), (2), and (3), we have $k = \ell = 3$ or $4$ and $\ell_i = 1 \forall i = 1, \dots, k$. This leads to Equations (1) and (2).

(B) Now, suppose $m \ge 3$. Since $m-1$ does not divide $m$, that yields $\ell_i \ge 2 \forall i = 1, \dots, k$. Consequently, we have

$$ (7) \quad \max \left[ \frac{m}{2}, 2 \right] \le \ell_i \le m - 1. $$

Hence, (2) becomes

$$ (8) \quad \max \left[ m + 1, \frac{km}{2}, 2k \right] \le \ell \le \min[2m, k(m-1)]. $$

So, we have $\frac{m+1}{m-1} \le k \le \min(m, \frac{2d}{m})$. And since $1 < \frac{m+1}{m-1} \le 2 \forall m \ge 3$, Relation (3) becomes

$$ (9) \quad 2 \le k \le \min \left[ m, \frac{2\ell}{m} \right] \le 4. $$
Now, by (7), we obtain $\frac{1}{m} \le \frac{m-\ell_i}{m} \le \min[\frac{m-2}{m}, \frac{1}{2}]$ i.e.

$$ (10) \quad \frac{1}{m} \le \frac{1}{u_i} \le \min \left[ \frac{m-2}{m}, \frac{1}{2} \right], $$

and by (5) and (8), we can derive

$$ (11) \quad \max \left[ \frac{1}{m}, \frac{k-2}{2}, \frac{2k-m}{m} \right] \le \frac{1}{u} \le \min \left[ 1, \frac{k(m-1)}{m} - 1 \right]. $$

(B1) Suppose $m=3$. By (7) and (8), we have $\ell_i = 1 \forall i = 1, \dots, k$ and $\max[4, \frac{3k}{2}, 2k] \le d \le \min[6, 2k]$. Consequently, either $\ell = 4$, $k=2$, or $\ell=6$, $k=3$. This leads to Equations (3) and (4).

(B2) $m \ge 4$. In that subcase, we have to consider three subcases.

(B2a) $k=4$. By (6), (9), (10), and (11), we obtain $\ell = 2m \ge 8$, $\frac{1}{m} \le \frac{1}{u_i} \le \frac{1}{2}$, $u = 1$, $\sum_{i=1}^4 \frac{1}{u_i} = 2$. Consequently, $u_i = 2$ and $\ell_i = \frac{m}{2} \forall i = 1, 2, 3, 4$. That leads to the equation $(y')^m = A(y-c_1)^{\frac{m}{2}}(y-c_2)^{\frac{m}{2}}(y-c_3)^{\frac{m}{2}}(y-c_4)^{\frac{m}{2}}$ with $m \ge 4$, $m$ multiple of 2.

That is the $m$th power of Equation (E2):

(B2b) $k=3$. Thanks to (6), (7), (8), and (10), we obtain

$$ \begin{gathered} \frac{m}{2} \le \ell_i \le m-1, \quad \frac{3m}{2} \le \ell \le 2m, \quad 2 \le u_i \le m, \\ 1 \le u \le 2, \quad \frac{1}{u} + \sum_{i=1}^{4} \frac{1}{u_i} = 2. \end{gathered} $$

We can assume that $2 \le u_1 \le u_2 \le u_3 \le m$ i.e. $\frac{1}{m} \le \frac{1}{u_3} \le \frac{1}{u_2} \le \frac{1}{u_1} \le \frac{1}{2}$.

(B2b)(i) Suppose first $u=1$. The only possible values for $\frac{1}{u_1}, \frac{1}{u_2}, \frac{1}{u_3}$ are

$$ \left(\frac{1}{2}, \frac{1}{3}, \frac{1}{6}\right), \left(\frac{1}{2}, \frac{1}{4}, \frac{1}{4}\right), \left(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\right). $$

Consequently, the only possible values for $(\ell_1, \ell_2, \ell_3)$ are respectively

$$ \left(\frac{m}{2}, \frac{3m}{2}, \frac{5m}{6}\right), \left(\frac{m}{2}, \frac{3m}{4}, \frac{3m}{4}\right), \left(\frac{2m}{3}, \frac{2m}{3}, \frac{2m}{3}\right). $$
That leads to the following equations, respectively:

$(y')^m = A(y - c_1)^{\frac{m}{2}} (y - c_2)^{\frac{m}{3}} (y - c_3)^{\frac{5m}{6}}$ with $m \ge 4$, $m$ multiple of 6.

$(y')^m = A(y - c_1)^{\frac{m}{2}} (y - c_2)^{\frac{3m}{4}} (y - c_3)^{\frac{3m}{4}}$ with $m \ge 4$, $m$ multiple of 4.

$(y')^m = A(y - c_1)^{\frac{2m}{3}} (y - c_2)^{\frac{2m}{3}} (y - c_3)^{\frac{2m}{3}}$ with $m \ge 4$, $m$ multiple of 3.

Such equations are respectively the $m$th power of $(\mathcal{E}_5)$, $(\mathcal{E}_6)$, $(\mathcal{E}_4)$.

(B2b) (ii) $u = 2$. The only possible value for $(\frac{1}{u_1}, \frac{1}{u_2}, \frac{1}{u_3})$ is $(\frac{1}{2}, \frac{1}{2}, \frac{1}{2})$. Consequently, the only possible value for $(\ell_1, \ell_2, \ell_3)$ is $(\frac{m}{2}, \frac{m}{2}, \frac{m}{2})$. That leads to the equation $(y')^m = A(y - c_1)^{\frac{m}{2}} (y - c_2)^{\frac{m}{2}} (y - c_3)^{\frac{m}{2}}$ with $m \ge 4$, $m$ multiple of 2.

Thus, that equation is the $m$th power of $(\mathcal{E}_1)$.

(B2c) $k = 2$. Thanks to (7), (8), (10), and (11), we can write

$$ \frac{m}{2} \le \ell_i \le m - 1, \quad m + 1 \le \ell \le 2m - 2, $$

$$ 2 \le u_i \le m, \quad (i = 1, 2) \quad \frac{m}{m-2} \le u \le m. $$

Particularly, $2 \le u \le m$. By (6), we then obtain $\frac{1}{u_1} + \frac{1}{u_2} = \frac{u-1}{u}$ i.e.:

$$ (12) \quad \frac{1}{u_1} + \frac{1}{u_2} + \frac{1}{u} = 1. $$

Suppose $u > 6$. By (12), we have $\frac{1}{u_1} + \frac{1}{u_2} > \frac{5}{6}$. But by (12), $u_1 = u_2 = 2$ is impossible. And if $(u_1, u_2) \ne (2, 2)$ we can see that $\frac{1}{u_1} + \frac{1}{u_2} \le \frac{5}{6}$. Consequently, $u \le 6$. Moreover, if $u = 5$, by (12), we have $\frac{1}{u_1} + \frac{1}{u_2} = \frac{4}{5}$. If $u_i \ge 3, i = 1, 2$, then $\frac{1}{u_1} + \frac{1}{u_2} \le \frac{2}{3}$. Suppose $u_1 = 2$. If $u_2 = 3$, $\frac{1}{u_1} + \frac{1}{u_2} = \frac{5}{6}$. And if $u_2 \ge 4$, then $\frac{1}{u_1} + \frac{1}{u_2} \le \frac{3}{4}$. Consequently, $u \ne 5$ and hence $u$ can only take the values 2, 3, 4, and 6. We consider those 4 cases.

(B2c) (i) $u = 2$. The only possible values for $(\frac{1}{u_1}, \frac{1}{u_2})$ are $(\frac{1}{3}, \frac{1}{6})$ and $(\frac{1}{4}, \frac{1}{4})$. Consequently, the only possible values for $(\ell_1, \ell_2)$
are respectively $(\frac{2m}{3}, \frac{5m}{6})$ and $(\frac{3m}{4}, \frac{3m}{4})$. They correspond to the equations

$(y')^m = A(y - c_1)^{\frac{2m}{3}} (y - c_2)^{\frac{5m}{6}}$ with $m \ge 4$ and $m$ multiple of 6,

$(y')^m = A(y - c_1)^{\frac{3m}{4}} (y - c_2)^{\frac{3m}{4}}$ with $m \ge 4$ and $m$ multiple of 4,

which are the $m$th power of $(\mathcal{E}_7)$ and $(\mathcal{E}_8)$.

(B2c) (ii) $u = 3$. The only possible values for $(\frac{1}{u_1}, \frac{1}{u_2})$ are $(\frac{1}{2}, \frac{1}{6})$ and $(\frac{1}{3}, \frac{1}{3})$. Consequently, the only possible values for $(\ell_1, \ell_2)$ are respectively $(\frac{m}{2}, \frac{5m}{6})$ and $(\frac{2m}{3}, \frac{2m}{3})$. They correspond to the equations

$(y')^m = A(y - c_1)^{\frac{m}{2}} (y - c_2)^{\frac{5m}{6}}$ with $m \ge 4$ and $m$ multiple of 6,

$(y')^m = A(y - c_1)^{\frac{2m}{3}} (y - c_2)^{\frac{2m}{3}}$ with $m \ge 4$ and $m$ multiple of 3,

which are the $m$th power of $(\mathcal{E}_9)$ and $(\mathcal{E}_3)$.

(B2c) (iii) $u = 4$. The only possible value for $(\frac{1}{u_1}, \frac{1}{u_2})$ is $(\frac{1}{2}, \frac{1}{4})$. Consequently, the only possible value for $(\ell_1, \ell_2)$ is $(\frac{m}{2}, \frac{3m}{4})$. This corresponds to the equation $(y')^m = A(y - c_1)^{\frac{m}{2}} (y - c_2)^{\frac{3m}{4}}$ with $m \ge 4$ and $m$ multiple of 4 and this is the $m$th power of $(\mathcal{E}_9)$ and $(\mathcal{E}_{10})$.

(B2c) (ii) $u = 6$. The only possible value for $(\frac{1}{u_1}, \frac{1}{u_2})$ is $(\frac{1}{2}, \frac{1}{3})$. Consequently, the only possible value for $(\ell_1, \ell_2)$ is $(\frac{m}{2}, \frac{2m}{3})$. This corresponds to the equation $(y')^m = A(y - c_1)^{\frac{m}{2}} (y - c_2)^{\frac{2m}{3}}$ with $m \ge 4$ and $m$ multiple of 6 and this is the $m$th power of $(\mathcal{E}_9)$ and $(\mathcal{E}_{11})$.

That finishes the proof of Theorem 62.6. □

**Corollary 62.7.** Suppose $\mathbb{K}$ is of characteristic 0. Let $F(Y) \in \mathbb{K}(Y)\backslash\mathbb{K}$ and let $m \in \mathbb{N}^*$. If one of the following hypotheses is satisfied, there exists no solution $f \in \mathcal{M}_u(da, R^{-})$ for the irreducible equation $(\mathcal{E})$ $(y')^m = F(y)$:

(1) $F(Y) \in \mathbb{K}[Y]$ and $\deg(F) \ge 2m + 1$,

(2) $F(Y) \in \mathbb{K}[Y]$ has at least five zeros, ignoring multiplicity,

(3) $m = 2$, $F(Y) \in \mathbb{K}[Y]$ has at least one multiple zero,
(4) $m = 3$, $F(Y) \in \mathbb{K}[Y]$ has at least four distinct zeros,

(5) $m = 3$, $F(Y) \in \mathbb{K}[Y]$ has at least one zero of order $s \neq 2$,

(6) $m = 4$, $F(Y) \in \mathbb{K}[Y]$ has four distinct zeros,

(7) $m = 5$,

(8) $m = 6$, $F(Y) \in \mathbb{K}[Y]$ has at least four distinct zeros,

(9) $m \geq 7$.
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# Definitions

## Chapter 1

centered similarity,
center of a centered similarity,
non-centered similarity,
translation,
affinely rigid set,
centered affinely rigid set,
non-centered affinely rigid analytic set,
Condition (F).

## Chapter 2

valuation ring, valuation ideal.
residue class, residue class field, residue characteristic,
circle of center *a*, or radius *r*, classes of a circle,
diameter of a set, codiameter of a set,
hole of a set,
infraconnected set, infraconnected component,
affinoid set,
empty annulus of a set.

## Chapter 3

filter thinner than another one,
sequence thinner than a filter,
filter secant with a set, with another filter,
increasing distances sequence, decreasing distances sequence,
monotonous distances sequence,
equal distances sequence,
increasing filter of center *a* and diameter *r*,
decreasing filter of center *a* and diameter *r*,
decreasing filter with no center of diameter *r*, of canonical
basis (*D*<sub>*n*</sub>),
monotonous filter,
body of a monotonous filter, beach of a monotonous filter,
spherically complete field,
pierced monotonous filter,
circular filter of center *a* and diameter *r*,
peripheral of a bounded set,
circular filter with no center, of diameter *r*,
of canonical basis (*D*<sub>*n*</sub>),
large circular filter, punctual circular filter,
*F*-affinoid.

**Chapter 4**

Gauss norm.

**Chapter 5**

quasi-monic polynomial.

**Chapter 6**

*p*-adic absolute value.

**Chapter 7**

Eisenstein polynomial,
uniformizer of an extension of $\mathbb{Q}_p$,
ramification index of an extension of $\mathbb{Q}_p$.

**Chapter 8**

principal ultrafilter,
incomplete ultrafilter,
immediate extension of an ultrametric field.

**Chapter 9**

transcendence order $\leq \tau$ over $\mathbb{Q}_p$,
weak transcendence order $\leq \tau$ over $\mathbb{Q}_p$, infinite order,
algebraic number in $\mathbb{C}_p$, transcendental number in $\mathbb{C}_p$,
denominator of an algebraic number.
**Chapter 10**

transcendence type $\leq \tau$ in $C_p$,
infinite transcendence type.

**Chapter 11**

analytic element on $D$,
invertible element of a space $H(D)$,
pole of order $q$ of an element $f \in H(D)$,
polynomial of poles of an element of $H(D)$ in $\bar{D} \setminus D$,
residue of an element $f \in H(D)$ at a pole $a$.

**Chapter 12**

bianalytic element from $A$ onto $B$.

**Chapter 13**

punctual semi-norms of $H(D)$.

**Chapter 14**

radius of convergence,
entire function,
power series, Laurent series,
zero of multiplicity order $q$.

**Chapter 15**

$f$-hole, Mittag-Leffler term of $f$ associated with a hole $T_n$,
principal term of $f$,
Mittag-Leffler series of $f$ on the infraconnected set $D$,
specific circular filter of a hole, of $\tilde{D}$,
residue of an analytic element on a hole.

**Chapter 16**

polynomial of zeros of an analytic element on a subset $A$,
semi-invertible analytic element,
quasi-invertible analytic elements,
quasi-minorated analytic element.

**Chapter 17**

analytic element vanishing along a filter $\mathcal{F}$,
analytic element properly vanishing along a filter $\tilde{\mathcal{F}}$.

**Chapter 18**

piercing of a subset,
well pierced subset.
**Chapter 20**
analytic element strictly vanishing along a monotonous filter $\mathcal{F}$,
analytic element collapsing along a monotonous filter $\mathcal{F}$.

**Chapter 21**
analytic set.

**Chapter 23**
strictly injective analytic element,
strictly injective analytic function,
bianalytic function.

**Chapter 28**
divisor on $K$, divisor on a disk $d(a, r^{-})$,
bounded divisor in a disk,
divisor of a function,
Euclidean division by a polynomial.

**Chapter 29**
index of an analytic element,
pure factor associated to a hole,
Motzkin factor in a hole,
Motzkin index,
$f$-supersequence,
Motzkin factorization,
principal factor.

**Chapter 30**
order of growth of an entire function,
cotype of growth of an entire function.

**Chapter 31**
type of growth of an entire function.

**Chapter 33**
order of growth of an analytic function inside a disk,
type of growth of an analytic function inside a disk,
cotype of growth of an analytic function inside a disk.

**Chapter 34**
contiguous filters,
quasi-invertible analytic function inside a disk.
**Chapter 35**

regular sequence,
regular ultrafilter,
coroner filter,
multib injective Banach algebra.

**Chapter 36**

meromorphic function in $K$, in a disk $d(a, R^{-})$,
pole of order $s$ of a meromorphic function,
meromorphic function holomorphic at a point $\alpha$,
divisor of a meromorphic function in $K$, in a disk $d(a, R^{-})$,
divisor of the poles of $f$,
exceptional value,
pseudo-exceptional value,
quasi-exceptional value.

**Chapter 37**

residue of a meromorphic function at a point $a$,
analytic element meromorphic in a hole,
singular part at a point $b$,
polynomial of the poles in a hole.

**Chapter 39**

exceptional value,
pseudo-exceptional value,
quasi-exceptional value,
Motzkin factor,
Motzkin index.

**Chapter 40**

circular filter surrounding another circular filter,
circular filter surrounding a monotonous filter,
monotonous surrounding a circular filter,
monotonous surrounding another monotonous filter,
*D*-bordering filter,
strictly *D*-bordering filter,
$\delta$-topology,
Shilov set,
Shilov boundary.
**Chapter 41**

degree of a rational function,

absolute degree of a rational function,

segment of the tree $\Phi(D)$.

**Chapter 42**

shortest path.

**Chapter 43**

counting function of zeros $Z(r, f)$, counting multiplicity,

counting function of zeros $\bar{Z}(r, f)$, ignoring multiplicity,

counting function of poles $N(r, f)$, counting multiplicity,

counting function of poles $\bar{N}(r, f)$, ignoring multiplicity,

characteristic function or Nevanlinna function $T(r, f)$.

**Chapter 47**

function of uniqueness for a family of functions, polynomial
of uniqueness,

function of strong uniqueness for a family of functions,
polynomial of strong uniqueness.

**Chapter 48**

singular point of a curve,

degenerate curve,

elliptic curve, hyperelliptic curve.

**Chapter 49**

perfectly branched value for $f$,

totally branched value for $f$.

**Chapter 51**

small function with respect to $f$,

perfectly branched small function with respect to $f$.

**Chapter 52**

functions $f, g$ sharing a function $h$ ignoring multiplicity.

**Chapter 53**

Laurent series of a function at a point,

regular point,

singular point,

point of high singularity,

Laurent series of $h \circ f$ at $\alpha$.
## Chapter 56

unique range set counting multiplicity, urscm for $\mathcal{F}$,
unique range set ignoring multiplicity, ursim for $\mathcal{F}$,
$f, g$ share $S$, counting multiplicity,
$f, g$ share $S$, ignoring multiplicity.

## Chapter 58

ramification index.

## Chapter 62

$q$th power of an equation,
irreducible equation.
# Notations

## Chapter 2

$\mathbb{N}, \mathbb{Z}, \mathbb{Q}, \mathbb{R}, \mathbb{C}$

$\bar{S}, \hat{S}$

$\bar{a} (a \in \mathbb{L}), \bar{P} (P \in \mathbb{L}[x])$

$|x|, (x \in E)$

$|t|_{\infty}, (t \in \mathbb{R})$

$\log, \theta, v, \Psi, |E|$

$U_E, M_E, \mathcal{L}$

$d(a, r), d(a, r^{-}), C(a, r)$

$\Delta(a, r', r''), \Gamma(a, r', r''), \delta(a, D), \delta(D, E)$

$\text{diam}(D), \text{codiam}(D)$

$\hat{L}, \hat{d}(a, r), \hat{d}(a, r^{-}), \hat{C}(a, r))$

$\hat{D}, \tilde{D}$

$\mathcal{J}(\Lambda), \mathcal{E}(\Lambda), \chi(D)$

## Chapter 3

$\mathcal{C}_D(\mathcal{F})$

$\mathcal{B}_D(\mathcal{F})$

$\text{diam}(\mathcal{F})$

$\hat{G}$

$\gamma$

$\hat{\mathcal{F}}, \hat{\mathcal{G}}$
## Chapter 4

$f^r, f^l$
$\| \cdot \|$
$\Phi(D), \Phi^0(D)$
$\nu(h, \mu)$
$\Psi(h, \mu)$
$\Psi_a(h, \mu)$
$h_a(u)$
$\Psi_a(h, \mu)$
$\| \cdot \|_0$
$|P|(r), P \in \mathbb{K}[x], |h|(r), h \in \mathbb{K}(x)$
$\varphi_{\mathcal{F}}, \varphi_{a,r}$
$\varphi_a$
$\nu^+(f, \mu), \nu^-(f, \mu), \nu(f, \mu).$
$\frac{d_r f}{d\mu}, \frac{d_l f}{d\mu}$

## Chapter 5

$(g, h)$
$B(f, g)$

## Chapter 6

irr$(a, L)$
$\mathcal{N}$
$|n|_p$
$\Psi_p$
$\mathbb{Z}_p, \mathbb{Q}_p, \Omega_p, \mathbb{C}_p, \mathbb{F}_p$

## Chapter 7

$u_s, r_s$
$A_s, B_s$
$F_s, G_s$
$W_s, W$
$\mathcal{H}$
## Chapter 8

$\mathcal{U}, \mathcal{R}$

$\hat{\mathbb{K}}$

$\hat{d}(a, r), \hat{d}(a, r^{-})$

$\hat{C}(a, r)$

## Chapter 9

$\mathcal{S}(\tau)$

$\mathcal{S}(\tau^{+})$

$\mathrm{Card}(\mathcal{S})$

$|z|_{\infty}, z \in \mathbb{C}$

$\mathrm{den}(a), a \in \Omega$

$\mathcal{T}(a), a \in \Omega$

$|a| = \max_{j=1,...,n}(|a_j|_{\infty}), a \in \Omega$

$s(a) = \max(\log(|a|, \mathrm{den}(a)))$

## Chapter 10

$\mathcal{U}_D$

$R(D), R_b(D), R_0(D)$

$\| \cdot \|_D$

$\mathrm{Mult}(A)$

$\mathrm{Mult}(A, \| \cdot \|)$

$\mathrm{Mult}_m(A, \| \cdot \|)$, $\mathrm{Mult}_1(A, \| \cdot \|)$

$\mathrm{Ker}(\phi)$

$\| \cdot \|_{si}$

## Chapter 11

$H(D), H_b(D), H_0(D)$

$\mathcal{U}_D$

Alg

$\mathrm{res}(f, a)$

## Chapter 12

$\Xi(D', D)$