Subfield: Algebra


Question: Let $P(x), Q(x), R(x)$ and $S(x)$ be non-constant polynomials with real coefficients such that $P(Q(x))=R(S(x))$. Suppose that the degree of $P(x)$ is divisible by the degree of $R(x)$.



Prove that there is a polynomial $T(x)$ with real coefficients such that $P(x)=R(T(x))$.

Solution:
Degree comparison of $P(Q(x))$ and $R(S(x))$ implies that $q=\operatorname{deg} Q \mid \operatorname{deg} S=s$. We will show that $S(x)=T(Q(x))$ for some polynomial $T$. Then $P(Q(x))=R(S(x))=R(T(Q(x)))$, so the polynomial $P(t)-R(T(t))$ vanishes upon substitution $t=S(x)$; it therefore vanishes identically, as desired.



Choose the polynomials $T(x)$ and $M(x)$ such that



$$

S(x)=T(Q(x))+M(x) \tag{*}

$$



where $\operatorname{deg} M$ is minimised; if $M=0$, then we get the desired result. For the sake of contradiction, suppose $M \neq 0$. Then $q \nmid m=\operatorname{deg} M$; otherwise, $M(x)=\beta Q(x)^{m / q}+M_{1}(x)$, where $\beta$ is some number and $\operatorname{deg} M_{1}<\operatorname{deg} M$, contradicting the choice of $M$. In particular, $0<m<s$ and hence $\operatorname{deg} T(Q(x))=s$.



Substitute now (*) into $R(S(x))-P(Q(x))=$ 0 ; let $\alpha$ be the leading coefficient of $R(x)$ and let $r=\operatorname{deg} R(x)$. Expand the brackets to get a sum of powers of $Q(x)$ and other terms including powers of $M(x)$ as well. Amongst the latter, the unique term of highest degree is $\operatorname{ar} M(x) T(Q(x))^{r-1}$. So, for some polynomial $N(x), \quad N(Q(x))=\operatorname{arM}(x) T(Q(x))^{r-1}+$ a polynomial of lower degree.



This is impossible, since $q$ divides the degree of the left-hand member, but not that of the righthand member.
All polynomials in the solution have real coefficients. As usual, the degree of a polynomial $f(x)$ is denoted $\operatorname{deg} f(x)$.



Of all pairs of polynomials $P(x), R(x)$, satisfying the conditions in the statement, choose one, say, $P_{0}(x), R_{0}(x)$, so that $P_{0}(Q(x))=R_{0}(S(x))$ has a minimal (positive) degree. We will show that $\operatorname{deg} R_{0}(x)=1$, say, $R_{0}(x)=\alpha x+\beta$ for some real numbers $\alpha \neq 0$ and $\beta$, so $P_{0}(Q(x))=\alpha S(x)+\beta$. Hence $S(x)=T(Q(x))$ for some polynomial $T(x)$.



Now, if $P(x)$ and $R(x)$ are polynomials satisfying $P(Q(x))=R(S(x))$, then $P(Q(x))=$ $R(T(Q(x)))$. Since $Q(x)$ is not constant, it takes infinitely many values, so $P(x)$ and $R(T(x))$ agree at infinitely many points, implying that $P(x)=$ $R(T(x))$, as required.



It is therefore sufficient to solve the problem in the particular case where $F(x)=P(Q(x))=$ $R(S(x))$ has a minimal degree. Let $d=$ $\operatorname{gcd}(\operatorname{deg} Q(x), \operatorname{deg} S(x))$ to write $\operatorname{deg} Q(x)=a d$ and $\operatorname{deg} S(x)=b d$, where $\operatorname{gcd}(a, b)=1$. Then $\operatorname{deg} P(x)=b c, \operatorname{deg} R(x)=a c$ and $\operatorname{deg} F(x)=a b c d$ for some positive integer $c$. We will show that minimality of $\operatorname{deg} F(x)$ forces $c=1$, so $\operatorname{deg} P(x)=b$, $\operatorname{deg} R(x)=a$ and $\operatorname{deg} F(x)=a b d$. The conditions $a=\operatorname{deg} R(x) \mid \operatorname{deg} P(x)=b$ and $\operatorname{gcd}(a, b)=1$ then force $a=1$, as stated above.



Consequently, the only thing we are left with is the proof of the fact that $c=1$. For convenience, we may and will assume that $P(x), Q(x), R(x), S(x)$ are all monic; hence so is $F(x)$. The argument hinges on the lemma below.



Lemma. If $f(x)$ is a monic polynomial of degree $m n$, then there exists a degree $n$ monic polynomial $g(x)$ such that $\operatorname{deg}\left(f(x)-g(x)^{m}\right)<(m-1) n$. (If $m=0$ or 1 , or $n=0$, the conclusion is still consistent with the usual convention that the identically zero polynomial has degree $-\infty$.)



Proof. Write $f(x)=\sum_{k=0}^{m n} \alpha_{k} x^{k}, \alpha_{m n}=1$, and seek $g(x)=\sum_{k=0}^{n} \beta_{k} x^{k}, \beta_{n}=1$, so as to fit the bill. To this end, notice that, for each positive integer $k \leq n$, the coefficient of $x^{m n-k}$ in the expansion of $g(x)^{m}$ is of the form $m \beta_{n-k}+\varphi_{k}\left(\beta_{n}, \ldots, \beta_{n-k+1}\right)$, where $\varphi_{k}\left(\beta_{n}, \ldots, \beta_{n-k+1}\right)$ is an algebraic expression in $\beta_{n}, \ldots, \beta_{n-k+1}$. Recall that $\beta_{n}=1$ to determine the $\beta_{n-k}$ recursively by requiring $\beta_{n-k}=$ $\frac{1}{m}\left(a_{m n-k}-\varphi_{k}\left(\beta_{n}, \ldots, \beta_{n-k+1}\right)\right), k=1, \ldots, n$.



The outcome is then the desired polynomial $g(x)$.



We are now in a position to prove that $c=$ 1. Suppose, if possible, that $c>1$. By the lemma, there exist monic polynomials $U(x)$ and $V(x)$ of degree $b$ and $a$, respectively, such that $\operatorname{deg}\left(P(x)-U(x)^{c}\right)<(c-1) b$ and $\operatorname{deg}(R(x)-$ $\left.V(x)^{c}\right)<(c-1) a$. Then $\operatorname{deg}\left(F(x)-U(Q(x))^{c}\right)=$ $\operatorname{deg}\left(P(Q(x))-U(Q(x))^{c}\right)<(c-1) a b d, \operatorname{deg}(F(x)-$ $\left.V(S(x))^{c}\right)=\operatorname{deg}\left(R(S(x))-V(S(x))^{c}\right)<(c-1) a b d$, so $\operatorname{deg}\left(U(Q(x))^{c}-V(S(x))^{c}\right)=\operatorname{deg}((F(x)-$ $\left.\left.V(S(x))^{c}\right)-\left(F(x)-U(Q(x))^{c}\right)\right)<(c-1) a b d$.



On the other hand, $U(Q(x))^{c}-V(S(x))^{c}=$ $(U(Q(x))-V(S(x)))\left(U(Q(x))^{c-1}+\cdots+\right.$ $\left.V(S(x))^{c-1}\right)$.



By the preceding, the degree of the left-hand member is (strictly) less than $(c-1) a b d$ which is precisely the degree of the second factor in the righthand member. This forces $U(Q(x))=V(S(x))$, so $U(Q(x))=V(S(x))$ has degree $a b d<a b c d=$ $\operatorname{deg} F(x)$ - a contradiction. Consequently, $c=1$. This completes the argument and concludes the proof.