non-trivial, in which case it must be carefully chosen. To understand the
best case scenario, we introduce the notion of *helper ideals*.

**Definition 20:** A primitive ideal I is a helper ideal if the norm $\mathcal{N}(I^c)$ of its complementary ideal $I^c$ is squarefree.

These ideals are interesting because they do not impact the value of $t$ as
explained by the following lemma.

**Lemma 21:** Let $p$ be a prime number not dividing $\tilde{t}qN\mathcal{N}(\mathfrak{b})$. Suppose that $h = mh'$ is an admissible vector such that the ideal generated by $h'$ has $p$-part a helper ideal dividing $p\mathcal{O}_{\mathbb{K}}$. This means that there is a prime ideal $\mathfrak{P}$ of norm $p$ such that $h'\mathfrak{P} \in p\mathcal{L}$ and $h' \notin p\mathcal{L}$. Then under the assumptions $(*)$, $v_p(l) = 1$ and $v_p(t) = 0$.

**Proof :**

We use once again formula (21). Under the assumptions $(*)$, we have $\tilde{\lambda} = 1$. Here, by assumption we also have $v_p(\tilde{t}) = 0$, $v_p(\mathcal{N}(h')/\mathcal{N}(L)) = d-1$. Note that $\mathfrak{P}$ is a degree one prime so all elements in $\mathcal{O}_{\mathbb{K}}$ are congruent to some integer mod $\mathfrak{P}$ via the isomorphism $\mathcal{O}_{\mathbb{K}}/\mathfrak{P} \simeq \mathbb{Z}/p\mathbb{Z}$. Hence, there are integers $r_j$ such that $u_j - r_j \in \mathfrak{P}$ for all $1 \le j \le r$ and therefore $(u_j - r_j)h' \in p\mathcal{L}$. Using lemma 17 successively for all indices $1 \le j \le r$ with integers $n_0 = -r_j$ and $n_i = 0$ for $1 \le i \le j-1$ we get that $p$ divides $c_{jj0}$. Therefore, $v_p(\lambda) \ge r = d-2$ and using formula (21) we get $v_p(lt) \le 1$. We now want to prove that $v_p(l) = 1$ so that $v_p(t) = 0$. This is essentially the same proof as for target ideals. There exists $0 \le j \le r$ such that $p$ does not divide $\alpha_j - nh'$ for all $n \in \mathbb{Z}$ (in fact this is true for all $j$ here). Take $m_j$ an integer such that $\tilde{\alpha}_j - m_j \in \mathfrak{P}$. There are an integer $k_j$ and a sign $\epsilon = \pm 1$ such that $\epsilon l \alpha_j = (\tilde{\alpha}_j - m_j)h' + k_j h'$. Suppose that $p$ does not divide $l$. Then there is an integer $n$ such that $\epsilon l n - k_j \in p\mathbb{Z}$ and $\epsilon l(\alpha_j - nh') = (\tilde{\alpha}_j - m_j)h' - (\epsilon l n - k_j)h' \in p\mathcal{L}$. This is once again a contradiction and we must conclude that $p$ divides $l$ and $p$ does not divide $t$. $\square$

### 4.3.3 Using target and helper ideals to choose $(h_\rho)_{\rho \in \mathfrak{S}_r}$

We now use both target ideals and helper ideals to describe how we choose
the vectors $h_\rho$ in Conjecture 11.

**Proposition 22:** Suppose that $\varepsilon_1, \dots, \varepsilon_r$ is a set of fundamental units for $\mathcal{O}_j^{+\times}$ satisfying (*) and (**). For all $\rho \in \mathfrak{S}_r$, write $p_{\rho,j}$ for the prime numbers dividing $\tilde{t}_{\rho}$ for $1 \le j \le j_{\rho}$ and $I_{\rho,j}$ for any corresponding target ideal in (**). Put $D = q \prod_{\rho} \prod_{j=1}^{j_{\rho}} p_{\rho,j}$. Choose for the smoothing ideal $\mathfrak{a}$ a degree one prime
$v_\ell(\mathcal{N}(h'_\rho)/\mathcal{N}(L)) = kd - k$ such that using formula (19) we get $v_\ell(t_\rho) = 0$. The value of $\mathcal{N}(\mathfrak{b})$ is completely invisible in formula (19) because

$$ \mathcal{N}(h'_{\rho})/\mathcal{N}(L) = q^{d-1}N^{d-1}\prod_{j=1}^{j_{\rho}}p_{\rho,j}^{2k_{\rho,j}-w_{\rho,j}}n_{\rho}^{d-1} $$

Lastly, $N/\mathfrak{a}$ is in fact a helper ideal for which we have $v_N(l_\rho) = 1$ and $v_N(t_\rho) = 0$. We use formula (19) and we obtain the values

$$ l_{\rho} = qNn_{\rho}\prod_{j=1}^{j_{\rho}}p_{\rho,j}^{k_{\rho,j}} $$

for the levels in the computations of the values $G_{r,f,\mathfrak{b},\mathfrak{a}}^{\pm}([\varepsilon_{\rho(1)}|\dots|\varepsilon_{\rho(r)}];h_\rho)$, as well as $t_\rho = 1$, which implies that the value $I_{r,f,\mathfrak{b},\mathfrak{a}}(\varepsilon_1, \dots, \varepsilon_r; \underline{h}, \underline{\mu}, \underline{\nu})$ is a product of $r!$ smoothed ordinary elliptic $G_r$ functions.

<figure><img src="image_5.png" /></figure>

**Remark:** Our choice of vectors $h_\rho$ are made to satisfy two important conditions. The main focus is to produce values for $I_{r,f,\mathfrak{b},\mathfrak{a}}(\varepsilon_1, \dots, \varepsilon_r; \underline{h}, \underline{\mu}, \underline{\nu})$ with arithmetical properties. It is plainly not true that all such products of $G_r$ functions yield algebraic integers. A key point in computing interesting $I_{r,f,\mathfrak{b},\mathfrak{a}}(\varepsilon_1, \dots, \varepsilon_r; \underline{h}, \underline{\mu}, \underline{\nu})$ is to make sure that there is some uniformisation on the choice of $h_\rho$ for $\rho \in \mathfrak{S}_r$. We believe that this uniformisation property has to do with the lattices $M_\rho$ defined in section 4.1. The condition $t_\rho = 1$ for all $\rho$ guarantees that $M_\rho = L$ for all $\rho$, which can be seen as the strongest form of uniformisation possible. On the other hand, this condition is useful to perform quick computations.

Before giving numerical evidence to support our conjecture, we explain how proposition 22 is used to compute the arithmetic $G_r$ functions.

1. Fix $\mathbb{K}$ and $f \neq \mathcal{O}_{\mathbb{K}}$. Compute fundamental units for $\mathcal{O}_{\mathbb{K}}^{\times}$ (as given by Pari/GP for example). Use these units to compute a set of fundamental units for $\mathcal{O}_{f}^{+,-\times}$.

2. Compute the values $\tilde{t}_\rho$ and $\lambda_\rho$ for all $\rho$. Check conditions (*) and (**) and compute target ideals. Note that when $\tilde{t}_\rho = 1$ there are no target ideals for the permutation $\rho$ and the empty product $\prod_{j=1}^{j_\rho} I_{\rho,j}$ is equal to the trivial ideal $\mathcal{O}_{\mathbb{K}}$. If the conditions (*) and (**) are not satisfied, try small base changes for the units.

3. Choose $\mathfrak{a}$ and $\mathfrak{b}$ as in proposition 22, with the smallest possible norms.
ideal of norm N coprime to D. Choose a representative integral ideal b for a class in $Cl^+(f)$ coprime to DN. Then there is an ideal $\mathfrak{J}$ independent of $\rho$ which is either $O_K$ or a helper ideal coprime to $DNN(b)$ and a positive integer $m_\rho$ coprime to $q$ such that for all $\rho \in S_r$ the ideal

$$q \frac{N}{ab} m_{\rho} \mathfrak{J} \prod_{j=1}^{j_{\rho}} I_{\rho,j}$$

is principal, with an admissible generator $h_{\rho}$. For all orientations $\mu, \nu$, the value $I_{r,f,b,a}(\epsilon_1, \dots, \epsilon_r; \underline{h}, \underline{\mu}, \underline{\nu})$ is a product of $r!$ smoothed ordinary elliptic $G_r$ functions.

**Proof:**

We want to pick an admissible generator of the ideal $q \frac{N}{ab} \prod_{j=1}^{j_{\rho}} I_{\rho,j}$ but this ideal may not be principal nor possess an admissible generator. We use helper ideals to deal with those as follows. By Cebotarev's density theorem (see [[6], Theorem 8.17]), there are infinitely many degree one prime ideals in each class of the wide ideal class group $Cl(f)$. For any prime integer $p$ coprime to $q$ there exists an integer $m(p)$ coprime to $q$ such that $pm(p) \equiv 1 \mod q$. Conditions (**) states that there exist integers $k_{\rho}$ such that all ideals $k_{\rho} \frac{N}{ab} \prod_{j=1}^{j_{\rho}} I_{\rho,j}$ belong to the same class $c$ in $Cl(f)$. Consider a degree one prime $\mathfrak{P}$ of norm $p$ coprime to $DNN(b)$ in the class $c$. Then we may choose $m_{\rho} = k_{\rho}m(p)$ and $\mathfrak{J} = p/\mathfrak{P}$. One notices that if the class $c$ is already the trivial class, then one may take $m_{\rho} = k_{\rho}$ and $\mathfrak{J} = O_K$. In either case, we may find a generator $g_{\rho}$ for the ideal $\frac{N}{ab}m_{\rho}\mathfrak{J} \prod_{j=1}^{j_{\rho}} I_{\rho,j}$ which is congruent to $1 \mod \mathfrak{f}\mathfrak{b}^{-1}$. We then obtain an admissible vector $h_{\rho} = qg_{\rho}$ because $h_{\rho}/q \equiv 1 \mod L$ and $h_{\rho}/N \in \mathfrak{a}^{-1}L - L$ is a generator of the cyclic group $\mathfrak{a}^{-1}L/L$ of prime order $N$.

Using lemma 19 we get $v_{p_{\rho,j}}(\lambda_{\rho}) = v_{p_{\rho,j}}(\tilde{s}_{\rho}), v_{p_{\rho,j}}(l_{\rho}) = v_{p_{\rho,j}}(\tilde{t}_{\rho})$ and $v_{p_{\rho,j}}(t_{\rho}) = 0$ for all $\rho, j$. Using lemma 21 for all prime divisors of $n$ such that $N(\mathfrak{J}) = n^{d-1}$ we get $l_{\rho} = nl'_{\rho}$ with $gcd(n, l'_{\rho}) = 1$ and $gcd(n, t_{\rho}) = 1$. Now we have to understand what happens for the $\ell$-adic valuations in formula (21) where $N(f) = q = \ell^k$. In this setting, we get $v_{\ell}(\lambda_{\rho}) = rk = (d-2)k$, $v_{\ell}(l_{\rho}) = k$, and $v_{\ell}(N(h'_{\rho})/N(L)) = kd - k$ such that using formula (21) we get $v_{\ell}(t_{\rho}) = 0$. The value of $N(b)$ is completely invisible in formula (21) because

$$\mathcal{N}(h'_{\rho})/\mathcal{N}(L) = q^{d-1}N^{d-1} \prod_{j=1}^{j_{\rho}} p_{\rho,j}^{v_{p_{\rho,j}}(\tilde{s}_{\rho})} n^{d-1}$$

Lastly, $N/a$ behaves in fact as a helper ideal for which we have $v_N(l_{\rho}) = 1$
and $v_N(t_\rho) = 0$. We use formula (21) and we obtain the values

$$l_{\rho} = qNn \prod_{j=1}^{j_{\rho}} p_{\rho,j}^{v_{\rho,j}(\tilde{t}_{\rho})}$$

for the levels in the computations of the values $G_{r,f,b,a}^{\pm}([\varepsilon_{\rho(1)}|\dots|\varepsilon_{\rho(r)}]; h_{\rho})$, as well as $t_{\rho} = 1$, which implies that the value $I_{r,f,b,a}(\varepsilon_1, \dots, \varepsilon_r; \underline{h}, \underline{\mu}, \underline{\nu})$ is a product of $r!$ smoothed ordinary elliptic $G_r$ functions. $\square$

**Remark:** Our choices for the vectors $h_\rho$ are made to satisfy two important conditions. The main focus is to produce values for $I_{r,f,b,a}(\varepsilon_1, \dots, \varepsilon_r; \underline{h}, \underline{\mu}, \underline{\nu})$ with arithmetical properties. It is plainly not true that all such products of $G_r$ functions yield algebraic numbers. A key point in computing interesting $I_{r,f,b,a}(\varepsilon_1, \dots, \varepsilon_r; \underline{h}, \underline{\mu}, \underline{\nu})$ is to make sure that the vectors $h_\rho$ share some common property as $\rho$ varies in $\mathfrak{S}_r$. We believe that this property has to do with the lattices $M_\rho$ defined in section 4.1 being somewhat “compatible” if not equal. The condition $t_\rho = 1$ for all $\rho$ guarantees that $M_\rho = L$ for all $\rho$, which can be seen as the strongest form of this common property. On the other hand, this condition is useful to perform quick computations.

### 4.3.4 Invariance under the choice of vectors $(h_\rho)_{\rho \in \mathfrak{S}_r}$

In this section we define the set $\mathcal{Z}_f^1$ appearing in Conjecture 11 and explain why it is necessary to compute an average on this set. We first begin by two simple lemmas.

**Lemma 23:** Take $h$ an admissible vector for $f, b, a$. Fix a system $u_1, \dots, u_r$ of linearly independent units. Consider $n$ a positive integer congruent to $1 \mod q$. Then

$$G_{r,f,b,a}^{\mu}(u_1, \dots, u_r; nh) = G_{r,f,b,a}^{\mu}(u_1, \dots, u_r; h)$$

**Proof:**

Take $\underline{\alpha}, \gamma, M$ as defined in section 4.1 for the orientation $\mu$. Write $h = m'\gamma$ for some integer $m' \in \mathbb{Z} - \{0\}$. Then

$$G_{r,f,b,a}^{\mu}(u_1, \dots, u_r; h) = \prod_{\delta \in L/M} \frac{G_r\left(\frac{m'}{q} + \frac{\delta}{\gamma}, \frac{\alpha_0}{\gamma}, \dots, \frac{\alpha_r}{\gamma}\right)}{G_r\left(\frac{Nm'}{q} + \frac{N\delta}{\gamma}, \frac{N\alpha_0}{\gamma}, \dots, \frac{N\alpha_r}{\gamma}\right)}$$
## 5.1 Cubic examples

Here we present six examples in the cubic case. The first example showcases the work on target ideal with $\tilde{t} = 131$ in the spirit of (*) and (**). The second example is one of the simplest ones, with $\tilde{t} = 1$ and class number one. The third example is a case where $q = 5$. The fourth and fifth examples showcase the work on helper ideals when the class number gets larger, for $q = 3$ and $q = 7$. The sixth example shows what can be done outside (*), (**), in the cubic case.

### 5.1.1 Example 1

We first discuss in full detail a cubic example. Let $z$ be the complex root of the polynomial $x^3 - 13$ lying in the upper half-plane. Then $K_1 = \mathbb{Q}(z)$ has class number 3. We choose $f^3 = (3)$. The unit group $O_f^{+, \times}$ is generated by $\varepsilon = 2z^2 - 3z - 4$. We compute $\tilde{\lambda} = 1$, $\tilde{s} = \tilde{t} = 131$. Then, we search for target ideals above the prime number 131 and we find the unique degree one prime $\mathfrak{P}_{131}$ above 131. Choose $b = (1)$ and $a$ the unique degree one prime above 5 in $K_1$. Then the ideal $qN\mathfrak{P}_{131}/a$ has an admissible generator $h = -21z^2 - 42z - 114$ so that we don't need to find a helper ideal. The corresponding level is $l = 3 \cdot 5 \cdot 131 = 1965$. The value

$$ \Gamma_{f,b,a}^{+}(\varepsilon, h)^{-1} = \frac{\Gamma\left(\frac{-1}{3}, \frac{-\varepsilon^{-1}-5348}{1965}, \frac{-\varepsilon-467}{1965}\right)^{-5}}{\Gamma\left(\frac{-5}{3}, \frac{-\varepsilon^{-1}-5348}{393}, \frac{-\varepsilon-467}{393}\right)^{-1}} \approx -0.0660917... + i \cdot 0.0932299... $$

is close to a root of the palindromic polynomial $x^{18} + 384x^{17} + 2310x^{16} - 10646490x^{15} + 1596241353x^{14} + 18608357181x^{13} + 156933809421x^{12} + 215098256580x^{11} + 381407365338x^{10} + 338205493469x^9 + \dots$ which defines $K_1^+(f)$. The computation time for 1000 digits is 11 seconds. Take another admissible vector $h = 3z^2 + 6z + 12$. The valuation of $h$ at any prime ideal dividing 131 is 0. This is one of the simplest admissible vectors we could find. Using this vector $h$ we have to compute $t = 131$ ordinary elliptic Gamma functions and the computation time is 30 seconds for 1000 digits. In the cubic case, because the computation time for the precision $\delta$ is $O(l|\log(\delta)|t)$, reducing the value of $t$ by raising the level $l$ may not result in a massive time gain. In the cases $r \ge 2$, the gain will be more substantial.

### 5.1.2 Example 2

We now discuss one of the simplest cubic examples which was presented in the introduction. Let $z$ be the complex root of the polynomial $x^3 - 2$ in the
Because $n > 0$ the same $\underline{\alpha}, \gamma, M$ are used to compute

$$G_{r,f,b,a}^{\mu}(u_1, \dots, u_r; nh) = \prod_{\delta \in L/M} \frac{G_r \left( \frac{m'}{q} + \frac{m'(n-1)}{q} + \frac{\delta}{\gamma}, \frac{\alpha_0}{\gamma}, \dots, \frac{\alpha_r}{\gamma} \right)}{G_r \left( \frac{Nm'}{q} + \frac{Nm'(n-1)}{q} + \frac{N\delta}{\gamma}, \frac{N\alpha_0}{\gamma}, \dots, \frac{N\alpha_r}{\gamma} \right)}$$

Because the function $G_r$ is 1-periodic in its first argument and $(n-1)/q \in \mathbb{Z}$ we have the desired equality. $\square$

**Lemma 24:** Take $h$ an admissible vector for $f, b, a$. Fix a system $u_1, \dots, u_r$ of linearly independent units. Then for all orientations $\mu, \nu$ and for all unit $\varepsilon \in \mathcal{O}_{\mathbb{K}}^{\times}$:

$$G_{r,f,b,a}^{\mu}(u_1, \dots, u_r; \varepsilon h)^{\nu} = G_{r,f,b,a}^{\mu \mathcal{N}(\varepsilon)}(u_1, \dots, u_r; h)^{\nu \mathcal{N}(\varepsilon)}$$

**Proof :**

Let us write $\eta = \mathcal{N}(\varepsilon) = \pm 1$. We analyse how the construction in section 4.1 changes when the vector $h$ is replaced with $\varepsilon h$. First, the linear forms $a_h$ and $a_{\varepsilon h}$ are related by:

$$\lambda a_{\varepsilon h} = \det_{B_L}(\varepsilon h, u_1 \varepsilon h, \dots, u_r \varepsilon h, \cdot)$$

$$\lambda a_{\varepsilon h} = \eta \det_{B_L}(h, u_1 h, \dots, u_r h, \varepsilon^{-1} \cdot)$$

$$\lambda a_{\varepsilon h} = \lambda \eta a_h (\varepsilon^{-1} \cdot)$$

This implies directly that for all $0 \le j \le r$ we have $\alpha_{\varepsilon h, j} = \eta \varepsilon \alpha_{h, j}$ and

$$s \gamma_{\varepsilon h} = \det_{B_{\Lambda}} (a_{\varepsilon h}, u_1 a_{\varepsilon h}, \dots, u_r a_{\varepsilon h}, \cdot) = s \eta^r \varepsilon \gamma_h$$

This also gives $\eta \varepsilon L/M_h = L/M_{\varepsilon h}$. Thus, using $\gamma = \gamma_h$, $\alpha_j = \alpha_{h, j}$, $M = M_h$ we get:

$$G_{r,f,b,a}^{\mu}(u_1, \dots, u_r; \varepsilon h)^{\nu} = \prod_{\delta \in L/M} \frac{G_r \left( \frac{\varepsilon h + q \eta \varepsilon \delta}{q \eta^r \varepsilon \gamma}, \frac{\eta \varepsilon \alpha_0}{\eta^r \varepsilon \gamma}, \dots, \frac{\eta \varepsilon \alpha_r}{\eta^r \varepsilon \gamma} \right)^{\nu}}{G_r \left( \frac{N(\varepsilon h + q \eta \varepsilon \delta)}{q \eta^r \varepsilon \gamma}, \frac{N \eta \varepsilon \alpha_0}{\eta^r \varepsilon \gamma}, \dots, \frac{N \eta \varepsilon \alpha_r}{\eta^r \varepsilon \gamma} \right)^{\nu}}$$

It then follows from formula (8) that

$$G_{r,f,b,a}^{\mu}(u_1, \dots, u_r; \varepsilon h)^{\nu} = \prod_{\delta \in L/M} \frac{G_r \left( \frac{\eta h + q \delta}{q \gamma}, \frac{\alpha_0}{\gamma}, \dots, \frac{\alpha_r}{\gamma} \right)^{\nu \eta^{r+1}}}{G_r \left( \frac{N(\eta h + q \delta)}{q \gamma}, \frac{N \alpha_0}{\gamma}, \dots, \frac{N \alpha_r}{\gamma} \right)^{\nu \eta^{r+1}}}$$
upper half-plane. Then $K_2 = \mathbb{Q}(z)$ has class number 1. We choose $f^3 = (3)$. Then $\epsilon = z - 1$ is a generator for $O_f^{+,\times}$ such that $\tilde{\lambda} = \tilde{t} = 1$. The ideal $qN/\mathfrak{a}$ has an admissible generator $h = 9z^2 - 3z - 9$. The corresponding level is $l = 15$. The value

$$ \Gamma_{f, \mathfrak{b}, \mathfrak{a}}^{+}(\epsilon, h)^{-1} = \frac{\Gamma\left(\frac{-1}{3}, \frac{\epsilon^{-1}+2}{15}, \frac{\epsilon-7}{15}\right)^{-5}}{\Gamma\left(\frac{-5}{3}, \frac{\epsilon^{-1}+2}{3}, \frac{\epsilon-7}{3}\right)^{-1}} \approx -1.2937005\dots + i \cdot 1.4743341\dots $$

is close to a root of the polynomial $x^6 + 3x^5 + 6x^4 + 5x^3 + 6x^2 + 3x + 1$ which defines $K_2^+(f)$. The computation time for 1000 digits is 1 second.

### 5.1.3 Example 3

Here we explain how we have obtained the example in section 3.2 with $q = 5$. Let $z = e^{2i\pi/3}10^{1/3}$ be the root of the polynomial $x^3 - 10$ in the upper half-plane. Then $K_3 = \mathbb{Q}(z)$ has class number 1. We choose $f^3 = (5)$. Then $\epsilon = (2z^2 - z - 7)/3$ is a generator for $O_f^{+,\times}$ such that $\tilde{\lambda} = 1$ and $\tilde{t} = 9$. We may choose $\mathfrak{b} = (1)$ and $\mathfrak{a}$ the unique degree one prime above 11 in $K_3$. We find the target ideal $\mathfrak{P}_3$ above 3 where $\mathfrak{P}_3\mathfrak{P}_3' = (3)$. Then, $h = -(35z^2 + 20z + 35)/3$ is an admissible generator for the ideal $qN\mathfrak{P}_3^2/\mathfrak{a}$. The value

$$ \Gamma_{f, \mathfrak{b}, \mathfrak{a}}^{+}(\epsilon, h) = \frac{\Gamma\left(\frac{1}{5}, \frac{-\epsilon^{-1}+1751}{495}, \frac{-\epsilon+776}{495}\right)^{11}}{\Gamma\left(\frac{11}{5}, \frac{-\epsilon^{-1}+1751}{45}, \frac{-\epsilon+776}{45}\right)} \approx -27.5333588\dots - i \cdot 32.7146180\dots $$

is close to a root of the polynomial $x^{12} + 57x^{11} + 1956x^{10} + 4640x^9 + 35415x^8 - 109818x^7 + 150139x^6 - 109818x^5 + 35415x^4 + 4640x^3 + 1956x^2 + 57x + 1$ which defines $K_3^+(f)$. The computation time for 1000 digits is 7 seconds.

### 5.1.4 Example 4

Let $z$ be the complex root of the polynomial $x^3 - 65$ in the upper half-plane. Then $K_4 = \mathbb{Q}(z)$ has class number 18. This means that most ideals won't be principal ideals and we will need to use helper ideals to build an admissible vector $h$. We choose $f^3 = (3)$. Then $\epsilon = z - 4$ is a generator for $O_f^{+,\times}$ such that $\tilde{\lambda} = \tilde{t} = 1$. We choose $\mathfrak{b} = (1)$, $\mathfrak{b} = \mathfrak{P}_{59}, \dots$ representatives for the 36 classes in $Cl^+(f)$, and $\mathfrak{a}$ the unique degree one prime above 5 in $K_4$. The ideals $qN/(\mathfrak{a}\mathfrak{b})$ are unfortunately not principal in general so we use helper ideals. In the case $\mathfrak{b} = (1)$ we find the helper ideal $\mathfrak{P}_{157}\mathfrak{P}_{157}'$ and in the case $\mathfrak{b} = \mathfrak{P}_{59}$ we find the helper ideal $\mathfrak{P}_{101}'$. The corresponding admissible generators we found for $qN\mathfrak{P}_{101}'/\mathfrak{a}$ and $qN\mathfrak{P}_{157}\mathfrak{P}_{157}'/(\mathfrak{a}\mathfrak{P}_{59})$ are $h_{\mathfrak{b}=(1)} = 3z^2 + 600z - 2460$
Using formula (23) we recognize the right-hand side as the expression of

$$G_{r,f,b,a}^{\mu\eta}(u_1, \dots, u_r; h)^{\nu\eta} = G_{r,f,b,a}^{\mu N(\epsilon)}(u_1, \dots, u_r; h)^{\nu N(\epsilon)}$$

which gives the desired result.

<figure><img src="image_3.png" /></figure>

Consider now two possible vectors $g, h$ associated to $[\epsilon_{\rho(1)}| \dots | \epsilon_{\rho(r)}]$ in Proposition 22, for the same choice of helper ideal $\mathfrak{I}$. This means that there is a fractional ideal $J$ of $\mathbb{K}$ such that $(g) = mJ$ and $(h) = m'J$ for some integers $m, m' \in \mathbb{Z}_{>0}$ coprime to $q$. Then the ideal $mJ$ is generated by $mh/m'$ so there exists a unit $\epsilon \in \mathcal{O}_{\mathbb{K}}^{\times}$ such that $g = m\epsilon h/m'$. Moreover, because $g, h \equiv 1 \mod \mathfrak{f}$ we get $m\epsilon/m' \equiv 1 \mod \mathfrak{f}$. This leads us to define

$$\mathbb{Z}_{\mathfrak{f}}^{1} = \{(n, \epsilon) \in \mathbb{Z}/q\mathbb{Z}^{\times} \times \mathcal{O}_{\mathbb{K}}^{\times}, n\epsilon \equiv 1 \mod \mathfrak{f}\} \quad (24)$$

In addition, if $n\epsilon \equiv n\epsilon' \equiv 1 \mod \mathfrak{f}$ then $\epsilon\epsilon'^{-1} \equiv 1 \mod \mathfrak{f}$ and we work under the assumption that all units congruent to $1 \mod \mathfrak{f}$ have norm +1. Therefore, $\mathcal{N}(\epsilon) = \mathcal{N}(\epsilon')$. We may then consider the following quotient space.

**Definition 25:** The elements $(n, \epsilon)$ and $(n', \epsilon')$ in $\mathbb{Z}_{\mathfrak{f}}^{1}$ are considered equivalent if $n = n'$ (remember that $n$ and $n'$ represent classes in $\mathbb{Z}/q\mathbb{Z}^{\times}$). We write $\mathbb{Z}_{\mathfrak{f}}^{1} = \mathbb{Z}_{\mathfrak{f}}^{1}/\sim$ for the corresponding equivalence classes. This set forms a group which is naturally embedded in $\mathbb{Z}/q\mathbb{Z}^{\times}$.

It is easy to see that $\mathbb{Z}_{\mathfrak{f}}^{1}$ contains at least $(1, 1)$ and $(q-1, -1)$ and that these two elements differ unless $\mathbb{K}$ is of even degree and $q = 2$. It follows from lemma 24 that the product

$$\prod_{(n, \epsilon) \in \mathbb{Z}_{\mathfrak{f}}^{1}} G_{r, f, b, a}^{\mu}(u_1, \dots, u_r; n\epsilon h)^{\nu}$$

is well-defined for any orientations $\mu, \nu$ where it is understood that $(n, \epsilon) \in \mathbb{Z}_{\mathfrak{f}}^{1}$ always comes with $n > 0$. We now want to prove the following invariance proposition:

**Proposition 26:** Consider two vectors $g, h$ associated to the unit system $[\epsilon_{\rho(1)}| \dots | \epsilon_{\rho(r)}]$ in Proposition 22 with the same helper ideal $\mathfrak{I}$. Fix the orientations $\mu, \nu$. Then:

1. There exists $(n, \epsilon) \in \mathbb{Z}_{\mathfrak{f}}^{1}$ such that

$$G_{r,f,b,a}^{\mu}(u_1, \dots, u_r; g)^{\nu} = G_{r,f,b,a}^{\mu}(u_1, \dots, u_r; n\epsilon h)^{\nu}$$
and $h_{b=\mathfrak{P}_{59}} = (-222z^2 + 4710z - 15375)/59$ respectively. The corresponding levels will be $l_{(1)} = 3 \cdot 5 \cdot 157$, $l_{\mathfrak{P}_{59}} = 3 \cdot 5 \cdot 101$. Then we compute

$$ \Gamma_{f,(1),\mathfrak{a}}^{+}(\varepsilon, h)^{-1} = \frac{\Gamma\left(\frac{-1}{3}, \frac{-\varepsilon^{-1}+781}{2355}, \frac{-\varepsilon-32774}{2355}\right)^{-5}}{\Gamma\left(\frac{-5}{3}, \frac{-\varepsilon^{-1}+781}{471}, \frac{-\varepsilon-32774}{471}\right)^{-1}} \approx -1.6691052\dots + i \cdot 5.7493283\dots $$

$$ \Gamma_{f,\mathfrak{P}_{59},\mathfrak{a}}^{+}(\varepsilon, h)^{-1} = \frac{\Gamma\left(\frac{1}{3}, \frac{-\varepsilon^{-1}+676}{1515}, \frac{-\varepsilon-26954}{1515}\right)^{-5}}{\Gamma\left(\frac{5}{3}, \frac{-\varepsilon^{-1}+676}{303}, \frac{-\varepsilon-26954}{303}\right)^{-1}} \approx 0.0344135\dots - i \cdot 0.0123218\dots $$

and the remaining 34 out of 36 values $I_b = \Gamma_{f,b,\mathfrak{a}}(\varepsilon, h, \sigma_C)$ attached to the 36 classes in $Cl^+(f)$. We may then compute the polynomial

$$ \prod_{b \in Cl^+(f)} (X - I_b) = X^{36} - (u + vz + wz^2)X^{35} + \dots \in \mathcal{O}_{\mathbb{K}_4}[X] $$

where $u = 93377174024326$, $v = 769211619985$, $w = -5967373310133$. This palindromic polynomial defines a relative equation of the class field $\mathbb{K}_4^+(f)$ above $\mathbb{K}_4$ and we identify the rest of its coefficients in $\mathcal{O}_{\mathbb{K}_4}$. The computation time for 1000 digits for each of the 36 computations is 1 minute and 20 seconds on average.

### 5.1.5 Example 5

Here we give an example with $q = 7$. Let $z$ be the complex root of the polynomial $x^3 - 14$ in the upper half-plane. Then $\mathbb{K}_5 = \mathbb{Q}(z)$ has class number 3. We choose $f^3 = (7)$. Then $\varepsilon = -z^2 + 2z + 1$ is a generator for $\mathcal{O}_f^{+\times}$ such that $\tilde{\lambda} = 1$, $\tilde{t} = 2 \cdot 11$. We choose $\mathfrak{b} = (1)$ and $\mathfrak{a}$ the unique degree one prime above 5 in $\mathbb{K}_5$. Computing target ideals gives $\mathfrak{P}_2$ and $\mathfrak{P}_{11}$ the degree one primes above 2 and 11. The ideal $qN\mathfrak{P}_2\mathfrak{P}_{11}/(\mathfrak{a}\mathfrak{b})$ is unfortunately not principal so we need to look for a helper ideal. One of the helper ideals with smallest norm we find is $\mathfrak{P}_{53}$ the degree two prime above 53 in $\mathbb{K}_5$. The associated admissible generator is $h = 392z^2 + 1043z + 2212$. The corresponding level will be $l = 2 \cdot 5 \cdot 7 \cdot 11 \cdot 53$. The value

$$ \Gamma_{f,b,\mathfrak{a}}^{+}(\varepsilon, h)^{-1} = \frac{\Gamma\left(\frac{-1}{7}, \frac{-\varepsilon^{-1}-445073}{40810}, \frac{-\varepsilon-17177}{40810}\right)^{-5}}{\Gamma\left(\frac{-5}{7}, \frac{-\varepsilon^{-1}-445073}{8162}, \frac{-\varepsilon-17177}{8162}\right)^{-1}} \approx -0.1700923\dots + i \cdot 3.8609499\dots $$

is close to a root of the palindromic polynomial $x^{54} - 4167x^{53} + 7931535x^{52} - 259219286x^{51}\dots$ which defines $\mathbb{K}_5^+(f)$ over $\mathbb{Q}$. This polynomial has very large coefficients, and we could alternatively compute the remaining 17 out of 18
2. The following equality holds:

$$ \prod_{(n,\varepsilon) \in \mathcal{Z}_f^1} G_{r,f,b,a}^{\mu}(u_1, \dots, u_r; n\varepsilon g)^{\nu} = \prod_{(n,\varepsilon) \in \mathcal{Z}_f^1} G_{r,f,b,a}^{\mu}(u_1, \dots, u_r; n\varepsilon h)^{\nu} $$

**Proof :**

1. Write once again $g = m\varepsilon h/m'$ where $m, m' > 0$ are coprime to $q$ and $m\varepsilon \equiv m' \mod f$. This gives:

$$ G_{r,f,b,a}^{\mu}(u_1, \dots, u_r; g)^{\nu} = G_{r,f,b,a}^{\mu}(u_1, \dots, u_r; m\varepsilon h/m')^{\nu} $$

Consider a positive integer $m''$ such that $m'm'' \equiv 1 \mod q$. Then $mm''\varepsilon \equiv m'm'' \equiv 1 \mod f$, so $(mm'', \varepsilon) \in \mathcal{Z}_f^1$ and it follows from lemma 23 that

$$ G_{r,f,b,a}^{\mu}(u_1, \dots, u_r; m\varepsilon h/m')^{\nu} = G_{r,f,b,a}^{\mu}(u_1, \dots, u_r; mm''\varepsilon h)^{\nu} $$

2. Write once again $g = m\varepsilon h/m'$ with $m\varepsilon \equiv m' \mod f$ and fix a positive integer $m''$ such that $mm'' \equiv 1 \mod q$. Then for all $(n', \varepsilon') \in \mathcal{Z}_f^1$, we have $n'\varepsilon'g = (mn')(\varepsilon\varepsilon')h/m'$ and

$$ G_{r,f,b,a}^{\mu}(u_1, \dots, u_r; n'\varepsilon'g)^{\nu} = G_{r,f,b,a}^{\mu}(u_1, \dots, u_r; mm''n'\varepsilon\varepsilon'h)^{\nu} $$

Then we get for the whole product:

$$ \prod_{(n', \varepsilon') \in \mathcal{Z}_f^1} G_{r,f,b,a}^{\mu}(u_1, \dots, u_r; n'\varepsilon'g)^{\nu} = \prod_{(n', \varepsilon') \in \mathcal{Z}_f^1} G_{r,f,b,a}^{\mu}(u_1, \dots, u_r; mm''n'\varepsilon\varepsilon'h)^{\nu} $$

with $mm''n'\varepsilon\varepsilon' \equiv mm''\varepsilon \equiv m'm'' \equiv 1 \mod f$. We put $n'' = mm''n', \varepsilon'' = \varepsilon\varepsilon'$ and we get $(n'', \varepsilon'') \in \mathcal{Z}_f^1$. The map $(n', \varepsilon') \to (n'', \varepsilon'')$ defines a bijection of $\mathcal{Z}_f^1$ and we get:

$$ \prod_{(n', \varepsilon') \in \mathcal{Z}_f^1} G_{r,f,b,a}^{\mu}(u_1, \dots, u_r; n'\varepsilon'g)^{\nu} = \prod_{(n'', \varepsilon'') \in \mathcal{Z}_f^1} G_{r,f,b,a}^{\mu}(u_1, \dots, u_r; n''\varepsilon''h)^{\nu} $$

□

We have thus successfully proved that for a fixed helper ideal $\mathfrak{I}$ the product in the right-hand side of Conjecture 11 is independent of the choice of vectors $h_\rho$ in Proposition 22. We now want to give another formulation for the term

$$ \prod_{(n,\varepsilon) \in \mathcal{Z}_f^1} G_{r,f,b,a}^{\mu}(u_1, \dots, u_r; n\varepsilon h)^{\nu} $$
values associated to the 18 classes in $Cl^+(\mathfrak{f})$ to identify a relative polynomial in $O_{K_5}[x]$ instead, as we did in the previous example. The computation time for the above value and for 1000 digits of precision is 6 minutes and 50 seconds, because the value of the level $l$ is quite high.

### 5.1.6 Example 6

This example falls outside of the conditions (*) and (**) because we will have $\tilde{\lambda} = 2$. Yet, it shows that these conditions are a set of p-adic conditions which can be satisfied for most $p$. In other words, the unique prime number for which (*) and (**) are not true in the present case is 2. Because of this, we can try to reduce the number of computations outside of the prime number 2.

Let $z$ be the complex root of the polynomial $x^3 - 5$ in the upper half-plane. Then $K_6 = \mathbb{Q}(z)$ has class number 1. We choose $f^3 = (3)$. Then $\varepsilon = 2z^2 - 4z + 1$ is a generator for $O_f^{+,*}$ such that $\tilde{\lambda} = 2$ and $\tilde{t} = 2 \times 13$. We choose $b = 1$ and $a$ the unique degree one prime above 5 in $K_6$. We find $\mathfrak{P}_{13}$ a target ideal above 13 and we find an admissible generator $h = 6z^2 + 15$ for the ideal $qN\mathfrak{P}_{13}/a$. The corresponding level will be $l = 3 \cdot 5 \cdot 13$. Then $\Gamma_{f,b,a}^+(\varepsilon, h)^{-1}$ is a product of two ordinary smoothed elliptic Gamma functions with parameters

$$ \tau = \frac{-\varepsilon^{-1} - 119}{2 \times 195}, \quad \sigma = \frac{-\varepsilon - 59}{2 \times 195} $$

More precisely, put $\delta = -2z^2 - z - 5$. Then the set $L/M$ in formula (21) is equal to $\{0, \delta\}$ and $\delta/q\gamma = (-4z^2 - 5z - 55)/195$. The product

$$ \frac{\Gamma\left(\frac{1}{3}, \frac{-\varepsilon^{-1}-119}{390}, \frac{-\varepsilon-59}{390}\right)^{-5}}{\Gamma\left(\frac{5}{3}, \frac{-\varepsilon^{-1}-119}{78}, \frac{-\varepsilon-59}{78}\right)^{-1}} \cdot \frac{\Gamma\left(\frac{1}{3} + \frac{-4z^2-5z-55}{195}, \frac{-\varepsilon^{-1}-119}{390}, \frac{-\varepsilon-59}{390}\right)^{-5}}{\Gamma\left(\frac{5}{3} + \frac{-4z^2-5z-55}{39}, \frac{-\varepsilon^{-1}-119}{78}, \frac{-\varepsilon-59}{78}\right)^{-1}} $$

is close to the root $\approx 9.8439696... + i \cdot 5.1060682...$ of the palindromic polynomial $x^6 - 21x^5 + 150x^4 - 185x^3 + 150x^2 - 21x + 1$ which defines $K_6^+(\mathfrak{f})$. The computation time for 1000 digits is 11 seconds. Note that in the present case, we could have chosen a different target ideal, namely $\mathfrak{P}_{13}^2\mathfrak{P}_{13}'$ where $\mathfrak{P}_{13}'$ is one of the other primes above 13, but in this case the level $l$ would be $169 \times 195$ instead of 195.

## 5.2 Quartic examples

We now present four examples in the quartic case. The first example is one of the simplest cases, for which $\tilde{t}_1 = \tilde{t}_2 = 1$. The second example
which is useful for computations. Let us define the following character on $\mathbb{Z}_f^1$:

$$ \chi_f := \begin{cases} \mathbb{Z}_f^1 \to \{-1, 1\} \\ (n, \varepsilon) \to \mathcal{N}(\varepsilon) \end{cases} $$

This character is well-defined because units congruent to 1 mod $f$ all have norm +1 and it only depends on $n$, thus we will often write $\chi_f(n)$ for $\chi_f(n, \varepsilon)$. This specific character is closely related to norm-type characters of the class group $Cl^+(f)$ as follows. Consider any such character $\chi$. This means that there exists a character $\chi_{finite}$ of $(\mathcal{O}_K/f)^\times$ such that for all element $a \in K$ coprime to $f$, $\chi((a)) = \chi_{finite}(a)sg(\mathcal{N}(a))$. Using this for $(n, \varepsilon) \in \mathbb{Z}_f^1$ we have

$$ \chi((n)) = \chi((n\varepsilon)) = \chi_{finite}(n\varepsilon)\mathcal{N}(\varepsilon) = \mathcal{N}(\varepsilon) $$

because $n\varepsilon \equiv 1 \mod f$ so $\chi((n)) = \chi_f(n)$. Therefore $\chi_f(n)$ is the common value of all norm-type characters of the class group $Cl^+(f)$ at $(n)$. Note that the functional equation of Hecke $L$-functions in ATR fields implies that if $L'(0, \chi) \neq 0$ for a character $\chi$ of the class group $Cl^+(f)$ then $\chi$ is of norm-type (see for instance [15] for the functional equation). It is worth noting that $\chi_f(1) = 1$ and $\chi_f(q-1) = (-1)^d$ where $d$ is the degree of $K$. We use this character to write using lemma 24:

$$ \prod_{(n,\varepsilon) \in \mathbb{Z}_f^1} G_{r,f,b,a}^{\mu}(u_1, \dots, u_r; n\varepsilon h)^{\nu} = \prod_{(n,\varepsilon) \in \mathbb{Z}_f^1} G_{r,f,b,a}^{\mu\chi_f(n)}(u_1, \dots, u_r; nh)^{\chi_f(n)\nu} \quad (25) $$

The right-hand side of formula (25) may be explicitly written as

$$ \prod_{(n,\varepsilon) \in \mathbb{Z}_f^1} G_{r,f,b,a}^{\mu\chi_f(n)}(u_1, \dots, u_r; nh)^{\chi_f(n)\nu} = \prod_{(n,\varepsilon) \in \mathbb{Z}_f^1} \prod_{\delta \in L/M} \left( \frac{G_r\left(\frac{\mu\chi_f(n)nh+q\delta}{q\gamma}, \frac{\alpha_0}{\gamma}, \dots, \frac{\alpha_r}{\gamma}\right)^N}{G_r\left(\frac{\mu\chi_f(n)nNh+Nq\delta}{q\gamma}, \frac{N\alpha_0}{\gamma}, \dots, \frac{N\alpha_r}{\gamma}\right)} \right)^{\nu\mu^r\chi_f(n)^{r+1}} \quad (26) $$

where $\alpha, \gamma, M$ are defined in section 4.1. Most of the computation of this double product can be done at once for all terms since the parameters $\alpha_0/\gamma, \dots, \alpha_r/\gamma$ are fixed. Remember that in the context of $(\ast)$, $(\ast\ast)$ this
showcases the work on target ideals with $\tilde{t}_1 = 7, \tilde{t}_2 = 7129$. The third example shows that we can sometimes hope to find better units to gain time in the computations. The fourth example is a case where the quartic field $\mathbb{K}$ contains a real quadratic field and the set of units falls just outside $(*)$, $(**)$.
In all examples, we use the ordering $\{Id, (21)\}$ of $\mathfrak{S}_2$ and when we say that we “choose” fundamental units for $\mathcal{O}_f^{+,\times}$, we let Pari/GP compute a set of fundamental units for $\mathcal{O}_{\mathbb{K}}^{\times}$, which we use to compute a set of fundamental units for $\mathcal{O}_f^{+,\times}$. Most of the time Pari/GP does a great job in finding small fundamental units, and the few operations required to compute $\mathcal{O}_f^{+,\times}$ keep these units relatively small.

### 5.2.1 Example 7

We first discuss one of the simplest quartic examples mentioned in section 3.2. Let $z$ be the complex root of the polynomial $x^4 - 6x^3 - x^2 - 3x + 1$ lying in the upper half-plane. Then $\mathbb{K}_7 = \mathbb{Q}(z)$ has class number 1. We choose $f = \mathfrak{P}_2$ the degree one prime above 2. We choose the fundamental units

$$ \varepsilon_1 = \frac{-2z^3 + 13z^2 - z + 3}{7}, \quad \varepsilon_2 = \frac{-5z^3 + 29z^2 + 15z + 18}{7} $$

for $\mathcal{O}_f^{+,\times}$. We compute $\tilde{\lambda}_1 = \tilde{\lambda}_2 = 1$, $\tilde{t}_1 = \tilde{t}_2 = 1$. We may choose $\mathfrak{b} = (1)$ and $\mathfrak{a}$ the unique degree one prime above 13 in $\mathbb{K}_7$. The ideal $qN/\mathfrak{a}$ has an admissible generator $h_1 = h_2 = (-170z^3 + 972z^2 + 468z + 640)/7$ so that we don't have to look for helper ideals. The corresponding levels will be $l_1 = l_2 = 2 \cdot 13$. Computations show that the correct orientations should be $\underline{\mu} = [1, -1]$ and $\underline{\nu} = [-1, 1]$. Let us write the parameters

$$ \begin{aligned} \tau &= \varepsilon_2 + 11, & \tau' &= 20 + \frac{1}{\varepsilon_1} + \frac{1}{\varepsilon_1 \varepsilon_2} \\ \sigma &= -7 + \frac{1}{\varepsilon_2}, & \sigma' &= -\varepsilon_2 - 11 \\ \rho &= -\varepsilon_1 + 23, & \rho' &= 4\varepsilon_1 - 85 - \frac{1}{\varepsilon_2} \end{aligned} $$

Then

$$ \frac{G_2\left(\frac{1}{2}, \frac{\tau}{26}, \frac{\sigma}{26}, \frac{\rho}{26}\right)^{-13}}{G_2\left(\frac{13}{2}, \frac{\tau}{2}, \frac{\sigma}{2}, \frac{\rho}{2}\right)^{-1}} \times \frac{G_2\left(\frac{-1}{2}, \frac{\tau'}{26}, \frac{\sigma'}{26}, \frac{\rho'}{26}\right)^{13}}{G_2\left(\frac{-13}{2}, \frac{\tau'}{2}, \frac{\sigma'}{2}, \frac{\rho'}{2}\right)} $$

is close to the root $\approx 4.1210208... - i \cdot 5.0617720...$ of the polynomial $x^8 - 7x^7 + 33x^6 + 49x^5 + 17x^4 + 49x^3 + 33x^2 - 7x + 1$ which defines $\mathbb{K}_7^+(f)$. The
computation time for 1000 digits is 6 seconds. We compare this to a simple choice for the admissible vectors $h_1, h_2$ given by Pari/GP. Namely for $h_1 = h_2 = (86z^3 - 496z^2 - 286z - 206)/7$ we compute $t_1 = t_2 = 31$ and $l = 26$. The computation time for this pair and for 1000 digits is 2 minutes, which is roughly 20 times longer than for the previous pair.

### 5.2.2 Example 8

We now discuss in full detail a quartic example with specific work on target ideals. Let $z$ be the complex root of the polynomial $x^4 - 19x^3 + 18x^2 + 8x + 1$ lying in the upper half-plane. Then $K_8 = \mathbb{Q}(z)$ has class number 1. We choose $f = \mathfrak{P}_3$ such that $(3) = \mathfrak{P}_3^2\mathfrak{P}_3'^2$. We choose the fundamental units

$$ \varepsilon_1 = \frac{19z^3 - 366z^2 + 438z + 44}{9}, \quad \varepsilon_2 = -z^3 + 19z^2 - 18z - 8 $$

for $O_f^{+, \times}$. We compute $\tilde{\lambda}_1 = \tilde{\lambda}_2 = 1$, $\tilde{t}_1 = 7$, $\tilde{t}_2 = 7129$ where $7129$ is a prime number. We may choose $b = (1)$ and $a$ the unique degree one prime above 13 in $K_8$. We search for target ideals and find $\mathfrak{P}_7$ in the first case and $\mathfrak{P}_{7129}$ in the second case. Both ideals $qN\mathfrak{P}_7/a$ and $qN\mathfrak{P}_{7129}/a$ are principal with admissible generators $h_1 = 43z^3 - 820z^2 + 830z + 313$ and $h_2 = -230z^3 + 4406z^2 - 4825z - 1208$. The corresponding levels will be $l_1 = 3 \cdot 13 \cdot 7$ and $l_2 = 3 \cdot 13 \cdot 7129$. Computations show that the correct orientations should be $\mu = [1, 1]$ and $\nu = [1, -1]$. Let us write the parameters

$$ \begin{aligned} \tau &= \frac{2\varepsilon_1\varepsilon_2 + 454\varepsilon_1 - 3}{\varepsilon_1}, & \tau' &= \frac{57488\varepsilon_1\varepsilon_2 + 3\varepsilon_2 + 16}{\varepsilon_1\varepsilon_2} \\ \sigma &= \frac{-3443\varepsilon_1\varepsilon_2 + 367\varepsilon_1 - 47}{\varepsilon_1\varepsilon_2}, & \sigma' &= \frac{47\varepsilon_1\varepsilon_2 - 328\varepsilon_2 + 485399}{1} \\ \rho &= \frac{-\varepsilon_1\varepsilon_2 - 3\varepsilon_1 + 40}{1}, & \rho' &= \frac{-57\varepsilon_1\varepsilon_2 + 188027\varepsilon_2 - 2}{\varepsilon_2} \end{aligned} $$

Then

$$ \frac{G_2\left(\frac{1}{3}, \frac{\tau}{273}, \frac{\sigma}{273}, \frac{\rho}{273}\right)^{13}}{G_2\left(\frac{13}{3}, \frac{\tau}{21}, \frac{\sigma}{21}, \frac{\rho}{21}\right)} \times \frac{G_2\left(\frac{1}{3}, \frac{\tau'}{278031}, \frac{\sigma'}{278031}, \frac{\rho'}{278031}\right)^{-13}}{G_2\left(\frac{13}{3}, \frac{\tau'}{21387}, \frac{\sigma'}{21387}, \frac{\rho'}{21387}\right)^{-1}} $$

is close to the root $\approx 10.6409709... - i \cdot 5.9332732...$ of the polynomial $x^8 - 18x^7 + 83x^6 + 396x^5 + 597x^4 + 396x^3 + 83x^2 - 18x + 1$ which defines a subextension of $K_8^+(f)/K_8$. The computation time for 1000 digits is 2 minutes and 15 seconds.
## 5 Numerical evidence to support the conjecture

In this section, we provide numerical examples to support our conjecture. They may be computed with high precision in a low amount of time. In what follows, we will give computation times for 1000 digits precision on a personal computer. Computations were carried out using number fields found in the LMFDB database [13] as well as the computer algebra system PARI/GP [24], making extensive use of algebraic number theory tools it provides. To understand the result of the computations, we use the commands `lindep` to test if the value obtained from our computations is close to the expected value for $N(a)ζ'_f([b], 0) - ζ'_f([ab], 0)$ and `algdep` to test if the value obtained is close to some algebraic integer in an abelian extension of $K$. Note that if $[K^+(f) : K] = m$ and $[K : \mathbb{Q}] = d$ then we may compute using our method $2m$ roots of a polynomial defining a subfield of $K^+(f)$ over $\mathbb{Q}$ out of a maximal amount of $dm$. This is achieved by varying the class $[b]$ and using both complex embeddings $\sigma_C$ and $\bar{\sigma}_C$. This way, we may also compute all the roots of a polynomial in $K[x]$ which defines a subextension of $K^+(f)/K$ and we can thus compute the elementary symmetric polynomials in these roots to recover the polynomial.

In what follows, we will define our fields as $K = \mathbb{Q}(z)$ where $z$ is the complex root of some polynomial $P = (X-z)(X-\bar{z}) \prod_{j=1}^r (X-z_j) \in \mathbb{Q}[X]$ lying in the upper half-plane (here the $z_j$ are real numbers with $z_1 < \cdots < z_r$). Thus, to define our orientations we will fix the ordering on the real embeddings of $K$ such that $\sigma_j(z) = z_j$ for $1 \le j \le r$. Throughout this section, we focus mainly on computations for $\mathfrak{b} = (1)$ because most of the work on target ideals is independent of the choice for the ideal $\mathfrak{b}$. For ease of presentation, we will name prime ideals above a prime $p$ as $\mathfrak{P}_p, \mathfrak{P}'_p, \mathfrak{P}''_p, \dots$ in their order of appearance using the commands `idealprimedec` or `idealfactor` in Pari/GP version 2.15.4. We provide examples in the cubic, quartic and quintic cases with $\kappa = \#\mathfrak{Z}_f^1$ first and then quartic examples with $\kappa < \#\mathfrak{Z}_f^1$. In all examples, the orientations have been computed as in Definition 10 and all values $v$ computed will satisfy directly formula (16) or

$$ N(a)ζ'_f([b], 0) - ζ'_f([ab], 0) = \frac{\kappa}{\#\mathfrak{Z}_f^1} \log|v|^2 $$

which is an avatar of formula (16) with $\kappa$ defined in section 3.2. For ease of presentation we will not write the partition sets $U_{\rho,j}$ (see section 3.2) in the examples below.
### 5.2.3 Example 9

Keep $K_8$, $f$, $b$, $a$ as in example 8 and let us change our choice of fundamental units. We fix another set of fundamental units for $O_f^{+, \times}$:

$$ \varepsilon_1 = \frac{z^3 - 21z^2 + 54z + 11}{9}, \quad \varepsilon_2 = -z^3 + 19z^2 - 18z - 8 $$

Then, we compute $\tilde{\lambda}_1 = \tilde{\lambda}_2 = 1$, $\tilde{t}_1 = 1$, $\tilde{t}_2 = 25$. In the second computation, a target ideal is $\mathfrak{P}_5^2$ the square of the unique degree one prime above 5 and the ideals $qN/a$ and $qN\mathfrak{P}_5^2/a$ are generated by the admissible vectors $h_1 = (-44z^3 + 843z^2 - 927z - 232)/3$, $h_2 = (76z^3 - 1449z^2 + 1470z + 344)/3$. The corresponding levels will be $l_1 = 3 \cdot 13$ and $l_2 = 3 \cdot 5^2 \cdot 13$. Computations show that the correct orientations should be $\underline{\mu} = [1, 1]$ and $\underline{\nu} = [-1, 1]$. Let us write the parameters

$$ \begin{aligned} \tau &= \frac{2\varepsilon_1\varepsilon_2 + 109\varepsilon_1 - 3}{\varepsilon_1}, & \tau' &= \frac{842\varepsilon_1\varepsilon_2 + \varepsilon_2 + 3}{\varepsilon_1\varepsilon_2} \\ \sigma &= \frac{-321\varepsilon_1\varepsilon_2 - 16\varepsilon_1 + 7}{\varepsilon_1\varepsilon_2}, & \sigma' &= \frac{-7\varepsilon_1\varepsilon_2 + 4\varepsilon_2 - 63}{1} \\ \rho &= \frac{-\varepsilon_1 - 47}{1}, & \rho' &= \frac{-3\varepsilon_1\varepsilon_2 + 566\varepsilon_2 - 2}{\varepsilon_2} \end{aligned} $$

Then

$$ \frac{G_2\left(\frac{1}{3}, \frac{\tau}{39}, \frac{\sigma}{39}, \frac{\rho}{39}\right)^{-13}}{G_2\left(\frac{13}{3}, \frac{\tau}{3}, \frac{\sigma}{3}, \frac{\rho}{3}\right)^{-1}} \times \frac{G_2\left(\frac{1}{3}, \frac{\tau'}{975}, \frac{\sigma'}{975}, \frac{\rho'}{975}\right)^{13}}{G_2\left(\frac{13}{3}, \frac{\tau'}{75}, \frac{\sigma'}{75}, \frac{\rho'}{75}\right)} $$

is close to the same root $\approx 10.6409709... - i \cdot 5.9332732...$ of the polynomial $x^8 - 18x^7 + 83x^6 + 396x^5 + 597x^4 + 396x^3 + 83x^2 - 18x + 1$. The computation time for 1000 digits is 11 seconds. We see that a good choice for the fundamental units can have a strong impact on the computation time.

### 5.2.4 Example 10

This example falls outside of (*) and (**) because $gcd(\tilde{t}_1, \tilde{t}_2, q = 2) = 2$. Still, we can find a pair of vectors $h_1, h_2$ for which the computations give interesting results. Let $z$ be the complex root of the polynomial $x^4 - 12$ lying in the upper half-plane. Then $K_{10} = \mathbb{Q}(z)$ has class number 1 and contains the real quadratic field $\mathbb{Q}(\sqrt{3})$. Remember that this means that we must carefully choose the fundamental units in this case to avoid totally real elements. We choose $f^4 = (2)$. We choose the fundamental units

$$ \varepsilon_1 = \frac{z^2 + 2z + 2}{4}, \quad \varepsilon_2 = \frac{-z^3 + z^2 + 4z - 2}{4} $$
## 5.1 Cubic examples

Here we present six examples in the cubic case. The first example showcases the work on target ideals with $\tilde{t} = 131$ in the spirit of (*) and (**). The second example is one of the simplest ones, with $\tilde{t} = 1$ and class number one. The third example is a case where $q = 5$. The fourth and fifth examples showcase the work on helper ideals when the class number gets larger, for $q = 3$ and $q = 7$. The sixth example shows what can be done outside (*), (***) in the cubic case. We also provide a table in Appendix A giving the values of $\tilde{\lambda}$ and $\tilde{t}$ for the pure cubic fields defined by $X^3 - m$ for $2 \le m \le 50$ up to isomorphism.

### 5.1.1 Example 1

We first discuss in full detail a cubic example. Let $z$ be the complex root of the polynomial $x^3 - 13$ lying in the upper half-plane. Then $\mathbb{K}_1 = \mathbb{Q}(z)$ has class number 3. We choose the ideal $f$ above $q = 3$ such that $f^3 = (3)$. The corresponding narrow ray class group is $Cl^+(f) \simeq \mathbb{Z}/6\mathbb{Z}$. The unit group $O_f^{+,\times}$ of totally positive units congruent to 1 mod $f$ is generated by $\varepsilon = 2z^2 - 3z - 4$.

* We compute $\tilde{\lambda} = 1$, $\tilde{s} = \tilde{t} = 131$ as defined in section 4.1.

* Then, we search for target ideals above the prime number 131 (see Definition 18) and we find the unique degree one prime $\mathfrak{P}_{131}$ above 131.

* We choose for the smoothing ideal $\mathfrak{a}$ the unique degree one prime above $N = 5$ in $\mathbb{K}_1$. The ideals $\mathfrak{P}_2^k$ for $0 \le k \le 5$ represent all classes in $Cl^+(f)$, where $\mathfrak{P}_2$ is the only prime ideal of norm 2 in $\mathbb{K}$. These integral ideals satisfying the conditions in Proposition 22.

* Using Proposition 22 for $0 \le k \le 5$ we obtain the following ideals and admissible generators (see Definition 7):

<table><thead><tr><th>k</th><th>m</th><th>J</th><th>Ideal in Prop. 22</th><th>Admissible generator h<sub>k</sub></th></tr></thead><tbody><tr><td>0</td><td>1</td><td>(1)</td><td>qN<sub>Cl</sub><sup>131</sup>/a</td><td>-21z<sup>2</sup> - 42z - 114</td></tr><tr><td>1</td><td>1</td><td><b>P</b>'<sub>7</sub><b>P</b>''<sub>7</sub></td><td>qN<sub>Cl</sub><sup>131</sup><b>P</b>'<sub>7</sub><b>P</b>''<sub>7</sub>/(a<sub>Cl</sub><sup>2</sup>)</td><td>(87z<sup>2</sup> + 9z - 477)/2</td></tr><tr><td>2</td><td>1</td><td><b>P</b>'<sub>11</sub></td><td>qN<sub>Cl</sub><sup>131</sup><b>P</b>'<sub>11</sub>/(a<sub>Cl</sub><sup>2</sup>)</td><td>-(3z<sup>2</sup> + 309z - 807)/4</td></tr><tr><td>3</td><td>1</td><td>(1)</td><td>qN<sub>Cl</sub><sup>131</sup>/(a<sub>Cl</sub><sup>3</sup>)</td><td>(-9z<sup>2</sup> + 27z + 159)/8</td></tr><tr><td>4</td><td>1</td><td><b>P</b>'<sub>7</sub><b>P</b>''<sub>7</sub></td><td>qN<sub>Cl</sub><sup>131</sup><b>P</b>'<sub>7</sub><b>P</b>''<sub>7</sub>/(a<sub>Cl</sub><sup>4</sup>)</td><td>(-159z<sup>2</sup> - 603z - 1431)/16</td></tr><tr><td>5</td><td>1</td><td><b>P</b>'<sub>11</sub></td><td>qN<sub>Cl</sub><sup>131</sup><b>P</b>'<sub>11</sub>/(a<sub>Cl</sub><sup>5</sup>)</td><td>(921z<sup>2</sup> + 2157z + 4209)/32</td></tr></tbody></table>

The corresponding levels will be $l_0 = l_3 = 3 \cdot 5 \cdot 131 = 1965$, $l_1 = l_4 = 7l_0$, $l_2 = l_5 = 11l_0$ as defined in section 4.1.
for $O_f^{+, \times}$. We compute $\tilde{\lambda}_1 = \tilde{\lambda}_2 = 1$, $\tilde{t}_1 = 2 \cdot 3$, $\tilde{t}_2 = 2 \cdot 11$. We may choose $\mathfrak{b} = (1)$ and $\mathfrak{a} = \mathfrak{P}_{23}$ a degree one prime above 23 in $\mathbb{K}_{10}$. We search for target ideals and find $\mathfrak{P}_3$ the degree one prime ideal above 3 in the first case and $\mathfrak{P}_{11}$ one of the degree one primes above 11 in the second case. Then $qN\mathfrak{P}_3/\mathfrak{a}$ and $qN\mathfrak{P}_{11}/\mathfrak{a}$ are generated by the admissible vectors $h_1 = (-4z^3 + 11z^2 + 10z + 30)/2$ and $h_2 = (9z^3 + 4z^2 - 34z - 56)/2$. The corresponding levels will be $l_1 = 2 \cdot 3 \cdot 23$ and $l_2 = 2 \cdot 11 \cdot 23$. Fixing the orientations $\mu = [1, -1]$, $\nu = [-1, 1]$ we compute $I_{2, f, b, \mathfrak{a}}(\varepsilon_1, \varepsilon_2, \underline{h}, \mu, \nu, \sigma_{\mathbb{C}}) \approx 13.9102308... - i \cdot 24.0932265...$ with $t_1 = t_2 = 2$, i.e. we compute a product of 8 ordinary elliptic smoothed $G_2$ functions. The result is close to a root of the polynomial $x^8 - 28x^7 + 778x^6 - 112x^5 - 749x^4 - 112x^3 + 778x^2 - 28x + 1$ which defines $\mathbb{K}_{10}^+(f)$. The computation time for 1000 digits is 17 seconds.

## 5.3 One Quintic example

We have already successfully produced a dozen of quintic examples and we give here a detailed presentation of our simplest one. Let $z$ be the complex root of the polynomial $x^5 - x^4 - x^3 - 2x^2 + x + 1$ lying in the upper half-plane. Then $\mathbb{K}_{11} = \mathbb{Q}(z)$ has class number 1. We choose $f = \mathfrak{P}_3$ the degree one prime above 3. We choose the fundamental units (once again take the fundamental units provided by Pari/GP and then compute the simplest totally positive ones):

$$ \varepsilon_1 = z^4 - 2z^3 - z + 3, \quad \varepsilon_2 = 2z^4 - 2z^3 - 3z + 3, \quad \varepsilon_3 = 2z^4 - 3z^3 - 4z + 4 $$

for $O_f^{+, \times}$. We fix the order $\{\mathrm{Id}, (32), (21), (231), (312), (31)\}$ of $\mathfrak{S}_3$ given by Pari/GP. We compute $\tilde{\lambda}_1 = \tilde{\lambda}_2 = \tilde{\lambda}_3 = \tilde{\lambda}_4 = \tilde{\lambda}_5 = \tilde{\lambda}_6 = 1$, $\tilde{t}_1 = 7 \cdot 37 \cdot 137$, $\tilde{t}_2 = 31 \cdot 53$, $\tilde{t}_3 = 491$, $\tilde{t}_4 = 107$, $\tilde{t}_5 = 1$ and $\tilde{t}_6 = 145637$ where we have written the prime number decomposition of all the overflows $\tilde{t}$. We may choose $\mathfrak{b} = (1)$ and $\mathfrak{a}$ the unique degree one prime above 11 in $\mathbb{K}_{11}$. We look for target ideals above the prime numbers listed above and find the ideals $\mathfrak{P}_7, \mathfrak{P}'_{37}, \mathfrak{P}'_{137}, \mathfrak{P}_{31}, \mathfrak{P}_{53}, \mathfrak{P}'_{491}, \mathfrak{P}'_{107}, \mathfrak{P}_{145637}$ in the order provided by Pari/GP version 2.15.4. The ideals $qN\mathfrak{P}_7\mathfrak{P}'_{37}\mathfrak{P}'_{137}/\mathfrak{a}, qN\mathfrak{P}_{31}\mathfrak{P}_{53}/\mathfrak{a}, qN\mathfrak{P}'_{491}/\mathfrak{a}, qN\mathfrak{P}'_{107}/\mathfrak{a}, qN/\mathfrak{a}, qN\mathfrak{P}_{145637}/\mathfrak{a}$ admit the following admissible generators:

$$ h_1 = 147z^4 - 135z^3 - 90z^2 - 234z + 72 $$

$$ h_2 = -15z^4 - 30z^3 + 90z^2 + 36z + 60 $$

$$ h_3 = 42z^4 - 15z^3 - 87z^2 - 48z + 30 $$

$$ h_4 = -21z^4 + 57z^3 - 6z^2 - 9z - 48 $$

$$ h_5 = 3z^4 - 6z^3 + 18z^2 - 6z + 12 $$

$$ h_6 = 108z^4 - 246z^3 + 111z^2 - 81z + 195 $$
• We compute the orientations $\mu = \nu = -1$ as in Definition 10.

• We compute $Z_f^1 = \{(1, 1), (2, -1)\}$ (see Definition 25) and we check that the equality $\Gamma_{f,b,a}^-(\varepsilon, -2h)^{-1} = \Gamma_{f,b,a}^-(\varepsilon, h)^{-1}$ holds. This gives $\kappa = 2$ as defined in section 3.2.

We may compute the values

$$ \Gamma_{f,(1),a}^{-}(\varepsilon, h_0)^{-1} = \frac{\Gamma\left(\frac{-1}{3}, \frac{\varepsilon^{-1}+5348}{1965}, \frac{\varepsilon+467}{1965}\right)^{-5}}{\Gamma\left(\frac{-5}{3}, \frac{\varepsilon^{-1}+5348}{393}, \frac{\varepsilon+467}{393}\right)^{-1}} \approx -0.0660917\dots + i \cdot 0.0932299\dots $$

$$ \Gamma_{f,\mathfrak{P}_2,a}^{-}(\varepsilon, h_1)^{-1} = \frac{\Gamma\left(\frac{1}{3}, \frac{\varepsilon^{-1}+1418}{13755}, \frac{\varepsilon-3463}{13755}\right)^{-5}}{\Gamma\left(\frac{5}{3}, \frac{\varepsilon^{-1}+1418}{2751}, \frac{\varepsilon-3463}{2751}\right)^{-1}} \approx 0.0059953\dots + i \cdot 0.0047179\dots $$

$$ \Gamma_{f,\mathfrak{P}_2,a}^{-}(\varepsilon, h_2)^{-1} = \frac{\Gamma\left(\frac{-1}{3}, \frac{\varepsilon^{-1}-547}{21615}, \frac{\varepsilon+6362}{21615}\right)^{-5}}{\Gamma\left(\frac{-5}{3}, \frac{\varepsilon^{-1}-547}{4323}, \frac{\varepsilon+6362}{4323}\right)^{-1}} \approx -289.3045814\dots - i \cdot 127.6382732\dots $$

$$ \Gamma_{f,\mathfrak{P}_2,a}^{-}(\varepsilon, h_3)^{-1} = \frac{\Gamma\left(\frac{1}{3}, \frac{\varepsilon^{-1}+5348}{1965}, \frac{\varepsilon+467}{1965}\right)^{-5}}{\Gamma\left(\frac{5}{3}, \frac{\varepsilon^{-1}+5348}{393}, \frac{\varepsilon+467}{393}\right)^{-1}} \approx -5.0606452\dots - i \cdot 7.1386178\dots $$

$$ \Gamma_{f,\mathfrak{P}_2,a}^{-4}(\varepsilon, h_4)^{-1} = \frac{\Gamma\left(\frac{-1}{3}, \frac{\varepsilon^{-1}+1418}{13755}, \frac{\varepsilon-3463}{13755}\right)^{-5}}{\Gamma\left(\frac{-5}{3}, \frac{\varepsilon^{-1}+1418}{2751}, \frac{\varepsilon-3463}{2751}\right)^{-1}} \approx 103.0063956\dots - i \cdot 81.0605592\dots $$

$$ \Gamma_{f,\mathfrak{P}_2,a}^{-5}(\varepsilon, h_5)^{-1} = \frac{\Gamma\left(\frac{1}{3}, \frac{\varepsilon^{-1}-547}{21615}, \frac{\varepsilon+6362}{21615}\right)^{-5}}{\Gamma\left(\frac{5}{3}, \frac{\varepsilon^{-1}-547}{4323}, \frac{\varepsilon+6362}{4323}\right)^{-1}} \approx -0.0028933\dots + i \cdot 0.0012765\dots $$

Using all 6 values $I_k = \Gamma_{f,\mathfrak{P}_2,a}^-(\varepsilon, h_k)^{-1}$ we may compute and identify the relative polynomial:

$$ \prod_{k=0}^{5} (X - I_k) = (X^6 + 1) + (-34z^2 + 26z + 128)(X^5 + X) \\ + (1127z^2 + 8879z - 27106)(X^4 + X^2) \\ + (40740z^2 - 16965z - 185350)X^3 $$

Alternatively, all values $I_k$ are close to roots of the absolute palindromic polynomial $x^{18}+384x^{17}+2310x^{16}-10646490x^{15}+1596241353x^{14}+18608357181x^{13}+156933809421x^{12}+215098256580x^{11}+381407365338x^{10}+338205493469x^9+\dots+384x+1$ which defines the degree 6 extension $K_1^+(f)/K$. The computation time for 1000 digits is 15 seconds for $I_0 = I_3^{-1}$, 65 seconds for $I_1 = I_4^{-1}$
The corresponding levels are $l_1 = 3 \cdot 11 \cdot 7 \cdot 37 \cdot 137$, $l_2 = 3 \cdot 11 \cdot 491$, $l_3 = 3 \cdot 11 \cdot 107$, $l_4 = 3 \cdot 11 \cdot 31 \cdot 53$, $l_5 = 3 \cdot 11$, $l_6 = 3 \cdot 11 \cdot 145637$. Computations give the orientations $\mu = [1, -1, -1, 1, 1, -1]$, $\nu = [1, -1, -1, 1, 1, -1]$. We compute the six quotients:

$$
\begin{align*}
& \frac{G_3\left(\frac{-1}{3}, \frac{\tau_1}{l_1}, \frac{\sigma_1}{l_1}, \frac{\rho_1}{l_1}, \frac{\varpi_1}{l_1}\right)^{11}}{G_3\left(\frac{-11}{3}, \frac{11\tau_1}{l_1}, \frac{11\sigma_1}{l_1}, \frac{11\rho_1}{l_1}, \frac{11\varpi_1}{l_1}\right)} & \approx -6.9846353\dots10^{-18} + i \cdot 6.2764063\dots10^{-17} \\
& \frac{G_3\left(\frac{-1}{3}, \frac{\tau_2}{l_2}, \frac{\sigma_2}{l_2}, \frac{\rho_2}{l_2}, \frac{\varpi_2}{l_2}\right)^{-11}}{G_3\left(\frac{-11}{3}, \frac{11\tau_2}{l_2}, \frac{11\sigma_2}{l_2}, \frac{11\rho_2}{l_2}, \frac{11\varpi_2}{l_2}\right)^{-1}} & \approx 1.7434761\dots10^{11} - i \cdot 1.7914686\dots10^{11} \\
& \frac{G_3\left(\frac{1}{3}, \frac{\tau_3}{l_3}, \frac{\sigma_3}{l_3}, \frac{\rho_3}{l_3}, \frac{\varpi_3}{l_3}\right)^{-11}}{G_3\left(\frac{11}{3}, \frac{11\tau_3}{l_3}, \frac{11\sigma_3}{l_3}, \frac{11\rho_3}{l_3}, \frac{11\varpi_3}{l_3}\right)^{-1}} & \approx 0.2815024\dots - i \cdot 0.0665163\dots \\
& \frac{G_3\left(\frac{1}{3}, \frac{\tau_4}{l_4}, \frac{\sigma_4}{l_4}, \frac{\rho_4}{l_4}, \frac{\varpi_4}{l_4}\right)^{11}}{G_3\left(\frac{11}{3}, \frac{11\tau_4}{l_4}, \frac{11\sigma_4}{l_4}, \frac{11\rho_4}{l_4}, \frac{11\varpi_4}{l_4}\right)} & \approx 9.8457700\dots - i \cdot 2.3506772\dots \\
& \frac{G_3\left(\frac{-1}{3}, \frac{\tau_5}{l_5}, \frac{\sigma_5}{l_5}, \frac{\rho_5}{l_5}, \frac{\varpi_5}{l_5}\right)^{11}}{G_3\left(\frac{-11}{3}, \frac{11\tau_5}{l_5}, \frac{11\sigma_5}{l_5}, \frac{11\rho_5}{l_5}, \frac{11\varpi_5}{l_5}\right)} & \approx -0.2518907\dots - i \cdot 0.2274847\dots \\
& \frac{G_3\left(\frac{-1}{3}, \frac{\tau_6}{l_6}, \frac{\sigma_6}{l_6}, \frac{\rho_6}{l_6}, \frac{\varpi_6}{l_6}\right)^{-11}}{G_3\left(\frac{-11}{3}, \frac{11\tau_6}{l_6}, \frac{11\sigma_6}{l_6}, \frac{11\rho_6}{l_6}, \frac{11\varpi_6}{l_6}\right)^{-1}} & \approx 100864.0193260\dots - i \cdot 767246.5816458\dots
\end{align*}
$$

where the parameters are given in the table below. The product of these six quotients is close to the root $\approx -11.6360077\dots + i \cdot 3.4634701\dots$ of the polynomial $x^{10}+24x^9+164x^8+99x^7-62x^6-89x^5-62x^4+99x^3+164x^2+24x+1$ which defines $K_{11}^+(f)$. The computation time for 1000 digits is 1 minute and 35 seconds, but the computation time for each of the individual computations is not uniform at all. The fifth computation requires 1 second whereas the sixth computation requires 58 seconds because of the level difference $l_5 = 33$ versus $l_6 = 4806021$.
and 105 seconds for $I_2 = I_5^{-1}$. We may also check formula (16) as

$$ \mathcal{N}(\mathfrak{a})\zeta'_{f}([1], 0) - \zeta'_{f}([\mathfrak{a}.(1)], 0) = \frac{1}{2} \log \left| \prod_{(n, \varepsilon') \in \mathbb{Z}_{f}^{1}} \Gamma_{f, \mathfrak{b}, \mathfrak{a}}^{-}(\varepsilon, n\varepsilon'h)^{-1} \right|^{2} \approx -4.3382052... $$

and similarly for the other classes. Consider now another admissible vector $h = 3z^2 + 6z + 12$ for $\mathfrak{b} = (1)$. The valuation of $h$ at any prime ideal dividing 131 is 0. This is one of the simplest admissible vectors we could find. Using this vector $h$ we have to compute $t = 131$ ordinary elliptic Gamma functions and the computation time is 30 seconds for 1000 digits. In the cubic case, because the computation time for the precision $\delta$ is $O(l|\log(\delta)|t)$, reducing the value of $t$ by raising the level $l$ may not result in a massive time gain. In the cases $r \ge 2$, the gain will be more substantial.

### 5.1.2 Example 2

We now discuss one of the simplest cubic examples which was presented in the introduction. Let $z$ be the complex root of the polynomial $x^3 - 2$ in the upper half-plane. Then $\mathbb{K}_2 = \mathbb{Q}(z)$ has class number 1. We choose the ideal $f$ above $q = 3$ such that $f^3 = (3)$. The corresponding narrow ray class group is $Cl^+(f) \simeq \mathbb{Z}/2\mathbb{Z}$. The unit $\varepsilon = z - 1$ is a generator for $O_f^{+, \times}$ such that $\tilde{\lambda} = \tilde{t} = 1$. We choose $\mathfrak{b} = (1)$ and $\mathfrak{a}$ the degree one prime above $N = 5$ in $\mathbb{K}_2$. The ideal $qN/\mathfrak{a}$ has an admissible generator $h = -3z^2 + 6z + 3$. The corresponding level is $l = 15$. Here we have $Z_f^1 = \{(1, 1), (2, -1)\}$ and $\kappa = 2$. The value

$$ \Gamma_{f, \mathfrak{b}, \mathfrak{a}}^{-}(\varepsilon, h)^{-1} = \frac{\Gamma\left(\frac{-1}{3}, \frac{\varepsilon^{-1}+2}{15}, \frac{\varepsilon-7}{15}\right)^{-5}}{\Gamma\left(\frac{-5}{3}, \frac{\varepsilon^{-1}+2}{3}, \frac{\varepsilon-7}{3}\right)^{-1}} \approx -1.2937005... + i \cdot 1.4743341... $$

is close to a root of the polynomial $x^6 + 3x^5 + 6x^4 + 5x^3 + 6x^2 + 3x + 1$ which defines $\mathbb{K}_2^+(f)$. The computation time for 1000 digits is 1 second.

### 5.1.3 Example 3

Here we explain how we have obtained the example in section 3.2 with $q = 5$. Let $z = e^{2i\pi/3}10^{1/3}$ be the root of the polynomial $x^3 - 10$ in the upper half-plane. Then $\mathbb{K}_3 = \mathbb{Q}(z)$ has class number 1. We choose the ideal $f$ above $q = 5$ such that $f^3 = (5)$. The corresponding narrow ray class group is $Cl^+(f) \simeq \mathbb{Z}/4\mathbb{Z}$. The unit $\varepsilon = (2z^2 - z - 7)/3$ is a generator for $O_f^{+, \times}$ such that $\tilde{\lambda} = 1$ and $\tilde{t} = 9$. We may choose $\mathfrak{b} = (1)$ and $\mathfrak{a}$ the unique degree
<table>
  <tbody>
    <tr>
      <td>ε<sub>1</sub>τ<sub>1</sub> =</td>
      <td>229ε<sub>1</sub>ε<sub>2</sub>ε<sub>3</sub></td>
      <td>-1727ε<sub>1</sub>ε<sub>2</sub></td>
      <td>-796386ε<sub>1</sub></td>
      <td>+3385</td>
    </tr>
    <tr>
      <td>ε<sub>1</sub>ε<sub>2</sub>σ<sub>1</sub> =</td>
      <td>-114ε<sub>1</sub>ε<sub>2</sub>ε<sub>3</sub></td>
      <td>-987551ε<sub>1</sub>ε<sub>2</sub></td>
      <td>+718ε<sub>1</sub></td>
      <td>+949</td>
    </tr>
    <tr>
      <td>ε<sub>1</sub>ε<sub>2</sub>ε<sub>3</sub>ρ<sub>1</sub> =</td>
      <td>-595138ε<sub>1</sub>ε<sub>2</sub>ε<sub>3</sub>+</td>
      <td>12ε<sub>1</sub>ε<sub>2</sub></td>
      <td>+1445ε<sub>1</sub></td>
      <td>-907</td>
    </tr>
    <tr>
      <td>ω<sub>1</sub> =</td>
      <td>-52ε<sub>1</sub>ε<sub>2</sub>ε<sub>3</sub></td>
      <td>+857ε<sub>1</sub>ε<sub>2</sub></td>
      <td>-295ε<sub>1</sub></td>
      <td>+2067162</td>
    </tr>
    <tr>
      <td>ε<sub>1</sub>τ<sub>2</sub> =</td>
      <td>99ε<sub>1</sub>ε<sub>2</sub>ε<sub>3</sub></td>
      <td>-356ε<sub>1</sub>ε<sub>3</sub></td>
      <td>+19304ε<sub>1</sub></td>
      <td>+57</td>
    </tr>
    <tr>
      <td>ε<sub>1</sub>ε<sub>3</sub>σ<sub>2</sub> =</td>
      <td>-17ε<sub>1</sub>ε<sub>2</sub>ε<sub>3</sub></td>
      <td>-70870ε<sub>1</sub>ε<sub>3</sub></td>
      <td>-145ε<sub>1</sub></td>
      <td>+40</td>
    </tr>
    <tr>
      <td>ε<sub>1</sub>ε<sub>2</sub>ε<sub>3</sub>ρ<sub>2</sub> =</td>
      <td>324170ε<sub>1</sub>ε<sub>2</sub>ε<sub>3</sub></td>
      <td>-85ε<sub>1</sub>ε<sub>3</sub></td>
      <td>+197ε<sub>1</sub></td>
      <td>-111</td>
    </tr>
    <tr>
      <td>ω<sub>2</sub> =</td>
      <td>91ε<sub>1</sub>ε<sub>2</sub>ε<sub>3</sub></td>
      <td>-460ε<sub>1</sub>ε<sub>3</sub></td>
      <td>+3ε<sub>1</sub></td>
      <td>-28725</td>
    </tr>
    <tr>
      <td>ε<sub>1</sub>ε<sub>2</sub>τ<sub>4</sub> =</td>
      <td>-62ε<sub>1</sub>ε<sub>2</sub>ε<sub>3</sub></td>
      <td>-12636ε<sub>1</sub>ε<sub>2</sub></td>
      <td>+20ε<sub>2</sub></td>
      <td>+3</td>
    </tr>
    <tr>
      <td>ε<sub>1</sub>ε<sub>2</sub>ε<sub>3</sub>σ<sub>4</sub> =</td>
      <td>327915ε<sub>1</sub>ε<sub>2</sub>ε<sub>3</sub></td>
      <td>-37ε<sub>1</sub>ε<sub>2</sub></td>
      <td>-1101ε<sub>2</sub></td>
      <td>+154</td>
    </tr>
    <tr>
      <td>ρ<sub>4</sub> =</td>
      <td>-44ε<sub>1</sub>ε<sub>2</sub>ε<sub>3</sub></td>
      <td>+314ε<sub>1</sub>ε<sub>2</sub></td>
      <td>-65ε<sub>2</sub></td>
      <td>-52915</td>
    </tr>
    <tr>
      <td>ε<sub>2</sub>ω<sub>4</sub> =</td>
      <td>-262ε<sub>1</sub>ε<sub>2</sub>ε<sub>3</sub></td>
      <td>-5ε<sub>1</sub>ε<sub>2</sub></td>
      <td>-768869ε<sub>2</sub></td>
      <td>-19</td>
    </tr>
    <tr>
      <td>ε<sub>2</sub>ε<sub>3</sub>τ<sub>3</sub> =</td>
      <td>-27ε<sub>1</sub>ε<sub>2</sub>ε<sub>3</sub></td>
      <td>-84ε<sub>2</sub>ε<sub>3</sub></td>
      <td>+11ε<sub>2</sub></td>
      <td>+1</td>
    </tr>
    <tr>
      <td>ε<sub>1</sub>ε<sub>2</sub>ε<sub>3</sub>σ<sub>3</sub> =</td>
      <td>2555ε<sub>1</sub>ε<sub>2</sub>ε<sub>3</sub></td>
      <td>+19ε<sub>2</sub>ε<sub>3</sub></td>
      <td>-391ε<sub>2</sub></td>
      <td>+52</td>
    </tr>
    <tr>
      <td>ρ<sub>3</sub> =</td>
      <td>-11ε<sub>1</sub>ε<sub>2</sub>ε<sub>3</sub></td>
      <td>+49ε<sub>2</sub>ε<sub>3</sub></td>
      <td>-51ε<sub>2</sub></td>
      <td>-491</td>
    </tr>
    <tr>
      <td>ε<sub>2</sub>ω<sub>3</sub> =</td>
      <td>-80ε<sub>1</sub>ε<sub>2</sub>ε<sub>3</sub></td>
      <td>-23ε<sub>2</sub>ε<sub>3</sub></td>
      <td>-1402ε<sub>2</sub></td>
      <td>-1</td>
    </tr>
    <tr>
      <td>ε<sub>1</sub>ε<sub>2</sub>ε<sub>3</sub>τ<sub>5</sub> =</td>
      <td>-456ε<sub>1</sub>ε<sub>2</sub>ε<sub>3</sub></td>
      <td>-7ε<sub>1</sub>ε<sub>3</sub></td>
      <td>-2ε<sub>3</sub></td>
      <td>+3</td>
    </tr>
    <tr>
      <td>σ<sub>5</sub> =</td>
      <td>-4ε<sub>1</sub>ε<sub>2</sub>ε<sub>3</sub></td>
      <td>+17ε<sub>1</sub>ε<sub>3</sub></td>
      <td>+12ε<sub>3</sub></td>
      <td>+566</td>
    </tr>
    <tr>
      <td>ε<sub>3</sub>ρ<sub>5</sub> =</td>
      <td>9ε<sub>1</sub>ε<sub>2</sub>ε<sub>3</sub></td>
      <td>-39ε<sub>1</sub>ε<sub>3</sub></td>
      <td>+1265ε<sub>3</sub></td>
      <td>+13</td>
    </tr>
    <tr>
      <td>ε<sub>1</sub>ε<sub>3</sub>ω<sub>5</sub> =</td>
      <td>ε<sub>1</sub>ε<sub>2</sub>ε<sub>3</sub></td>
      <td>-252ε<sub>1</sub>ε<sub>3</sub></td>
      <td>-3ε<sub>3</sub></td>
      <td>+2</td>
    </tr>
    <tr>
      <td>ε<sub>1</sub>ε<sub>2</sub>ε<sub>3</sub>τ<sub>6</sub> =</td>
      <td>9785210ε<sub>1</sub>ε<sub>2</sub>ε<sub>3</sub></td>
      <td>+2609ε<sub>2</sub>ε<sub>3</sub></td>
      <td>-523ε<sub>3</sub></td>
      <td>-729</td>
    </tr>
    <tr>
      <td>σ<sub>6</sub> =</td>
      <td>479ε<sub>1</sub>ε<sub>2</sub>ε<sub>3</sub></td>
      <td>+987ε<sub>2</sub>ε<sub>3</sub></td>
      <td>-13874ε<sub>3</sub></td>
      <td>-19886676</td>
    </tr>
    <tr>
      <td>ε<sub>3</sub>ρ<sub>6</sub> =</td>
      <td>1713ε<sub>1</sub>ε<sub>2</sub>ε<sub>3</sub></td>
      <td>-8328ε<sub>2</sub>ε<sub>3</sub></td>
      <td>-13204073ε<sub>3</sub></td>
      <td>+764</td>
    </tr>
    <tr>
      <td>ε<sub>2</sub>ε<sub>3</sub>ω<sub>6</sub> =</td>
      <td>4002ε<sub>1</sub>ε<sub>2</sub>ε<sub>3</sub></td>
      <td>+7773526ε<sub>2</sub>ε<sub>3</sub></td>
      <td>+1749ε<sub>3</sub></td>
      <td>-2296</td>
    </tr>
  </tbody>
</table>

References

[1] N. Bergeron, P. Charollois, and L. E. García. Elliptic units for complex cubic fields. https://arxiv.org/abs/2311.04110, 2023.

[2] M. Bhargava. Higher composition laws III: the parametrization of quadratic rings. *Annals of Mathematics*, 159:1329–1360, 2004.

[3] P. Charollois and H. Darmon. *Arguments des unités de Stark et périodes de séries d'Eisenstein*. *Algebra & Number Theory*, 2.6:655–688, 2008.

[4] H. Cohen. *A Course in Computational Algebraic Number Theory*. Springer-Verlag, 1993.

[5] D. A. Cox. *Primes of the form $x^2 + ny^2$*. John Wiley & Sons, Ltd, 2013.
one prime above $N = 11$ in $\mathbb{K}_3$. We find the target ideal $\mathfrak{P}_3^2$ above 3 where $\mathfrak{P}_3\mathfrak{P}_3^2 = (3)$. Then, $h = -(35z^2 + 20z + 35)/3$ is an admissible generator for the ideal $qN\mathfrak{P}_3^2/\mathfrak{a}$. Here we have $\mathcal{Z}_f^1 = \{(1, 1), (4, -1)\}$ and $\kappa = 2$. The value

$$ \Gamma_{\mathfrak{f},\mathfrak{b},\mathfrak{a}}^{-}(\varepsilon, h)^{-1} = \frac{\Gamma\left(\frac{-1}{5}, \frac{\varepsilon^{-1}-1751}{495}, \frac{\varepsilon-776}{495}\right)^{-11}}{\Gamma\left(\frac{-11}{5}, \frac{\varepsilon^{-1}-1751}{45}, \frac{\varepsilon-776}{45}\right)^{-1}} \approx -27.5333588\dots - i \cdot 32.7146180\dots $$

is close to a root of the polynomial $x^{12}+57x^{11}+1956x^{10}+4640x^9+35415x^8-109818x^7+150139x^6-109818x^5+35415x^4+4640x^3+1956x^2+57x+1$ which defines $\mathbb{K}_3^+(\mathfrak{f})$. The computation time for 1000 digits is 7 seconds.

### 5.1.4 Example 4

Let $z$ be the complex root of the polynomial $x^3 - 65$ in the upper half-plane. Then $\mathbb{K}_4 = \mathbb{Q}(z)$ has class number 18. This means that most ideals won't be principal ideals and we will need to use helper ideals to build an admissible vector $h$. We choose the ideal $\mathfrak{f}$ above $q = 3$ such that $\mathfrak{f}^3 = (3)$. The corresponding narrow ray class group is $Cl^+(\mathfrak{f}) \simeq \mathbb{Z}/6\mathbb{Z} \times \mathbb{Z}/6\mathbb{Z}$. The unit $\varepsilon = z - 4$ is a generator for $\mathcal{O}_f^{+,\times}$ such that $\tilde{\lambda} = \tilde{t} = 1$. We choose $\mathfrak{b} = (1), \mathfrak{b} = \mathfrak{P}_{59}, \dots$ representatives for the 36 classes in $Cl^+(\mathfrak{f})$, and $\mathfrak{a}$ the unique degree one prime above $N = 5$ in $\mathbb{K}_4$. The ideals $qN/(\mathfrak{a}\mathfrak{b})$ are unfortunately not principal in general so we use helper ideals. In the case $\mathfrak{b} = (1)$ we find the helper ideal $\mathfrak{P}_{13}^2$ and in the case $\mathfrak{b} = \mathfrak{P}_{59}$ we find the helper ideal $\mathfrak{P}_{41}$. The corresponding admissible generators we found for $qN\mathfrak{P}_{13}^2/\mathfrak{a}$ and $qN\mathfrak{P}_{41}'/(\mathfrak{a}\mathfrak{P}_{59})$ are $h_{\mathfrak{b}=(1)} = 3z^2$ and $h_{\mathfrak{b}=\mathfrak{P}_{59}} = -(72z^2 + 75z + 1590)/59$ respectively. The corresponding levels will be $l_{(1)} = 3 \cdot 5 \cdot 13$, $l_{\mathfrak{P}_{59}} = 3 \cdot 5 \cdot 41$. Here we have $\mathcal{Z}_f^1 = \{(1, 1), (2, -1)\}$ and $\kappa = 2$. We may compute

$$ \Gamma_{\mathfrak{f},(1),\mathfrak{a}}^{-}(\varepsilon, h_{\mathfrak{b}=(1)})^{-1} = \frac{\Gamma\left(\frac{-1}{3}, \frac{\varepsilon^{-1}-211}{195}, \frac{\varepsilon-61}{195}\right)^{-5}}{\Gamma\left(\frac{-5}{3}, \frac{\varepsilon^{-1}-211}{39}, \frac{\varepsilon-61}{39}\right)^{-1}} \approx -1.6691052\dots + i \cdot 5.7493283\dots \\ \Gamma_{\mathfrak{f},\mathfrak{P}_{59},\mathfrak{a}}^{-}(\varepsilon, h_{\mathfrak{b}=\mathfrak{P}_{59}})^{-1} = \frac{\Gamma\left(\frac{1}{3}, \frac{\varepsilon^{-1}+1034}{615}, \frac{\varepsilon-91}{615}\right)^{-5}}{\Gamma\left(\frac{5}{3}, \frac{\varepsilon^{-1}+1034}{123}, \frac{\varepsilon-91}{123}\right)^{-1}} \approx 0.0344135\dots - i \cdot 0.0123218\dots $$

and the remaining 34 out of 36 values $I_{\mathfrak{b}} = \Gamma_{\mathfrak{f},\mathfrak{b},\mathfrak{a}}^{-}(\varepsilon, h_{\mathfrak{b}})^{-1}$ attached to the 36 classes in $Cl^+(\mathfrak{f})$. We may then compute the polynomial

$$ \prod_{\mathfrak{b} \in Cl^+(\mathfrak{f})} (X - I_{\mathfrak{b}}) = X^{36} - (u + vz + wz^2)X^{35} + \dots + 1 \in \mathcal{O}_{\mathbb{K}_4}[X] $$
[6] S. Dasgupta and M. Greenberg. Lecture notes on the Rank One Abelian Stark Conjecture, Arizona Winter School, 2011.

[7] S. Dasgupta and M. Kakde. Brumer-Stark Units and Explicit Class Field Theory. *Duke Math.*, 2023.

[8] G. Eisenstein. Beiträge zur Theorie der elliptischen Functionen, part VI. *Crelles Journal*, 35:153–274, 1847.

[9] G. Felder, A. Henriques, C. A. Rossi, and C. Zhu. A gerbe for the elliptic gamma function. *Duke Math.*, 141:1–74, 2008.

[10] G. Felder and A. Varchenko. The elliptic gamma function and SL(3, Z) ⋉ Z³. *Adv. Math.*, 156:44–76, 2000.

[11] The LMFDB Collaboration. The L-functions and modular forms database. https://www.lmfdb.org, 2024.

[12] A. Narukawa. The modular properties and the integral representations of the multiple elliptic gamma functions. *Adv. Math.*, 189(2):247–267, 2004.

[13] M. Nishizawa. An elliptic analogue of the multiple gamma function. *J. Phys. A: Math. Gen.*, 34:7411–7421, 2001.

[14] Vicençiu Paşol and Wadim Zudilin. A study of elliptic gamma function and allies. *Research in the Mathematical Sciences*, 5(4), 2018.

[15] A. R. Rankin. *Modular Forms and Functions*. Cambridge University Press, 1977.

[16] T. Ren and R. Sczech. A refinement of Stark’s conjecture over complex cubic number fields. *Journal of number theory*, 129:831–857, 2009.

[17] G. Robert. Unités elliptiques. *Bull. Soc. Math. France*, Mémoire 36, 1973.

[18] S. N. M. Ruijsenaars. First order analytic difference equations and integrable quantum systems. *J. Math. Phys*, 38(2):1069–1146, 1997.

[19] Sczech. Eisenstein group cocycles for GL$_n$ and values of l-functions. *Inventiones mathematicae*, 113:581–616, 1993.

[20] H. Stark. L-functions at s = 1. IV. First Derivatives at s = 0. *Advances in Mathematics*, 35:197–235, 1980.
where $u = 93377174024326, v = 769211619985, w = -5967373310133$. This palindromic polynomial defines a relative equation of the class field $K_4^+(f)$ above $K_4$ and we identify the rest of its coefficients in $O_{K_4}$. The computation time for 1000 digits and for all of the 36 computations is 8 minutes, which gives 13 seconds per individual computation on average.

It is interesting to note that this field as well as the field defined by $x^3 - 2$ belong to a special family parametrized by $x^3 - 2^{3k} - (-1)^k$ for $k \ge 0$ which behaves nicely compared to other pure cubic fields with respect to the values $\tilde{\lambda}, \tilde{t}$. Generally speaking, the values of these parameters vary significantly with the discriminant, as shown in the table in Appendix A. Putting $\mathbb{L}_k = \mathbb{Q}(z_k)$ where $z_k$ is the complex root of the polynomial $x^3 - 2^{3k} - (-1)^k$ lying in the upper half-plane, it seems that the positive fundamental unit in $\mathbb{L}_k$ is given by $\varepsilon_k = (-1)^k(z_k - 2^k)$ and its inverse by $\varepsilon_k^{-1} = z_k^2 + 2^k z_k + 2^{2k}$. When $D_k = 2^{3k} + (-1)^k$ is cube-free (at least for $0 \le k \le 100$ except for $k = 49, 50$), because $D_k \ne \pm 1 \pmod 9$, the field $\mathbb{L}_k$ is a so-called pure cubic field of the first kind and an integral basis of $O_{\mathbb{L}_k}$ is given by $(1, z_k, z_k^2/s_k)$ where $D_k = r_k s_k^2$ with $r_k, s_k$ square-free and relatively coprime. In that case, we may compute $\tilde{\lambda} = 1$ and $\tilde{t} = s_k$. When $D_k$ is also square-free (at least for $0 \le k \le 100$ except for $k = 7, 10, 21, 26, 30, 35, 63, 68, 70, 77, 78, 90, 91$) we get $\tilde{\lambda}_k = \tilde{t}_k = 1$. The class group grows very rapidly in this family (ex: $h(\mathbb{L}_{11}) = 2^2 \cdot 3^5 \cdot 5 \cdot 3191$ and $h(\mathbb{L}_{20}) = 2^6 \cdot 3^4 \cdot 5 \cdot 11 \cdot 19 \cdot 79 \cdot 863 \cdot 18047$). It would be interesting to understand if for $k \ge 100$ there are infinitely many fields in this family for which the conditions $\tilde{\lambda} = \tilde{t} = 1$ are satisfied for the fundamental unit. Our construction would then conjecturally allow to construct algebraic units of very high degree.

### 5.1.5 Example 5

Here we give an example with $q = 7$. Let $z$ be the complex root of the polynomial $x^3 - 14$ in the upper half-plane. Then $K_5 = \mathbb{Q}(z)$ has class number 3. We choose the ideal $f$ above $q = 7$ such that $f^3 = (7)$. The corresponding narrow ray class group is $Cl^+(f) \simeq \mathbb{Z}/18\mathbb{Z}$. The unit $\varepsilon = -z^2 + 2z + 1$ is a generator for $O_f^{+,\times}$ such that $\tilde{\lambda} = 1, \tilde{t} = 2 \cdot 11$. We choose $b = (1)$ and $a$ the unique degree one prime above 5 in $K_5$. Computing target ideals gives $\mathfrak{P}_2$ and $\mathfrak{P}_{11}$ the degree one primes above 2 and 11. The ideal $qN\mathfrak{P}_2\mathfrak{P}_{11}/(ab)$ is unfortunately not principal so we need to look for a helper ideal. In Proposition 22 we may take $m_1 = 3$ and $\mathfrak{J} = \mathfrak{P}_3^2$ so that the ideal $3qN\mathfrak{P}_2\mathfrak{P}_{11}\mathfrak{P}_3^2/(ab)$ is principal with an admissible generator $h = -21z^2 + 21z - 336$. The corresponding level will be $l = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11$.
[21] The PARI Group, Univ. Bordeaux. *PARI/GP version 2.15.4*, 2023. available from http://pari.math.u-bordeaux.fr/.

[22] A. Weil. *Elliptic functions according to Eisenstein and Kronecker*, volume 88 of *Ergebnisse der Mathematik und ihrer Grenzgebiete*. Springer-Verlag, first edition, 1976.

[23] Jacob Winding. Multiple elliptic gamma functions associated to cones. *Adv. Math.*, 325:56-86, 2018.
Here we have $\mathcal{Z}_f^1 = \{(1, 1), (6, -1)\}$ and $\kappa = 2$. The value

$$ \Gamma_{f,b,a}^{-}(\varepsilon, h)^{-1} = \frac{\Gamma\left(\frac{3}{7}, \frac{\varepsilon^{-1}-3067}{2310}, \frac{\varepsilon+1007}{2310}\right)^{-5}}{\Gamma\left(\frac{15}{7}, \frac{\varepsilon^{-1}-3067}{462}, \frac{\varepsilon+1007}{462}\right)^{-1}} \approx -0.1700923\dots + i \cdot 3.8609499\dots $$

is close to a root of the palindromic polynomial $x^{54} - 4167x^{53} + 7931535x^{52} - 259219286x^{51}\dots - 4167x + 1$ which defines $K_5^+(f)$ over $\mathbb{Q}$. This polynomial has very large coefficients, and we could alternatively compute the remaining 17 out of 18 values associated to the 18 classes in $Cl^+(f)$ to identify a relative polynomial in $O_{K_5}[x]$ instead, as we did in the previous example. The computation time for 1000 digits and for all 18 values is 18 minutes, which gives 1 minute per individual computation on average.

### 5.1.6 Example 6

This example falls outside of the conditions (*) and (**) because we will have $\tilde{\lambda} = 2$. Yet, it shows that these conditions are a set of p-adic conditions which can be satisfied for most $p$. In other words, the unique prime number for which (*) and (**) are not true in the present case is 2. Because of this, we can try to reduce the number of computations outside of the prime number 2.

Let $z$ be the complex root of the polynomial $x^3 - 5$ in the upper half-plane. Then $K_6 = \mathbb{Q}(z)$ has class number 1. We choose the ideal $f$ above $q = 3$ such that $f^3 = (3)$. The corresponding narrow ray class group is $Cl^+(f) \simeq \mathbb{Z}/2\mathbb{Z}$. The unit $\varepsilon = 2z^2 - 4z + 1$ is a generator for $O_f^{+,\times}$ such that $\tilde{\lambda} = 2$ and $\tilde{t} = 2 \times 13$. We choose $b = 1$ and $a$ the unique degree one prime above $N = 5$ in $K_6$. We find $\mathfrak{P}_{13}$ a target ideal above 13 and we find an admissible generator $h = 6z^2 + 15$ for the ideal $qN\mathfrak{P}_{13}/a$. The corresponding level will be $l = 3 \cdot 5 \cdot 13$. Here we have $\mathcal{Z}_f^1 = \{(1, 1), (2, -1)\}$ and $\kappa = 2$. Then $\Gamma_{f,b,a}^{-}(\varepsilon, h)^{-1}$ is a product of two ordinary smoothed elliptic Gamma functions with parameters

$$ \tau = \frac{\varepsilon^{-1} + 119}{2 \times 195}, \quad \sigma = \frac{\varepsilon + 59}{2 \times 195} $$

More precisely, put $\delta = -2z^2 - z - 5$. Then the set $L/M$ in formula (23) is exactly $\{0, \delta\}$ with $\delta/\gamma = (-4z^2 - 5z - 55)/195$. The product

$$ \frac{\Gamma\left(\frac{1}{3}, \frac{\varepsilon^{-1}+119}{390}, \frac{\varepsilon+59}{390}\right)^{-5}}{\Gamma\left(\frac{5}{3}, \frac{\varepsilon^{-1}+119}{78}, \frac{\varepsilon+59}{78}\right)^{-1}} \cdot \frac{\Gamma\left(\frac{1}{3} + \frac{-4z^2-5z-55}{195}, \frac{\varepsilon^{-1}+119}{390}, \frac{\varepsilon+59}{390}\right)^{-5}}{\Gamma\left(\frac{5}{3} + \frac{-4z^2-5z-55}{39}, \frac{\varepsilon^{-1}+119}{78}, \frac{\varepsilon+59}{78}\right)^{-1}} $$
is close to the root ≈ 9.8439696... + i · 5.1060682... of the palindromic polynomial $x^6 - 21x^5 + 150x^4 - 185x^3 + 150x^2 - 21x + 1$ which defines $K_6^+(f)$. The computation time for 1000 digits is 11 seconds.

## 5.2 Quartic examples

We now present four examples in the quartic case. The first example is one of the simplest cases, for which $\tilde{t}_1 = \tilde{t}_2 = 1$. The second example showcases the work on target ideals with $\tilde{t}_1 = 7, \tilde{t}_2 = 7129$. The third example shows that we can sometimes hope to find better units to gain time in the computations. The fourth example is a case where the quartic field $\mathbb{K}$ contains a real quadratic field and the set of units falls just outside $(*)$, $(**)$.

In all examples, we use the ordering $\{Id, (21)\}$ of $\mathfrak{S}_2$ and when we say that we “choose” fundamental units for $O_f^{+, \times}$, we let Pari/GP compute a set of fundamental units for $O_{\mathbb{K}}^{\times}$, which we use to compute a set of fundamental units for $O_f^{+, \times}$. Most of the time Pari/GP does a great job in finding small fundamental units, and the few operations required to compute $O_f^{+, \times}$ keep these units relatively small.

### 5.2.1 Example 7

We first discuss one of the simplest quartic examples, mentioned in section 3.2, this time with a “big smoothing”. Let $z$ be the complex root of the polynomial $x^4 - 6x^3 - x^2 - 3x + 1$ lying in the upper half-plane. Then $\mathbb{K}_7 = \mathbb{Q}(z)$ has class number 1. We choose $f = \mathfrak{P}_2$ the degree one prime above $q = 2$. The corresponding narrow ray class group is $Cl^+(f) \simeq \mathbb{Z}/2\mathbb{Z}$. We choose the fundamental units

$$ \varepsilon_1 = \frac{-2z^3 + 13z^2 - z + 3}{7}, \quad \varepsilon_2 = \frac{-5z^3 + 29z^2 + 15z + 18}{7} $$

for the set $O_f^{+, \times}$ of totally positive units congruent to 1 mod $f$. We compute the contents $\tilde{\lambda}_1 = \tilde{\lambda}_2 = 1$, the overflows $\tilde{t}_1 = \tilde{t}_2 = 1$ as defined in section 4.1. We may choose $b = (1)$ and $a$ the unique degree one prime above $N = 13$ in $\mathbb{K}_7$. Since $\tilde{t}_1 = \tilde{t}_2 = 1$, there are no target ideals to be found here. The ideal $qN/a$ has an admissible generator $h_1 = h_2 = (44z^3 - 258z^2 - 104z - 80)/7$ (see Definition 7). The corresponding levels will be $l_1 = l_2 = 2 \cdot 13$. We compute the orientations $\mu = \nu = [-1, 1]$ as given in Definition 10. Here, we have $Z_f^1 = \{(1, 1)\}$ as in Definition 25 and $\kappa = 1$. Let us write the parameters
$$ \tau = \varepsilon_2 - 15, \qquad \tau' = -6 + \frac{1}{\varepsilon_1} + \frac{1}{\varepsilon_1 \varepsilon_2} $$

$$ \sigma = -7 + \frac{1}{\varepsilon_2}, \qquad \sigma' = -\varepsilon_2 + 15 $$

$$ \rho = -\varepsilon_1 - 3, \qquad \rho' = 4\varepsilon_1 + 19 - \frac{1}{\varepsilon_2} $$

Then

$$ \frac{G_{2,f,b,a}^{+}(\varepsilon_2, \varepsilon_1\varepsilon_2; h_2)}{G_{2,f,b,a}^{-}(\varepsilon_1, \varepsilon_1\varepsilon_2; h_1)} = \frac{G_2\left(\frac{-1}{2}, \frac{\tau}{26}, \frac{\sigma}{26}, \frac{\rho}{26}\right)^{-13}}{G_2\left(\frac{-13}{2}, \frac{\tau}{2}, \frac{\sigma}{2}, \frac{\rho}{2}\right)^{-1}} \times \frac{G_2\left(\frac{1}{2}, \frac{\tau'}{26}, \frac{\sigma'}{26}, \frac{\rho'}{26}\right)^{13}}{G_2\left(\frac{13}{2}, \frac{\tau'}{2}, \frac{\sigma'}{2}, \frac{\rho'}{2}\right)} $$

is close to the root $\approx 4.1210208... - i \cdot 5.0617720...$ of the polynomial $x^8 - 7x^7 + 33x^6 + 49x^5 + 17x^4 + 49x^3 + 33x^2 - 7x + 1$ which defines $K_7^+(f)$. The computation time for 1000 digits is 6 seconds. We may also check formula (16) up to 1000 digits as:

$$ \mathcal{N}(\mathfrak{a})\zeta'_{f}([\mathfrak{b}], 0) - \zeta'_{f}([\mathfrak{ab}], 0) \approx \log \left| \frac{G_{2,f,b,a}^{+}(\varepsilon_2, \varepsilon_1\varepsilon_2; h_2)}{G_{2,f,b,a}^{-}(\varepsilon_1, \varepsilon_1\varepsilon_2; h_1)} \right|^2 \approx 3.7519563... $$

We compare the above computation to a simple choice for the admissible vectors $h_1, h_2$ given by Pari/GP. Namely for $h_1 = h_2 = (86z^3 - 496z^2 - 286z - 206)/7$ we compute $t_1 = t_2 = 31$ and $l = 26$. The computation time for this pair and for 1000 digits is 2 minutes, which is roughly 20 times longer than for the previous pair.

### 5.2.2 Example 8

We now discuss in full detail a quartic example with specific work on target ideals. Let $z$ be the complex root of the polynomial $x^4 - 19x^3 + 18x^2 + 8x + 1$ lying in the upper half-plane. Then $K_8 = \mathbb{Q}(z)$ has class number 1. We choose $f = \mathfrak{P}_3$ a prime ideal above $q = 3$ such that $(3) = \mathfrak{P}_3^2\mathfrak{P}_3'^2$. The corresponding narrow ray class group is $Cl^+(f) \simeq \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$. We choose the fundamental units

$$ \varepsilon_1 = \frac{19z^3 - 366z^2 + 438z + 44}{9}, \quad \varepsilon_2 = -z^3 + 19z^2 - 18z - 8 $$

for $O_f^{+,*}$. We compute $\tilde{\lambda}_1 = \tilde{\lambda}_2 = 1$, $\tilde{t}_1 = 7$, $\tilde{t}_2 = 7129$ where 7129 is a prime number. We may choose $\mathfrak{b} = (1)$ and $\mathfrak{a}$ the unique degree one prime
above $N = 13$ in $K_8$. We search for target ideals and find $\mathfrak{P}_7$ in the first case and $\mathfrak{P}_{7129}$ in the second case. Both ideals $qN\mathfrak{P}_7/\mathfrak{a}$ and $qN\mathfrak{P}_{7129}/\mathfrak{a}$ are principal with admissible generators $h_1 = (-32z^3 + 606z^2 - 543z - 112)/3$ and $h_2 = (124z^3 - 2319z^2 + 1563z + 200)/3$. The corresponding levels will be $l_1 = 3 \cdot 7 \cdot 13$ and $l_2 = 3 \cdot 13 \cdot 7129$. Here we have $Z_f^1 = \{(1, 1), (2, 1)\}$ and $\kappa = 2$. Let us write the parameters

$$ 
\begin{aligned}
\tau &= \frac{2\varepsilon_1\varepsilon_2 - 92\varepsilon_1 - 3}{\varepsilon_1}, & \tau' &= \frac{57488\varepsilon_1\varepsilon_2 + 3\varepsilon_2 + 16}{\varepsilon_1\varepsilon_2} \\
\sigma &= \frac{925\varepsilon_1\varepsilon_2 + 367\varepsilon_1 - 47}{\varepsilon_1\varepsilon_2}, & \sigma' &= \frac{47\varepsilon_1\varepsilon_2 - 328\varepsilon_2 - 348694}{1} \\
\rho &= \frac{-\varepsilon_1\varepsilon_2 - 3\varepsilon_1 + 40}{1}, & \rho' &= \frac{-57\varepsilon_1\varepsilon_2 - 90004\varepsilon_2 - 2}{\varepsilon_2}
\end{aligned}
$$

Then

$$ \frac{G_{2,f,b,a}^{+}(\varepsilon_1, \varepsilon_1\varepsilon_2; h_1)}{G_{2,f,b,a}^{-}(\varepsilon_2, \varepsilon_1\varepsilon_2; h_2)} = \frac{G_2\left(\frac{1}{3}, \frac{\tau}{273}, \frac{\sigma}{273}, \frac{\rho}{273}\right)^{13}}{G_2\left(\frac{13}{3}, \frac{\tau}{21}, \frac{\sigma}{21}, \frac{\rho}{21}\right)} \times \frac{G_2\left(\frac{-2}{3}, \frac{\tau'}{278031}, \frac{\sigma'}{278031}, \frac{\rho'}{278031}\right)^{-13}}{G_2\left(\frac{-26}{3}, \frac{\tau'}{21387}, \frac{\sigma'}{21387}, \frac{\rho'}{21387}\right)^{-1}} $$

is close to the root $\approx 10.6409709... - i \cdot 5.9332732...$ of the polynomial $x^8 - 18x^7 + 83x^6 + 396x^5 + 597x^4 + 396x^3 + 83x^2 - 18x + 1$ which defines a subextension of $K_8^+(f)/K_8$. The computation time for 1000 digits is 3 minutes and 20 seconds.

### 5.2.3 Example 9

Keep $K_8$, $f$, $b$, $a$ as in example 8 and let us change our choice of fundamental units. We fix another set of fundamental units for $O_f^{+, \times}$:

$$ \varepsilon_1 = \frac{z^3 - 21z^2 + 54z + 11}{9}, \quad \varepsilon_2 = -z^3 + 19z^2 - 18z - 8 $$

Then, we compute $\tilde{\lambda}_1 = \tilde{\lambda}_2 = 1$, $\tilde{t}_1 = 1$, $\tilde{t}_2 = 25$. In the second computation, a target ideal is $\mathfrak{P}_5^2$ the square of the unique degree one prime above 5 and the ideals $qN/\mathfrak{a}$ and $qN\mathfrak{P}_5^2/\mathfrak{a}$ are generated by the admissible vectors $h_1 = (-44z^3 + 843z^2 - 927z - 232)/3$, $h_2 = (76z^3 - 1449z^2 + 1470z + 344)/3$. The corresponding levels will be $l_1 = 3 \cdot 13$ and $l_2 = 3 \cdot 5^2 \cdot 13$. Let us write the parameters
$$
\begin{align*}
\tau &= \frac{2\varepsilon_1\varepsilon_2 + 109\varepsilon_1 - 3}{\varepsilon_1}, \\
\sigma &= \frac{-321\varepsilon_1\varepsilon_2 - 16\varepsilon_1 + 7}{\varepsilon_1\varepsilon_2}, \\
\rho &= \frac{-\varepsilon_1 - 47}{1},
\end{align*}
$$

$$
\begin{align*}
\tau' &= \frac{842\varepsilon_1\varepsilon_2 + \varepsilon_2 + 3}{\varepsilon_1\varepsilon_2} \\
\sigma' &= \frac{-7\varepsilon_1\varepsilon_2 + 4\varepsilon_2 - 63}{1} \\
\rho' &= \frac{-3\varepsilon_1\varepsilon_2 + 566\varepsilon_2 - 2}{\varepsilon_2}
\end{align*}
$$

Then

$$
\frac{G_{2,f,b,a}^{+}(\varepsilon_2, \varepsilon_1\varepsilon_2; h_2)}{G_{2,f,b,a}^{-}(\varepsilon_1, \varepsilon_1\varepsilon_2; h_1)} = \frac{G_2\left(\frac{-1}{3}, \frac{\tau}{39}, \frac{\sigma}{39}, \frac{\rho}{39}\right)^{-13}}{G_2\left(\frac{-13}{3}, \frac{\tau}{3}, \frac{\sigma}{3}, \frac{\rho}{3}\right)^{-1}} \times \frac{G_2\left(\frac{2}{3}, \frac{\tau'}{975}, \frac{\sigma'}{975}, \frac{\rho'}{975}\right)^{13}}{G_2\left(\frac{26}{3}, \frac{\tau'}{75}, \frac{\sigma'}{75}, \frac{\rho'}{75}\right)}
$$

is close to the same root ≈ 10.6409709... − i · 5.9332732... of the polynomial
x⁸ − 18x⁷ + 83x⁶ + 396x⁵ + 597x⁴ + 396x³ + 83x² − 18x + 1. The compu-
tation time for 1000 digits is 11 seconds. This value is the same as for the
previous example in which the choice of fundamental units was different,
which supports the idea that the product in Conjecture 11 is independent of
said choice, even though computations may be longer when a poor choice of
fundamental units is made.

5.2.4 Example 10

This example falls outside of (*) and (**) because gcd(̃t₁, ̃t₂, q = 2) = 2. Still,
we can find a pair of vectors h₁, h₂ for which the computations give interesting
results. Let z be the complex root of the polynomial x⁴ − 12 lying in the
upper half-plane. Then K₁₀ = Q(z) has class number 1 and *contains the real
quadratic field Q(√3).* Remember that this means that we must carefully
choose the fundamental units in this case to avoid totally real elements (see
(*) conditions in section 4.3). We choose the ideal *f* above q = 2 such that
*f*⁴ = (2). The corresponding narrow ray class group is *Cl*⁺(*f*) ≃ Z/2Z. We
choose the fundamental units

$$
\epsilon_1 = \frac{z^2 + 2z + 2}{4}, \quad \epsilon_2 = \frac{-z^3 + z^2 + 4z - 2}{4}
$$

for $O_f^{+ , \times}$. We compute $\tilde{\lambda}_1 = \tilde{\lambda}_2 = 1$, $\tilde{t}_1 = 2 \cdot 3$, $\tilde{t}_2 = 2 \cdot 11$. We may choose $\mathfrak{b} = (1)$ and $\mathfrak{a} = \mathfrak{P}_{23}$ a degree one prime above $N = 23$ in $\mathbb{K}_{10}$. We search for target ideals and find $\mathfrak{P}_3$ the degree one prime ideal above 3 in the first case and $\mathfrak{P}_{11}$ one of the degree one primes above 11 in the second case. Then $qN\mathfrak{P}_3/\mathfrak{a}$ and $qN\mathfrak{P}_{11}/\mathfrak{a}$ are generated by the admissible vectors
$h_1 = (-4z^3 + 11z^2 + 10z + 30)/2$ and $h_2 = (9z^3 + 4z^2 - 34z - 56)/2$. The corresponding levels will be $l_1 = 2 \cdot 3 \cdot 23$ and $l_2 = 2 \cdot 11 \cdot 23$. Here we have $Z_f^1 = \{(1, 1)\}$ and $\kappa = 1$. We compute the orientations $\underline{\mu} = \underline{\nu} = [1, -1]$ and the value $I_{2,f,b,a}(\varepsilon_1, \varepsilon_2, \underline{h}, \underline{\mu}, \underline{\nu}) \approx 13.9102308... - i \cdot 24.0932265...$ with $t_1 = t_2 = 2$, i.e. we compute a product of 8 ordinary elliptic smoothed $G_2$ functions. The result is close to a root of the polynomial $x^8 - 28x^7 + 778x^6 - 112x^5 - 749x^4 - 112x^3 + 778x^2 - 28x + 1$ which defines $K_{10}^+(f)$. The computation time for 1000 digits is 8 seconds.

Now we would like to discuss the case where some of the $\tilde{\alpha}_j$ belong to $\mathbb{R}$. Choose the fundamental units:

$$ \varepsilon_1 = \frac{z^2 + 2z + 2}{4}, \quad \varepsilon_2 = \frac{-z^2 + 4}{2} = 2 + \sqrt{3} $$

Then we may compute $\tilde{\lambda}_1 = \tilde{\lambda}_2 = 1$, $\tilde{t}_1 = \tilde{t}_2 = 1$. Keeping the same ideals $\mathfrak{b} = (1)$, $\mathfrak{a} = \mathfrak{P}_{23}$, we find the admissible vectors $h_1 = h_2 = (-3z^3 - 9z^2 - 4z + 34)/2$ for $qN/\mathfrak{P}_{23}$. We write the obtained parameters as:

$$
\begin{align*}
\tau &= \varepsilon_2 + 175 = 177 + \sqrt{3}, & \tau' &= 39 + \frac{1}{\varepsilon_1} \\
\sigma &= -342 + \frac{3}{\varepsilon_2} - \frac{1}{\varepsilon_1\varepsilon_2}, & \sigma' &= \varepsilon_2\varepsilon_1 + \varepsilon_2 - 352 \\
\rho &= -\varepsilon_1 + 217, & \rho' &= 179 - \frac{1}{\varepsilon_2} = 177 + \sqrt{3}
\end{align*}
$$

We get in particular $\tau, \rho' \in \mathbb{R}$, $\mathfrak{F}(\tau')$, $\mathfrak{F}(\rho) < 0$ and $\mathfrak{F}(\sigma)$, $\mathfrak{F}(\sigma') > 0$. In this setting, as suggested by my advisor P. Charollois, it is possible to make sense of formula (6). Indeed, using Liouville's theorem on the bad approximation of real irrational algebraic numbers by rational numbers, we get that

$$ \frac{1}{\sin \pi j \tau} = O(j^2), \quad \frac{1}{\sin \pi j \rho'} = O(j^2) $$

since $\tau, \rho'$ are algebraic numbers of degree 2. This guarantees that formula (6) converges in this setting for $z \in \mathbb{Q}$. The computation of the product

$$ \frac{G_{2,f,b,a}^{+}(\varepsilon_1, \varepsilon_1\varepsilon_2; h_2)}{G_{2,f,b,a}^{-}(\varepsilon_2, \varepsilon_1\varepsilon_2; h_1)} = \frac{G_2\left(\frac{1}{2}, \frac{\tau}{46}, \frac{\sigma}{46}, \frac{\rho}{46}\right)^{23}}{G_2\left(\frac{23}{2}, \frac{\tau}{2}, \frac{\sigma}{2}, \frac{\rho}{2}\right)} \times \frac{G_2\left(\frac{-1}{2}, \frac{\tau'}{46}, \frac{\sigma'}{46}, \frac{\rho'}{46}\right)^{-23}}{G_2\left(\frac{-23}{2}, \frac{\tau'}{2}, \frac{\sigma'}{2}, \frac{\rho'}{2}\right)} $$

using formula (6) yields a number close to the same root $\approx 13.9102308... - i \cdot 24.0932265...$ of the polynomial $x^8 - 28x^7 + 778x^6 - 112x^5 - 749x^4 - 112x^3 + 778x^2 - 28x + 1$. It is expected to be true generally that formula (6) may always be used to compute the right-hand side of formula (16) when some of the $\tilde{\alpha}_j$ are real.
## 5.3 One Quintic example

We have already successfully produced a dozen of quintic examples and we give here a detailed presentation of our simplest one. Let $z$ be the complex root of the polynomial $x^5 - x^4 - x^3 - 2x^2 + x + 1$ lying in the upper half-plane. Then $K_{11} = \mathbb{Q}(z)$ has class number 1. We choose $f = \mathfrak{P}_3$ the degree one prime above $q = 3$. The corresponding narrow ray class group is $Cl^+(f) \simeq \mathbb{Z}/2\mathbb{Z}$. We choose the fundamental units (once again take the fundamental units provided by Pari/GP version 2.15.4 and then compute the simplest totally positive ones):

$$ \epsilon_1 = z^4 - 2z^3 - z + 3, \quad \epsilon_2 = 2z^4 - 2z^3 - 3z + 3, \quad \epsilon_3 = 2z^4 - 3z^3 - 4z + 4 $$

for $O_f^{+, \times}$. We fix the order $\{Id, (32), (21), (231), (312), (31)\}$ of $\mathfrak{S}_3$ given by Pari/GP. We compute $\tilde{\lambda}_1 = \tilde{\lambda}_2 = \tilde{\lambda}_3 = \tilde{\lambda}_4 = \tilde{\lambda}_5 = \tilde{\lambda}_6 = 1$, $\tilde{t}_1 = 7 \cdot 37 \cdot 137$, $\tilde{t}_2 = 31 \cdot 53$, $\tilde{t}_3 = 491$, $\tilde{t}_4 = 107$, $\tilde{t}_5 = 1$ and $\tilde{t}_6 = 145637$ where we have written the prime number decomposition of all the overflows $\tilde{t}$. We may choose $\mathfrak{b} = (1)$ and $\mathfrak{a}$ the unique degree one prime above $N = 11$ in $K_{11}$. We look for target ideals above the prime numbers listed above and find the ideals $\mathfrak{P}_7, \mathfrak{P}'_{37}, \mathfrak{P}'_{137}, \mathfrak{P}_{31}, \mathfrak{P}_{53}, \mathfrak{P}'_{491}, \mathfrak{P}'_{107}, \mathfrak{P}_{145637}$ in the order provided by Pari/GP version 2.15.4. The ideals $qN\mathfrak{P}_7\mathfrak{P}'_{37}\mathfrak{P}'_{137}/\mathfrak{a}, qN\mathfrak{P}_{31}\mathfrak{P}_{53}/\mathfrak{a}, qN\mathfrak{P}'_{491}/\mathfrak{a}, qN\mathfrak{P}'_{107}/\mathfrak{a}, qN/\mathfrak{a}, qN\mathfrak{P}_{145637}/\mathfrak{a}$ admit the following admissible generators:

$$ h_1 = 147z^4 - 135z^3 - 90z^2 - 234z + 72 $$

$$ h_2 = -15z^4 - 30z^3 + 90z^2 + 36z + 60 $$

$$ h_3 = 42z^4 - 15z^3 - 87z^2 - 48z + 30 $$

$$ h_4 = -21z^4 + 57z^3 - 6z^2 - 9z - 48 $$

$$ h_5 = -3z^4 - 6z^3 + 18z^2 - 6z + 12 $$

$$ h_6 = 108z^4 - 246z^3 + 111z^2 - 81z + 195 $$

The corresponding levels are $l_1 = 3 \cdot 7 \cdot 11 \cdot 37 \cdot 137$, $l_2 = 3 \cdot 11 \cdot 491$, $l_3 = 3 \cdot 11 \cdot 107$, $l_4 = 3 \cdot 11 \cdot 31 \cdot 53$, $l_5 = 3 \cdot 11$, $l_6 = 3 \cdot 11 \cdot 145637$. Here we have $Z_f^1 = \{(1, 1), (2, -1)\}$ and $\kappa = 2$. We compute the six quotients:

$$
\begin{aligned}
& \frac{G_3\left(\frac{-1}{3}, \frac{\tau_1}{l_1}, \frac{\sigma_1}{l_1}, \frac{\rho_1}{l_1}, \frac{\varpi_1}{l_1}\right)^{11}}{G_3\left(\frac{-11}{3}, \frac{11\tau_1}{l_1}, \frac{11\sigma_1}{l_1}, \frac{11\rho_1}{l_1}, \frac{11\varpi_1}{l_1}\right)} & \approx -6.9846353\dots10^{-18} + i \cdot 6.2764063\dots10^{-17} \\[1.5em]
& \frac{G_3\left(\frac{-1}{3}, \frac{\tau_2}{l_2}, \frac{\sigma_2}{l_2}, \frac{\rho_2}{l_2}, \frac{\varpi_2}{l_2}\right)^{-11}}{G_3\left(\frac{-11}{3}, \frac{11\tau_2}{l_2}, \frac{11\sigma_2}{l_2}, \frac{11\rho_2}{l_2}, \frac{11\varpi_2}{l_2}\right)} & \approx 1.7434761\dots10^{11} - i \cdot 1.7914686\dots10^{11}
\end{aligned}
$$
$$
\begin{align*}
& \frac{G_3\left(\frac{1}{3}, \frac{\tau_3}{l_3}, \frac{\sigma_3}{l_3}, \frac{\rho_3}{l_3}, \frac{\varpi_3}{l_3}\right)^{-11}}{G_3\left(\frac{11}{3}, \frac{11\tau_3}{l_3}, \frac{11\sigma_3}{l_3}, \frac{11\rho_3}{l_3}, \frac{11\varpi_3}{l_3}\right)^{-1}} & \approx 0.2815024\dots - i \cdot 0.0665163\dots \\
& \frac{G_3\left(\frac{1}{3}, \frac{\tau_4}{l_4}, \frac{\sigma_4}{l_4}, \frac{\rho_4}{l_4}, \frac{\varpi_4}{l_4}\right)^{11}}{G_3\left(\frac{11}{3}, \frac{11\tau_4}{l_4}, \frac{11\sigma_4}{l_4}, \frac{11\rho_4}{l_4}, \frac{11\varpi_4}{l_4}\right)} & \approx 9.8457700\dots - i \cdot 2.3506772\dots \\
& \frac{G_3\left(\frac{-1}{3}, \frac{\tau_5}{l_5}, \frac{\sigma_5}{l_5}, \frac{\rho_5}{l_5}, \frac{\varpi_5}{l_5}\right)^{11}}{G_3\left(\frac{-11}{3}, \frac{11\tau_5}{l_5}, \frac{11\sigma_5}{l_5}, \frac{11\rho_5}{l_5}, \frac{11\varpi_5}{l_5}\right)} & \approx -0.2518907\dots - i \cdot 0.2274847\dots \\
& \frac{G_3\left(\frac{-1}{3}, \frac{\tau_6}{l_6}, \frac{\sigma_6}{l_6}, \frac{\rho_6}{l_6}, \frac{\varpi_6}{l_6}\right)^{-11}}{G_3\left(\frac{-11}{3}, \frac{11\tau_6}{l_6}, \frac{11\sigma_6}{l_6}, \frac{11\rho_6}{l_6}, \frac{11\varpi_6}{l_6}\right)^{-1}} & \approx 100864.0193260\dots - i \cdot 767246.5816458\dots
\end{align*}
$$

where the parameters are given in the table below. The product of these six quotients is close to the root $\approx -11.6360077\dots + i \cdot 3.4634701\dots$ of the polynomial $x^{10}+24x^9+164x^8+99x^7-62x^6-89x^5-62x^4+99x^3+164x^2+24x+1$ which defines $K_{11}^+(f)$. The computation time for 1000 digits is 1 minute and 35 seconds, but the computation time for each of the individual computations is not uniform at all. The fifth computation requires 1 second whereas the sixth computation requires 58 seconds because of the level difference $l_5 = 33$ versus $l_6 = 4806021$.
<table>
  <tbody>
    <tr>
      <td>ε<sub>1</sub>τ<sub>1</sub> =</td>
      <td>229ε<sub>1</sub>ε<sub>2</sub>ε<sub>3</sub></td>
      <td>-1727ε<sub>1</sub>ε<sub>2</sub></td>
      <td>-796386ε<sub>1</sub></td>
      <td>+3385</td>
    </tr>
    <tr>
      <td>ε<sub>1</sub>ε<sub>2</sub>σ<sub>1</sub> =</td>
      <td>-114ε<sub>1</sub>ε<sub>2</sub>ε<sub>3</sub></td>
      <td>-987551ε<sub>1</sub>ε<sub>2</sub></td>
      <td>+718ε<sub>1</sub></td>
      <td>+949</td>
    </tr>
    <tr>
      <td>ε<sub>1</sub>ε<sub>2</sub>ε<sub>3</sub>ρ<sub>1</sub> =</td>
      <td>-595138ε<sub>1</sub>ε<sub>2</sub>ε<sub>3</sub>+</td>
      <td>12ε<sub>1</sub>ε<sub>2</sub></td>
      <td>+1445ε<sub>1</sub></td>
      <td>-907</td>
    </tr>
    <tr>
      <td>ω<sub>1</sub> =</td>
      <td>-52ε<sub>1</sub>ε<sub>2</sub>ε<sub>3</sub></td>
      <td>+857ε<sub>1</sub>ε<sub>2</sub></td>
      <td>-295ε<sub>1</sub></td>
      <td>+2067162</td>
    </tr>
    <tr>
      <td>ε<sub>1</sub>τ<sub>2</sub> =</td>
      <td>99ε<sub>1</sub>ε<sub>2</sub>ε<sub>3</sub></td>
      <td>-356ε<sub>1</sub>ε<sub>3</sub></td>
      <td>+19304ε<sub>1</sub></td>
      <td>+57</td>
    </tr>
    <tr>
      <td>ε<sub>1</sub>ε<sub>3</sub>σ<sub>2</sub> =</td>
      <td>-17ε<sub>1</sub>ε<sub>2</sub>ε<sub>3</sub></td>
      <td>-70870ε<sub>1</sub>ε<sub>3</sub></td>
      <td>-145ε<sub>1</sub></td>
      <td>+40</td>
    </tr>
    <tr>
      <td>ε<sub>1</sub>ε<sub>2</sub>ε<sub>3</sub>ρ<sub>2</sub> =</td>
      <td>324170ε<sub>1</sub>ε<sub>2</sub>ε<sub>3</sub></td>
      <td>-85ε<sub>1</sub>ε<sub>3</sub></td>
      <td>+197ε<sub>1</sub></td>
      <td>-111</td>
    </tr>
    <tr>
      <td>ω<sub>2</sub> =</td>
      <td>91ε<sub>1</sub>ε<sub>2</sub>ε<sub>3</sub></td>
      <td>-460ε<sub>1</sub>ε<sub>3</sub></td>
      <td>+3ε<sub>1</sub></td>
      <td>-28725</td>
    </tr>
    <tr>
      <td>ε<sub>1</sub>ε<sub>2</sub>τ<sub>4</sub> =</td>
      <td>-62ε<sub>1</sub>ε<sub>2</sub>ε<sub>3</sub></td>
      <td>-12636ε<sub>1</sub>ε<sub>2</sub></td>
      <td>+20ε<sub>2</sub></td>
      <td>+3</td>
    </tr>
    <tr>
      <td>ε<sub>1</sub>ε<sub>2</sub>ε<sub>3</sub>σ<sub>4</sub> =</td>
      <td>327915ε<sub>1</sub>ε<sub>2</sub>ε<sub>3</sub></td>
      <td>-37ε<sub>1</sub>ε<sub>2</sub></td>
      <td>-1101ε<sub>2</sub></td>
      <td>+154</td>
    </tr>
    <tr>
      <td>ρ<sub>4</sub> =</td>
      <td>-44ε<sub>1</sub>ε<sub>2</sub>ε<sub>3</sub></td>
      <td>+314ε<sub>1</sub>ε<sub>2</sub></td>
      <td>-65ε<sub>2</sub></td>
      <td>-52915</td>
    </tr>
    <tr>
      <td>ε<sub>2</sub>ω<sub>4</sub> =</td>
      <td>-262ε<sub>1</sub>ε<sub>2</sub>ε<sub>3</sub></td>
      <td>-5ε<sub>1</sub>ε<sub>2</sub></td>
      <td>-768869ε<sub>2</sub></td>
      <td>-19</td>
    </tr>
    <tr>
      <td>ε<sub>2</sub>ε<sub>3</sub>τ<sub>3</sub> =</td>
      <td>-27ε<sub>1</sub>ε<sub>2</sub>ε<sub>3</sub></td>
      <td>-84ε<sub>2</sub>ε<sub>3</sub></td>
      <td>+11ε<sub>2</sub></td>
      <td>+1</td>
    </tr>
    <tr>
      <td>ε<sub>1</sub>ε<sub>2</sub>ε<sub>3</sub>σ<sub>3</sub> =</td>
      <td>2555ε<sub>1</sub>ε<sub>2</sub>ε<sub>3</sub></td>
      <td>+19ε<sub>2</sub>ε<sub>3</sub></td>
      <td>-391ε<sub>2</sub></td>
      <td>+52</td>
    </tr>
    <tr>
      <td>ρ<sub>3</sub> =</td>
      <td>-11ε<sub>1</sub>ε<sub>2</sub>ε<sub>3</sub></td>
      <td>+49ε<sub>2</sub>ε<sub>3</sub></td>
      <td>-51ε<sub>2</sub></td>
      <td>-491</td>
    </tr>
    <tr>
      <td>ε<sub>2</sub>ω<sub>3</sub> =</td>
      <td>-80ε<sub>1</sub>ε<sub>2</sub>ε<sub>3</sub></td>
      <td>-23ε<sub>2</sub>ε<sub>3</sub></td>
      <td>-1402ε<sub>2</sub></td>
      <td>-1</td>
    </tr>
    <tr>
      <td>ε<sub>1</sub>ε<sub>2</sub>ε<sub>3</sub>τ<sub>5</sub> =</td>
      <td>-456ε<sub>1</sub>ε<sub>2</sub>ε<sub>3</sub></td>
      <td>-7ε<sub>1</sub>ε<sub>3</sub></td>
      <td>-2ε<sub>3</sub></td>
      <td>+3</td>
    </tr>
    <tr>
      <td>σ<sub>5</sub> =</td>
      <td>-4ε<sub>1</sub>ε<sub>2</sub>ε<sub>3</sub></td>
      <td>+17ε<sub>1</sub>ε<sub>3</sub></td>
      <td>+12ε<sub>3</sub></td>
      <td>+566</td>
    </tr>
    <tr>
      <td>ε<sub>3</sub>ρ<sub>5</sub> =</td>
      <td>9ε<sub>1</sub>ε<sub>2</sub>ε<sub>3</sub></td>
      <td>-39ε<sub>1</sub>ε<sub>3</sub></td>
      <td>+1265ε<sub>3</sub></td>
      <td>+13</td>
    </tr>
    <tr>
      <td>ε<sub>1</sub>ε<sub>3</sub>ω<sub>5</sub> =</td>
      <td>ε<sub>1</sub>ε<sub>2</sub>ε<sub>3</sub></td>
      <td>-252ε<sub>1</sub>ε<sub>3</sub></td>
      <td>-3ε<sub>3</sub></td>
      <td>+2</td>
    </tr>
    <tr>
      <td>ε<sub>1</sub>ε<sub>2</sub>ε<sub>3</sub>τ<sub>6</sub> =</td>
      <td>9785210ε<sub>1</sub>ε<sub>2</sub>ε<sub>3</sub></td>
      <td>+2609ε<sub>2</sub>ε<sub>3</sub></td>
      <td>-523ε<sub>3</sub></td>
      <td>-729</td>
    </tr>
    <tr>
      <td>σ<sub>6</sub> =</td>
      <td>479ε<sub>1</sub>ε<sub>2</sub>ε<sub>3</sub></td>
      <td>+987ε<sub>2</sub>ε<sub>3</sub></td>
      <td>-13874ε<sub>3</sub></td>
      <td>-19886676</td>
    </tr>
    <tr>
      <td>ε<sub>3</sub>ρ<sub>6</sub> =</td>
      <td>1713ε<sub>1</sub>ε<sub>2</sub>ε<sub>3</sub></td>
      <td>-8328ε<sub>2</sub>ε<sub>3</sub></td>
      <td>-13204073ε<sub>3</sub></td>
      <td>+764</td>
    </tr>
    <tr>
      <td>ε<sub>2</sub>ε<sub>3</sub>ω<sub>6</sub> =</td>
      <td>4002ε<sub>1</sub>ε<sub>2</sub>ε<sub>3</sub></td>
      <td>+7773526ε<sub>2</sub>ε<sub>3</sub></td>
      <td>+1749ε<sub>3</sub></td>
      <td>-2296</td>
    </tr>
  </tbody>
</table>

## 5.4 Quartic examples with $\kappa < \#\mathcal{Z}_f^1$

We now present three quartic examples with $\kappa < \#\mathcal{Z}_f^1$. The first example is a case where $q = 5$, $\kappa = 2$ and $\mathcal{Z}_f^1 \simeq \mathbb{Z}/5\mathbb{Z}^\times$. The second example is a case where $q = 5$, $\kappa = 1$ and $\mathcal{Z}_f^1 \simeq \mathbb{Z}/5\mathbb{Z}^\times$. The third example is a case where $q = 7$, $\kappa = 1$ and $\mathcal{Z}_f^1 = \{(1, 1), (6, -1)\}$.

### 5.4.1 Example 12

We first discuss a quartic example with $q = 5$, $\kappa = 2$ and $\mathcal{Z}_f^1 \simeq \mathbb{Z}/5\mathbb{Z}^\times$, where the character $\chi_f$ is $(\frac{1}{5})$. Let $z$ be the complex root of the polynomial $x^4 - 3x + 1$ lying in the upper half-plane. Then $K_{12} = \mathbb{Q}(z)$ has class number 1. We choose $f = \mathfrak{P}_5$ the degree one prime above $q = 5$. The corresponding narrow ray class group is $Cl^+(f) \simeq \mathbb{Z}/2\mathbb{Z}$. We choose the fundamental units

$$ \varepsilon_1 = -2z^3 + z^2 + 3z - 1, \quad \varepsilon_2 = z^2 - 2z + 1 $$
for $O_f^{+, \times}$. We compute $\tilde{\lambda}_1 = \tilde{\lambda}_2 = 1$, $\tilde{t}_1 = 25561$, $\tilde{t}_2 = 59$ where 25561 is a prime number. We may choose $\mathfrak{b} = (1)$ and $\mathfrak{a}$ the unique degree one prime above $N = 7$ in $\mathbb{K}_{12}$. We search for target ideals and find $\mathfrak{P}_{25561}$ in the first case and $\mathfrak{P}_{59}$ in the second case. Both ideals $qN\mathfrak{P}_{25561}/\mathfrak{a}$ and $qN\mathfrak{P}_{59}/\mathfrak{a}$ are principal with admissible generators $h_1 = 130z^3 - 40z^2 - 125z - 225$ and $h_2 = 65z^3 + 50z^2 + 60z - 95$. The corresponding levels will be $l_1 = 5 \cdot 7 \cdot 25561$ and $l_2 = 5 \cdot 7 \cdot 59$. Let us write the parameters

$$
\begin{align*}
\tau &= \frac{9\varepsilon_1\varepsilon_2 + 694144\varepsilon_1 + 70}{\varepsilon_1}, &
\tau' &= \frac{-19730\varepsilon_2\varepsilon_1 + 11\varepsilon_2 - 4}{\varepsilon_1\varepsilon_2} \\
\sigma &= \frac{-3353996\varepsilon_1\varepsilon_2 + 504\varepsilon_1 - 215}{\varepsilon_1\varepsilon_2}, &
\sigma' &= \frac{-215\varepsilon_2\varepsilon_1 - 2436\varepsilon_2 + 64874}{1} \\
\rho &= \frac{32\varepsilon_1\varepsilon_2 + 11\varepsilon_1 + 130894}{1}, &
\rho' &= \frac{-5\varepsilon_2\varepsilon_1 + 5654\varepsilon_2 + 9}{\varepsilon_2}
\end{align*}
$$

Then the two products

$$
\begin{align*}
v_1 &= \prod_{k=1,4} \frac{G_2\left(\frac{k}{5}, \frac{\tau}{894635}, \frac{\sigma}{894635}, \frac{\rho}{894635}\right)^7}{G_2\left(\frac{7k}{5}, \frac{\tau}{127805}, \frac{\sigma}{127805}, \frac{\rho}{127805}\right)} \times \frac{G_2\left(\frac{k}{5}, \frac{\tau'}{2065}, \frac{\sigma'}{2065}, \frac{\rho'}{2065}\right)^{-7}}{G_2\left(\frac{7k}{5}, \frac{\tau'}{295}, \frac{\sigma'}{295}, \frac{\rho'}{295}\right)^{-1}} \\
v_2 &= \prod_{k=2,3} \frac{G_2\left(\frac{-k}{5}, \frac{\tau}{894635}, \frac{\sigma}{894635}, \frac{\rho}{894635}\right)^{-7}}{G_2\left(\frac{-7k}{5}, \frac{\tau}{127805}, \frac{\sigma}{127805}, \frac{\rho}{127805}\right)^{-1}} \times \frac{G_2\left(\frac{-k}{5}, \frac{\tau'}{2065}, \frac{\sigma'}{2065}, \frac{\rho'}{2065}\right)^7}{G_2\left(\frac{-7k}{5}, \frac{\tau'}{295}, \frac{\sigma'}{295}, \frac{\rho'}{295}\right)^{1}}
\end{align*}
$$

coincide up to 1000 digits and are close to the root $\approx -33.4405123... - i \cdot 62.3162955...$ of the polynomial $x^8 + 68x^7 + 5078x^6 + 5703x^5 + 7945x^4 + 5703x^3 + 5078x^2 + 68x + 1$ which defines $K_{12}^+(f)$. The computation time for 1000 digits is 48 minutes. The fact that the two products seem to be equal suggests that the integer $\kappa$ defined in section 3.2 is equal to 2 in this case. We then check formula (16) as:

$$
\mathcal{N}(\mathfrak{a})\zeta'_{f}([\mathfrak{b}], 0) - \zeta'_{f}([\mathfrak{ab}], 0) \approx \frac{1}{4} \log |v_1 v_2|^2 \approx 4.2587554\dots
$$

5.4.2 Example 13

We now discuss a case where $q = 5$, $\kappa = 1$ and $\mathcal{Z}_f^1 \simeq \mathbb{Z}/5\mathbb{Z}^\times$ with $\chi_f = 1$. In this case all $L$-functions attached to $\mathbb{K}^+(f)$ have a zero of order at least 2 at 0 and $\mathbb{K}^+(f)$ is the narrow Hilbert class field of $\mathbb{K}$, therefore the Stark unit we aim to compute is just 1. We consider $z$ the complex root of the polynomial $x^4 - x^3 - 2x^2 - 3x - 2$ lying in the upper half-plane. Then $\mathbb{K}_{13} = \mathbb{Q}(z)$ has class number 1. We choose $f = \mathfrak{P}_5$ the degree one prime above $q = 5$.
The corresponding narrow ray class group is $Cl^+(\mathfrak{f}) \simeq \mathbb{Z}/2\mathbb{Z}$. We choose the fundamental units

$$ \varepsilon_1 = -z^2 + 3z + 3, \quad \varepsilon_2 = z^3 + 2z^2 - 8z - 7 $$

for $O_f^{+,\times}$. We compute $\tilde{\lambda}_1 = \tilde{\lambda}_2 = 1$, $\tilde{t}_1 = 2 \cdot 12907$, $\tilde{t}_2 = 61 \cdot 73$ where 12907 is a prime number. We may choose $\mathfrak{b} = (1)$ and $\mathfrak{a}$ the unique degree one prime above $N=7$ in $\mathbb{K}_{13}$. We search for target ideals and find $\mathfrak{P}_2$ a degree one prime above 2 and $\mathfrak{P}'_{12907}$ a degree one prime above 12907 in the first case. In the second case, we find $\mathfrak{P}_{61}$ a degree one prime ideal above 61 and $\mathfrak{P}_{73}$ a degree one prime above 73. Both ideals $qN\mathfrak{P}_2\mathfrak{P}'_{12907}/\mathfrak{a}$ and $qN\mathfrak{P}_{61}\mathfrak{P}_{73}/\mathfrak{a}$ are principal with admissible generators $h_1 = 40z^3 - 210z^2 + 95z + 80$ and $h_2 = 360z^3 - 525z^2 - 160z - 785$. The corresponding levels will be $l_1 = 2 \cdot 5 \cdot 7 \cdot 12907$ and $l_2 = 5 \cdot 7 \cdot 61 \cdot 73$. Let us write the parameters

$$ \begin{aligned} \tau &= \frac{-59\varepsilon_1\varepsilon_2 - 194580149\varepsilon_1 + 3255}{\varepsilon_1}, & \tau' &= \frac{-159315\varepsilon_2\varepsilon_1 + 1189\varepsilon_2 - 76}{\varepsilon_1\varepsilon_2} \\ \sigma &= \frac{44607037\varepsilon_1\varepsilon_2 + 135\varepsilon_1 - 49}{\varepsilon_1\varepsilon_2}, & \sigma' &= \frac{-49\varepsilon_2\varepsilon_1 - 108\varepsilon_2 - 106940}{1} \\ \rho &= \frac{95\varepsilon_1\varepsilon_2 + 1189\varepsilon_1 + 3636283}{1}, & \rho' &= \frac{-108\varepsilon_2\varepsilon_1 + 36840\varepsilon_2 - 59}{\varepsilon_2} \end{aligned} $$

Then

$$ \prod_{k=1}^{4} \frac{G_2\left(\frac{-k}{5}, \frac{\tau}{903490}, \frac{\sigma}{903490}, \frac{\rho}{903490}\right)^7}{G_2\left(\frac{-7k}{5}, \frac{\tau}{129070}, \frac{\sigma}{129070}, \frac{\rho}{129070}\right)} \times \frac{G_2\left(\frac{-k}{5}, \frac{\tau'}{155855}, \frac{\sigma'}{155855}, \frac{\rho'}{155855}\right)^{-7}}{G_2\left(\frac{-7k}{5}, \frac{\tau'}{22265}, \frac{\sigma'}{22265}, \frac{\rho'}{22265}\right)^{-1}} $$

is $\approx 1$ which is a correct Stark unit when all $L$ functions have a zero of order 2 at 0. The computation time for 1000 digits is 70 minutes. Computations show that the above product is not a power of one of its subproducts, therefore the integer $\kappa$ defined in section 3.2 is equal to 1. Amongst our examples, this one exhibits one of the largest computed value for $\#\mathcal{Z}_f^1/\kappa = 4$.

### 5.4.3 Example 14

We now discuss a case where $q = 7$, $\kappa = 1$ and $\mathcal{Z}_f^1 = \{(1,1), (6,-1)\}$. Consider $z$ the complex root of the polynomial $x^4 - x^3 - x^2 - 5x + 1$ lying in the upper half-plane. Then $\mathbb{K}_{14} = \mathbb{Q}(z)$ has class number 1. We choose $\mathfrak{f} = \mathfrak{P}_7$ the degree one prime above $q = 7$ in $\mathbb{K}_{14}$. The corresponding narrow ray class group is $Cl^+(\mathfrak{f}) \simeq \mathbb{Z}/6\mathbb{Z}$. We choose the fundamental units

$$ \varepsilon_1 = 2z^3 - 5z^2 + z, \quad \varepsilon_2 = -z^3 + z^2 + 3z $$
for $\mathcal{O}_f^{+,\times}$. We compute $\tilde{\lambda}_1 = \tilde{\lambda}_2 = 1$, $\tilde{t}_1 = 13 \cdot 1069$, $\tilde{t}_2 = 227257$ where 1069, 227257 are prime numbers. We may choose $\mathfrak{b} = (1)$ and $\mathfrak{a} = \mathfrak{P}_{11}$ the degree one prime above $N = 11$ in $\mathbb{K}_{14}$. We search for target ideals and find $\mathfrak{P}_{13}$ the degree one prime above 13 and $\mathfrak{P}'_{1069}$ a degree one prime above 1069 in the first case. In the second case, we find $\mathfrak{P}_{227257}$ a degree one prime above 227257 in the second case. In Proposition 22 we may choose $m_1 = 1$, $m_2 = 2$ and $\mathfrak{J} = \mathcal{O}_{\mathbb{K}}$ such that both ideals $qN\mathfrak{P}_{13}\mathfrak{P}'_{1069}/\mathfrak{a}$ and $2qN\mathfrak{P}_{227257}/\mathfrak{a}$ are principal with admissible generators $h_1 = 182z^3 + 35z^2 - 84z - 714$ and $h_2 = 532z^3 - 490z^2 + 406z - 1554$. The corresponding levels will be $l_1 = 7 \cdot 11 \cdot 13 \cdot 1069$ and $l_2 = 7 \cdot 11 \cdot 227257$. Let us write the parameters

$$
\begin{align*}
\tau &= \frac{5\varepsilon_1\varepsilon_2 - 5863655\varepsilon_1 + 51}{\varepsilon_1}, &
\tau' &= \frac{11916518\varepsilon_2\varepsilon_1 + 276\varepsilon_2 - 191}{\varepsilon_1\varepsilon_2} \\
\sigma &= \frac{18641035\varepsilon_1\varepsilon_2 + 692\varepsilon_1 - 197}{\varepsilon_1\varepsilon_2}, &
\sigma' &= \frac{197\varepsilon_2\varepsilon_1 + 4832\varepsilon_2 + 7904541}{1} \\
\rho &= \frac{-49\varepsilon_1\varepsilon_2 - 276\varepsilon_1 + 2381446}{1}, &
\rho' &= \frac{216\varepsilon_2\varepsilon_1 - 1644334\varepsilon_2 - 5}{\varepsilon_2}
\end{align*}
$$

For $k = 1, 2, 3, 4, 5, 6$, write $I_k$ for the value

$$
I_k = \frac{G_2\left(\frac{k}{7}, \frac{\tau}{1070069}, \frac{\sigma}{1070069}, \frac{\rho}{1070069}\right)^{11}}{G_2\left(\frac{11k}{7}, \frac{\tau}{97279}, \frac{\sigma}{97279}, \frac{\rho}{97279}\right)} \times \frac{G_2\left(\frac{-2k}{7}, \frac{\tau'}{17498789}, \frac{\sigma'}{17498789}, \frac{\rho'}{17498789}\right)^{-11}}{G_2\left(\frac{-22k}{7}, \frac{\tau'}{1590799}, \frac{\sigma'}{1590799}, \frac{\rho'}{1590799}\right)^{-1}}
$$

On the one hand, $I_1I_6 \approx -1197160922532.7652149... - i \cdot 455052155255.6575547...$
is the product in the right-hand side of formula (16), which we check as:

$$
\mathcal{N}(\mathfrak{a})\zeta'_{f}([\mathfrak{b}], 0) - \zeta'_{f}([\mathfrak{ab}], 0) \approx \frac{1}{2} \log |I_1 I_6|^2 \approx 27.8784505\dots
$$

On the other hand, the values $I_1 I_6$, $I_2 I_5 \approx 10^{-4}(6.0177934... - i \cdot 6.7709930...)$
and $I_3 I_4 \approx 10^{-10}(-3.0632199... - i \cdot 8.0567091...)$ seem to be algebraically
dependent and we may identify the polynomial

$$
\begin{align*}
& \prod_{k=1,2,3} (X - I_k I_{7-k})(X - I_k^{-1} I_{7-k}^{-1}) = X^6 + 1 \\
& \qquad + (-153504836673z^3 + 136635341749z^2 \\
& \qquad \phantom{+ (-153504836673z^3 +} + 544161498400z - 108155523221)(X^5 + X) \\
& \qquad \phantom{+ (-153504836673z^3 + 544161498400z -} + (45202964386965263585z^3 - 350943983942803910564z^2 \\
& \qquad \phantom{+ (-153504836673z^3 + 45202964386965263585z^3 -} + 613535010394689372486z - 104952772799918427079)(X^4 + X^2) \\
& \qquad \phantom{+ (-153504836673z^3 + 45202964386965263585z^3 - 613535010394689372486z -} + (138218090908063915784215z^3 + 7152666826995541968563z^2)
\end{align*}
$$
$$ -814169082025891017196278z + 154702240787650756485938)X^3 $$

as a relative polynomial defining the extension $K_{14}^{+}(f)/K_{14}$. Alternatively, there are roots $u_k^{\pm 1}$ of the palindromic polynomial

$$ \begin{aligned} & x^{24} + 1 + 1292(x^{23} + x) + 610850(x^{22} + x^2) - 324070132(x^{21} + x^3) \\ & + 43597528236(x^{20} + x^4) - 641906806311(x^{19} + x^5) \\ & + 3661996071192(x^{18} + x^6) - 9521279678257(x^{17} + x^7) \\ & + 11598437454383(x^{16} + x^8) - 3951777294409(x^{15} + x^9) \\ & - 4643997885968(x^{14} + x^{10}) + 3477156436009(x^{13} + x^{11}) - 43787701643x^{12} \end{aligned} $$

defining $K_{14}^{+}(f)$ such that for $k = 1, 2, 3$, $(I_k I_{7-k})^{\pm 1} \approx u_k^{\pm 4}$ and it seems that $I_k I_{7-k}$ is already a 4-th power inside $K_{14}^{+}(f)$.

## A Appendix: Pure cubic table for $2 \le m \le 50$

The following table gives one of the positive fundamental units $\varepsilon$ of the pure cubic field defined by $X^3 - m$ in terms of the root $z = e^{2i\pi/3}m^{1/3}$ for cube-free integers $2 \le m \le 50$ up to isomorphism. It also gives the corresponding contents $\tilde{\lambda}$ and the overflows $\tilde{t}$ computed regardless of whether or not there is a modulus $f$ satisfying the conditions in Conjecture 11 with $O_f^{+, \times}$ generated by $\varepsilon$. In particular, it is clear that the parameters $\tilde{\lambda}, \tilde{t}$ vary significantly with the value of $m$. In general the value of $\tilde{\lambda}$ stays rather small while the value of $\tilde{t}$ may become very large. Our work shows that the computation time for the right-hand side of formula (16) is at least $O(qN\tilde{\lambda}\tilde{t}|\log \delta|)$ for the precision $\delta$ so computations can quickly become overwhelming, in particular for the cases $m = 23, 29, 33, 41, 46, 47$.
<table><thead><tr><th>X<sup>3</sup> - m</th><th>ε</th><th>λ</th><th>τ̃</th></tr></thead><tbody><tr><td>X<sup>3</sup> - 2</td><td>z - 1</td><td>1</td><td>1</td></tr><tr><td>X<sup>3</sup> - 3</td><td>z<sup>2</sup> - 2</td><td>1</td><td>3</td></tr><tr><td>X<sup>3</sup> - 5</td><td>2z<sup>2</sup> - 4z + 1</td><td>2</td><td>2 · 13</td></tr><tr><td>X<sup>3</sup> - 6</td><td>-3z<sup>2</sup> + 6z - 1</td><td>3</td><td>2 · 3 · 7</td></tr><tr><td>X<sup>3</sup> - 7</td><td>-z + 2</td><td>1</td><td>1</td></tr><tr><td>X<sup>3</sup> - 10</td><td>2/3z<sup>2</sup> - 1/3z - 7/3</td><td>1</td><td>3<sup>2</sup></td></tr><tr><td>X<sup>3</sup> - 11</td><td>-2z<sup>2</sup> + 4z + 1</td><td>2</td><td>2 · 19</td></tr><tr><td>X<sup>3</sup> - 12</td><td>-3/2z<sup>2</sup> + 3z + 1</td><td>3</td><td>3 · 5</td></tr><tr><td>X<sup>3</sup> - 13</td><td>2z<sup>2</sup> - 3z - 4</td><td>1</td><td>131</td></tr><tr><td>X<sup>3</sup> - 14</td><td>-z<sup>2</sup> + 2z + 1</td><td>1</td><td>2 · 11</td></tr><tr><td>X<sup>3</sup> - 15</td><td>-12z<sup>2</sup> + 30z - 1</td><td>6</td><td>2 · 3 · 5 · 7<sup>2</sup></td></tr><tr><td>X<sup>3</sup> - 17</td><td>-7z + 18</td><td>7</td><td>3 · 7</td></tr><tr><td>X<sup>3</sup> - 19</td><td>1/3z<sup>2</sup> - 2/3z - 2/3</td><td>1</td><td>3</td></tr><tr><td>X<sup>3</sup> - 20</td><td>-1/2z<sup>2</sup> + z + 1</td><td>1</td><td>7</td></tr><tr><td>X<sup>3</sup> - 21</td><td>-4z<sup>2</sup> - 6z + 47</td><td>2</td><td>2 · 3 · 47</td></tr><tr><td>X<sup>3</sup> - 22</td><td>4z<sup>2</sup> - 3z - 23</td><td>1</td><td>5 · 7 · 41</td></tr><tr><td>X<sup>3</sup> - 23</td><td>6230z<sup>2</sup> - 3160z - 41399</td><td>10</td><td>2 · 5 · 31 · 2659 · 67853</td></tr><tr><td>X<sup>3</sup> - 26</td><td>z - 3</td><td>1</td><td>3</td></tr><tr><td>X<sup>3</sup> - 28</td><td>1/6z<sup>2</sup> - 1/3z - 1/3</td><td>1</td><td>1</td></tr><tr><td>X<sup>3</sup> - 29</td><td>370284z<sup>2</sup> + 103819462z - 322461439</td><td>2</td><td>2 · 7 · 1953569 · 16988921 · 602078573</td></tr><tr><td>X<sup>3</sup> - 30</td><td>-3z<sup>2</sup> + 9z + 1</td><td>3</td><td>3<sup>2</sup> · 19</td></tr><tr><td>X<sup>3</sup> - 31</td><td>20z<sup>2</sup> + 54z - 367</td><td>2</td><td>2 · 11317</td></tr><tr><td>X<sup>3</sup> - 33</td><td>-394098z<sup>2</sup> + 97392z + 3742201</td><td>6</td><td>2 · 3 · 21313 · 224363 · 1956481</td></tr><tr><td>X<sup>3</sup> - 34</td><td>51z<sup>2</sup> + 24z - 613</td><td>3</td><td>2 · 3<sup>2</sup> · 5 · 7 · 13 · 61</td></tr><tr><td>X<sup>3</sup> - 35</td><td>-1/3z<sup>2</sup> + 10/3z - 22/3</td><td>1</td><td>5 · 23</td></tr><tr><td>X<sup>3</sup> - 37</td><td>-3z + 10</td><td>3</td><td>3<sup>2</sup></td></tr><tr><td>X<sup>3</sup> - 38</td><td>3z<sup>2</sup> - 55z + 151</td><td>1</td><td>13 · 79 · 163</td></tr><tr><td>X<sup>3</sup> - 39</td><td>2z<sup>2</sup> - 23</td><td>2</td><td>2 · 3 · 13</td></tr><tr><td>X<sup>3</sup> - 41</td><td>-70761183382z<sup>2</sup> + 305478475184z - 211991370839</td><td>2</td><td>2 · 7 · 17 · 967 · 17467 · 810791 · 126525109 · 26087689931</td></tr><tr><td>X<sup>3</sup> - 42</td><td>-12z<sup>2</sup> + 42z - 1</td><td>6</td><td>2 · 3 · 7 · 97</td></tr><tr><td>X<sup>3</sup> - 43</td><td>2z - 7</td><td>2</td><td>2</td></tr><tr><td>X<sup>3</sup> - 44</td><td>17/6z<sup>2</sup> + 2/3z - 113/3</td><td>1</td><td>3<sup>2</sup> · 23 · 29</td></tr><tr><td>X<sup>3</sup> - 45</td><td>-104z<sup>2</sup> + 66z + 1081</td><td>6</td><td>2 · 3 · 37 · 97 · 197</td></tr><tr><td>X<sup>3</sup> - 46</td><td>-309z<sup>2</sup> - 48z + 4139</td><td>9</td><td>2 · 3<sup>2</sup> · 19 · 281 · 523</td></tr><tr><td>X<sup>3</sup> - 47</td><td>-64786z<sup>2</sup> + 69704z + 592199</td><td>2</td><td>2 · 17 · 37 · 71 · 2311 · 2791 · 5693</td></tr></tbody></table>
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