2·1. Properties of singular moduli

We collect here some well-known properties of singular moduli which we will need throughout the paper.

Let $j \colon {\mathbb{H}} \to {\mathbb{C}}$ denote the modular j-function, where ${\mathbb{H}}$ is the complex upper half plane. A singular modulus is a complex number $j(\tau)$ , where $\tau \in {\mathbb{H}}$ is such that $[{\mathbb{Q}}(\tau) \;:\; {\mathbb{Q}}] = 2$ . The discriminant $\Delta$ of a singular modulus $j(\tau)$ is given by $\Delta = b^2-4ac$ , where $a, b, c \in {\mathbb{Z}}$ , not all zero, are such that $a \tau^2 + b \tau + c = 0$ and $\gcd(a, b, c) = 1$ . In particular, $\Delta \lt 0$ and $\Delta \equiv 0, 1 \bmod 4$ . Hence, $\lvert \Delta \rvert \geq 3$ always.

Note that ${\mathbb{Q}}(\tau) = {\mathbb{Q}}(\sqrt{\Delta})$ . The fundamental discriminant D of $j(\tau)$ is defined to be the discriminant of the imaginary quadratic field ${\mathbb{Q}}(\tau) = {\mathbb{Q}}(\sqrt{\Delta})$ . One has that $\Delta = f^2 D$ for some $f \in {\mathbb{Z}}_{\gt0}$ .

Write $F_j$ for the standard fundamental domain for the action (by fractional linear transformations) of ${\mathrm{SL}_2({\mathbb{Z}})}$ on ${\mathbb{H}}$ , i.e.

\begin{align*} {}F_j = &\left\{ z \in {\mathbb{H}} \;:\; -\frac{1}{2} \leq {\mathrm{Re}} z \lt \frac{1}{2} \mbox{ and } \lvert z \rvert \gt 1\right\}\\ {}&\cup \left\{ z \in {\mathbb{H}} \;:\; -\frac{1}{2} \leq {\mathrm{Re}} z \leq 0 \mbox{ and } \lvert z \rvert = 1\right\}.\end{align*}

The j-function restricts to a bijection $F_j \to {\mathbb{C}}$ . Therefore, for $\Delta \lt 0$ such that $\Delta \equiv 0, 1 \bmod 4$ , the map

\[ (a, b, c) \mapsto j\left( \frac{-b + \lvert \Delta \rvert^{1/2} i}{2a}\right)\]

is a bijection between the set

\begin{align*} {}T_\Delta = \{&(a, b, c) \in {\mathbb{Z}}^3 \;:\; b^2 - 4 ac = \Delta, \, \gcd(a,b,c) = 1,\\ &\mbox{ and either } -a\lt b \leq a \lt c \mbox{ or } 0 \leq b \leq a =c\}\end{align*}

and the set of singular moduli of discriminant $\Delta$ . For each such $\Delta$ , there exists a unique triple $(a, b, c) \in T_\Delta$ with $a=1$ , given by

\[ (a, b, c) = \left(1, k, \frac{k^2-\Delta}{4}\right),\]

where $k \in \{0, 1\}$ is such that $k \equiv \Delta \bmod 2$ ; we call the corresponding singular modulus the dominant singular modulus of discriminant $\Delta$ .

If x is a singular modulus of discriminant $\Delta$ , then, see [ Reference Cox10 , lemma 9·3 and theorem 11·1], the field ${\mathbb{Q}}(\sqrt{\Delta}, x)$ is a Galois extension of both ${\mathbb{Q}}(\sqrt{\Delta})$ and ${\mathbb{Q}}$ and

\[ {\mathrm{Gal}}\left({\mathbb{Q}}\left(\sqrt{\Delta}, x\right) / {\mathbb{Q}}\left(\sqrt{\Delta}\right)\right) \cong {\mathrm{cl}} (\Delta),\]

where ${\mathrm{cl}}(\Delta)$ denotes the class group of the unique imaginary quadratic order of discriminant $\Delta$ .

Let $x_1, \ldots, x_n$ be all the distinct singular moduli of some discriminant $\Delta$ . Then, by [ Reference Cox10 , theorem 11·1 and proposition 13·2], the polynomial

\[ H_\Delta(z) = \prod_{i=1}^n (z- x_i)\]

has coefficients in ${\mathbb{Z}}$ and is irreducible over ${\mathbb{Q}}$ and over ${\mathbb{Q}}(\sqrt{\Delta})$ . Hence, for every $i \in \{1, \ldots, n\}$ , we have that

\[ [{\mathbb{Q}}(x_i) \;:\; {\mathbb{Q}}] = \left[{\mathbb{Q}}\left(\sqrt{\Delta}, x_i\right) \;:\; {\mathbb{Q}}\left(\sqrt{\Delta}\right)\right] = h(\Delta),\]

where $h(\Delta)$ denotes the class number of the imaginary quadratic order of discriminant $\Delta$ .

The class numbers of small discriminants may be computed straightforwardly in PARI. We will make use of the following consequence of this.

We will also need the following bound on singular moduli.

Proposition 2·2. Let x be a singular modulus of discriminant $\Delta$ which corresponds to a triple $(a, b, c) \in T_\Delta$ . Then

\[ \exp\left(\frac{\pi \lvert \Delta \rvert^{1/2}}{a}\right) - 2079 \leq \lvert x \rvert \leq \exp\left(\frac{\pi \lvert \Delta \rvert^{1/2}}{a}\right) + 2079.\]

Proof. Let $(a, b, c) \in T_\Delta$ be the triple corresponding to x. Then $x = j(\tau)$ , where

\[ \tau = \frac{-b + \lvert \Delta \rvert^{1/2} i}{2 a} \in F_j. \]

The result follows immediately, since Bilu, Masser and Zannier [ Reference Bilu, Masser and Zannier7 , lemma 1] proved that if $z \in F_j$ , then

\[ \left\lvert \left\lvert j(z) \right\rvert - \exp(2 \pi {\mathrm{Im}} z) \right\rvert \leq 2079.\]

This bound has the following consequence.

Lemma 2·3. Let x, y be distinct singular moduli of the same discriminant $\Delta$ . Suppose that x is dominant. Then

\[\lvert y \rvert \leq \frac{6 \lvert x \rvert}{\exp\left(\frac{\pi \lvert \Delta \rvert^{1/2}}{2}\right)}.\]

Proof. There are at least two distinct singular moduli of discriminant $\Delta$ , so $\lvert \Delta \rvert \geq 15$ by Lemma 2·1. Since x is dominant, y is not dominant. Thus, Proposition 2·2 implies that

\begin{align*} {} {}\frac{\lvert y \rvert}{\lvert x \rvert} &\leq \frac{1}{\exp\left(\frac{\pi \lvert \Delta \rvert^{1/2}}{2}\right)} \frac{1 + 2079 \exp\left(\frac{-\pi \lvert \Delta \rvert^{1/2}}{2}\right)}{1 - 2079 \exp\left(- \pi \lvert \Delta \rvert^{1/2}\right)}\\ {} {}&\leq \frac{1}{\exp\left(\frac{\pi \lvert \Delta \rvert^{1/2}}{2}\right)} \frac{1 + 2079 \exp\left(\frac{-\pi \sqrt{15}}{2}\right)}{1 - 2079 \exp\left(- \pi \sqrt{15}\right)}\\ {} {}&\leq \frac{6}{\exp\left(\frac{\pi \lvert \Delta \rvert^{1/2}}{2}\right)}. {}\end{align*}

2·2. Effective bounds for the class number

A theorem of Tatuzawa [ Reference Tatuzawa29 , theorem 1] and Dirichlet’s class number formula together imply the following result. It shows that Siegel’s [ Reference Siegel28 ] ineffective lower bound for the class number of imaginary quadratic fields may be made effective, apart from at most one possible exceptional imaginary quadratic field.

Theorem 2·4 ([ Reference Tatuzawa29 , theorem 1]). Let $\epsilon \in (0, 1/2)$ . Let $K_1, K_2$ be imaginary quadratic fields with respective discriminants $D_1, D_2$ . If

\[ h(D_1) \lt \frac{\epsilon}{10 \pi} \lvert D_1 \rvert^{\frac{1}{2} - \epsilon} \mbox{ and } h(D_2) \lt \frac{\epsilon}{10 \pi} \lvert D_2 \rvert^{\frac{1}{2} - \epsilon},\]

then $K_1 = K_2$ .

In other words, for a fixed $\epsilon \in (0, 1/2)$ , there is at most one imaginary quadratic field K for which the bound

(2·1) \begin{align}h(D) \geq \frac{\epsilon}{10 \pi} \lvert D \rvert^{\frac{1}{2} - \epsilon}\end{align}

is false, where D denotes the discriminant of K. It is possible (indeed, it would follow from GRH) that, for some $\epsilon \in (0, 1/2)$ , the bound (2·1) in fact holds for every imaginary quadratic field K. In such a case, the results of [ Reference Binyamini8 ] would imply an effective version of Theorem 1·1.

Throughout this paper, we fix $\epsilon_* = 1/12$ and denote by $K_*$ the unique imaginary quadratic field for which the corresponding bound (2·1) is false. Denote by $D_*$ the discriminant of $K_*$ . If no such field $K_*$ exists, then we adopt the notational convention that the inequalities $K \neq K_*$ and $D \neq D_*$ hold for every imaginary quadratic field K with discriminant D. We have the following consequence of Tatuzawa’s result.

Proposition 2·5 ([ Reference Bilu and Kühne4 , (17)]). Let $\Delta \lt 0$ be such that $\Delta \equiv 0, 1 \bmod 4$ and let $K = {\mathbb{Q}}(\sqrt{\Delta})$ . If $K \neq K_*$ , then

\[ h(\Delta) \geq \frac{37}{50000} \lvert \Delta \rvert^{5/12}.\]

For discriminants $\Delta$ with ${\mathbb{Q}}(\sqrt{\Delta}) = K_*$ , the known effective bounds for the class number are much weaker.

Proposition 2·6. Let $D \lt 0$ be a fundamental discriminant. Then

\[ h(D) \geq \frac{1}{\sqrt{42000}} (\log \lvert D \rvert)^{1/2}.\]

Proof. Oesterlé [ Reference Oesterlé20 , théorème 1 and section 5·1], building on work of Goldfeld [ Reference Goldfeld14 ] and Gross–Zagier [ Reference Gross and Zagier15 , theorem 8·1], proved that

\[ h(D) \geq \frac{\log \lvert D \rvert}{7000} F(D),\]

where

\[ F(D) = \prod \left( 1 - \frac{\lfloor 2 \sqrt{p} \rfloor}{p+1}\right)\]

and this product is taken over all the primes p which divide D except the largest. Let $\omega(D)$ denote the number of distinct prime divisors of D. So there are $\omega(D) - 1$ terms in the product defining F(D). By the theory of genera, see e.g. [ Reference Cox10 , proposition 3·11 and theorem 6·1], we have that

\[ \lvert {\mathrm{cl}}(D)[2] \rvert = 2^{\omega(D) -1}.\]

In particular, $2^{\omega(D) -1} \mid h(D)$ . So $\omega(D) - 1 \leq v_2(h(D))$ , where $ v_2(h(D))$ denotes the largest integer k such that $2^k \mid h(D)$ . Hence, there are at most $v_2(h(D))$ terms appearing in the product defining F(D).

Observe that

\[ 1 - \frac{\lfloor 2 \sqrt{p} \rfloor}{p+1} \geq \frac{1}{2}\]

if $p \geq 11$ . Hence,

\[F(D) \geq \frac{1}{3} \frac{1}{4} \frac{1}{3} \frac{3}{8} \left(\frac{1}{2}\right)^{v_2(h(D)) - 4} = \frac{1}{6} 2^{-v_2(h(D))}.\]

Since $h(D) \geq 2^{v_2(h(D))}$ by definition, we obtain that

\[ h(D)^2 \geq \frac{\log \lvert D \rvert}{42000}. \]

Proposition 2·7. Let $\Delta \lt 0$ be such that $\Delta \equiv 0, 1 \bmod 4$ and let $k \in {\mathbb{Z}}_{\gt 0}$ . If $h(\Delta) \leq k$ , then

\[ \lvert \Delta \rvert^{1/2} \leq 2k^2 e^{21000k^2}.\]

Proof. Write $\Delta = f^2 D$ , where D is the discriminant of ${\mathbb{Q}}(\sqrt{\Delta})$ . Suppose that $h(\Delta) \leq k$ . Assume first that $D \notin \{-3, -4\}$ . The class number formula in [ Reference Cox10 , theorem 7·24] gives that

\[ h(\Delta) = h(D) f \prod_{p \mid f} \left(1 - \left(\frac{D}{p}\right)\frac{1}{p}\right),\]

where $(D/p)$ denotes the Kronecker symbol. Hence, $h(D) \mid h(\Delta)$ and

\[ h(\Delta) \geq h(D) \varphi(\,f),\]

where $\varphi(\!\cdot\!)$ denotes Euler’s totient function. Hence,

\[h(D) \leq k \mbox{ and } \varphi(\,f) \leq k.\]

Proposition 2·6 then implies that

\[ \lvert D \rvert \leq e^{42000k^2}, \]

while the classical bound $\varphi(\,f) \geq \sqrt{f/2}$ implies that $f \leq 2k^2$ . Hence,

\[ \lvert \Delta \rvert^{1/2} \leq 2k^2 e^{21000k^2}.\]

Now assume $D \in \{-3, -4\}$ . Then $h(D) = 1$ . In this case, the class number formula from [ Reference Cox10 , theorem 7·24] implies that

\[ h(\Delta) \geq \frac{\varphi(\,f)}{3}.\]

Hence, $\varphi(\,f) \leq 3 k$ and so $f \leq 18 k^2$ . Thus, clearly,

\[ \lvert \Delta \rvert^{1/2} \leq 36k^2 \leq 2 k^2 e^{21000k^2}. \]

Finally, we will also need an upper bound for the class number. This bound is a straightforward consequence of Dirichlet’s class number formula.

Proof. By [ Reference Paulin21 , proposition 2·2], we have that

\[ h(\Delta) \leq \frac{1}{\pi} \lvert \Delta \rvert^{1/2} (2 + \log \lvert \Delta \rvert).\]

It then suffices to note that, for every $x\gt0$ ,

\[ \frac{2 + \log x}{\pi} \lt x^{1/6}. \]

3·1. Uniformity in Theorem 1·1

Recall, from Section 2·1, that the singular moduli of a given discriminant $\Delta$ form a complete set of Galois conjugates over ${\mathbb{Q}}$ . Moreover, by Proposition 2·7, the number of distinct singular moduli of discriminant $\Delta$ is $\geq c_1 (\log \lvert \Delta \rvert)^{c_2}$ for some absolute effective constants $c_1, c_2\gt0$ . Therefore, if m, n are fixed, then $c(m, n, d) \to \infty$ as $d \to \infty$ , where c(m, n, d) is the constant in Theorem 1·1. Similarly, if m, d are fixed, then $c(m, n, d) \to \infty$ as $n \to \infty$ .

On the other hand, if n, d are fixed, then it is not obvious what happens to the constant c(m, n, d) as $m \to \infty$ . Since singular moduli are algebraic, one cannot bound the discriminants associated to a special point lying on a general hypersurface in ${\mathbb{C}}^n$ solely in terms of n and the minimal degree of a field of definition of the hypersurface. However, it may be possible to obtain bounds that are uniform in m for the specific family of hypersurfaces considered in Theorem 1·1, i.e. those defined by equations of the form

\[ a_1 x_1^m + \cdots + a_n x_n^m = b.\]

Indeed, the constant c(m, n, d) in Theorem 1·1 may be taken to be uniform in m if either $n = 1$ or $(d, n) = (1, 2)$ . If x is a singular modulus of discriminant $\Delta$ such that

\[a x^m = b\]

for some $a, b \in {\overline{\mathbb{Q}}} \setminus \{0\}$ and $m \in {\mathbb{Z}}_{\gt0}$ , then $\lvert \Delta \rvert$ may be bounded solely in terms of $[{\mathbb{Q}}(a, b) \;:\; {\mathbb{Q}}]$ , thanks to Proposition 2·7 and the fact [ Reference Riffaut25 , lemma 2·6] that ${\mathbb{Q}}(x^m) = {\mathbb{Q}}(x)$ . If $x_1, x_2$ are distinct singular moduli of respective discriminants $\Delta_1, \Delta_2$ such that

\[ a_1 x_1^{m_1} + a_2 x_2^{m_2} \in {\mathbb{Q}}\]

for some $a_1, a_2 \in {\mathbb{Q}} \setminus \{0\}$ and $m_1, m_2 \in {\mathbb{Z}}_{\gt0}$ , then $\max \{\lvert \Delta_1 \rvert, \lvert \Delta_2 \rvert \} \leq 427$ by [ Reference Luca and Riffaut19 , theorem 1·3] (see also [ Reference Riffaut25 , theorem 1·5]). For $n = 3$ , we have the previous result of the author [ Reference Fowler13 , theorem 1·1]: if $x_1, x_2, x_3$ are pairwise distinct singular moduli of respective discriminants $\Delta_1, \Delta_2, \Delta_3$ such that

\[ a_1 x_1^m + a_2 x_2^m + a_3 x_3^m \in {\mathbb{Q}}\]

for some $m \in {\mathbb{Z}}_{\gt0}$ and $a_1, a_2, a_3 \in {\mathbb{Q}} \setminus \{0\}$ with $\lvert a_1 \rvert = \lvert a_2 \rvert = \lvert a_3 \rvert$ , then $\max \{\lvert \Delta_1 \rvert, \lvert \Delta_2 \rvert, \lvert \Delta_3 \rvert \} \leq 907$ .

6·1. Completing the proof of Theorem 1·2

In this section, we complete the proof of Theorem 1·2. We start by recalling Binyamini’s [ Reference Binyamini8 ] “almost” effective André–Oort result for ${\mathbb{C}}^n$ . For the definition of a special subvariety of ${\mathbb{C}}^n$ , see e.g. [ Reference Binyamini8 , (1·1)] or [ Reference Pila24 , definition 4·10]. For a subvariety $V \subset {\mathbb{C}}^n$ , we denote by $\deg V$ the degree of V with respect to the projective embedding ${\mathbb{C}}^n \subset \mathbb{P}^n$ . Binyamini proved the following theoremFootnote ² . (Recall that $K_*$ denotes the exceptional imaginary quadratic field that was defined above Proposition 2·5.)

Theorem 6·1 ([ Reference Binyamini8 , theorem 1]). Let $n, d, l \in {\mathbb{Z}}_{\gt0}$ . There exists an effective constant $c(n, d, l)\gt0$ with the following property:

Let $V \subset {\mathbb{C}}^n$ be a subvariety defined over some number field L with $[L \;:\; {\mathbb{Q}}] \leq d$ and $\deg V \leq l$ . Denote by $V^\mathrm{sp}$ the union of all the positive-dimensional special subvarieties of V. If $x_1, \ldots, x_n$ are singular moduli with respective discriminants $\Delta_1, \ldots, \Delta_n$ such that

\[(x_1, \ldots, x_n) \in V \setminus V^\mathrm{sp},\]

then, for every $i \in \{1, \ldots, n\}$ , either

\[ \lvert \Delta_i \rvert \leq c\left(n, d, l\right)\]

or

\[ {\mathbb{Q}}\left(\sqrt{\Delta_i}\right) = K_*.\]

Let $m, n \in {\mathbb{Z}}_{\gt0}$ . For $a = (a_1, \ldots, a_n) \in ({\mathbb{Q}} \setminus \{0\})^n$ and $b \in {\mathbb{Q}}$ , let

\[ V_{a, b} = \left\{\left(z_1, \ldots, z_n\right) \in {\mathbb{C}}^n \;:\; a_1 z_1^m + \cdots + a_n z_n^m = b \right\}.\]

Since the $V_{a, b}$ are all defined over ${\mathbb{Q}}$ and have degree m, the constant in Theorem 6·1 is uniform (for the given m, n) across the $V_{a, b}$ .

Proof of Theorem 1·2. Let $m, n \in {\mathbb{Z}}_{\gt0}$ . Let $c_1(m, n)\gt0$ be the effective constant given by Theorem 6·1 applied to the family of linear subvarieties $V_{a, b} \subset {\mathbb{C}}^n$ , where $a \in ({\mathbb{Q}} \setminus \{0\})^n$ and $b \in {\mathbb{Q}}$ (i.e. $c_1(m, n) = c(n, 1, m)$ for the constant c(n, d, l) in the statement of Theorem 6·1).

Suppose that $x_1, \ldots, x_n$ are pairwise distinct singular moduli such that

\[ a_1 x_1^m + \cdots + a_n x_n^m = b\]

for some $a_1, \ldots, a_n \in {\mathbb{Q}} \setminus \{0\}$ and $b \in {\mathbb{Q}}$ . Then, by Lemma 3·1,

\[(x_1, \ldots, x_n) \in V_{a, b} \setminus V_{a, b}^\mathrm{sp}.\]

Hence, by Theorem 6·1, for every $i \in \{1, \ldots, n\}$ , either

\[ \lvert \Delta_i \rvert \leq c_1(m, n)\]

or

\[ {\mathbb{Q}}\left(\sqrt{\Delta_i}\right) = K_*.\]

If there is no i such that ${\mathbb{Q}}(\sqrt{\Delta_i}) = K_*$ , then we are done. We may thus assume that there exists a discriminant $\Delta_*$ such that ${\mathbb{Q}}(\sqrt{\Delta_*}) = K_*$ and, for every $i \in \{1, \ldots, n\}$ , either $\Delta_i = \Delta_*$ or ${\mathbb{Q}}(\sqrt{\Delta_i}) \neq K_*$ . Relabelling as necessary, we may assume that there exists some $k \in \{1, \ldots, n\}$ such that

\[ \Delta_i = \Delta_* \iff i \in \{1, \ldots, k\}.\]

In particular, if $k+1 \leq i \leq n$ , then ${\mathbb{Q}}(\sqrt{\Delta_i}) \neq K_*$ and so

\[ \lvert \Delta_i \rvert \leq c_1(m, n).\]

Observe that

\[ \sum_{i=1}^k a_i x_i^m = b - \sum_{i = k+1}^n a_i x_i^m,\]

where the sum on the right-hand side may be empty. Hence,

\[ {\mathbb{Q}}\left(a_1 x_1^m + \cdots + a_k x_k^m \right) \subset {\mathbb{Q}}\left(\left\{x_i \;:\; k+1 \leq i \leq n\right\}\right).\]

Since the $\lvert \Delta_i \rvert$ are effectively bounded (solely in terms of m, n) for $i \geq k+1$ , there exists, by Proposition 2·8, an effective constant $c_2(m, n)\gt0$ such that

\[ \left[ {\mathbb{Q}}\left(\left\{x_i \;:\; k+1 \leq i \leq n\right\}\right)\;:\; {\mathbb{Q}}\right] \leq c_2(m, n).\]

Hence,

\[ \left [{\mathbb{Q}}\left(a_1 x_1^m + \cdots + a_k x_k^m \right) \;:\; {\mathbb{Q}} \right] \leq c_2(m, n).\]

Since $\Delta_1 = \cdots = \Delta_k = \Delta_*$ , we may apply Theorem 1·5 to see that there exists an effective constant $c_3(n)\gt0$ such that either

\[ \lvert \Delta_* \rvert \leq c_3(n)\]

or

\[ [{\mathbb{Q}}(x_1, a_1 x_1^m + \cdots + a_k x_k^m) \;:\; {\mathbb{Q}}(a_1 x_1^m + \cdots + a_k x_k^m)] \leq 2k \leq 2n.\]

In the first case, we are done. So assume then that

\[ [{\mathbb{Q}}(x_1, a_1 x_1^m + \cdots + a_k x_k^m) \;:\; {\mathbb{Q}}(a_1 x_1^m + \cdots + a_k x_k^m)] \leq 2n.\]

Then

\begin{align*} {}h(\Delta_*) &= {}[{\mathbb{Q}}(x_1) \;:\; {\mathbb{Q}}]\\ {}&\leq [{\mathbb{Q}}(x_1, a_1 x_1^m + \cdots + a_k x_k^m) \;:\; {\mathbb{Q}}]\\ {}&= [{\mathbb{Q}}(x_1, a_1 x_1^m + \cdots + a_k x_k^m) \;:\; {\mathbb{Q}}(a_1 x_1^m + \cdots + a_k x_k^m)]\\ {}&\quad \times [{\mathbb{Q}}(a_1 x_1^m + \cdots + a_k x_k^m) \;:\; {\mathbb{Q}}]\\ {}&\leq 2n c_2(m, n).\end{align*}

Thus, by Proposition 2·7, there exists an effective constant $c_4(m, n) \gt 0$ , which depends only on m, n, such that $\lvert \Delta_* \rvert \leq c_4(m, n)$ .

6·2. The proof of Theorem 1·3

The first part of Theorem 1·3 (i.e. the existence of the constant $c_1(n)$ ) is given by the following proposition. It follows from Theorem 1·5 by an argument of Bilu and Kühne [ Reference Bilu and Kühne4 , section 3, step 4], which in turn relies on an earlier result of Kühne [ Reference Kühne18 , corollary 1·2].

Proposition 6·2. Let $x_1, \ldots, x_n$ be pairwise distinct singular moduli such that

\[ a_1 x_1^m + \cdots + a_n x_n^m = b\]

for some $m \in {\mathbb{Z}}_{\gt0}$ and $a_1, \ldots, a_n \in {\mathbb{Q}} \setminus \{0\}$ and $b \in {\mathbb{Q}}$ . Write $\Delta_i$ for the discriminant of $x_i$ and $D_i$ for the fundamental discriminant. Suppose that $k \in \{1, \ldots, n\}$ is such that $D_k \neq D_*$ and

\[ \# \{\Delta_i \;:\; i \in \{1, \ldots, n\} \mbox{ such that } D_i = D_k\} = 1.\]

Then

\[ \lvert \Delta_k \rvert^{1/2} \leq (3.8 \times 10^{10})(2.1 \times 10^4)^n (n+1)^{4n+6}.\]

Proof. Let $k \in \{1, \ldots, n\}$ be such that $D_k \neq D_*$ and

\[ \# \{\Delta_i \;:\; i \in \{1, \ldots, n\} \mbox{ such that } D_i = D_k\} = 1.\]

Let

\[ r = \# \{D_1, \ldots, D_n\}.\]

Suppose first that $r=1$ . Then $\Delta_1 = \cdots = \Delta_n$ by assumption. Hence, by Theorem 1·4, either

\[ \lvert \Delta_k \rvert^{1/2} \leq \frac{1}{\pi} \left(\left(2n+3\right) \log\left(n+1\right) -2n +4\right),\]

or $h(\Delta) = n$ . In the latter case, since $D_k \neq D_*$ , Proposition 2·2 implies that

\[ \lvert \Delta_k \rvert^{1/2} \leq \left(\frac{50000 n}{37}\right)^{6/5}.\]

Thus, in either case, the desired bound on $\lvert \Delta_k \rvert^{1/2}$ evidently holds.

So we may assume that $r \geq 2$ . Relabelling as necessary, we may assume that $k = 1$ and there are $0 = n_0 \lt n_1 \lt \ldots \lt n_r = n$ such that

\begin{align*} {}D_{n_0 + 1} = &\ldots = D_{n_1},\\ {}D_{n_1 + 1} = &\ldots = D_{n_2},\\ {}&\ldots\\ {}D_{n_{r-1} + 1} = &\ldots = D_{n_r}.\end{align*}

For $i \in \{1, \ldots, r\}$ , let $K_i = {\mathbb{Q}}(\sqrt{D_{n_i}})$ and

\[ L_i = K_i(a_{n_{i-1} + 1} x_{n_{i-1} + 1}^m + \cdots + a_{n_i} x_{n_i}^m).\]

By assumption, $\Delta_{1} = \cdots = \Delta_{n_1}$ . We may also assume that

\[\lvert \Delta_1 \rvert^{1/2} \gt \frac{1}{\pi} \left(\left(4n+3\right) \log\left(2n+1\right) -4n + 4\right),\]

since otherwise the desired bound already holds. Hence, by Theorem 1·5,

\[ [L_1(x_1) \;:\; L_1] \leq n_1.\]

Since $D_1 \neq D_*$ , one may now argue exactly as below [ Reference Bilu and Kühne4 , (39)] to obtain that

\[ \lvert \Delta_1 \rvert^{1/2} \leq (3.8 \times 10^{10})(2.1 \times 10^4)^n (n+1)^{4n+6}. \]

The second part of Theorem 1·3 is an easy consequence of Theorem 1·5 and Proposition 6·2.

Proof of Theorem 1·3. Let $n \in {\mathbb{Z}}_{\gt0}$ . Let $x_1, \ldots, x_n$ be pairwise distinct singular moduli. Write $\Delta_i$ for the discriminant of $x_i$ and $D_i$ for the fundamental discriminant. Suppose that

\[ a_1 x_1^m + \cdots + a_n x_n^m = b\]

for some $m \in {\mathbb{Z}}_{\gt0}$ and $a_1, \ldots, a_n \in {\mathbb{Q}} \setminus \{0\}$ and $b \in {\mathbb{Q}}$ . Suppose that, for every $k \in \{1, \ldots, n\}$ ,

\[ \# \left \{ \Delta_i \;:\; \, i \in \left \{1, \ldots, n\right \} \mbox{ such that } D_i = D_k \right \} = 1.\]

Hence, by Proposition 6·2, if $i \in \{1, \ldots, n\}$ is such that $D_i \neq D_*$ , then

\[ \lvert \Delta_i \rvert^{1/2} \leq (3.8 \times 10^{10})(2.1 \times 10^4)^n (n+1)^{4n+6}.\]

So if $D_* \notin \{D_1, \ldots, D_n\}$ , then we are done.

Without loss of generality, assume then that $D_1 = D_*$ . For $i \geq 2$ , either $\Delta_i = \Delta_1$ , or $D_i \neq D_*$ and so $\lvert \Delta_i \rvert$ is bounded as above. Relabelling as necessary, there exists $k \in \{1, \ldots, n\}$ such that:

\[ \Delta_1 = \cdots = \Delta_{k}\]

and if $i \gt k$ , then $D_i \neq D_*$ and hence

\[ \lvert \Delta_i \rvert^{1/2} \leq (3.8 \times 10^{10})(2.1 \times 10^4)^n (n+1)^{4n+6}.\]

Hence, by Proposition 2·8, if $i \gt k$ , then

\[[{\mathbb{Q}}(x_i) \;:\; {\mathbb{Q}}] \leq \left((3.8 \times 10^{10})(2.1 \times 10^4)^n (n+1)^{4n+6}\right)^{4/3}.\]

Let

\[ L = {\mathbb{Q}}(a_1 x_1^m + \cdots + a_k x_k^m).\]

We may and do assume that

\[ \lvert \Delta_1 \rvert^{1/2} \gt \frac{1}{\pi} \left(\left(4k+3\right) \log\left(2k+1\right) -4k + 4\right).\]

Since $\Delta_1 = \cdots = \Delta_k$ , Theorem 1·5 implies that

\[ [L(\sqrt{\Delta_1}, x_1) \;:\; L(\sqrt{\Delta_1})] \leq k \leq n.\]

Thus,

\[[{\mathbb{Q}}(x_1) \;:\; {\mathbb{Q}}] \leq [L(\sqrt{\Delta_1}, x_1) \;:\; {\mathbb{Q}}] \leq 2n[L \;:\; {\mathbb{Q}}].\]

Since

\[ a_1 x_1^m + \cdots + a_k x_k^m = b - \sum_{i = k + 1}^n a_{i} x_{i}^m ,\]

we have that

\[ L \subset {\mathbb{Q}}(\{x_i \;:\; k \lt i \leq n\}).\]

Hence,

\[ [L \;:\; {\mathbb{Q}}] \leq \prod_{i=k+1}^n [{\mathbb{Q}}(x_i) \;:\; {\mathbb{Q}}] \leq \left( \left((3.8 \times 10^{10})(2.1 \times 10^4)^n (n+1)^{4n+6}\right)^{4/3} \right)^n,\]

and so

\[ [{\mathbb{Q}}(x_1) \;:\; {\mathbb{Q}}] \leq 2n\left((3.8 \times 10^{10})(2.1 \times 10^4)^n (n+1)^{4n+6}\right)^{4n/3}.\]

Hence, if we write

\[ g(n) = \left(2n\left((3.8 \times 10^{10})(2.1 \times 10^4)^n (n+1)^{4n+6}\right)^{4n/3}\right)^2\]

for convenience, then Proposition 2·7 implies that

\[ \lvert \Delta_1 \rvert^{1/2} \leq 2 g(n) e^{21000 g(n)}. \]

7·1. Fields generated by singular moduli

The fields generated by ${\mathbb{Q}}$ -linear combinations of two singular moduli may be described explicitly.

Theorem 7·4 ([ Reference Bilu, Faye and Zhu2 , theorem 1·5] and [ Reference Faye and Riffaut11 , theorem 4·1]). Let $\alpha \in {\mathbb{Q}} \setminus \{0\}$ . Let x, y be distinct singular moduli of respective discriminants $\Delta_x, \Delta_y$ . If

\[ {\mathbb{Q}}(x + \alpha y) \subsetneq {\mathbb{Q}}(x, y),\]

then one of the following holds:

1. (i) $\alpha = 1$ and $\Delta_x = \Delta_y$ . In this case,

   \[ [{\mathbb{Q}}(x, y) \;:\; {\mathbb{Q}}(x + y)] = 2;\]

2. (ii) $\Delta_x \neq \Delta_y$ , ${\mathbb{Q}}(x) = {\mathbb{Q}}(y)$ and this field has degree 2 over ${\mathbb{Q}}$ , and

   \[ \alpha = -\frac{x - x'}{y - y'},\]
   where x^′, y^′ are the non-trivial ${\mathbb{Q}}$ -conjugates of x, y respectively. In this case,
   \[ {\mathbb{Q}}(x + \alpha y) = {\mathbb{Q}}.\]

We also need the following results describing when singular moduli x, y satisfy either ${\mathbb{Q}}(x) = {\mathbb{Q}}(y)$ or ${\mathbb{Q}}(x) \subset {\mathbb{Q}}(y)$ with $[{\mathbb{Q}}(y) \;:\; {\mathbb{Q}}(x)] = 2$ .

Proposition 7·5 ([ Reference Allombert, Bilu and Pizarro-Madariaga1 , corollary 4·2]). Let x, y be singular moduli with discriminants $\Delta_x, \Delta_y$ and fundamental discriminants $D_x, D_y$ respectively. If

\[ D_x \neq D_y \mbox{ and } {\mathbb{Q}}(x) = {\mathbb{Q}}(y),\]

then all the possible fields ${\mathbb{Q}}(x) = {\mathbb{Q}}(y)$ and the corresponding possible discriminants $\Delta_x, \Delta_y$ are listed in [ Reference Allombert, Bilu and Pizarro-Madariaga1 , table 2].

Proposition 7·6. Let x, y be singular moduli with discriminants $\Delta_x, \Delta_y$ and fundamental discriminants $D_x, D_y$ respectively. If

\[ D_x \neq D_y, \, {\mathbb{Q}}(x) \subset {\mathbb{Q}}(y), \mbox{ and } [{\mathbb{Q}}(y) \;:\; {\mathbb{Q}}(x)] = 2,\]

then there are only finitely many possibilities for the pair $(\Delta_x, \Delta_y)$ and these may be listed explicitly.

This is an explicit version of [ Reference Bilu, Gun and Tron3 , corollary 2·13(2)]. To give a proof, we need the following terminology from [ Reference Bilu, Gun and Tron3 ]. Let G be a finite abelian group, written multiplicatively. Say that G is 2-elementary if every element of G has order $\leq 2$ . Say that G is almost 2-elementary if there exists a subgroup $H \leq G$ such that H is 2-elementary and $[G \;:\; H] = 2$ . Denote by $\rho_2(G)$ the dimension of $G / G^2$ as an $\mathbb{F}_2$ -vector space, where $G^2 = \{g^2 \;:\; g \in G\}$ .

Proof. Suppose that

(7·1) \begin{align} {}D_x \neq D_y, \, {\mathbb{Q}}(x) \subset {\mathbb{Q}}(y), \mbox{ and } [{\mathbb{Q}}(y) \;:\; {\mathbb{Q}}(x)] = 2.\end{align}

Then $h(\Delta_x) \leq 16$ and $h(\Delta_y) \leq 32$ by [ Reference Bilu, Gun and Tron3 , corollary 2·13]. So $\lvert \Delta_x \rvert, \lvert \Delta_y \rvert \leq 166147$ by Proposition 7·3. The proof of [ Reference Bilu, Gun and Tron3 , corollary 2·13] also shows that the class group ${\mathrm{cl}}(\Delta_x)$ is 2-elementary (and hence also almost 2-elementary) and that the class group ${\mathrm{cl}}(\Delta_y)$ is almost 2-elementary.

One may then compute in PARI all the almost 2-elementary discriminants $\Delta$ with $\lvert \Delta \rvert \leq 166147$ and $h(\Delta) \leq 32$ and then, among those, compute all the 2-elementary discriminants $\Delta$ with $h(\Delta) \leq 16$ . To do this, we use the following two characterisations [ Reference Bilu, Gun and Tron3 , (2·9), (2·10)]:

\begin{align*} {}\Delta \mbox{ is 2-elementary} &\iff h(\Delta) = 2^{\rho_2({\mathrm{cl}} (\Delta))}\\ {}\Delta \mbox{ is almost 2-elementary} &\iff h(\Delta) \mid 2^{\rho_2({\mathrm{cl}} (\Delta))+1}\end{align*}

and the classical fact (see e.g. [ Reference Cox10 , proposition 3·11] and the isomorphism in [ Reference Cox10 , (3·19)]) that

\[ \rho_2({\mathrm{cl}}(\Delta)) = \begin{cases} {}\omega(\Delta) & \mbox{if } \Delta \equiv 0 \bmod 32\\ {}\omega(\Delta)-2 & \mbox{if } \Delta \equiv 4 \bmod 16\\ {}\omega(\Delta) - 1 & \mbox{otherwise,}\end{cases} \]

where $\omega(\Delta)$ denotes the number of distinct prime divisors of $\Delta$ as usual.

Let $\Delta_y$ be an almost 2-elementary discriminant such that $h(\Delta_y)\leq 32$ and $\Delta_x$ a 2-elementary discriminant such that $h(\Delta_x) \leq 16$ . One may then check in PARI whether the conditions in (7·1) are satisfied by the pair $(\Delta_x, \Delta_y)$ . One finds in this way that there are 873 different possibilities for $(\Delta_x, \Delta_y)$ . The computations take about 30 minutes in total.

Proposition 7·8 ([ Reference Fowler12 , lemma 2·3]). Let x, y be singular moduli with respective discriminants $\Delta_x, \Delta_y$ and respective fundamental discriminants $D_x, D_y$ . Suppose that

\[x \notin {\mathbb{Q}}, \, D_x = D_y, \, {\mathbb{Q}}(x) \subset {\mathbb{Q}}(y), \mbox{ and } [{\mathbb{Q}}(y) \;:\; {\mathbb{Q}}(x)] = 2.\]

Then $\Delta_y/ \Delta_x \in \{9/4, 4, 9, 16\}$ . In addition, if $\Delta_y / \Delta_x = 9/4$ , then $\Delta_x \equiv 0, 4 \bmod 16$ .

The “in addition” part of Proposition 7·8 is not stated in [ Reference Fowler12 , lemma 2·3], but is a trivial consequence of the fact that $\Delta_x, \Delta_y \equiv 0, 1 \bmod 4$ .

8·1. Controlling the fields

Claim 8·3. Either:

\[ {\mathbb{Q}}(x) = {\mathbb{Q}}(y) = {\mathbb{Q}}(z),\]

or:

\[ {\mathbb{Q}}(x) \subsetneq {\mathbb{Q}}(y) = {\mathbb{Q}}(z) \mbox{ and } [{\mathbb{Q}}(y) \;:\; {\mathbb{Q}}(x)] = [{\mathbb{Q}}(z) \;:\; {\mathbb{Q}}(x)] = 2.\]

Proof. Observe that

\[ {\mathbb{Q}}(x) \subset {\mathbb{Q}}(y, z).\]

The same holds permuting x, y, z.

Suppose then that

\[ {\mathbb{Q}}(x) \subsetneq {\mathbb{Q}}(y, z).\]

Note that

\[ {\mathbb{Q}}\left(y + \frac{C}{B}z\right) = {\mathbb{Q}}(x).\]

We apply Theorem 7·4 to the field

\[{\mathbb{Q}}\left(y + \frac{C}{B} z\right).\]

Since $x \notin {\mathbb{Q}}$ by Claim 8·1, case (ii) of Theorem 7·4 is not possible. Thus, we must have that $B = C$ , $\Delta_y = \Delta_z$ , and

\[ [{\mathbb{Q}}(y, z) \;:\; {\mathbb{Q}}(x)] = 2.\]

Suppose, in addition, that

\[ {\mathbb{Q}}(y) \subsetneq {\mathbb{Q}}(x, z).\]

Then, by the same argument, we must have that $A = C$ and $\Delta_x = \Delta_z$ . Thus, $\Delta_x = \Delta_y = \Delta_z$ , which contradicts Claim 8·2. Similarly, if ${\mathbb{Q}}(z) \subsetneq {\mathbb{Q}}(x, y)$ .

We are thus left with the following two possibilities. Either:

\[ {\mathbb{Q}}(x) = {\mathbb{Q}}(y, z), \, {\mathbb{Q}}(y) = {\mathbb{Q}}(x, z), \mbox{ and } {\mathbb{Q}}(z) = {\mathbb{Q}}(x, y);\]

or, up to permuting x, y, z:

\[{\mathbb{Q}}(x) \subsetneq {\mathbb{Q}}(y, z), \, {\mathbb{Q}}(y) = {\mathbb{Q}}(x, z), \mbox{ and } {\mathbb{Q}}(z) = {\mathbb{Q}}(x, y). \]

Hence, either:

\[ {\mathbb{Q}}(x) = {\mathbb{Q}}(y) = {\mathbb{Q}}(z),\]

or:

\[ {\mathbb{Q}}(x) \subsetneq {\mathbb{Q}}(y) = {\mathbb{Q}}(z) \mbox{ and } [{\mathbb{Q}}(y) \;:\; {\mathbb{Q}}(x)] = [{\mathbb{Q}}(z) \;:\; {\mathbb{Q}}(x)] = 2. \]

Claim 8·4. If

\[ {\mathbb{Q}}(x) = {\mathbb{Q}}(y) = {\mathbb{Q}}(z),\]

then we may assume that this field has degree at least 3 over ${\mathbb{Q}}$ .

Proof. Suppose not. Then

\[ {\mathbb{Q}}(x) = {\mathbb{Q}}(y) = {\mathbb{Q}}(z)\]

and, by Claim 8·1, this field must be a degree 2 extension of ${\mathbb{Q}}$ . Write x ^′, y ^′, z ^′ for the unique non-trivial ${\mathbb{Q}}$ -conjugates of x, y, z respectively. Then

\[Ax + By + Cz = Ax' + By' + Cz'.\]

Hence,

\[A = -\frac{B(y-y')+C(z-z')}{x-x'},\]

and so we must be in case (iii) of Theorem 1·6.

Claim 8·5. If

\[ {\mathbb{Q}}(x) \subsetneq {\mathbb{Q}}(y) = {\mathbb{Q}}(z) \mbox{ and } [{\mathbb{Q}}(y) \;:\; {\mathbb{Q}}(x)] = [{\mathbb{Q}}(z) \;:\; {\mathbb{Q}}(x)] = 2,\]

then we may assume that

\[ [{\mathbb{Q}}(x) \;:\; {\mathbb{Q}}] \gt 2.\]

Proof. Suppose that

\[ {\mathbb{Q}}(x) \subsetneq {\mathbb{Q}}(y) = {\mathbb{Q}}(z) \mbox{ and } [{\mathbb{Q}}(y) \;:\; {\mathbb{Q}}(x)] = [{\mathbb{Q}}(z) \;:\; {\mathbb{Q}}(x)] = 2.\]

Suppose, in addition, that

\[ [{\mathbb{Q}}(x) \;:\; {\mathbb{Q}}] = 2.\]

We will show that we must then be in case (v) of Theorem 1·6.

Since ${\mathbb{Q}}(x) \subsetneq {\mathbb{Q}}(y, z)$ , we must have that $B=C$ and $\Delta_y = \Delta_z$ . Let x ^′ denote the unique non-trivial Galois conjugate of x over ${\mathbb{Q}}$ . Let v, w be the other (along with y, z) two singular moduli of discriminant $\Delta_y$ . Since

\[ Ax + B(y + z) \in {\mathbb{Q}},\]

there are then $y', z' \in \{y, z, v, w\}$ such that

\[ Ax' + B(y'+z') = Ax + B(y + z).\]

So

\[ A(x-x') = -B((y+z) - (y'+z')).\]

Suppose that $\{y, z\} \cap \{y', z'\} \neq \emptyset$ . Without loss of generality, we may assume that $y = y'$ . So

\[ A(x-x') = B(z-z')\]

and hence

\[ {\mathbb{Q}}(x-x') = {\mathbb{Q}}(z-z').\]

We thus have that

\[ {\mathbb{Q}}(z - z') = {\mathbb{Q}}(x - x') = {\mathbb{Q}}(x, x') = {\mathbb{Q}}(x) \neq {\mathbb{Q}},\]

where the second equality is by Theorem 7·4 and the third equality holds since $[{\mathbb{Q}}(x) \;:\; {\mathbb{Q}}] = 2$ . In particular, $z \neq z'$ , and so, by Theorem 7·4 again,

\[ {\mathbb{Q}}(z - z') = {\mathbb{Q}}(z, z').\]

Hence,

\[{\mathbb{Q}}(z, z') = {\mathbb{Q}}(x).\]

This is a contradiction, since $[{\mathbb{Q}}(x) \;:\; {\mathbb{Q}}] = 2$ , but

\[ [{\mathbb{Q}}(z, z') \;:\; {\mathbb{Q}}] \geq [{\mathbb{Q}}(z) \;:\; {\mathbb{Q}}] = [{\mathbb{Q}}(z) \;:\; {\mathbb{Q}}(x)] [{\mathbb{Q}}(x) \;:\; {\mathbb{Q}}] = 4.\]

Therefore, we must have that $\{y', z'\} = \{v, w\}$ and so

\[ \frac{A}{B} = - \frac{(y+z) - (v+w)}{x - x'}.\]

We are thus in case (v) of Theorem 1·6.

Proof. By Claim 8·3, either

\[ {\mathbb{Q}}(x) = {\mathbb{Q}}(y) = {\mathbb{Q}}(z),\]

or:

\[ {\mathbb{Q}}(x) \subsetneq {\mathbb{Q}}(y) = {\mathbb{Q}}(z) \mbox{ and } [{\mathbb{Q}}(y) \;:\; {\mathbb{Q}}(x)] = [{\mathbb{Q}}(z) \;:\; {\mathbb{Q}}(x)] = 2.\]

If

\[ {\mathbb{Q}}(x) = {\mathbb{Q}}(y) = {\mathbb{Q}}(z),\]

then, by Claim 8·4, this field has degree at least 3 over ${\mathbb{Q}}$ , and hence we are not in any of cases (i)–(v) of Theorem 1·6. (Recall that, by Claim 8·2, we are not in case (iv) of Theorem 1·6.) If

\[ {\mathbb{Q}}(x) \subsetneq {\mathbb{Q}}(y) = {\mathbb{Q}}(z) \mbox{ and } [{\mathbb{Q}}(y) \;:\; {\mathbb{Q}}(x)] = [{\mathbb{Q}}(z) \;:\; {\mathbb{Q}}(x)] = 2,\]

then, by Claim 8·5,

\[[{\mathbb{Q}}(x) \;:\; {\mathbb{Q}}] \gt 2,\]

and hence we are not in any of cases (i)–(v) of Theorem 1·6.

8·2. Controlling the discriminants

Proof. By Claim 8·3, either:

\[ {\mathbb{Q}}(x) = {\mathbb{Q}}(y) = {\mathbb{Q}}(z),\]

or:

\[ {\mathbb{Q}}(x) \subsetneq {\mathbb{Q}}(y) = {\mathbb{Q}}(z) \mbox{ and } [{\mathbb{Q}}(y) \;:\; {\mathbb{Q}}(x)] = [{\mathbb{Q}}(z) \;:\; {\mathbb{Q}}(x)] = 2.\]

In either case, if the fundamental discriminants $D_x, D_y, D_z$ are not all equal, then all the possibilities for $(\Delta_x, \Delta_y, \Delta_z)$ are given by Propositions 7·5 and 7·6. These possibilities will be dealt with in Section 8·4.

Proof. Suppose that

\[ {\mathbb{Q}}(x) = {\mathbb{Q}}(y) = {\mathbb{Q}}(z).\]

Without loss of generality, we assume that $\lvert \Delta_x \rvert \geq \lvert \Delta_y \rvert \geq \lvert \Delta_z \rvert$ . Hence, $\lvert \Delta_x \rvert \gt \lvert \Delta_z \rvert$ by Claim 8·2. By Claim 8·7, $D_x = D_y = D_z$ . Hence, Proposition 7·6 implies that there are the following two possibilities:

1. (1) $\Delta_x = 4 \Delta$ and $\Delta_y = \Delta_z = \Delta$ for some $\Delta \equiv 1 \bmod 8$ ;

2. (2) $\Delta_x = \Delta_y = 4 \Delta$ and $\Delta_z = \Delta$ for some $\Delta \equiv 1 \bmod 8$ .

Claim 8·9. If

\[{\mathbb{Q}}(x) \subsetneq {\mathbb{Q}}(y) = {\mathbb{Q}}(z),\]

then $\Delta_x = \Delta$ and $\Delta_y = \Delta_z = l^2 \Delta$ for some $\Delta$ , where $l \in \{3/2, 2, 3, 4\}$ . If $l = 3/2$ , then $\Delta \equiv 0, 4 \bmod 16$ .

Proof. Since

\[ {\mathbb{Q}}(B y + C z) = {\mathbb{Q}}(x) \subsetneq {\mathbb{Q}}(y) = {\mathbb{Q}}(z),\]

Theorem 7·4 and Claim 8·1 together imply that $\Delta_y = \Delta_z$ and $B=C$ . Proposition 7·7 then implies that we may write $\Delta_x = \Delta$ and $\Delta_y = \Delta_z = l^2 \Delta$ , where $l \in \{3/2, 2, 3, 4\}$ , and also that $\Delta \equiv 0, 4 \bmod 16$ if $l = 3/2$ .

8·3. Bounding the discriminants

Let

\[ \alpha = A x + B y + C z.\]

So $\alpha \in {\mathbb{Q}}$ by assumption. Suppose that $(x_i, y_i, z_i)$ , where $i=1, \ldots, 4$ , are Galois conjugates of (x, y, z) over ${\mathbb{Q}}$ . Then

\[ A x_i + B y_i + C z_i = \alpha\]

for $i=1, 2, 3, 4$ . In particular, we have that

(8·1) \begin{align}\begin{vmatrix} {}1 & 1 & 1 & 1\\ {}x_1 & x_2 & x_3 & x_4\\ {}y_1 & y_2 & y_3 & y_4\\ {}z_1 & z_2 & z_3 & z_4\end{vmatrix}= 0.\end{align}

In this section, we will establish effective bounds on $(\Delta_x, \Delta_y, \Delta_z)$ by showing that, if these discriminants are too large (in absolute value), then one may find conjugates $(x_i, y_i, z_i)$ such that the determinant on the left-hand side of (8·1) is non-zero. Typically, we will do this by showing that there exists a permutation $\tau \in S_4$ such that

\begin{align*}\lvert x_{\tau(2)} y_{\tau(3)} z_{\tau(4)} \rvert \gt \sum_{\sigma \in S_4 \setminus \{\tau\}} \lvert x_{\sigma(2)} y_{\sigma(3)} z_{\sigma(4)} \rvert.\end{align*}

We will make repeated use (without special reference) of the bound on singular moduli given in Proposition 2·2, the description of the conjugates of a singular modulus in Subsection 2·1, and the bound on the number of singular moduli of a given discriminant and denominator in Proposition 7·1. The bounds themselves were calculated in PARI; this took only a few seconds.

8·3·1. The case where ${\mathbb{Q}}(x) = {\mathbb{Q}}(y) = {\mathbb{Q}}(z)$

In this section, we assume that

\[ {\mathbb{Q}}(x) = {\mathbb{Q}}(y) = {\mathbb{Q}}(z).\]

By Claim 8·4, the field ${\mathbb{Q}}(x) = {\mathbb{Q}}(y) = {\mathbb{Q}}(z)$ has degree at least 3 over ${\mathbb{Q}}$ . Each singular modulus of discriminant $\Delta_x$ occurs precisely once among the first coordinates of the distinct ${\mathbb{Q}}$ -conjugates of (x, y, z). Correspondingly for the singular moduli of discriminant $\Delta_y$ among the second coordinates, and for the singular moduli of discriminant $\Delta_z$ among the third coordinates.

Claim 8·10. If $\Delta_x = 4 \Delta$ and $\Delta_y = \Delta_z = \Delta$ , where $\Delta \equiv 1 \bmod 8$ , then $h(\Delta) \leq 6$ .

Proof. Suppose that $\Delta_x = 4 \Delta$ and $\Delta_y = \Delta_z = \Delta$ , where $\Delta \equiv 1 \bmod 8$ . Assume that $h(\Delta) \geq 7$ . In particular, $\Delta \notin \{-7, -15\}$ by Lemma 2·1. Thus there are exactly two subdominant singular moduli of discriminant $\Delta$ and no subdominant singular moduli of discriminant $4 \Delta$ . Let $(x_1, y_1, z_1)$ be the unique conjugate of (x, y, z) with $x_1$ dominant. We distinguish two subcases.

Subcase 1: neither of $y_1, z_1$ is dominant. In this case, we may take $(x_2, y_2, z_2), (x_3, y_3, z_3)$ to be the unique conjugates such that $y_2, z_3$ are dominant. Then

(8·2) \begin{align} {}\lvert x_1 y_2 z_3 \rvert \geq (\!\exp(2 \pi \lvert \Delta \rvert^{1/2}) - 2079)(\!\exp(\pi \lvert \Delta \rvert^{1/2}) - 2079)^2.\end{align}

Since $h(\Delta) \geq 7$ , we may take $(x_4, y_4, z_4)$ to be a conjugate with none of $x_4, y_4, z_4$ dominant. Then, for $\sigma \in S_4$ such that $\sigma(2, 3, 4) \neq (1, 2, 3)$ , we have that either: $\sigma(2)=1$ and

(8·3) \begin{align} {}&\lvert x_{\sigma(2)} y_{\sigma(3)} z_{\sigma(4)} \rvert \nonumber \\ {}&\quad \leq (\!\exp(2 \pi \lvert \Delta \rvert^{1/2}) + 2079)(\!\exp(\pi \lvert \Delta \rvert^{1/2}) + 2079)\nonumber \\ {}&\quad \times \left(\exp\left(\frac{\pi \lvert \Delta \rvert^{1/2}}{2}\right) + 2079\right),\end{align}

or: $\sigma(2) \neq 1$ and

(8·4) \begin{align} {}&\lvert x_{\sigma(2)} y_{\sigma(3)} z_{\sigma(4)} \rvert \nonumber \\ {}&\quad \leq \left(\exp\left(\frac{2 \pi \lvert \Delta \rvert^{1/2}}{3}\right) + 2079\right)(\!\exp(\pi \lvert \Delta \rvert^{1/2}) + 2079)^2.\end{align}

The bounds (8·2), (8·3) and (8·4) are incompatible with (8·1) if $\lvert \Delta \rvert \geq 10$ . By Lemma 2·1, $\lvert \Delta \rvert \leq 9$ is impossible since $z \notin {\mathbb{Q}}$ .

Subcase 2: one of $y_1, z_1$ is dominant. Without loss of generality, suppose that $y_1$ is dominant. Let $(x_2, y_2, z_2)$ be the unique conjugate with $z_2$ dominant. There also exists a conjugate $(x_3, y_3, z_3)$ with $y_3$ subdominant and $x_3, z_3$ not dominant. There exists a conjugate $(x_4, y_4, z_4)$ with neither of $y_4, z_4$ dominant or subdominant. Note that $x_4$ is thus not dominant as well. Hence,

(8·5) \begin{align} {}&\lvert x_1 y_3 z_2 \rvert \nonumber\\ {}&\quad \geq (\!\exp(2 \pi \lvert \Delta \rvert^{1/2}) - 2079)\left(\exp\left(\frac{\pi \lvert \Delta \rvert^{1/2}}{2}\right) - 2079\right) \nonumber \\ {}&\qquad \times (\!\exp(\pi \lvert \Delta \rvert^{1/2}) - 2079).\end{align}

Now let $\sigma \in S_4$ be such that $\sigma(2, 3, 4) \neq (1, 3, 2)$ . If $\sigma(2) \neq 1$ , then

(8·6) \begin{align} {}&\lvert x_{\sigma(2)} y_{\sigma(3)} z_{\sigma(4)} \rvert \nonumber \\ {}&\quad \leq \left(\exp\left(\frac{2 \pi \lvert \Delta \rvert^{1/2}}{3}\right) + 2079\right)(\!\exp(\pi \lvert \Delta \rvert^{1/2}) + 2079)^2.\end{align}

If $\sigma(2) = 1$ and $\sigma(4) = 2$ , then

(8·7) \begin{align} {}&\lvert x_{\sigma(2)} y_{\sigma(3)} z_{\sigma(4)} \rvert \nonumber \\ {}&\quad \leq (\!\exp(2 \pi \lvert \Delta \rvert^{1/2}) + 2079)\left(\exp\left(\frac{\pi \lvert \Delta \rvert^{1/2}}{3}\right) + 2079\right) \nonumber \\ {}&\qquad \times (\!\exp(\pi \lvert \Delta \rvert^{1/2}) + 2079).\end{align}

If $\sigma(2) = 1$ and $\sigma(4) \neq 2$ , then

(8·8) \begin{align} {}&\lvert x_{\sigma(2)} y_{\sigma(3)} z_{\sigma(4)} \rvert \nonumber \\ {}&\quad \leq (\!\exp(2 \pi \lvert \Delta \rvert^{1/2}) + 2079)\left(\exp\left(\frac{\pi \lvert \Delta \rvert^{1/2}}{2}\right) + 2079\right)^2.\end{align}

The bounds (8·5), (8·6), (8·7) and (8·8) are incompatible with (8·1) if $\lvert \Delta \rvert \geq 32$ . By Lemma 2·1, $\lvert \Delta \rvert \leq 31$ is impossible since $h(\Delta) \geq 7$ by assumption.

We must therefore have that $h(\Delta) \leq 6$ .

Claim 8·11. If $\Delta_x = \Delta_y = 4 \Delta$ and $\Delta_z = \Delta$ , where $\Delta \equiv 1 \bmod 8$ , then $h(\Delta) \leq 5$ .

Proof. Suppose that $\Delta_x = \Delta_y = 4 \Delta$ and $\Delta_z = \Delta$ , where $\Delta \equiv 1 \bmod 8$ . Assume further that $h(\Delta) \geq 6$ . In particular, $\Delta \notin \{-7, -15\}$ by Lemma 2·1. Thus there are exactly two subdominant singular moduli of discriminant $\Delta$ and no subdominant singular moduli of discriminant $4 \Delta$ . Let $(x_1, y_1, z_1), (x_2, y_2, z_2)$ be the unique conjugates of (x, y, z) such that $x_1, y_2$ are dominant. There must also exist a conjugate $(x_3, y_3, z_3)$ such that $z_3$ is either dominant or subdominant and neither of $x_3, y_3$ is dominant.

Then

(8·9) \begin{align} {}\lvert x_1 y_2 z_3 \rvert \geq (\!\exp(2 \pi \lvert \Delta \rvert^{1/2}) - 2079)^2\left(\exp\left(\frac{\pi \lvert \Delta \rvert^{1/2}}{2}\right) - 2079\right).\end{align}

There exists a conjugate $(x_4, y_4, z_4)$ such that neither of $x_4, y_4$ is dominant and $z_4$ is neither dominant nor subdominant. Let $\sigma \in S_4$ be such that $\sigma(2, 3, 4) \neq (1, 2, 3)$ . If $\sigma(2)=1$ and $\sigma(3)=2$ , then

(8·10) \begin{align} {}&\lvert x_{\sigma(2)} y_{\sigma(3)} z_{\sigma(4)} \rvert \nonumber \\ {}&\quad \leq (\!\exp(2 \pi \lvert \Delta \rvert^{1/2}) + 2079)^2\left(\exp\left(\frac{\pi \lvert \Delta \rvert^{1/2}}{3}\right) + 2079\right).\end{align}

If either $\sigma(2) \neq 1$ or $\sigma(3) \neq 2$ , then

(8·11) \begin{align} {}&\lvert x_{\sigma(2)} y_{\sigma(3)} z_{\sigma(4)} \rvert \nonumber \\ {}&\quad \leq (\!\exp(2 \pi \lvert \Delta \rvert^{1/2}) + 2079)\left(\exp\left(\frac{2 \pi \lvert \Delta \rvert^{1/2}}{3}\right) + 2079\right) \nonumber \\ {}&\qquad \times (\!\exp(\pi \lvert \Delta \rvert^{1/2}) + 2079).\end{align}

The bounds (8·9), (8·10), and (8·11) are incompatible with (8·1) if $\lvert \Delta \rvert \geq 29$ . By Lemma 2·1, there are no discriminants $\Delta$ with $\lvert \Delta \rvert \leq 28$ and $h(\Delta) \geq 6$ . Hence, we must have that $h(\Delta) \leq 5$ .

8·3·2. The case where ${\mathbb{Q}}(x) \subsetneq {\mathbb{Q}}(y) = {\mathbb{Q}}(z)$

In this section, we assume that

\[ {\mathbb{Q}}(x) \subsetneq {\mathbb{Q}}(y) = {\mathbb{Q}}(z).\]

By Claim 8·9, we have that $\Delta_x = \Delta$ and $\Delta_y = \Delta_z = l^2 \Delta$ , where $l \in \{3/2, 2, 3, 4\}$ . Also, by Claims 8·3 and 8·5,

\[ [{\mathbb{Q}}(y) \;:\; {\mathbb{Q}}(x)] = [{\mathbb{Q}}(z) \;:\; {\mathbb{Q}}(x)] = 2 \mbox{ and } [{\mathbb{Q}}(x) \;:\; {\mathbb{Q}}] \geq 3.\]

So $h(\Delta) \geq 3$ .

Each singular modulus of discriminant $\Delta$ occurs exactly twice among the first coordinates of the distinct ${\mathbb{Q}}$ -conjugates of (x, y, z). Each singular modulus of discriminant $l^2 \Delta$ occurs exactly once among the second coordinates of the distinct ${\mathbb{Q}}$ -conjugates of (x, y, z) and exactly once among the third coordinates. Let $(x_1, y_1, z_1), (x_2, y_2, z_2)$ be the conjugates such that $y_1, z_2$ are dominant.

Claim 8·12. Both of $x_1, x_2$ are dominant.

Proof. Suppose that at least one of $x_1, x_2$ is not dominant. There is then a conjugate $(x_3, y_3, z_3)$ with $x_3$ dominant and neither of $y_3, z_3$ dominant. Then

(8·12) \begin{align} {}\lvert x_3 y_1 z_2 \rvert \geq (\!\exp(\pi \lvert \Delta \rvert^{1/2}) - 2079)(\!\exp(l \pi \lvert \Delta \rvert^{1/2}) - 2079)^2.\end{align}

Since $h(\Delta) \geq 3$ , there is a conjugate $(x_4, y_4, z_4)$ with none of $x_4, y_4, z_4$ dominant. Now let $\sigma \in S_4$ be such that $\sigma(2, 3, 4) \neq (3, 1, 2)$ . If $\sigma(3) = 1$ and $\sigma(4) = 2$ , then

(8·13) \begin{align} {}&\lvert x_{\sigma(2)} y_{\sigma(3)} z_{\sigma(4)} \rvert \nonumber \\ {}&\quad \leq \left(\exp\left(\frac{\pi \lvert \Delta \rvert^{1/2}}{2}\right) + 2079\right)(\!\exp(l \pi \lvert \Delta \rvert^{1/2}) + 2079)^2.\end{align}

If either $\sigma(3) \neq 1$ or $\sigma(4) \neq 2$ , then

(8·14) \begin{align} {}&\lvert x_{\sigma(2)} y_{\sigma(3)} z_{\sigma(4)} \rvert \nonumber \\ {}&\quad \leq (\!\exp(\pi \lvert \Delta \rvert^{1/2}) + 2079)(\!\exp(l \pi \lvert \Delta \rvert^{1/2}) + 2079) \nonumber \\ {}& \qquad \times \left(\exp\left(\frac{l \pi \lvert \Delta \rvert^{1/2}}{2}\right) + 2079\right).\end{align}

For every $l \in \{3/2, 2, 3, 4\}$ , the bounds (8·12), (8·13) and (8·14) are incompatible with (8·1) if $\lvert \Delta \rvert \geq 8$ . By Lemma 2·1, $\lvert \Delta \rvert \leq 7$ is impossible since $x \notin {\mathbb{Q}}$ .

Note that $x_1 = x_2$ , since $x_1, x_2$ are both dominant. We therefore rearrange (8·1) to obtain that

(8·15) \begin{align}y_1 z_2 (x_3 - x_4) = \sum_{\substack{\sigma \in S_4\\ \sigma(3, 4) \neq (1, 2)}} \mathrm{sgn}(\sigma) x_{\sigma(2)} y_{\sigma(3)} z_{\sigma(4)}.\end{align}

We may then use Proposition 7·2 to obtain the lower bound

(8·16) \begin{align}\lvert y_1 z_2 (x_3 - x_4) \rvert \geq 800 \frac{(\!\exp(l \pi \lvert \Delta \rvert^{1/2}) - 2079)^2}{\lvert \Delta \rvert^{4}}.\end{align}

Claim 8·13. We have that $l \in \{3/2, 2\}$ and:

1. (i) if $l = 3/2$ , then one of the following holds:

   1. (a) $h(\Delta_x) \leq 6$ ;

   2. (b) $h(\Delta_x) \in \{7, 8\}$ and $\lvert \Delta_x \rvert \leq 5879$ ;

   3. (c) $h(\Delta_x) \in \{9, 10\}$ and $\lvert \Delta_x \rvert \leq 1557$ ;

   4. (d) $h(\Delta_x) \in \{11, 12, 13, 14\}$ and $\lvert \Delta_x \rvert \leq 790$ .

2. (ii) if $l = 2$ , then one of the following holds:

   1. (a) $h(\Delta_x) \leq 4$ ;

   2. (b) $h(\Delta_x) \in \{5, 6\}$ and $\lvert \Delta_x \rvert \leq 304$ ;

Proof. For each $k \in \{3, 5, 7, 9, 11, 15\}$ , there is some $a_\mathrm{min}(k)$ , the value of which is given in Table 1, with the property that: for every $l \in \{3/2, 2, 3, 4\}$ , if $h(\Delta) \geq k$ , then there exist conjugates $(x_3, y_3, z_3), (x_4, y_4, z_4)$ such that $x_3 \neq x_4$ and

\[ a(y_3), a(y_4), a(z_3), a(z_4) \geq a_\mathrm{min}(k),\]

where $a(y_i)$ (respectively $a(z_i)$ ) denotes the denominator of $y_i$ (respectively $z_i$ ).

Fix some $l \in \{3/2, 2, 3, 4\}$ . Next, let $k \in \{3, 5, 7, 9, 11, 15\}$ , and take the corresponding value of $a_\mathrm{min}(k)$ from Table 1. Now let $\sigma \in S_4$ be such that $\sigma(3, 4) \neq (1, 2)$ . If either $\sigma(3) = 1$ or $\sigma(4) = 2$ (note there are 8 such $\sigma$ ), then

(8·17) \begin{align} {}&\lvert x_{\sigma(2)} y_{\sigma(3)} z_{\sigma(4)} \rvert \nonumber \\ {}&\quad \leq (\!\exp(\pi \lvert \Delta \rvert^{1/2}) + 2079)(\!\exp(l \pi \lvert \Delta \rvert^{1/2}) + 2079) \nonumber \\ {}&\qquad \times\left(\exp\left(\frac{l \pi \lvert \Delta \rvert^{1/2}}{a_\mathrm{min}(k)}\right) + 2079\right).\end{align}

If $\sigma(3) = 2$ and $\sigma(4) = 1$ (note there are 2 such $\sigma$ ), then

(8·18) \begin{align} {}&\lvert x_{\sigma(2)} y_{\sigma(3)} z_{\sigma(4)} \rvert \nonumber \\ {}&\quad \leq \left(\exp\left(\frac{\pi \lvert \Delta \rvert^{1/2}}{2}\right) + 2079\right)\left(\exp\left(\frac{l \pi \lvert \Delta \rvert^{1/2}}{2}\right) + 2079\right)^2.\end{align}

Otherwise (i.e. for the 12 remaining $\sigma$ ),

(8·19) \begin{align} {}&\lvert x_{\sigma(2)} y_{\sigma(3)} z_{\sigma(4)} \rvert \nonumber \\ {}&\quad \leq (\!\exp(\pi \lvert \Delta \rvert^{1/2}) + 2079)\left(\exp\left(\frac{l \pi \lvert \Delta \rvert^{1/2}}{2}\right) + 2079\right) \nonumber \\ {}&\qquad \times \left(\exp\left(\frac{l \pi \lvert \Delta \rvert^{1/2}}{a_\mathrm{min}(k)}\right) + 2079\right).\end{align}

If k is suitably large (for the given l), then the bounds (8·16), (8·17), (8·18) and (8·19) are incompatible with (8·15) for large enough $\lvert \Delta \rvert$ . The entries in Table 1 are the upper bounds on $\lvert \Delta \rvert$ which are obtained for a given l when $h(\Delta) \geq k$ by taking suitable conjugates $(x_3, y_3, z_3), (x_4, y_4, z_4)$ with $x_3 \neq x_4$ and

\[ a(y_3), a(y_4), a(z_3), a(z_4) \geq a_\mathrm{min}(k).\]

Note that for $l = 3/2$ (respectively, for $l=2$ ) no upper bound on $\lvert \Delta \rvert$ is obtained if $k \lt 7$ (respectively if $k \lt 5$ ). An entry b in a row k and column l of Table 1 is in a grey-shaded cell if there are no discriminants $\Delta$ satisfying:

1. (1) $h(\Delta) \geq k$ ;

2. (2) $\lvert \Delta \rvert \leq b$ ; and

3. (3) if $l=3/2$ , then $\Delta \equiv 0, 4 \bmod 16$ .

(These conditions are easily checked in PARI.) In particular, this implies that $l \in \{3, 4\}$ is impossible, since $h(\Delta) \geq 3$ by Claim 8·5. For $l \in \{3/2, 2\}$ , we must be in one of the cases stated in the claim.

Table 1.

Upper bounds on $\lvert \Delta \rvert$ when $h(\Delta) \geq k$ for a given $l \in \{3/2, 2, 3, 4\}$

8·4. Eliminating the discriminants

So far in Section 8, we have shown that if x, y, z are pairwise distinct singular moduli of respective discriminants $\Delta_x, \Delta_y, \Delta_z$ such that

\[ Ax + By + Cz \in {\mathbb{Q}}\]

for some $A, B, C \in {\mathbb{Q}} \setminus \{0\}$ , then either we are in one of cases (1)–(5) of Theorem 1·6 or (without loss of generality) one of the following situations occurs:

1. (1) ${\mathbb{Q}}(x) = {\mathbb{Q}}(y) = {\mathbb{Q}}(z)$ and this field, denote it L, is an extension of ${\mathbb{Q}}$ of degree at least 3 and one of the following holds:

   1. (a) $(\Delta_x, \Delta_y, \Delta_z) = (4 \Delta, \Delta, \Delta)$ , where $\Delta \equiv 1 \bmod 8$ and $h(\Delta) \leq 6$ ;

   2. (b) $(\Delta_x, \Delta_y, \Delta_z) = (4 \Delta, 4 \Delta, \Delta)$ , where $\Delta \equiv 1 \bmod 8$ and $h(\Delta) \leq 5$ ;

   3. (c) the field L and the discriminants $\Delta_x, \Delta_y, \Delta_z$ are given in [ Reference Allombert, Bilu and Pizarro-Madariaga1 , table 2];

2. (2) ${\mathbb{Q}}(x) \subset {\mathbb{Q}}(y) = {\mathbb{Q}}(z)$ , $[{\mathbb{Q}}(y) \;:\; {\mathbb{Q}}(x)] = 2$ , $[{\mathbb{Q}}(x) \;:\; {\mathbb{Q}}] \geq 3$ , $\Delta_y = \Delta_z$ , and one of the following holds:

   1. (a) $9 \Delta_x = 4 \Delta_y$ and $\Delta_x \equiv 0, 4 \bmod 16$ and one of the following holds:

      1. (i) $h(\Delta_x) \leq 6$ ;

      2. (ii) $h(\Delta_x) \in \{7, 8\}$ and $\lvert \Delta_x \rvert \leq 5879$ ;

      3. (iii) $h(\Delta_x) \in \{9, 10\}$ and $\lvert \Delta_x \rvert \leq 1557$ ;

      4. (iv) $h(\Delta_x) \in \{11, 12, 13, 14\}$ and $\lvert \Delta_x \rvert \leq 790$ ;

1. (b) $4 \Delta_x = \Delta_y$ and one of the following holds:

   1. (i) $h(\Delta_x) \leq 4$ ;

   2. (ii) $h(\Delta_x) \in \{5, 6\}$ and $\lvert \Delta_x \rvert \leq 304$ ;

2. (c) $(\Delta_x, \Delta_y)$ is one of the 873 pairs given in Proposition 7·5.

Here (1)(a) arises from Claim 8·10, (1)(b) arises from Claim 8·11, 1(c) and (2)(c) arise from Claim 8·7, and (2)(a) and (2)(b) arise from Claim 8·13.

The corresponding list of all such $(\Delta_x, \Delta_y, \Delta_z)$ may be generated straightforwardly in PARI. Here we make use of Proposition 7·3, whence $h(\Delta) \leq 6$ implies that $\lvert \Delta \rvert \leq 4075$ . For each triple of discriminants $(\Delta_x, \Delta_y, \Delta_z)$ belonging to this list, we then use PARI to show that there are no pairwise distinct singular moduli x, y, z of the corresponding discriminants such that the set $\{1, x, y, z\}$ is linearly dependent over ${\mathbb{Q}}$ . The necessary computations take about 40 minutes to run. This completes the proof of Theorem 1·6.

One might expect that, in analogy to case (iii) of Theorem 1·6, there would be an additional case in Theorem 1·6, namely one where x, y, z all generate the same number field of degree 3 over ${\mathbb{Q}}$ . In fact, this cannot occur, for reasons that we now explain.

Suppose that x, y are distinct singular moduli such that ${\mathbb{Q}}(x) = {\mathbb{Q}}(y)$ and this field has degree 3 over ${\mathbb{Q}}$ . In particular, by [ Reference Allombert, Bilu and Pizarro-Madariaga1 , corollary 3·3], the field ${\mathbb{Q}}(x)$ is not a Galois extension of ${\mathbb{Q}}$ . Propositions 7·5 and 7·7, together with the list of discriminants $\Delta$ with $h(\Delta) = 3$ , imply that either $\Delta_x = \Delta_y$ or $\{\Delta_x, \Delta_y\} \in \{ \{-23, -92\}, \{-31, -124\} \}$ .

Hence, if x, y, z are pairwise distinct singular moduli such that ${\mathbb{Q}}(x) = {\mathbb{Q}}(y) = {\mathbb{Q}}(z)$ and this field has degree 3 over ${\mathbb{Q}}$ , then at least two of x, y, z have the same discriminant. Suppose that x, y have the same discriminant. Then the equality ${\mathbb{Q}}(x) = {\mathbb{Q}}(y)$ implies that ${\mathbb{Q}}(x)$ is a Galois extension of ${\mathbb{Q}}$ , which is a contradiction.