Proof. By our definition of the degree, we have $\deg (V \cap W) = \deg (\overline {V \cap W})$, $\deg V = \deg \bar {V}$, and $\deg W = \deg \bar {W}$. As $V = \bar {V} \cap U$ and $W = \bar {W} \cap U$, we find that $V \cap W = \bar {V} \cap \bar {W} \cap U$. It follows that the irreducible components of $\overline {V \cap W}$ are precisely the irreducible components of $\bar {V} \cap \bar {W}$ that have non-empty intersection with $U$. We deduce that $\deg (\overline {V \cap W}) \leq \deg (\bar {V} \cap \bar {W})$, so it suffices to prove the theorem for $U = \mathbb {P}^{n_1}_K \times _K \cdots \times _K \mathbb {P}^{n_k}_K$, $V = \bar {V}$, and $W = \bar {W}$.

In the case where $k = 1$ and $V$ and $W$ are irreducible, this then follows from Example 8.4.6 in [Reference FultonFul98]. Using the Segre embedding, we directly deduce the general case.

Proof. We identify subvarieties of $\mathbb {P}^{n_1}_K \times _K \cdots \times _K \mathbb {P}^{n_k}_K$ with their images under the Segre embedding and iterate the following step: we first choose an irreducible component $V_0$ of $V$ that contains $W$. If $V_0 = W$, we stop. If $V_0 \neq W$, our hypothesis on $W$ implies that we can find a hypersurface of degree at most $d$ in $\mathbb {P}^{\prod _{i=1}^{k}{(n_i+1)}-1}_K$ that contains $W$, but not $V_0$. We then replace $V$ by the intersection of $V_0$ with such a hypersurface. After at most $\dim V - \dim W$ steps, we end up with a variety that contains $W$ as an irreducible component. The theorem then follows from Theorem 2.1.

Proof. In the case where $k = 1$ and $V$ is equidimensional, this follows from Proposition 2.1 in [Reference FaltingsFal91]. Using the Segre embedding, we then directly deduce the general case.

Proof. First of all, we have $\overline {\pi (U)} = \pi (V)$ since $\pi$ is closed. Since $\deg U = \deg V$ by our definition, we can assume without loss of generality that $U = V$. We can also assume that $K$ is algebraically closed and $V$ is irreducible.

We use the following fact: for a linear projection $p: \mathbb {P}^{n}_K\backslash L \to \mathbb {P}^{n-l}_K$ with center $L \subset \mathbb {P}^{n}_K$ of dimension $l-1$ and an irreducible subvariety $X \subset \mathbb {P}^{n}_K$, let $Y$ denote the Zariski closure of $p((\mathbb {P}^{n}_K\backslash L) \cap X)$. Then $\deg Y \leq \deg X$. For a proof of this fact, see [Reference HrushovskiHru01, p. 55]. Note that there $\bar {L}$ should be chosen such that $\bar {L} \cap \bar {V} \subset \theta (V\backslash C)$.

Let $S \subset \{1,\ldots,k\}$ be such that $\pi : \prod _{i=1}^{k}{\mathbb {P}^{n_i}_K} \to \prod _{i \in S}{\mathbb {P}^{n_i}_K}$. The lemma now follows from the above fact together with the commutative diagram

where the horizontal arrows are Segre embeddings and the right vertical arrow is a suitable linear coordinate projection, chosen such that the image of $V$ under the Segre embedding is not contained in its center.

Proof. Let $a, M \in \mathbb {N}$ with $\gcd (a,M) = 1$ be given and fix $\hat {a} \in \hat {\mathbb {Z}}^{\ast }$ such that $\hat {a} \equiv a \mod M$. By Théorème 3 in [Reference WintenbergerWin02], due to Serre [Reference SerreSer00, No. 136, Théorème $2'$], there exists a constant $c = c(A_0,K) \in \mathbb {N}$ such that there exists $\sigma _a \in \operatorname {\mathrm {Gal}}(\bar {\mathbb {Q}}/K)$ acting on $\varprojlim A_0[n] \simeq \hat {\mathbb {Z}}^{2g}$ as multiplication by $\hat {a}^{c}$.

Let $x \in \mathcal {A}(\bar {\mathbb {Q}})$ be a torsion point of order $M$ such that the fiber $\mathcal {A}_{\pi (x)}$ is isogenous to $A_0$. Let $\varphi : A_0 \to \mathcal {A}_{\pi (x)}$ be an isogeny and let $y \in A_0(\bar {\mathbb {Q}})$ be a torsion point such that $\varphi (y) = x$. Choose $N \in \mathbb {N}$ large enough so that $y$ belongs to $A_0[N]$ and $\ker \varphi \subset A_0[N]$. The order $M$ of $x$ is equal to the greatest common divisor of all $n \in \mathbb {N}$ with $ny \in \ker \varphi$. In particular, $M$ divides $N$.

We now choose $\tilde {a} \in \mathbb {N}$ such that $\tilde {a} \equiv \hat {a} \mod N$ and replace $a$ by $\tilde {a}$. The Galois automorphism $\sigma$ will not depend on the choice of $\tilde {a}$, but only on $\sigma _a$ and $\mathcal {A}$. We deduce that $\sigma _a(y) = a^{c(A_0,K)}y$.

If $\lambda _s: \mathcal {A}_s \to \widehat {\mathcal {A}_s}$ denotes the principal polarization on $\mathcal {A}_s$ ($s \in S$), then we have that $\widehat {\varphi } \circ \lambda _{\pi (x)} \circ \varphi$ is a polarization of $A_0$. By Théorème 1.2 in [Reference RémondRém20], which improves and optimizes earlier results by Silverberg in [Reference SilverbergSil92] and by Masser and Wüstholz in [Reference Masser and WüstholzMW93b], every homomorphism between $A_0$ and $\widehat {A_0}$ is defined over a Galois extension of $K$ of degree at most $F(g) = 4\beta (g)6^{2(g-1)}(g!)^{2}$, where $\beta (g) = 3$ if $g \not \in \{2,4,6\}$, while $\beta (2) = 8$, $\beta (4) = 75$, and $\beta (6) = \frac {49}{12}$. Hence, the polarization $\widehat {\varphi } \circ \lambda _{\pi (x)} \circ \varphi$ of $A_0$ is fixed by $\sigma _a^{\circ F(g)!}$.

Since $\sigma _a$ acts on $A_0[N] \supset \ker \varphi$ as multiplication by a natural number that is coprime to $N$, we know that $\sigma _a^{\circ F(g)!}(\ker \varphi ) = \ker \varphi$. Therefore, there is an isomorphism $\iota : \mathcal {A}_{\pi (x)} \to \mathcal {A}_{\sigma _a^{\circ F(g)!}(\pi (x))}$ such that $\sigma _a^{\circ F(g)!}(\varphi ) = \iota \circ \varphi$ (see Theorem 5.13 in [Reference MilneMil17]).

Since $\widehat {\varphi } \circ \lambda _{\pi (x)} \circ \varphi$ is fixed by $\sigma _a^{\circ F(g)!}$ and the polarization $\lambda$ of $\mathcal {A}$ is defined over $K$, we deduce that

\[ \widehat{\varphi} \circ \lambda_{\pi(x)} \circ \varphi = \widehat{\sigma_a^{\circ F(g)!}(\varphi)} \circ \lambda_{\sigma_a^{\circ F(g)!}(\pi(x))} \circ \sigma_a^{\circ F(g)!}(\varphi) = \widehat{\varphi} \circ \hat{\iota} \circ \lambda_{\sigma_a^{\circ F(g)!}(\pi(x))} \circ \iota \circ \varphi. \]

Note that $\widehat {\sigma _a^{\circ F(g)!}(\varphi )} = \sigma _a^{\circ F(g)!}(\widehat {\varphi })$ since dualizing commutes with extending the base field. As $\varphi$ and $\widehat {\varphi }$ are isogenies, it follows that $\lambda _{\pi (x)} = \hat {\iota } \circ \lambda _{\sigma _a^{\circ F(g)!}(\pi (x))} \circ \iota$, so $\mathcal {A}_{\pi (x)}$ and $\mathcal {A}_{\sigma _a^{\circ F(g)!}(\pi (x))}$ are isomorphic as polarized abelian varieties.

Hence, the point $\rho (\pi (x))$ is fixed by $\sigma _a^{\circ F(g)!}$. It follows that the finite set $\rho ^{-1}(\rho (\pi (x)))$ of cardinality at most $M_1$ is permuted by $\sigma _a^{\circ F(g)!}$ and therefore the Galois automorphism $\sigma _a^{\circ F(g)!M_1!}$ fixes $\pi (x)$.

By Théorème 1.2 in [Reference RémondRém20], the isogeny $\varphi$ is defined over a Galois extension of $K(\pi (x))$ of degree at most $F(g)$ with $F(g)$ as above. We deduce that $\sigma = \sigma _a^{\circ M_1!(F(g)!)^{2}}$ fixes $\varphi$ and

\[ a^{B}x = \varphi(a^{B}y) = \varphi(\sigma(y)) = \sigma(\varphi(y)) = \sigma(x) \]

with $B = c(A_0,K)M_1!(F(g)!)^{2}$.

Proof. Let $p \in V(\bar {F})$ be a torsion point that is not contained in any translate of a positive-dimensional abelian subvariety of $A_{\bar {F}}$ that is contained in $V_{\bar {F}}$, and let $N$ denote the order of $p$. We want to show that $N$ is bounded from above as in the proposition.

Let $Y \subset V$ be an equidimensional subvariety such that $p \in Y(\bar {F})$ and $Y$ is a union of irreducible components of $V$. For $d \in \mathbb {N}$ fixed and $t \in \mathbb {Z}$, $t \geq 0$, set (as in [Reference HindryHin88])

\[ Y_t = \bigcap_{j=0}^{t}{[d^{j}]^{-1}(Y)}, \]

where $[d^{j}]$ denotes the multiplication-by-$d^{j}$ morphism on $A$.

Suppose that $d = a^{c}$ for some $a \in \mathbb {N}$ that is coprime to $N$. Our hypothesis then implies that there exists $\sigma _{a,N} \in \operatorname {\mathrm {Gal}}(\bar {F}/F)$ such that $\sigma _{a,N}(q) = dq$ for all torsion points $q \in A(\bar {F})$ of order $N$. Hence, we have that $d^{j} p = \sigma _{a,N}^{\circ j}(p) \in Y(\bar {F})$ for all $j \in \mathbb {Z}$, $j \geq 0$, and therefore $p \in Y_t(\bar {F})$ for all $t \in \mathbb {Z}$, $t \geq 0$.

Suppose, furthermore, that $\dim _p (Y_s)_{\bar {F}} = \cdots = \dim _p (Y_{s+k})_{\bar {F}} = m'$ for some integers $s \geq 1$, $k \geq 0$, and $m' \geq 1$. It follows that there exists an irreducible component $C$ of $(Y_s)_{\bar {F}}$ that contains $p$, has dimension $m'$, and is contained in $(Y_{s+k})_{\bar {F}}$. Hence, we have $C' = [d^{k}](C) \subset (Y_s)_{\bar {F}}$. Since $p' = d^{k} p = \sigma _{a,N}^{\circ k}(p)$ is a Galois conjugate of $p$ over $F$, we have that $\dim _{p'} (Y_s)_{\bar {F}} = \dim _{p} (Y_s)_{\bar {F}}$. Therefore $C'$ is an irreducible component of $(Y_s)_{\bar {F}}$. We deduce from Theorems 2.2 and 2.3 that

(3.1)\begin{equation} \deg [d^{k}](C) = \deg C' \leq (\deg Y)\Big(\max_{j=1,\ldots,s}{\deg [d^{j}]^{-1}(Y)}\Big)^{\dim Y-m'}. \end{equation}

Recall that $\operatorname {\mathrm {Stab}}(C,A_{\bar {F}})$ denotes the stabilizer of $C$ in $A_{\bar {F}}$. As $p$ is not contained in a translate of a positive-dimensional abelian subvariety of $A_{\bar {F}}$ that is contained in $Y_{\bar {F}}$, we have $\dim \operatorname {\mathrm {Stab}}(C,A_{\bar {F}}) = 0$. Since $L$ is symmetric, we have $[d]^{\ast }L \simeq L^{\otimes d^{2}}$ by Proposition 8.7.1 in [Reference Bombieri and GublerBG06]. Hence, we can deduce from the projection formula that

\[ \deg [d^{j}](C) = d^{2jm'}|\operatorname{\mathrm{Stab}}(C,A_{\bar{F}}) \cap \ker [d^{j}]|^{-1}(\deg C) \]

as well as

(3.2)\begin{equation} \deg [d^{j}]^{-1}(Y) = (\deg Y)d^{2j(g-\dim Y)} \end{equation}

for $j \in \mathbb {Z}$, $j \geq 0$.

Combining these two equations with (3.1), we deduce that

(3.3)\begin{equation} d^{2km'}|\operatorname{\mathrm{Stab}}(C,A_{\bar{F}})|^{-1}(\deg C) \leq (\deg Y)^{\dim Y-m'+1}d^{2s(g-\dim Y)(\dim Y-m')}. \end{equation}

Since $C$ is an irreducible component of $\bigcap _{j=0}^{s}{([d^{j}]^{-1}(Y))_{\bar {F}}}$, Theorems 2.2 and 2.3 then imply together with (3.2) that

(3.4)\begin{align} \deg C &\leq (\deg Y)\Big(\max_{j=1,\ldots,s}{\deg [d^{j}]^{-1}(Y)}\Big)^{\dim Y-m'} \nonumber\\ &= (\deg Y)^{\dim Y-m'+1}d^{2s(g-\dim Y)(\dim Y-m')}. \end{align}

Now $\operatorname {\mathrm {Stab}}(C,A_{\bar {F}})$ is a union of irreducible components of

\[ (C-x_1) \cap \cdots \cap (C-x_{m'+1}) \]

for well-chosen $x_1, \ldots, x_{m'+1} \in C(\bar {F})$. Theorem 2.1 implies that

\[ |\operatorname{\mathrm{Stab}}(C,A_{\bar{F}})| = \deg \operatorname{\mathrm{Stab}}(C,A_{\bar{F}}) \leq (\deg C)^{m'+1}. \]

We deduce from this together with (3.3) and (3.4) that

\begin{align*} d^{2km'} &\leq (\deg C)^{m'}(\deg Y)^{\dim Y-m'+1}d^{2s(g-\dim Y)(\dim Y-m')} \\ &\leq (\deg Y)^{(\dim Y-m'+1)(m'+1)}d^{2s(g-\dim Y)(\dim Y-m')(m'+1)}. \end{align*}

We now assume that

(3.5)\begin{equation} d^{2(g-\dim Y)} \geq \deg Y. \end{equation}

Note that $g-\dim Y > 0$ as otherwise $p$ would be contained in a translate of a positive-dimensional abelian subvariety of $A_{\bar {F}}$ that is contained in $Y_{\bar {F}}$.

It then follows that

\[ k \leq \frac{(s+1)(\dim Y-m'+1)(m'+1)(g-\dim Y)}{m'} \leq \frac{(s+1)\Xi}{m'}, \]

where

\[ \Xi = \bigg(\frac{(\dim Y+2)^{2}(g-\dim Y)}{4}\bigg). \]

Induction now shows that $\dim _p (Y_t)_{\bar {F}} = 0$ for some $t \leq t_0$, where $t_0$ is effective and depends only on $g$ and $\dim Y$.

The same holds for any conjugate of $p$ over $F$. By our hypothesis, there are at least $\phi (N)/(2c^{\omega (N)})$ such conjugates. We therefore get a lower bound

(3.6)\begin{equation} \deg Y_t \geq \frac{\phi(N)}{2c^{\omega(N)}}. \end{equation}

Applying (3.2) together with Theorem 2.1 and $t \leq t_0$ to bound the degree of $Y_t = \bigcap _{j=0}^{t}{[d^{j}]^{-1}(Y)}$, we obtain an upper bound

(3.7)\begin{equation} \deg Y_t \leq (\deg Y)^{t_0+1}d^{(g-\dim Y)t_0(t_0+1)} \leq d^{(g-\dim Y)(t_0+1)(t_0+2)}. \end{equation}

By Théorème 11 in [Reference RobinRob83], we have $\omega (N) \leq ({7\log N})/({5 \log \log N})$ if $N \geq 3$. We can also estimate

\[ \phi(N) = N\prod_{p' | N}{\frac{p'-1}{p'}} \geq N \prod_{j=2}^{\omega(N)+1}{\frac{j-1}{j}} = \frac{N}{\omega(N)+1} \geq \frac{N}{2\log N +1}, \]

where the product runs over all primes $p' \in \mathbb {N}$ that divide $N$. For $N \geq 3$, it then follows from combining this with (3.6) and (3.7) that

\[ \frac{N^{1-({7\log c})/({5 \log \log N})}}{2(2\log N+1)} \leq d^{(g-\dim Y)(t_0+1)(t_0+2)}. \]

If $N > \exp (c^{ {8}/{5}}) \geq 2$, which we will assume from now on, this implies that

\[ \frac{N^{{1}/{8}}}{2(2\log N+1)} \leq d^{(g-\dim Y)(t_0+1)(t_0+2)}. \]

As the function $x \mapsto x^{ {1}/{16}} \cdot (2(2\log x+1))^{-1}$ ($x \geq 1$) attains its minimum at $x = \exp (31/2)$, we deduce that

\[ \frac{\exp({31}/{32})}{64}N^{{1}/{16}} \leq d^{(g-\dim Y)(t_0+1)(t_0+2)}. \]

We want to obtain a contradiction for $N$ large enough, but we have to make sure that (3.5) is satisfied. Recall that $d = a^{c}$ for $a \in \mathbb {N}$ coprime to $N$ and set $b = c(g-\dim Y) > 0$. In order to obtain a contradiction, it suffices to find $a \in \mathbb {N}$, coprime to $N$, such that

\[ (\deg Y)^{{(t_0+1)(t_0+2)}/{2}} \leq a^{b(t_0+1)(t_0+2)} < \frac{N^{{1}/{16}}}{25}. \]

Simplifying the problem, we may look for a prime number $a$ that does not divide $N$ and satisfies

\[ (\deg Y)^{{1}/{(2b)}} \leq a < \bigg(\frac{N^{{1}/{16}}}{25}\bigg)^{{1}/{(b(t_0+1)(t_0+2))}}. \]

By Corollary 1 of Theorem 2 in [Reference Rosser and SchoenfeldRS62], this is possible if $N \geq (25\times 17^{b(t_0+1)(t_0+2)})^{16}$, which we will assume from now on, and

\[ \omega(N) + (\deg Y)^{{1}/{(2b)}} < \bigg(\frac{N^{{1}/{16}}}{25}\bigg)^{{1}/{(b(t_0+1)(t_0+2))}} \frac{b(t_0+1)(t_0+2)}{{(\log N)}/{16}-\log 25}, \]

which is a consequence of the simpler inequality

\[ \frac{\log N}{16}(\omega(N) + (\deg Y)^{{1}/{(2b)}}) < \bigg(\frac{N^{{1}/{16}}}{25}\bigg)^{{1}/{(b(t_0+1)(t_0+2))}}. \]

Thanks to the above bound for $\omega (N)$ and $N \geq 25^{16} \geq e^{e}$, this inequality in turn follows from

\[ (\log N)^{2}(\deg Y)^{{1}/{(2b)}} < \bigg(\frac{N^{{1}/{16}}}{25}\bigg)^{{1}/{(b(t_0+1)(t_0+2))}}. \]

Recall that $Y$ is a union of irreducible components of $V$ and therefore $\deg Y \leq \deg V$. Furthermore, if $\epsilon \in (0,1)$ and $N \geq \exp ( {1}/{\epsilon ^{2}}) \geq \epsilon ^{-({1}/{\epsilon })}$, then we have

\[ \log N \leq \frac{N^{\epsilon}}{\epsilon} \leq N^{2\epsilon}. \]

This implies that we get a contradiction as soon as $N \geq \exp ((128b(t_0+1)(t_0+2))^{2})$ and

\[ N^{{1}/{(32b(t_0+1)(t_0+2))}}(\deg V)^{{1}/{(2b)}} < \bigg(\frac{N^{{1}/{16}}}{25}\bigg)^{{1}/{(b(t_0+1)(t_0+2))}}. \]

This last inequality is equivalent to $N > 25^{32}(\deg V)^{16(t_0+1)(t_0+2)}$.

Since $\dim Y \in \{0,\ldots,g\}$, all this together implies that there exists an effective constant $\gamma (g)$, depending only on $g$, such that we get a contradiction if $N > \max \{\exp (\gamma (g)c^{2}),(\deg V)^{\gamma (g)}\}$. Thus, we have established the proposition.

Proof. We first perform a quasi-finite dominant base change $S' \to S$ such that $S'$ is smooth and irreducible and every irreducible component of $\mathcal {V}_\xi$ is defined over $\bar {\mathbb {Q}}(S') \subset \overline {\bar {\mathbb {Q}}(S)}$. Set $\mathcal {A}' = \mathcal {A} \times _{S} S'$. Let $\mathcal {V}'$ be an irreducible component of $\mathcal {V} \times _{S} S' \hookrightarrow \mathcal {A}'$ that dominates $S'$ (and hence dominates $\mathcal {V}$).

If $\zeta$ is a geometric generic point of $S'$, then $\operatorname {\mathrm {Stab}}(\mathcal {V}'_{\zeta },\mathcal {A}'_{\zeta })$ must be finite. Otherwise it would contain a positive-dimensional abelian subvariety of $\mathcal {A}'_{\zeta }$, which we identify with $\mathcal {A}_{\xi }$, but as all abelian subvarieties of $\mathcal {A}'_{\zeta }$ are defined over $\bar {\mathbb {Q}}(S)$, this abelian subvariety would be contained in the stabilizer of $\mathcal {V}_\xi$, which could therefore not be finite. Furthermore, the generic fiber of $\mathcal {V}'$ is irreducible by § 2.1.8 of Chapter 0 of [Reference GrothendieckEGA1] and hence also $\mathcal {V}'_\zeta$ is irreducible by our choice of $S'$.

If the union of all translates of positive-dimensional abelian subvarieties of $\mathcal {A}_s$ that are contained in $\mathcal {V}_s$ for some $s \in S(\bar {\mathbb {Q}})$ is Zariski dense in $\mathcal {V}$, then the union of all translates of positive-dimensional abelian subvarieties of $\mathcal {A}'_{s'}$ that are contained in $\mathcal {V}'_{s'}$ for some $s' \in S'(\bar {\mathbb {Q}})$ is Zariski dense in $\mathcal {V}'$. So we can replace $\mathcal {A}$ and $\mathcal {V}$ by $\mathcal {A}'$ and $\mathcal {V}'$ and assume without loss of generality that $\mathcal {V}_\xi$ is irreducible.

Let $N \in \mathbb {N}$ be a natural number that is larger than the order of $\operatorname {\mathrm {Stab}}(\mathcal {V}_\xi,\mathcal {A}_\xi )$. There are finitely many irreducible subvarieties $\mathcal {T}_1, \ldots, \mathcal {T}_R \subset \mathcal {A}$ such that each $\mathcal {T}_i$ dominates $S$ and the union of the $\mathcal {T}_i$ ($i=1,\ldots,R$) is equal to the set of torsion points of order $N$ on $\mathcal {A}$: first of all, every irreducible component of the pre-image of $\epsilon (S)$ under the multiplication-by-$N$ morphism $[N]$ dominates $S$ by Proposition 2.3.4(iii) in [Reference GrothendieckEGA4] since $[N]$ is étale, so flat (see [Reference MilneMil86, Proposition 20.7]). Therefore, every irreducible component of $[N]^{-1}(\epsilon (S))$ is of dimension $\dim S$. The same holds for any $M \in \mathbb {N}$ that divides $N$. Furthermore, $[N]^{-1}(\epsilon (S))$ is smooth as $[N]$ is étale and $S$ is smooth. Hence, no two distinct irreducible components of $[N]^{-1}(\epsilon (S))$ intersect each other. So every irreducible component of $[N]^{-1}(\epsilon (S))$ is either contained in $\bigcup _{M\,|\,N, M \neq N}[M]^{-1}(\epsilon (S))$ or disjoint from it and our claim follows.

We now consider $\mathcal {W}_i = \mathcal {V} \cap (\mathcal {V}+\mathcal {T}_i)$ for $i \in \{1,\ldots,R\}$. If this variety were equal to $\mathcal {V}$, then we would have $\mathcal {V} \subset \mathcal {V}+\mathcal {T}_i$ and so $\mathcal {V}_\xi \subset \mathcal {V}_\xi +(\mathcal {T}_i)_\xi$. For dimension reasons and thanks to the irreducibility of $\mathcal {V}_\xi$, we would get that $\mathcal {V}_\xi = t+\mathcal {V}_\xi$ for a torsion point $t \in \mathcal {A}_\xi (\overline {\bar {\mathbb {Q}}(S)})$ of order $N$. This contradicts our choice of $N$. So $\mathcal {W}_i \subsetneq \mathcal {V}$.

On the other hand, each positive-dimensional abelian variety contains a point of order $N$, so the union of all translates of positive-dimensional abelian subvarieties of $\mathcal {A}_s$ that are contained in $\mathcal {V}_s$ for some $s \in S(\bar {\mathbb {Q}})$ is contained in $\bigcup _{i=1}^{R}{\mathcal {W}_i}$. As every $\mathcal {W}_i$ is a proper closed subset of $\mathcal {V}$ and $\mathcal {V}$ is irreducible, the lemma follows.

Proof. Suppose that $\sigma \in \operatorname {\mathrm {Gal}}(\bar {K}/K)$ acts as a homothety on the torsion of $(E_0)_{\bar {K}}$ and let $s \in Y(2)(\bar {K})$ be such that $\mathcal {E}_s$ is isogenous to $(E_0)_{\bar {K}}$. Let $\varphi : (E_0)_{\bar {K}} \to \mathcal {E}_{s}$ be any isogeny.

Since $\sigma$ acts as a homothety on the torsion of $(E_0)_{\bar {K}}$, it fixes $\ker \varphi$. By Theorem 5.13 in [Reference MilneMil17], there exists an isomorphism $\psi : \mathcal {E}_s \to \mathcal {E}_{\sigma (s)}$ such that $\sigma (\varphi ) = \psi \circ \varphi$. Let $p_0 \in (\mathcal {E}_{s})_{\operatorname {\mathrm {tors}}}$ be a point of order $2$ and let $q_0 \in (E_0)_{\operatorname {\mathrm {tors}}}$ be a pre-image of $p_0$ under $\varphi$. Since $\sigma$ acts as a homothety on the torsion of $(E_0)_{\bar {K}}$, we have $\sigma (q_0) = b_0q_0$ for some odd $b_0 \in \mathbb {N}$. Since $p_0$ has order $2$, it follows that

\[ \psi(p_0) = \psi(b_0p_0) = (\psi \circ \varphi)(b_0q_0) = \sigma(\varphi)(\sigma(q_0)) = \sigma(p_0). \]

Because of the nature of the Legendre family, we deduce that $\sigma (s) = s$, which implies that $\psi$ is an automorphism of $\mathcal {E}_s$ that restricts to the identity on the $2$-torsion of $\mathcal {E}_s$. It follows that $\psi = \pm \operatorname {id}_{\mathcal {E}_s}$ and $\sigma ^{\circ 2}(\varphi ) = \varphi$.

Suppose now that $\sigma$ acts on the torsion of $(E_0)_{\bar {K}}$ as multiplication by $b \in \hat {\mathbb {Z}}^{\ast }$ and let $p \in (\mathcal {E}_{s})_{\operatorname {\mathrm {tors}}}$. There exists $q \in (E_0)_{\operatorname {\mathrm {tors}}}$ such that $\varphi (q) = p$. By the above, we know that $\sigma ^{\circ 2}(s) = s$, $\sigma ^{\circ 2}(\varphi ) = \varphi$, and

\begin{align*} \sigma^{\circ 2}(p) &= \sigma^{\circ 2}(\varphi)(\sigma^{\circ 2}(q)) \\ &= \varphi(b^{2}q) = b^{2}\varphi(q) = b^{2}p. \end{align*}

The lemma follows.

Proof. In the non-CM case, the theorem follows from the improved version of Corollary 9.3 in [Reference LombardoLom15] that is mentioned in Remark 9.4 in the same paper. The result by Gaudron and Rémond that is needed for this improvement and that was still unpublished when [Reference LombardoLom15] appeared is Corollaire 17.5 in [Reference Gaudron and RémondGR20].

In the CM case, let $L \subset \bar {K}$ denote the imaginary quadratic field such that $(\operatorname {\mathrm {End}} (E_0)_{\bar {K}}) \otimes _{\mathbb {Z}} \mathbb {Q} \simeq L$. Since $[KL:L(j(E_0))] \leq [K:\mathbb {Q}(j(E_0))]$, the theorem follows from Corollary 1.5 in [Reference Bourdon and ClarkBC20].

Proof. For (1), let $l_i$ denote the class modulo rational equivalence of the pull-back of a line in $\mathbb {P}^{2}_K$ to $(\mathbb {P}^{2}_K)^{g}$ under projection to the $i$th factor ($i=1,\ldots,g$). If $[B]$ denotes the class modulo rational equivalence of $B \subset (\mathbb {P}_K^{2})^{g}$, then $[B] \cdot l_i^{\cdot 2} = 0$ for all $i$ and so

\[ \deg B = [B] \cdot (l_1+\cdots+l_g)^{\cdot (g-k)} = (g-k)!\sum_{I \subset \{1,\ldots,g\},|I| = g-k}{[B]\cdot\prod_{i \in I}{l_i}}. \]

If $\pi _I: E^{g} \to E^{g-k}$ denotes the projection associated to $I \subset \{1,\ldots,g\}$, then the projection formula implies that

\[ [B]\cdot\prod_{i \in I}{l_i} = \begin{cases} 3^{g-k}\#(\ker \pi_I|_B) & \text{if $\pi_I|_B$ is finite},\\ 0 & \text{otherwise.} \end{cases} \]

The formula from the lemma now follows.

For (2), set

\[ \Lambda = \{(a_1,\ldots,a_g) \in \mathbb{Z}^{g}; a_1x_1+\cdots+a_gx_g = 0_E \mbox{ for all } (x_1,\ldots,x_g) \in B(\bar{K})\}. \]

Since $\operatorname {\mathrm {End}} E_{\bar {K}} \simeq \mathbb {Z}$, the free abelian group $\Lambda$ has rank $k$. Since $B$ is irreducible, we have $(\mathbb {Q} \cdot \Lambda ) \cap \mathbb {Z}^{g} = \Lambda$. Choosing a basis of $\Lambda$ yields the rows of a $k \times g$ matrix $M_B$ such that $B$ is an irreducible component of the kernel of the homomorphism from $E^{g}$ to $E^{k}$ induced by $M_B$. Since $(\mathbb {Q} \cdot \Lambda ) \cap \mathbb {Z}^{g} = \Lambda$, the rows of $M_B$ can be completed to a basis of $\mathbb {Z}^{g}$. It follows that the kernel of the homomorphism induced by $M_B$ is isomorphic to $E^{g-k}$ and hence equal to $B$.

Let $D$ denote the ($k$-dimensional) volume of the parallelotope spanned in $\mathbb {R} \cdot \Lambda$ by the rows of $M_B$. Let $M_B^{t}$ denote the transpose of $M_B$. Then $D^{2}$ is equal to the Gram determinant $\det (M_BM_B^{t})$. The Cauchy–Binet formula then implies together with part (1) that

\[ D = \sqrt{\sum_{\Delta}{\Delta^{2}}} \leq \sqrt{\deg B}, \]

where the sum runs over the maximal minors of $M_B$.

Let $\lambda _i$ denote the $i$th successive minimum of $\Lambda$ in $\mathbb {R} \cdot \Lambda$ with respect to the distance function induced by the Euclidean distance on $\mathbb {R}^{g}$ ($i=1,\ldots,k$). We have $\lambda _i \geq 1$ for all $i$ and therefore Theorem V in § VIII.4.3 of [Reference CasselsCas97] implies that $\lambda _i \leq 2^{k}\nu _k^{-1}D$ for all $i$, where $\nu _k$ denotes the volume of the unit ball in $\mathbb {R}^{k}$.

By the corollary in § VIII.5.2 of [Reference CasselsCas97], we can then find a basis $\{v_1,\ldots,v_k\}$ of $\Lambda$ such that the Euclidean norm $\lVert v_i \rVert$ of $v_i$ satisfies

\[ \lVert v_i \rVert \leq k2^{k}\nu_k^{-1}D, \quad i=1,\ldots,k. \]

Thus, we can choose $M_B$ such that all its entries are at most $c(g)\sqrt {\deg B}$ in absolute value. The lemma follows.

Proof. We fix an algebraic closure $\bar {K}$ of $K$.

For (1), if $[X_i:Y_i:Z_i]$ are projective coordinates on the $i$th $\mathbb {P}^{2}_K$-factor ($i=1,2,3$), then the polynomial is equal to

\[ \begin{vmatrix} X_1 & Y_1 & Z_1 \\ X_2 & Y_2 & Z_2 \\ X_3 & -Y_3 & Z_3 \end{vmatrix} = X_1Y_2Z_3-Y_1X_2Z_3+X_1Z_2Y_3-Z_1X_2Y_3+Y_1Z_2X_3-Z_1Y_2X_3. \]

For (2), let $[\Lambda :M]$ be projective coordinates on $\mathbb {P}^{1}_K$ and let $[X_i:Y_i:Z_i]$ be projective coordinates on the $i$th $\mathbb {P}^{2}_K$-factor ($i=1,2$). If $n = 0$, the polynomial $Z_2$ has the desired property. Hence, we can assume without loss of generality that $n \neq 0$. Since changing the sign of $Y_2$ sends $\Gamma _n$ to $\Gamma _{-n}$, we can even assume that $n > 0$.

We know that there exist polynomials $A_n, B_n \in \mathbb {Q}[X,\Lambda ]$ such that for $(\lambda,[x:y:1]) \in \mathcal {E}(\bar {K}) \subset \mathbb {A}^{1}(\bar {K}) \times \mathbb {P}^{2}(\bar {K})$, we have $B_n(x,\lambda ) = 0$ if and only if $x$ is the abscissa of a non-zero torsion point of $\mathcal {E}_{\lambda }$ of order dividing $n$, and

\[ [n](\lambda,[x:y:1]) = (\lambda,[A_n(x,\lambda)/B_n(x,\lambda):\ast:1]) \]

if $B_n(x,\lambda ) \neq 0$. Furthermore, we have $\deg _X A_n = n^{2}$, $\deg _X B_n = n^{2}-1$, and $A_n$ is monic as a polynomial in $X$. For all of this, see [Reference Masser and ZannierMZ13].

It follows that $\Gamma _n$ is an irreducible component of the intersection in $\mathbb {P}^{1}_K \times _K (\mathbb {P}^{2}_K)^{2}$ of $\mathcal {E}^{(2)}_K$ with the zero locus of

\[ Z_2\tilde{A}_n(X_1,Z_1,\Lambda,M) - X_2\tilde{B}_n(X_1,Z_1,\Lambda,M), \]

where

\[ \tilde{A}_n(X_1,Z_1,\Lambda,M) = Z_1^{d}M^{e}A_n\bigg(\frac{X_1}{Z_1},\frac{\Lambda}{M}\bigg) \]

and

\[ \tilde{B}_n(X_1,Z_1,\Lambda,M) = Z_1^{d}M^{e}B_n\bigg(\frac{X_1}{Z_1},\frac{\Lambda}{M}\bigg) \]

for $d = \max \{\deg _X A_n, \deg _X B_n\} = n^{2}$ and $e = \max \{\deg _\Lambda A_n, \deg _\Lambda B_n\}$.

By Lemma 2.2 in [Reference Masser and ZannierMZ13], we have $\Lambda ^{n^{2}}A_n(\Lambda ^{-1}X,\Lambda ^{-1}) = A_n(X,\Lambda )$ and $\Lambda ^{n^{2}-1}B_n(\Lambda ^{-1}X, \Lambda ^{-1})= B_n(X,\Lambda )$. This implies that $e \leq n^{2}$. The lemma follows.