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$\newcommand{\End}{\operatorname{End}}$Consider an elliptic curve E defined over $\mathbb Q$ such that $\End(E_{\bar{\mathbb Q}})\neq \mathbb Z$ (i.e. with CM).

Let $F/\mathbb Q$ be finite Galois extension such that $\End(E_F)=\End(E_{\bar{\mathbb Q}})$ (i.e. all endomorphisms of $E$ are defined over $F$).

Let $A=\operatorname{Res}_{F/\mathbb Q}(E_F)$ be the abelian variety over $\mathbb Q$ given by the Weil restriction of the elliptic curve $E_F$ over $F$, and write $\End^0(-)$ for $\End(-)\otimes \mathbb Q$.

What is $\End^0(A)$?

Can one compute $\End^0(A)$ in terms of the imaginary quadratic field $K=\End^0(E_F)$?

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By the universal property of the Weil restriction for abelian varieties, we have \begin{equation*} \mathrm{Hom}_{\mathbb{Q}}(A,A) = \mathrm{Hom}_{\mathbb{Q}}(A,\mathrm{Res}_{F/\mathbb{Q}}(E_F)) \cong \mathrm{Hom}_F(A_F,E_F). \end{equation*} A general property of the Weil restriction is that it commutes with base change in the following sense: \begin{equation*} \mathrm{Res}_{S'/S}(X) \times_S T \cong \mathrm{Res}_{(S' \times_S T)/T}(X \times_{S'} (S' \times_S T)) \end{equation*} where $S'$ and $T$ are $S$-schemes. This can be checked for the functors of points, thus it holds when the above Weil restrictions exist as schemes (or group schemes, or abelian schemes). In the present situation, this gives \begin{equation*} A_F \cong \mathrm{Res}_{(F \otimes F) /F}(E_{F \otimes F}). \end{equation*} Now let $G=\mathrm{Gal}(F/\mathbb{Q})$. The algebra $F \otimes F$ is isomorphic to $\prod_G F$ via the map $x \otimes y \mapsto (x \sigma(y))_{\sigma \in G}$. This means that $E_{F \otimes F}$ is just the disjoint union of finitely many copies of $E_F$ indexed by $G$, and consequently that $A_F$ is isomorphic to $\prod_G E_F$.

It follows that $\mathrm{End}^0(A)$ is isomorphic to $\prod_G K$ as a $\mathbb{Q}$-vector space.

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    $\begingroup$ Thanks so much! One thing that confuses me: Given this structure of the Q-algebra End^0(A) as a product of [F:Q] copies of the field K/Q, it now follows (since A has dimension [F:Q]) that A/Q is isogenous to a product of elliptic curves E_i over Q with End^0(E_i)=K? (I think I make a wrong argument here, because End(B)=Z for any elliptic curve B/Q) $\endgroup$
    – Hensel
    Commented Jan 1, 2020 at 12:35
  • $\begingroup$ @Hensel Thank you for your comment, indeed there is something wrong in the last paragraph. $\mathrm{End}^0(A)$ is isomorphic to $\prod_G K$ as an abelian group (or $\mathbb{Q}$-vector space) but not as a ring, otherwise $A$ would split (up to isogeny) as a product of elliptic curves over $\mathbb{Q}$, which is impossible as you point out. One needs to work out how the ring structure transports in $\mathrm{Hom}(A_F,E_F)$. My (naïve) guess is that one gets the group algebra $K[G]$ but I'm not sure how to prove it. $\endgroup$ Commented Jan 1, 2020 at 14:02
  • $\begingroup$ One thing one should be careful is what the first equation in my answer means. In the context of schemes, the Weil restriction is defined as the scheme representing a certain contravariant functor $\mathrm{Sch}/S \to \mathrm{Set}$. In the special case of abelian varieties, the Weil restriction also represents the analogous functor from the category of abelian $S$-schemes to abelian groups. The first equation in my answer holds in both categories. $\endgroup$ Commented Jan 1, 2020 at 14:10
  • $\begingroup$ Thank you very much for your clarifications and excellent remarks. You now completely answered my question. $\endgroup$
    – Hensel
    Commented Jan 2, 2020 at 10:50
  • $\begingroup$ Just to add a standard reference for these kind of arguments (and for the first comment by Hensel), one can look at the paper of Milne, "On the arithmetic of abelian varieties" (jmilne.org/math/articles/1972a.pdf) $\endgroup$ Commented Jan 7, 2020 at 10:59

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