Restriction of Scalars

Let $E$ be an elliptic curve over  $\mathbb{Q}$, and let $K$ be Galois over  $\mathbb{Q}$. We recall the definition of restriction of scalars, and prove that the kernel of a certain morphism induced by $\tr_{K/\mathbb{Q}}$ is geometrically connected.

The restriction of scalars $R=\Res_{K/\mathbb{Q}}(E_K)$ is an abelian variety over  $\mathbb{Q}$ of dimension $[K:\mathbb{Q}]$, which is characterized by the following universal property: There is a functorial group isomorphism $R(S) \cong E_K(S_K)$, where $S$ varies over all $\mathbb{Q}$-schemes. More explicitly, as $\Gal(\overline{\mathbb{Q}}/\mathbb{Q})$-modules we have

$$R(\overline{\mathbb{Q}}) = E(\overline{\mathbb{Q}}\otimes K) \cong E(\overline{\mathbb{Q}})\otimes _{\mathbb{Z}} \mathbb{Z}[\Gal(K/\mathbb{Q})],$$

where $\tau\in \Gal(\overline{\mathbb{Q}}/\mathbb{Q})$ acts on $\sum P_\sigma\otimes \sigma \in E(\overline{\mathbb{Q}})\otimes _{\mathbb{Z}}\mathbb{Z}[\Gal(K/\mathbb{Q})]$ by

$$\tau\left(\sum P_\sigma\otimes \sigma\right) = \sum \tau(P_\sigma)\otimes \sigma\tau_{\vert K}.$$

Moreover, the $L$-series of $R$ is $\prod_{a=1}^{n} L(E,\chi^a,s)$, and $R$ has good reduction at all $p\nmid \ell\cdot N$.

Proposition 2.1   The identity map induces a closed immerion $\iota: E\hookrightarrow R$, and the trace $\tr:K\rightarrow \mathbb{Q}$ induces a surjection $\tr:R\rightarrow E$ whose kernel is geometrically connected.

Proof. The existence of $\iota$ and $\tr$ follows from Yoneda's lemma. The map $\iota$ is induced by the functorial inclusion $E(S)\hookrightarrow E_K(S_K)=R(S)$, so $\iota$ is injective. The $\tr$ map is induced by the usual functorial trace map on points $R(S)=E_K(S_K)\xrightarrow{\tr} E(S)$.

To verify that $\ker(\tr)$ is geometrically connected, we base extend to  $\overline{\mathbb{Q}}$. First, note that

$$R_{\overline{\mathbb{Q}}} \approx E_{\overline{\mathbb{Q}}}\times \cdots \times E_{\overline{\mathbb{Q}}}.$$

After this base extension, the trace map is the summation map:

$$+: E_{\overline{\mathbb{Q}}} \times \cdots \times E_{\overline{\mathbb{Q}}} \longrightarrow E_{\overline{\mathbb{Q}}}.$$

Let $n=[K:\mathbb{Q}]$. The map defined by

$$\left(a_1,\ldots a_{n-1}\right) \mapsto \left(a_1,a_2,\ldots a_{n-1},-\sum_{i=1}^{n-1} a_i\right),$$

is an isomorphism from $E_{\overline{\mathbb{Q}}}^{\times (n-1)}$ to $\ker(+)=\ker(\tr_{\overline{\mathbb{Q}}})$. Thus $\ker(\tr_{\overline{\mathbb{Q}}})$ is a product of connected varieties, hence connected. $\qedsymbol$

Corollary 2.2   Let $n=[K:\mathbb{Q}]$. Then

$$(\iota(E)\cap \ker(\tr))(\overline{\mathbb{Q}}) \cong E[n](\overline{\mathbb{Q}})\approx (\mathbb{Z}/n\mathbb{Z})^2.$$

(The rightmost map is an isomorphism of groups, not Galois modules.)

Proof. Since the map

$$\mathbb{Q}\hookrightarrow K\xrightarrow{\tr} \mathbb{Q}$$

is multiplication by $n$, the composite map

$$E \hookrightarrow R \longrightarrow E$$

is also multiplication by $n$. The corollary now follows since $\iota(E)\cap \ker(\tr)$ is the kernel of  $\tr\circ\iota$, which equals $[n]$. It is elementary that $E[n](\overline{\mathbb{Q}})\approx (\mathbb{Z}/n\mathbb{Z})^2$, where we have, of course, forgotten the action of $\Gal(\overline{\mathbb{Q}}/\mathbb{Q})$. $\qedsymbol$