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Let $E$ be an elliptic curves defined over a number field $K$. Consider the $\ell$-adic representation attached to $E$

$$ \rho_{\ell}:\mathrm{Gal}(\overline{K}/K) \longrightarrow \mathrm{Aut}(V_{\ell})$$

With $\overline{K}$ an algebraic closure of $K$ and $V_{\ell}=T_{\ell} \otimes_{\mathbb{Z}_{\ell}} \mathbb{Q}_{\ell}$ with $T_{\ell}$ the Tate module of $E$.

My question is: why the representation $\rho_{\ell}$ is rational?

By criterion of Néron-Ogg-Shafarevich I understand that for all finite place $v$ such that $v$ not divides $\ell$ and $E$ has a good reduction at $v$, then $\rho_{\ell}$ is unramified. In particular the set of finite place of $K$ where $E$ has a bad reduction is finite and we denote it with $S_{\mathrm{bad}}$. So, if $S_{\ell}$ denote the set of finite place that not divide $\ell$, then for all finite place $v\notin S_{\mathrm{bad}}\cup S_{\ell}$ we have $\rho_{\ell}$ is unramified and $S_{\mathrm{bad}} \cup S_{\ell}$ is finite.

But I don't understand why the coefficients of the polynomial $\mathrm{det}(1-F_{v,\rho_{\ell}}T)$ are rationals, where $T$ is an indeterminate and $F_{v,\rho_{\ell}}$ is the conjugacy class in $\mathrm{Aut}(V_{\ell})$ of $\rho_{\ell}(F_{w})$, $F_w$ the Frobenius element corresponding to a place $w$ of $\overline{K}$ extending $v$.

I write a reasoning that I have done:

If E has a good reduction at $v$ and $v \nmid \ell$ ,then I have the isomorphism $E[\ell](K) \simeq \widetilde{E}[\ell](k_v)$ where $k_v$ is the residue field of the completion of $K$ with respect to $v$. So the action of $\rho_{\ell}(F_w)$ on $E[\ell](K)$ correspond to the Frobenius endomorphism $\phi$ on $\widetilde{E}[\ell](k_v)$.

Since $\mathrm{det}(1-F_{v,\rho_{\ell}})=\mathrm{det}(1-\rho_{\ell}(F_{w}))= 1-\mathrm{Tr}(\rho_{\ell}(F_{w}))T + \mathrm{det}(\rho_{\ell}(F_{w}))T^2$, can I deduce from my reasoning that $\mathrm{Tr}(\rho_{\ell}(F_{w}))=\mathrm{Tr}(\phi)$ and $\mathrm{det}(\rho_{\ell}(F_w))=\mathrm{det}(\phi)$?

If it is true then we are done because the characteristic polynomial would be $1-\mathrm{Tr}(\phi)T+\mathrm{det}(\phi)T^2$ so would be independent from $\ell$ and it is known that $\mathrm{Tr}(\phi),\mathrm{det}(\phi)$ are integers (cf. Silverman ''the arithmetic of elliptic curves", chap. V, theorem 2.3.1)

wythagoras
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user253407
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1 Answers1

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I think I can help clear up two points of confusion in your question.

At one point you switched from working with $T_\ell(E)$ to working with $E[\ell]$. So your Frobenius is actually acting on a characteristic $\ell$ object, so its characteristic polynomial cannot possibly be a characteristic $0$ thing. You really need to work $\ell$-adically if you want the rationality statement to even make sense.

The first part of what you're trying to do to identify the characteristic polynomial of a Frobenius at an unramified prime $p$ acting on $V_\ell(E)$ with the characteristic polynomial of the (finite field!) Frobenius acting on $V_\ell(\overline{E})$, where $\overline{E}$ is the reduction of $E$ mod $p$. Remark that now it makes sense to claim that these two characteristic polynomials are equal because they are both polynomials in $\mathbb Q_\ell[T]$. Showing that these two polynomials are equal is not hard; it essentially boils down to the fact that $V_\ell(E) \cong V_\ell(\overline E)$ as modules over the decomposition group at $p$ (because the representation is unramified at $p$).

The hard part is actually then proving (as in Silverman) that the characteristic polynomial of Frobenius on $V_\ell(\overline{E})$ is rational. This is a statement purely about $\overline E$ and having nothing to do with global fields.

Bruno Joyal
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