Suppose that $K/\mathbb{Q}_p$ is a finite extension and $k_K$ the residue field of $K$. Let $A/K$ be an abelian variety with good reduction. Suppose that $E\to\mathrm{End}^0_K(A)$ is an inclusion of a number field that sends $1$ to the identity.

Denote by $T_\ell A$ the Tate module for a prime $\ell\neq p$ and let $V_\ell A=T_\ell A\otimes_{\mathbb{Z}_\ell}\mathbb{Q}_\ell$. The Galois representation $$\rho_\ell:\mathrm{Gal(\bar{K}/K)}\to \mathrm{Aut}(V_\ell A)$$ is then $E_\ell=E\otimes \mathbb{Q}_\ell$-linear.

We can decompose $E_\ell=\prod_\lambda E_\lambda$ where $\lambda$ run through the places of $E$ dividing $\ell$ and $E_\lambda$ is the corresponding completion of $E$. Then by $E_\ell$-linearity, $V_\ell A=\prod_\lambda V_\lambda$ as $E_\ell[\mathrm{Gal}(\bar{K}/K)]$-modules.

Let $\mathrm{Frob}_K$ be a lift of the Frobenius element. I want to prove that each $E_\lambda$-characteristic polynomial of $\rho_\lambda(\mathrm{Frob}_K)$ is the image of some common polynomial $P_0\in E[X]$ via $E[X]\hookrightarrow E_\lambda[X]$.

I believe this follows from a result by Shimura from "Algebraic Number Fields and Symplectic Discontinuous Groups", Prop. 11.09, but I want to understand the argument of his proof.

It is well-known that the action of the lift of the Frobenius element is obtained via the Frobenius endomorphism $\pi$ of the reduction $\tilde{A}/k_K$, so the $\mathbb{Q}_\ell$-characteristic polynolmial of $\rho_\ell(\mathrm{Frob}_K)$ has rational coefficients. Then, we need to find an $E[\pi]$-module $U$ so that $V_\ell A\simeq U\otimes \mathbb{Q}_\ell$ as $E[\pi]\otimes \mathbb{Q}_\ell$-modules. But how to get this $U$?


Let's instead decompose the $\ell$-adic Tate module, tensored up with $\overline{\mathbb Q}_\ell$, as a product over embeddings of $E$ into $\overline{\mathbb Q}_\ell$ of a $ \overline{\mathbb Q}_\ell$-vector space with an action of $\pi$. On each such vector space $\pi$ admits a characteristic polynomial, and you want to know that these characteristic polynomials are all conjugate under the action of $\operatorname{Gal}(\overline{\mathbb Q}_{\ell}|\mathbb Q)$ permuting the embeddings.

This follows from two facts:

  1. The determinant of every element of $E[\pi]$ acting on this vector space is Galois-invariant, because it equals the degree of the corresponding rational endomorphism and hence is a rational number.

  2. The determinant of $\pi$ minus an element of $E$ can be calculated as the product of these different characteristic polynomials applied to the different embeddings of the element of $E$.

  3. From the determinant, we can recover the individual factors, because we can write the determinant as a polynomial in the element of $E$ under different embeddings into $\overline{\mathbb Q_\ell}$, which we treat as independent random variables, and then factor this polynomial.

Because the determinant is Galois-invariant, and we can use it to calculate the individual polynomials, the polynomials must be appropriately equivariant under the action of Galois.

Constructing a module like you suggest might be difficult, because, as Serre's example of a supersingular elliptic curve shows, it is impossible to do so canonically.

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  • $\begingroup$ in the first line, when you say $\text{tensored up with $\bar{\mathbb{Q}_l}$}$, what does mean it? I am new guest in this topic and have a fear with tensoring. Can you please the explain essence and necessity when we are tensoring with Tate module or something else like above case? $\endgroup$ – M. A. SARKAR Jan 20 at 11:56
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    $\begingroup$ @M.A.SARKAR It just means we find a basis of the Tate module and make a vector space with the same basis over $\overline{\mathbb Q}_\ell$. $\endgroup$ – Will Sawin Jan 20 at 13:21
  • $\begingroup$ Thank you very much for your nice explanation. You mean we construct a new vector space through the tensor product keeping same basis but replacing the "previous field" with a "new field" containing the "previous field". In this case the new field is $\overline{ \mathbb{Q}_l}$. Is this what you are saying? $\endgroup$ – M. A. SARKAR Jan 20 at 14:36
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    $\begingroup$ @M.A.SARKAR Correct. $\endgroup$ – Will Sawin Jan 20 at 16:33
  • $\begingroup$ thank you very much $\endgroup$ – M. A. SARKAR Jan 20 at 16:48

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