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 $U$?