# Local polynomial of Galois representation restricted to subextension

Let $$K$$ be an extension of $$\mathbb{Q}_p$$. Consider an Galois representation $$\rho: G_K \to GL_2(\mathbb{C})$$ and for any finite extension $$M/K$$ we call

$$P(\rho|_M,T) = \det(1-\operatorname{Frob}_{M}^{-1}T \, | \rho^{I_M})$$

the local polynomial of $$\rho$$ over $$M$$. Here, $$\operatorname{Frob}_{M}$$ is an Frobenius element of $$G_M$$, i.e. a lift of the automorphism $$x \mapsto x^{q_M} \in G_{\mathbb{F}_{q_M}}$$ (where $$q_M$$ denotes the cardinality of the residue field of $$M$$). Note that the local polynomial is independent of the choice of the Frobenius element.

Now we make the following assumptions:

• $$L/K$$ is a cyclic and finite extension,
• $$M$$ is a subextension of $$L/K$$ such that $$L/M$$ is unramified,
• $$\rho = A \otimes \psi$$ where $$A: G_K \to GL_{2}(\mathbb{C})$$ factors through $$\operatorname{Gal}(L/K)$$ (i.e. is an Artin representation) and $$\psi: G_K \to \mathbb{C}^*$$ is an unramified character (i.e. $$\psi(I_K) = 1$$),
• $$P(\rho|_L,T) = (1-\mu^{-f}T)^2$$ where $$\mu = \psi(\operatorname{Frob}_K) \in \mathbb{C}^*$$ for a fixed Frobenius element $$\operatorname{Frob}_K \in G_K$$ and $$f$$ is the inertial degree of $$L/K$$.

Question: Can we compute $$P(\rho|_M,T)$$ by using the result $$P(\rho|_L,T)$$?

Because $$L/M$$ is unramified, a Frobenius element in $$G_M$$ is given by $$\operatorname{Frob}_L^\ell$$ where $$\ell$$ is the degree of $$L/M$$. Now I am not sure how to proceed - a next step would be to consider the vector spaces $$\rho^{I_L}$$ and $$\rho^{I_M}$$ in the definition of the local polynomials.

• The notation $\operatorname{Frob}_M^{-1} T | \rho^{I_M}$ does not make sense to me. I assume $M/K$ is Galois, and $\rho$ factors through $\operatorname{Gal}(M/K)$? In that case, is $I_M$ the inertia group of $\operatorname{Gal}(M/K)$?
– D_S
Commented Apr 19, 2021 at 1:59
• $I_M=Gal(\overline{M}/M^{ur})$ and $\rho^{I_M}$ is the subspace of $\Bbb{C}^2$ where $I_M$ acts trivially, so that $\rho(Frob_M)$ doesn't depend on a choice of $Frob_M$ @D_S Commented Apr 19, 2021 at 2:01

This is what I understand of your "setting".

Optional:

• $$P(\rho|_L,T) = (1-\mu^{-f}T)^2$$ gives that $$P(A|_L,T) = (1-\psi(Frob|_L)\mu^{-f} T)^2$$.

Since $$A$$ has finite image it means that on $$A$$, $$Frob_L$$ is the multiplication by $$\psi(Frob_L)^{-1}\mu^f$$ and thus on $$\rho|_L$$ it is the multiplication by $$\mu^f$$.

The main point:

• Since $$L/M$$ is unramified of degree $$\ell$$ then for any $$Frob_M$$ you'll get that $$Frob_M^\ell$$ is a $$Frob_L$$. Also $$M^{ur}=L^{ur}\implies Gal(\overline{L}/L^{ur})=I_M=I_L$$, therefore $$P(\rho|_M,T) = (1- \zeta_{\ell}^a\mu^{-f/\ell}T)(1- \zeta_{\ell}^b\mu^{-f/\ell}T)$$ for some integers $$a,b$$, with in general no restriction on $$a,b$$.
• Thank you for your response! Could you explain to me why these $\zeta_\ell^a$ and $\zeta_\ell^b$ occur in $P(\rho_M,T)$ in the first place? This is the only part which is unclear to me. Commented Apr 19, 2021 at 12:10
• Eigenvalues of a matrix whose $\ell$ power is $\mu^{-f}$ (or if you skip the first part so you don't know that it is diagonalizable, the diagonal entries of its jordan normal form whose $\ell$ power has $\mu^{-f}$ diagonal entries) Commented Apr 19, 2021 at 16:23