Local factors determine Weil representations - proof of the cyclic case I already created this post on Math Stack Exchange but I was not so sure if this question fits better here. If it is not, I want to apologize in advance and feel free to delete my post.
I want to understand the proof of the following Theorem from "Euler Factors determine Weil Representations" by Tim and Vladimir Dokchitser:

Theorem 1 Every Frobenius-semisimple Weil representation $\rho$ is uniquely determined by its local polynomials $P(\rho/F,T)$ over finite separable extensions $F/K$.

Before we talk about the cyclic case of the proof, let us recall some definitions first:


*

*Let $K$ be a local field and $G_K = \operatorname{Gal}(\bar{K}/K)$ be the absolute Galois group of $K$. An (arithmetic) Frobenius element is any element $\operatorname{Frob}_K \in G_K$ that acts as $x \mapsto x^{|k|}$ on $\bar{k}$, the algebraic closure of the residue field $k$ of $K$.

*The Weil group $W_K$ is the subgroup of $G_K$ of all automorphisms that act as an integral power of Frobenius on the residue field.

*A Weil representation is a representation $\rho: W_K \to \operatorname{GL}_n(\mathbb{C})$ such that $\rho(I_K)$ is finite. It is called Frobenius-semisimple if the image of some (equivalently, any) Frobenius element is diagonalizable.

*The local polynomial $P(\rho,T)$ is the inverse characteristic polynomial of $\operatorname{Frob}_K^{-1}$ on the inertia invariants of $\rho$, i.e. $$P(\rho,T) = \det(1-T \cdot \operatorname{Frob_K^{-1}}).$$
Similarly, for a finite extension $F/K$, we write $P(\rho/F)$ for the local polynomial of the restriction of $\rho$ to $W_F$, i.e. $$P(\rho/F,T) = P(\rho|_{W_F},T).$$
Now I would like to talk about the proof of the cyclic case which is given in the paper:

Step 1: Cyclic. Suppose $\rho$ factors through a finite cyclic group $G = \operatorname{Gal}(F/K) \simeq C_n$ and $F/K$ has ramification degree $e$. By Lemma 2 (cf. below), there is a cyclic totally ramified extension $K/K$ of degree $e$ such that $FL/L$ is unramified of degree $n$. The restriction map $\operatorname{Gal}(FL/L) \to \operatorname{Gal}(F/K)$ is an isomorphism, as it is clearly injective and both groups have order $n$. So $\rho/L$ determines $\rho$, and $\rho/L$ can be recovered from its local polynomial $P(\rho/L,T)$.

In our proof we used the following Lemma which we shall take for granted in this post:

Lemma 2 Let $F/K$ be a cyclic extension of degree $n$ and ramification degree $e$. Then there exists a cyclic totally ramified extension $L/K$ of degree $e$ such that $FL/L$ is unramified of degree $n$.

Now I have the following specific questions about the proof above:


*

*Could you give me a good argument why the restriction map $\operatorname{Gal}(FL/L) \to \operatorname{Gal}(F/K)$ is injective? Let us say we have $\sigma, \sigma' \in \operatorname{Gal}(FL/L)$ which are not equal. Then there exists an $x \in FL \setminus L$ such that $\sigma(x) \neq \sigma'(x)$. If $x \in F$ then the restrictions $\bar{\sigma}, \bar{\sigma}$ are obviously not equal. But what happens if $x$ is not in $F$?

*Why does $\rho/L$ determine $\rho$ then? And how can we recover $\rho/L$ from its local polynomial $P(\rho/L,T)$? What do "determining" and "recovering" even mean in these cases?


Additional Remark:
I would also like to add a diagram which my professor drew to explain me something relating to this Theorem (or Lemma?):
$\require{AMScd}\DeclareMathOperator\Gal{Gal}$
\begin{CD}
@. \Gal(L F/L) @>{\text{$\chi|_L$, unramified}}>> \Gal(K^{\mathrm{ur}}/K) \\
@. @V{\simeq}VV @| \\
1 @>>> \Gal(F/K^{\mathrm{ur}}) @>>> \Gal(F/K) @>>> \Gal(K^{\mathrm{ur}}/K) @>>> 1 \\
@. @. @V{\chi}VV \\
@. @. \mathbb C^\times
\end{CD}
And next to this diagram, he also wrote $P(\chi|_L,T) = 1 - \operatorname{Frob}_L \cdot T$.
To this remark, I have the following questions (if it is really related to the proof):


*

*What is $\chi$ supposed to mean?

*Is the diagram commutative?

*Why all of the sudden does the maximal unramified extension (I think?) $K^{ur}$ of $K$ appear here?

*If or how is this related to the proof of Theorem 1 (resp. Lemma 2)?


Could you please help me answering my questions? I feel like I lack a lot of background knowledge which is required here, so it would be really nice if you could explain them to me carefully. Thank you in advance!
 A: Why is the map ${\rm Gal}(FL/L)\to {\rm Gal}(F/K)$ injective?
Elements of ${\rm Gal}(FL/L)$ are automorphisms of $FL$ that act trivially on $L$. To be in the kernel of the above map means to also act trivially on $F$. A field automorphism that acts trivially on $F$ and $L$ acts trivially on $FL$ (by definition of compositum and of field automorphism).
Why does $\rho|_L$ determine $\rho$? And what does "determine" mean?
Firstly, what does $\rho|_L$ mean? $\rho$ is a representation of the absolute Galois group of $K$, and $\rho|_L$ is really shorthand for the restriction of $\rho$ to the subgroup ${\rm Gal}(\bar{K}/L)$ of ${\rm Gal}(\bar{K}/K)$. "Determine" means that if you have a second representation $\rho'$ satisfying all the hypotheses such that $\rho|_L=\rho'|_L$, then in fact $\rho=\rho'$. So why is this the case? Well, firstly, $\rho$ is trivial on ${\rm Gal}(\bar{K}/LF)$, since this is a subgroup of ${\rm Gal}(\bar{K}/F)$, and you already know that $\rho$ is trivial on the latter (that's what it means to say that $\rho$ factors through $F/K$). So we can just think of $\rho$ as a representation of ${\rm Gal}(LF/K)$ that is trivial on ${\rm Gal}(LF/F)$. So the value of $\rho$ on $\sigma\in {\rm Gal}(LF/K)$ only depends on the image of $\sigma$ under the quotient map ${\rm Gal}(LF/K)\to {\rm Gal}(F/K)$. But we have just observed that this quotient map is an isomorphism on the subgroup ${\rm Gal}(LF/L)$ (i.e. different elements of that subgroup give different cosets in the quotient, and every element of the quotient has a representative in that subgroup), so $\rho$ is determined by its values on ${\rm Gal}(LF/L)$. It's really just a matter of unravelling the definitions, and has not much to do with Galois, it's just a fact about homomorphisms from arbitrary groups to anywhere that factor through a quotient, when the extension of groups splits (the diagram I would draw here is the diamond with $K$, $F$, $L$, and $FL$; just think about what $\rho$ and $\rho'$ would look like on the quotient ${\rm Gal}(F/K)$ if they agreed on ${\rm Gal}(LF/L)$).
How to recover $\rho$ from its local polynomial?
Now you are left with a representation of the Galois group of an unramified extension. That Galois group is generated by Frobenius, let's call him $\phi$ (that's what being unramified has bought you), so the representation $\rho$ is determined by $\rho(\phi)$. By assumption, the representation is Frobenius-semisimple, so $\rho(\phi)$ is just determined by the eigenvalues of $\rho(\phi)$, and these, in turn, can be read off from the characteristic polynomial.
I do not understand your professor's diagram ($K^{\rm ur}$ cannot denote the maximal unramified extension of $K$, since that would be infinite, so certainly not the same degree as $F/K$; it also cannot denote the maximal unramified extension contained within $F$, since that claimed equality would still not hold - $F/K$ was not assumed to be unramified), and I suggest that you ask your professor rather than us.
