Let $\overline{\rho}: \text{Gal}(\overline{K}/K)\to \text{GL}_2(\mathbb{F}_q)$ be an unramified Galois representation at a place $v$ of $K$. Since $\ker(\overline{\rho})$ is closed, it corresponds to a subextension $\overline{K}/L/K$, wherefore $|\text{GL}_2(\mathbb{F}_q)|<\infty$ implies that $L/K$ is finite, and hence $\ker(\overline{\rho})$ is open. Therefore, $\ker(\overline{\rho})$ being open normal implies $L/K$ is finite Galois.
Let $\mathfrak{p}',\mathfrak{p}$ be primes of $\overline{K},L$ respectively lying over $v$. Since $\overline{\rho}$ is unramified at $v$, $I(\mathfrak{p}'|v)\leq \ker\overline{\rho}$ or equivalently $I(\mathfrak{p}|v)=1$. Therefore, the Frobenius element $\text{Frob}_v\in \text{Gal}(L/K) = \text{im}(\overline{\rho})$ is well-defined up to conjugation by an element of $\text{im}(\overline{\rho})$.
But often times in papers, I see authors talk about $“\text{Frob}_v"\in \text{Gal}(\overline{K}/K)$ or $“\text{Frob}_v"\in\text{Gal}(\overline{K}/L)$. Is this simply abuse of notation for any lift of $\text{Frob}_v$ under the surjection $\text{Gal}(\overline{K}/K)\twoheadrightarrow\text{Gal}(L/K)$? In other words, $“\text{Frob}_v"\in\text{Gal}(\overline{K}/K)$ is well-defined only up to both a conjugation and an element of $\text{Gal}(\overline{K}/L)$?
Best Answer
You are right that $\operatorname{Frob}_v$ is well-defined only up to a conjugation, but you don't really need to choose an $L$. This is because the Frobenius for different $L$ can be packed together (as a projective limit) to $\overline K$. Let me explain this below.
Let $K$ be a global field (number field or function field). If $v$ is a finite place of $K$, then we may form the $v$-adic completion $K_v$, so that $K$ is canonically embedded in $K_v$.
Let $w$ be any lift of $v$ to $\overline K$. The choice of $w$ is equivalent to choosing an embedding $\overline K \hookrightarrow \overline {K_v}$. This induces an injection of groups $\operatorname{Gal}(\overline {K_v}/K_v) \hookrightarrow \operatorname{Gal}(\overline K / K)$, whose image is the decomposition group $D_{w/v}$. Note that two different choices of embeddings $\overline K \hookrightarrow \overline {K_v}$ differ by an automorphism of $\overline K$, hence the two resulting decomposition groups differ by a conjugation.
Now in the case of a local field $K_v$ with residue field $\Bbb F_q$, the residue field of $\overline {K_v}$ is simply $\overline{ \Bbb F_q}$. Thus any automorphism $\phi\in \operatorname{Gal}(\overline {K_v}/K_v)$ induces an automorphism $\psi \in \operatorname{Gal}(\overline {\Bbb F_q}/\Bbb F_q)$.
This gives rise to a surjective homomorphism of groups $\tau: \operatorname{Gal}(\overline {K_v}/K_v) \rightarrow \operatorname{Gal}(\overline{\Bbb F_q}/\Bbb F_q)$.
We know that $\operatorname{Gal}(\overline{\Bbb F_q}/\Bbb F_q)$ is canonically isomorphic to $\widehat{\Bbb Z}$, the profinite completion of $\Bbb Z$, and we call the inverse image of $1\in \widehat{\Bbb Z}$ in $\operatorname{Gal}(\overline{\Bbb F_q}/\Bbb F_q)$ the "Frobenius", denoted $\operatorname{Frob}_v$.
The inverse image $\tau^{-1}(\operatorname{Frob}_v) \subseteq \operatorname{Gal}(\overline {K_v}/K_v)$ is a coset of the kernel $I_v$, the inertia group. But by abuse of notation, it is also called "the" Frobenius, and by even more abuse of notation, any element in this coset is called "the" Frobenius.
To add even more confusion, there is a difference of arithmetic vs geometric Frobenius, which is we choose the inverse image of $-1\in \widehat{\Bbb Z}$ above.
Back in the global situation, we have the embedding $\operatorname{Gal}(\overline{K_v}/K_v)\hookrightarrow \operatorname{Gal}(\overline K/K)$ and the image of a Frobenius is again called a "Frobenius".
In view of this unfortunate situation, one should always pay attention to what is meant by "the" Frobenius.