Let $K$ be number field and $\rho:G_K\rightarrow \text{Gl}(V)$ a Galois representation. Let $\nu$ be a place of $K$ (non-archimedean if it helps/is necessary). We say that $\rho$ is unramified at $\nu$ if $\rho(I_\nu)$ is trivial. My question is if this can be tested galois-locally, i.e. if $L$ is a finite Galois extension of $K$ and $\rho\vert_{G_L}$ is unramified at all primes $\omega\vert \nu$, does it follow that $\rho$ is unramified at $\nu$?
Galois representation being unramified is Galois local
algebraic-number-theorygalois-extensionsgalois-representationsnumber theoryramification
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$\require{AMScd}$ If $\rho$ factors through some $\overline{\rho}\colon \mathrm{Gal}(L/K) \to \mathrm{GL}_n(F)$ with $L/K$ unramified at $\mathfrak{p}$, then the image of $I_{\mathfrak{p}}$ under $\rho$ is \begin{equation*} \rho(I_{\mathfrak{p}}) = \overline{\rho}\circ \pi(I_{\mathfrak{p}}) = \overline{\rho}(I_{\mathfrak{p}}(L/K)) = \overline{\rho}(1)=1 \end{equation*} because $\mathfrak{p}$ is unramified in $L/K$ so it must have trivial inertia group there, and the image of the absolute inertia group in some intermediate extension is the inertia group of that extension.
Conversely, if $I_{\mathfrak{p}}$ is trivial then the statement is trivial as well. If on the contrary $I_{\mathfrak{p}}\neq 1$ then $\rho$ factors through $\mathrm{Gal}(L/K)$ where $K\subseteq L\subseteq \overline{K}$ is the maximal extension where $\mathfrak{p}$ is unramified. Indeed, the kernel of the projection is exactly $I_{\mathfrak{p}}$ which lies in the kernel of $\rho$.
Potential purity implies purity. Fix a prime $\mathfrak p$ of $K$ and let $\mathfrak q$ be a prime of $L$ above it. Assume that $\mathfrak q/\mathfrak p$ is unramified, with residue degree $f$.
Then $\mathrm{Frob}_{\mathfrak q}\in G_L$ is exactly $\mathrm{Frob}_{\mathfrak p}^f$. Indeed, by definition, $\mathrm{Frob}_{\mathfrak p}$ is a pre-image of $x\mapsto x^{N\mathfrak p}$ in $\mathrm{Gal}(\overline{\mathbb F}_{\mathfrak p}/\mathbb F_{\mathfrak p})$ and $\mathrm{Frob}_{\mathfrak q}$ is a pre-image of $x\mapsto x^{N\mathfrak q}$ in $\mathrm{Gal}(\overline{\mathbb F}_{\mathfrak p}/\mathbb F_{\mathfrak q})$. But $N\mathfrak q = (N\mathfrak p)^f$ .
Now, by assumption, $\rho|_{G_L}$ is pure of some weight $w$. That means that the eigenvalues of $\rho(\mathrm{Frob}_{\mathfrak q})$ have absolute value $(N\mathfrak q)^{w/2}$.
But that means that the eigenvalues of $\rho(\mathrm{Frob}_{\mathfrak p}^f)$ have absolute value $(N\mathfrak p)^{fw/2}$. Hence the eigenvalues of $\rho(\mathrm{Frob}_{\mathfrak p})$ have absolute value $(N\mathfrak p)^{w/2}$, i.e. $\rho$ is pure of weight $w$.
For your second question, I think you've misunderstood what pure of weight $0$ means - it does not mean that Frobenii act as the identity. For example, let $\rho$ be the $\ell$-adic Galois representation attached to a weight $3$ modular form, and let $\epsilon$ be the $\ell$-adic cyclotomic character. Then $\rho$ is pure of weight $2$, and $\rho\otimes\epsilon^{-1}$ is pure of weight $0$. But it has infinite image, and its Frobenii do not act trivially.
Best Answer
No, I don't think so.
For example, take an elliptic curve $E/K$ which has potential good reduction at $\nu$ but not good reduction. Let $L$ be some finite extension over which $E$ achieves good reduction at all $\omega|\nu$. Then Ogg-Neron-Shafarevic tells us that the action on the Tate module of $E$ at some prime not divisible by $\nu$ is not unramified at $\nu$ (since we don't have good reduction) but is unramified at all those $\omega$ (since we get good reduction at all those places).
I think whenever $I_\nu$ has finite image you can come up with examples like that - the issue is that if the action factors through a finite quotient then you can find some finite extension $L$ which ``eats up'' that image and so the restriction to $L$ will always look unramified. Probably (?) the only way you can always guarantee that what you want holds is if you require that $L/K$ is unramified (in which case $I_\nu = I_\omega$).