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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 $\nu$?

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  • $\begingroup$ No. Consider any Galois representation $\rho$ with finite image (e.g. a Dirichlet character). Its image is isomorphic to $\mathrm{Gal}(L/K)$ for some $L$. But $\rho|_L$ is trivial and hence unramified for all primes. A similar idea works whenever the image of inertia is finite. $\endgroup$ – Mathmo123 Oct 3 '20 at 18:45
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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 $I_\nu = I_\omega$).

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    $\begingroup$ The last sentence is correct (so you can delete the "probably"). If $L/K$ ramifies at $v$, then the regular representation of $Gal(L / K)$ is ramified at $v$, but its restriction to $G_L$ is trivial, so certainly unramified at the primes above $v$. $\endgroup$ – David Loeffler Oct 2 '20 at 7:15
  • $\begingroup$ Thanks for confirming - I usually only think about these things locally and wanted to make sure I didn't say anything that would be trivially false if there were some splitting :) $\endgroup$ – user208649 Oct 17 '20 at 3:56

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