Let $K/ \mathbb Q$ be a Galois extension, and $\rho$ be an irreducible representation of the Galois group $Gal(K/ \mathbb Q)$. Consider an integer prime $p$ which doesn't ramify in $K$, and let $\sigma_p$ denote the conjugacy class of the Frobenius element. Is the linear transformation $\rho(\sigma_p)$ diagonalizable in general? If so, I'd really appreciate a proof or reference (preferably, as elementary as possible). If not, what other conditions are required and what is the best result we can say in that direction?
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1$\begingroup$ You have almost no hypotheses, so the answer is clearly no. Just take $K/\mathbb{Q}$ to be finite Galois, e.g. with Galois group $S_3$, and take $\rho$ to be irreducible over a finite field such that some element of $S_3$ is sent to a non-diagonalisable matrix. For $S_3$, the standard representation arising from the isomorphism $S_3\cong {\rm GL}_2(\mathbb{F}_2)$ will do, where the involutions are not diagonalisable. By Chebotarev, every element of your Galois group will be a Frobenius at some $p$. $\endgroup$– Alex B.Commented May 25, 2020 at 12:36
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2$\begingroup$ There is no "best result in that direction", since you have too few hypotheses. For example if $K/\mathbb{Q}$ is finite and the representation is over a field of characteristic $0$, then everything is diagonalisable, but that has nothing to do with Galois representations. In general, (Weil-Deligne) representations for which Frobenii are diagonalisable are called "Frobenius semisimple". You can google that term for lots of literature, but almost none of it will be "elementary". $\endgroup$– Alex B.Commented May 25, 2020 at 12:39
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