$\DeclareMathOperator{\char}{char}\DeclareMathOperator{\gal}{Gal}$
Let $P$ be a smooth projective variety over a field $K$ (one may certainly assume that $K$ is perfect; the case $K=\mathbb{Q}$ already seems to be interesting enough). For some $\ell\neq \char K$, $n>0$, should the $n$-th $\mathbb Q\_\ell$-adic Galois cohomology of $X\_{K^{sep}}$ be semi-simple as a $\gal(K)$-representation? Certainly; no proof of this fact is known, so I would rather like to know whether it is related with some 'motivic' conjectures.


Some remarks:


1. For a finite $K$ one can consider the 'motivic' Frobenius; thus the conjecture follows from standard (motivic) ones. Yet this argument does not seem to work already for $K=\mathbb Q$.
2. It is certainly tempting to apply some polarizability argument. Yet my impression is that polarizability can only be applied to Hodge structures (in general); is this true?


Upd. It seems (see the comment of Ulrich) that 'my conjecture' is wrong for $K= \mathbb Q\_\ell$; this settles my question. Yet I wonder where I can find the details for this example (when is the representation corresponding to an elliptic curve with multiplicative reduction indecomposable). 



Should the etale cohomology of a smooth projective variety (over rationals) be semi-simple; why?
For a base field finitely generated over the prime field, the Tate conjecture implies that the $l$-adic realization gives an equivalence from the category of pure motives tensor $\mathbb{Q}\_l$ to the category of $l$-adic Galois representations generated by the cohomology of smooth projective algebraic varieties over the field. The standard conjectures imply that the category of pure motives is semisimple, and hence also the category of Galois representations. (Actually, all you need for this is the Tate conjecture plus the conjecture that numerical equivalence and l-adic homological equivalence coincide, which are both in Tate's original Woods Hole article.)


For other base fields, the Galois representations need not be semisimple -- as ulrich explains in his comments.


