Jim has already given the correct answer "no".
Here is a hopefully instructive example. Let be an alg. closed field of pos. char , and let . Write for the "natural" 2-dimensional representation of say with basis . Let be the -th symmetric power of . Then contains a 2 dimensional submodule spanned by the -th powers and ; the module is isom. to the "first Frobenius twist" of .
It is an exercise to check that there is no -stable complement to in ; i.e. the SES
is not split.
Thus
is not completely reducible. Evidently there is a surjective mapping
, thus also the -th tensor power is not completely
reducible.
But is a simple (hence completely reducible) -module; thus tensor powers of a completely reducible module are not in general completely reducible. In fact,
the -th tensor power is completely reducible;
arguing as before, one sees that is not completely
reducible; thus in general the tensor product of two completely reducible modules
is not completely reducible.
I gave some further remarks about semisimplicity of tensor products in an answer to
this question.
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