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Why a projective module is a projective cover for its largest semisimple quotient? That is - why the projection on the quotient is an essential morphism in this case?

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This question is not quite on-topic on this site, as explained in the FAQ. The FAQ suggests a few other places where your question will be much more at ease. Good luck! –  Mariano Suárez-Alvarez Oct 28 '12 at 4:14
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You should probably accept your answer: otherwise, this question is going to be bumped automatically on the question lists periodically for ever! –  Mariano Suárez-Alvarez Nov 11 '12 at 4:27
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Answer: if Q is a submodule of a projective module P which projects surjectively on the largest semisimple quotient of P, then Q projects surjectively on each simple quotient of P, and hence Q lies outside of any maximal submodule of P - contradiction.

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In other words, because the kernel is contained in the radical (because the map is surjective and the quotient has zero radical!) –  Mariano Suárez-Alvarez Oct 28 '12 at 4:13
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