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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?

flag	asked Oct 28 at 0:28 George 53●2

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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 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 at 4:27

George · Accepted Answer · 2012-10-28 03:57:59Z UTC

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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.

answered Oct 28 at 3:57

George
53●2

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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 at 4:13

why a projective module is a projective cover for its largest semisimple quotient?

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