Talk:Projective module

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Section and split exact sequences

There was a recent concern about an equivalence of sections and direct sum decompositions for modules. First I'll describe the equivalence, and then describe what probably went wrong with the proposed example (an edit summary is not long enough to specify such an example, so one must interpret).

If f:P->M is surjective, and h:M->P is such that f(h(m))=m for all m in M, and both f,h are R-module homomorphisms, then P is the direct sum of ker(f) and im(h), and the image of h is isomorphic to M. An easy proof is that (1) P=ker(f)+im(h) since x = (x - h(f(x))) + h(f(x)), and f( x - h(f(x)) ) = f(x) - f(h(f(x))) = f(x) - f(x) = 0, so the first summand is in the kernel of f, and the second summand is in the image of h. If x is in the intersection of ker(f) and im(h), then x=h(m), but then 0 = f(x) = f(h(m)) = m, and x=h(0)=0. Finally, since f and h are R-module homomorphisms, im(h) and ker(f) are R-submodules of P.

The group ring example might have a few different interpretations, but either the implicit h is not a G-homomorphism, or the composition is not the identity. Since there is a copy of the trivial module Z contained in the center of ZG and a G-homomorphism from Z to that copy, I'll assume we are in the latter case: h:Z->Z(G):n->n*Sum(g,g in G). But then f(h(1)) = |G|, rather than 1. The composition is not the identity, but rather |G| times the identity. Now if |G| is invertible in the ring (changing Z to just some ring R), then one can fix this little snag, but then the trivial module R *is* a direct summand of RG. JackSchmidt (talk) 14:21, 26 March 2008 (UTC)[reply]

Locally free redirects here

Clearly there should be an extra page with at least the definition.T3kcit (talk) 20:40, 16 August 2008 (UTC)[reply]