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This is an old revision of this page, as edited by KonradVoelkel (talk | contribs) at 14:48, 5 April 2012 (Diagram suggested). The present address (URL) is a permanent link to this revision, which may differ significantly from the current revision.
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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]

Examples

Could someone please provide a few examples of non-free projective modules? —Preceding unsigned comment added by 128.40.136.156 (talk) 13:08, 3 March 2009 (UTC)[reply]

Done. Non-principal ideals of Dedekind domains are popular examples, but simpler examples are direct factor of direct product rings, or the natural module of a full matrix ring. JackSchmidt (talk) 14:32, 3 March 2009 (UTC)[reply]

Diagram suggested

From the german wikipedia I translated a diagram which helped me many times in my studies: Module properties in commutative algebra

I consider adding it here (and as well on the free, flat and torsion-free module pages). What do you think? Konrad (talk) 13:42, 5 April 2012 (UTC)[reply]

I've started comments on a duplicate of this post at Talk:Flat_module, if anyone cares to respond there. Rschwieb (talk) 14:35, 5 April 2012 (UTC)[reply]
I'm sorry for the double/triple-posting. I just see no other way of getting all people concerned about these three pages into this discussion. If one puts a diagram like this on wikipedia, I think having it on all three pages will be better. I will watch the Talk:Flat_module now. Konrad (talk) 14:48, 5 April 2012 (UTC)[reply]