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A submodule of a free module need not be free (for instance, in the free Z[X]-module Z[X] the submodule generated by 2 and X is not free). But over a principal ideal domain, submodules of free modules are free.

I was wondering about the center of a free (as a module) algebra. Is it always free? or are there weaker conditions on the ground ring which guarantee it?

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up vote 4 down vote accepted

Choose a ring R of characteristic not 2 which does not satisfy the condition "every projective is free" (e.g. R is not local). Pick a nonfree projective R-module M and make M into a commutative R-algebra in some way. Pick a complement N such that M + N (direct sum) is free and make N into an anticommutative R-algebra via a nondegnereate skew-symmetric form N x N -> R such that MN = NM = 0. I think this should yield an example, though I'm not sure on all the details.

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Thanks. I was going to complain that this algebra is not simple but by the Artin-Wedderburn theorem, it cannot be! (at least if associative and finitely generated) – Benoit Jubin Oct 29 '09 at 21:13

Following up on Akhil's plan, a good source of projective modules which are not free is via nontrivial vector bundles: smooth sections of the bundle form a projective module over the ring R of smooth functions, but a non-free module if the bundle is nontrivial.

So, you could do something simple-minded like this: let P be the module of sections of the tangent bundle of RP^2; put a commutative algebra structure on M = R + P where all products of elements of P are zero. As an R-module, M is not free. Let N be the module of sections of the exterior algebra bundle of the normal bundle to an embedding of RP^2 in R^4. The Whitney sum of the tangent bundle and the normal bundle is trivial; therefore the algebra M x N forms a free R-module, whose center M is not free.

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Thanks. I should have thought of the Serre-Swan theorem! – Benoit Jubin Oct 29 '09 at 21:17

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