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?



Is the center of a free (as a module) algebra free?
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.


