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Hi, I know that if $R$ is a ring such that every projective $R$-module finitely generated is free then $R$ has the unimodular column property. I would like to know if there is a ring $R$ that doesn't satisfy the unimodular column property but such that every finitely generated stably free $R$-module is free.


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Let me talk about unimodular rows instead of unimodular columns (it's obviously equivalent, because you can take the transpose). An unimodular row is (using the standard identification between matrices and linear maps) a surjective homomorphism $R^k\to R$ for some $k$. Since surjective homomorphisms between free $R$-modules always split, this homomorphism splits, and thus it writes $R^k$ as a direct sum $R\oplus T$, where $T$ is the kernel of that homomorphism and the projection from $R^k$ to $R$ is your unimodular row. ... – darij grinberg Feb 5 '11 at 19:39
... Now $T$ is stably free, thus free (by your condition), and this yields that your unimodular row can be completed to an invertible matrix. The only thing we don't know is whether this invertible matrix is going to be a square matrix (things can go wrong here if the ring $R$ does not satisfy the invariant basis property). – darij grinberg Feb 5 '11 at 19:40
Whether this is an answer to your question depends on how you define the "unimodular column property". – darij grinberg Feb 5 '11 at 19:41
up vote 4 down vote accepted

If "unimodular column property" means that every unimodular column can be completed to an invertible matrix, then this is equivalent to the statement that every finitely generated stably free module is free.

For finitely generated modules "Stably free implies free" is equivalent to "$P\oplus R$ free implies $P$ free" by an easy induction. The latter is equivalent to "Every unimodular column can be completed to an invertible matrix" because "$P\oplus R$ free" is equivalent to "$P$ is the cokernel of a unimodular column", and then $P$ is free if and only if the column can be completed.

I see that darij grinberg said most of this in comments already.

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