Let $R$ be a Noetherian local commutative ring, $F$ a finitely generated free $R$-module and $A,B$ some arbitrary $R$-modules. Consider a short exact sequence $0 \to A \to F \to B \to 0$.

In [Bruns, Herzog - Cohen-Mayaulay rings] Proposition 1.4.3 (d) (see here) the authors are arguing that under some special circumstances we have that $A$ is free if and only if $B$ is free.

I am wondering why here are any further assumptions necessary. Can't we simply argue as follows: Since $F$ is free the sequence splits, i.e. we have $F \cong A \oplus B$. As direct summands of a free module $A$ and $B$ are projective. But (finitely generated) projective modules over local rings are free.

Where is the mistake?


1 Answer 1


Your mistake is in believing that freeness of $F$ implies that the sequence splits. It is projectivity of $B$ that ensures the sequence splits. Projectivity of $A$ doesn't obviously relate to the sequence splitting without some additional hypotheses (presumably what's in the reference you give).

  • $\begingroup$ Of course you are right. Thank you very much! $\endgroup$
    – Dune
    Jul 18, 2014 at 16:00

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