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Let $A$ be a (commutative) noetherian ring, and let $I \subseteq A$ be an ideal. It is not hard to show that if $I$ is generated by a length $n$ regular sequence, then the $A/I$-module $I/I^2$ is a free module of rank $n$. Is the converse true? if we know that $I/I^2$ is free, does it follows that $I$ is generated by a regular sequence?


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Although this isn't quite the right notion, you are on to something by trying to relate these two. It turns out that $A$ is LCI if and only if the cotangent complex is concentrated in degree -1, which is related to the cotangent module. – Matt Feb 13 '11 at 0:18
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No. Here are two counterexamples:

It is possible to have $I=I^2$, in which case $I$ is generated by an idempotent, which is not a regular element.

If $A$ is local, and $I$ the maximal ideal, then $I/I^2$ is always free over the residue field $A/I$, but in general $I$ is not generated by a regular sequence. (If this is true, then $A$ is called a regular local ring.)

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so easy... don't know how I missed this. Thanks! – the L Feb 12 '11 at 17:52

Actually, your guess is right with a little more assumption. Namely, in a Noetherian local ring $A$, TFAE for an ideal $I$:

1) $I$ has finite projective dimension and $I/I^2$ is a free $A/I$-module.

2) $I$ is generated by a regular sequence.

(this is a result by Vasconcelos, I just found it as Exercise 20.23 in Eisenbud's book)

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