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?
Thanks!