Lemma 10.68.4. Let $R$ be a local Noetherian ring. Let $M$ be a finite $R$-module. Let $x_1, \ldots , x_ c$ be an $M$-regular sequence. Then any permutation of the $x_ i$ is a regular sequence as well.

**Proof.**
First we do the case $c = 2$. Consider $K \subset M$ the kernel of $x_2 : M \to M$. For any $z \in K$ we know that $z = x_1 z'$ for some $z' \in M$ because $x_2$ is a nonzerodivisor on $M/x_1M$. Because $x_1$ is a nonzerodivisor on $M$ we see that $x_2 z' = 0$ as well. Hence $x_1 : K \to K$ is surjective. Thus $K = 0$ by Nakayama's Lemma 10.20.1. Next, consider multiplication by $x_1$ on $M/x_2M$. If $z \in M$ maps to an element $\overline{z} \in M/x_2M$ in the kernel of this map, then $x_1 z = x_2 y$ for some $y \in M$. But then since $x_1, x_2$ is a regular sequence we see that $y = x_1 y'$ for some $y' \in M$. Hence $x_1 ( z - x_2 y' ) =0$ and hence $z = x_2 y'$ and hence $\overline{z} = 0$ as desired.

For the general case, observe that any permutation is a composition of transpositions of adjacent indices. Hence it suffices to prove that

is an $M$-regular sequence. This follows from the case we just did applied to the module $M/(x_1, \ldots , x_{i-2})$ and the length $2$ regular sequence $x_{i-1}, x_ i$. $\square$

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