Lemma 15.30.6. An $M$-$H_1$-regular sequence is $M$-quasi-regular.

**Proof.**
Let $R$ be a ring and let $M$ be an $R$-module. Let $f_1, \ldots , f_ r$ be an $M$-$H_1$-regular sequence. Denote $J = (f_1, \ldots , f_ r)$. The assumption means that we have an exact sequence

where the first arrow is given by $e_ i \wedge e_ j \otimes m \mapsto (f_ ie_ j - f_ je_ i) \otimes m$. In particular this implies that

is a finite free module. To finish the proof we have to prove for every $n \geq 2$ the following: if

then $m_ I \in JM$ for all $I$. Note that $f_1, \ldots , f_{r - 1}, f_ r^ n$ is an $M$-$H_1$-regular sequence by Lemma 15.30.4. Hence we see that the required result holds for the multi-index $I = (0, \ldots , 0, n)$. It turns out that we can reduce the general case to this case as follows.

Let $S = R[x_1, x_2, \ldots , x_ r, 1/x_ r]$. The ring map $R \to S$ is faithfully flat, hence $f_1, \ldots , f_ r$ is an $M$-$H_1$-regular sequence in $S$, see Lemma 15.30.5. By Lemma 15.28.4 we see that

is an $M$-$H_1$-regular sequence in $S$. Finally, note that our element $\xi $ can be rewritten

and the coefficient of $g_ r^ n$ in this expression is

Since the monomials $x_1^{i_1} \ldots x_ r^{i_ r}$ form part of an $R$-basis of $S$ over $R$ we conclude that $m_ I \in J$ for all $I$ as desired. $\square$

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## Comments (1)

Comment #6253 by Johan on

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