Lemma 15.29.6. 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 15.29.4. 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 15.29.5. 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. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)

There are also: