Lemma 10.68.9. Let $R$ be a ring. Let $M$ be an $R$-module. Let $f_1, \ldots , f_ r \in R$ and $e_1, \ldots , e_ r > 0$ integers. Then $f_1, \ldots , f_ r$ is an $M$-regular sequence if and only if $f_1^{e_1}, \ldots , f_ r^{e_ r}$ is an $M$-regular sequence.

**Proof.**
We will prove this by induction on $r$. If $r = 1$ this follows from the following two easy facts: (a) a power of a nonzerodivisor on $M$ is a nonzerodivisor on $M$ and (b) a divisor of a nonzerodivisor on $M$ is a nonzerodivisor on $M$. If $r > 1$, then by induction applied to $M/f_1M$ we have that $f_1, f_2, \ldots , f_ r$ is an $M$-regular sequence if and only if $f_1, f_2^{e_2}, \ldots , f_ r^{e_ r}$ is an $M$-regular sequence. Thus it suffices to show, given $e > 0$, that $f_1^ e, f_2, \ldots , f_ r$ is an $M$-regular sequence if and only if $f_1, \ldots , f_ r$ is an $M$-regular sequence. We will prove this by induction on $e$. The case $e = 1$ is trivial. Since $f_1$ is a nonzerodivisor under both assumptions (by the case $r = 1$) we have a short exact sequence

Suppose that $f_1, f_2, \ldots , f_ r$ is an $M$-regular sequence. Then by induction the elements $f_2, \ldots , f_ r$ are $M/f_1M$ and $M/f_1^{e - 1}M$-regular sequences. By Lemma 10.68.8 $f_2, \ldots , f_ r$ is $M/f_1^ eM$-regular. Hence $f_1^ e, f_2, \ldots , f_ r$ is $M$-regular. Conversely, suppose that $f_1^ e, f_2, \ldots , f_ r$ is an $M$-regular sequence. Then $f_2 : M/f_1^ eM \to M/f_1^ eM$ is injective, hence $f_2 : M/f_1M \to M/f_1M$ is injective, hence by induction(!) $f_2 : M/f_1^{e - 1}M \to M/f_1^{e - 1}M$ is injective, hence

is a short exact sequence by Lemma 10.4.1. This proves the converse for $r = 2$. If $r > 2$, then we have $f_3 : M/(f_1^ e, f_2)M \to M/(f_1^ e, f_2)M$ is injective, hence $f_3 : M/(f_1, f_2)M \to M/(f_1, f_2)M$ is injective, and so on. Some details omitted. $\square$

## 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Your email address will not be published.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 (4)

Comment #787 by Keenan Kidwell on

Comment #805 by Johan on

Comment #3793 by Kestutis Cesnavicius on

Comment #3915 by Johan on