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One of the first results showed when studying regular sequences is that we are allowed to shuffle the elements of the sequence if the ring is noetherian, local, and the module is finite (see Proposition 2 in here for a proof).

I know the ring being local and the module being finite are required to use Nakayama's lemma, but I can't spot where is noetherianity used!

Edit: By MooS' comment I realize Akhil and probably all the other sources are using Krull's Intersection theorem, which I thought I didn't need for 'my' proof (I learned it from Bruns-Herzog). So now I wonder what is wrong with the following:

We want to show that under all the hypothesis, if $x,y$ is an $M$-sequence, then $y,x$ is an $M$-sequence. Let's show that $y$ is not a zero-divisor.

Denote the kernel of multiplication by $y$ on $M$ by $K$. Assume $m\in K$. Then, by the regularity of the sequence $m\in xK$ (use $\overline{ym}=0\in xM$) and we can write $m=xm'$ for some $m'\in K$, i.e., $K\subseteq xK$. Conversely, if $xym'=0$ by regularity $ym'=0$ and $m'\in K$. Therefore $K=xK$, and by Nakayama's lemma (version from AM) $K=0$. Thus, $y$ is not a zero-divisor. QED

What is wrong with this proof?

user347489
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1 Answers1

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The assumption that the ring be noetherian is used when Krull's intersection theorem is applied. And the assumption is necessary, as Stacks Project's tag 00LH shows, for example: consider $k[x,y,w_1,w_2,\ldots]/(yw_1,yw_2,\ldots,w_1-xw_2,w_2-xw_3,\ldots)$ and localise in the maximal ideal generated by $x,y$ and all the $w_i$, $i\in\mathbb{N}$. Then $x,y$ is a regular sequence, but $y$ is a zero divisor.

We can detect the issue with the OP's proof in the example: $K=\mathrm{Ann}(y) = (w_1,w_2,\ldots)$ does satisfy $xK=K$, but we can't conclude $K=0$ from Nakayama's lemma since $K$ is not finitely generated. So it seems the proof is okay if the ring is noetherian and it does not use Krull's intersection theorem.

Ben
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  • Hi Ben, thanks for your reply! MooS had already made me realize that KIT is being used in the proof, so I edited my answer to show my work, for which I can't find a mistake. If you can spot the mistake in it I'll be happy to accept your answer. – user347489 Apr 12 '17 at 21:12
  • Dear @user347489, there you go. – Ben Apr 12 '17 at 21:23
  • Awesome! And it's great to see how it fails in the example you mention. Good to see my proof is correct but Noetherianity is needed :) thanks. – user347489 Apr 12 '17 at 21:29