Lemma 15.29.15. Let $R$ be a ring. Let $I$ be an ideal generated by $f_1, \ldots , f_ r \in R$.

1. If $I$ can be generated by a quasi-regular sequence of length $r$, then $f_1, \ldots , f_ r$ is a quasi-regular sequence.

2. If $I$ can be generated by an $H_1$-regular sequence of length $r$, then $f_1, \ldots , f_ r$ is an $H_1$-regular sequence.

3. If $I$ can be generated by a Koszul-regular sequence of length $r$, then $f_1, \ldots , f_ r$ is a Koszul-regular sequence.

Proof. If $I$ can be generated by a quasi-regular sequence of length $r$, then $I/I^2$ is free of rank $r$ over $R/I$. Since $f_1, \ldots , f_ r$ generate by assumption we see that the images $\overline{f}_ i$ form a basis of $I/I^2$ over $R/I$. It follows that $f_1, \ldots , f_ r$ is a quasi-regular sequence as all this means, besides the freeness of $I/I^2$, is that the maps $\text{Sym}^ n_{R/I}(I/I^2) \to I^ n/I^{n + 1}$ are isomorphisms.

We continue to assume that $I$ can be generated by a quasi-regular sequence, say $g_1, \ldots , g_ r$. Write $g_ j = \sum a_{ij}f_ i$. As $f_1, \ldots , f_ r$ is quasi-regular according to the previous paragraph, we see that $\det (a_{ij})$ is invertible mod $I$. The matrix $a_{ij}$ gives a map $R^{\oplus r} \to R^{\oplus r}$ which induces a map of Koszul complexes $\alpha : K_\bullet (R, f_1, \ldots , f_ r) \to K_\bullet (R, g_1, \ldots , g_ r)$, see Lemma 15.28.3. This map becomes an isomorphism on inverting $\det (a_{ij})$. Since the cohomology modules of both $K_\bullet (R, f_1, \ldots , f_ r)$ and $K_\bullet (R, g_1, \ldots , g_ r)$ are annihilated by $I$, see Lemma 15.28.6, we see that $\alpha$ is a quasi-isomorphism.

Now assume that $g_1, \ldots , g_ r$ is a $H_1$-regular sequence generating $I$. Then $g_1, \ldots , g_ r$ is a quasi-regular sequence by Lemma 15.29.6. By the previous paragraph we conclude that $f_1, \ldots , f_ r$ is a $H_1$-regular sequence. Similarly for Koszul-regular sequences. $\square$

Comment #37 by Matthias Kümmerer on

Hi!

I think the second part of this lemma ("In other words...") is not correct. Only generating sets of the minimal possible length are Koszul-regular. An Example:

Consider the Ring $R=\mathbb{Z}$ and the ideal $(2)$. As it is generated by a non-zero-divisor, it is Koszul-regular. The Elements $4$ and $6$ form a minimal generating system. But of course, they do not form a Koszul-regular sequence (both are zero divisors in the factor ring of the other one).

Matthias

Comment #38 by Johan on

Ok, thanks! I fixed this by simply removing the entire second part, even though your formulation is correct. The reason is that it doesn't add much to the precise statement as given in the lemma anyway.

Comment #39 by on

And when I say something is fixed it means you can find a commit in the stacks project repository at github https://github.com/stacks/stacks-project/commits/master that addresses the issue. Most of the time the actual change will be propagated to the stacks project website in a few hours.

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• 2 comment(s) on Section 15.29: Koszul regular sequences

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