**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.
$\square$

## Comments (3)

Comment #37 by Matthias Kümmerer on

Comment #38 by Johan on

Comment #39 by Johan on

There are also: