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 15.30.6. 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
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