The Stacks project

Proof. First assume that $\mathcal{J}$ is quasi-regular. We may choose an open neighbourhood $U \subset X$ of $x$ and a quasi-regular sequence $g_1, \ldots , g_ s \in \mathcal{O}_ X(U)$ which generates $\mathcal{J}|_ U$. Note that this implies that $\mathcal{J}/\mathcal{J}^2$ is free of rank $s$ over $\mathcal{O}_ U/\mathcal{J}|_ U$ (see Lemma 31.20.4 and its proof) and hence $r = s$. We may shrink $U$ and assume $f_1, \ldots , f_ r \in \mathcal{J}(U)$. Thus we may write

for some $a_{ij} \in \mathcal{O}_ X(U)$. By assumption the matrix $A = (a_{ij})$ maps to an invertible matrix over $\kappa (x)$. Hence, after shrinking $U$ once more, we may assume that $(a_{ij})$ is invertible. Thus we see that $f_1, \ldots , f_ r$ give a basis for $(\mathcal{J}/\mathcal{J}^2)|_ U$ which proves that $f_1, \ldots , f_ r$ is a quasi-regular sequence over $U$.

Note that in order to prove (2) and (3) we may, because the assumptions of (2) and (3) are stronger than the assumption in (1), already assume that $f_1, \ldots , f_ r \in \mathcal{J}(U)$ and $f_ i = \sum a_{ij}g_ j$ with $(a_{ij})$ invertible as above, where now $g_1, \ldots , g_ r$ is a $H_1$-regular or Koszul-regular sequence. Since the Koszul complex on $f_1, \ldots , f_ r$ is isomorphic to the Koszul complex on $g_1, \ldots , g_ r$ via the matrix $(a_{ij})$ (see More on Algebra, Lemma 15.28.4) we conclude that $f_1, \ldots , f_ r$ is $H_1$-regular or Koszul-regular as desired. $\square$