Lemma 15.31.4. Let $A \to B$ and $A \to A'$ be ring maps. Set $B' = B \otimes _ A A'$. Let $f_1, \ldots , f_ r \in B$. Assume $B/(f_1, \ldots , f_ r)B$ is flat over $A$
If $f_1, \ldots , f_ r$ is a quasi-regular sequence, then the image in $B'$ is a quasi-regular sequence.
If $f_1, \ldots , f_ r$ is a $H_1$-regular sequence, then the image in $B'$ is a $H_1$-regular sequence.
Proof. Assume $f_1, \ldots , f_ r$ is quasi-regular. Set $J = (f_1, \ldots , f_ r)$. By assumption $J^ n/J^{n + 1}$ is isomorphic to a direct sum of copies of $B/J$ hence flat over $A$. By induction and Algebra, Lemma 10.39.13 we conclude that $B/J^ n$ is flat over $A$. The ideal $(J')^ n$ is equal to $J^ n \otimes _ A A'$, see Algebra, Lemma 10.39.12. Hence $(J')^ n/(J')^{n + 1} = J^ n/J^{n + 1} \otimes _ A A'$ which clearly implies that $f_1, \ldots , f_ r$ is a quasi-regular sequence in $B'$.
Assume $f_1, \ldots , f_ r$ is $H_1$-regular. By Lemma 15.31.3 the vanishing of the Koszul homology group $H_1(K_\bullet (B, f_1, \ldots , f_ r))$ implies the vanishing of $H_1(K_\bullet (B', f'_1, \ldots , f'_ r))$ and we win. $\square$
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