The Stacks project

Proof. We fix the prime $\mathfrak p \supset I$ throughout the proof. Assume $IB$ is quasi-regular. By Lemma 15.32.2 $IB$ is a finite module, hence $I$ is a finite $A$-module by Algebra, Lemma 10.83.2. As $A \to B$ is flat we see that

As $IB$ is quasi-regular, the $B/IB$-module $IB/(IB)^2$ is finite locally free. Hence $I/I^2$ is finite projective, see Algebra, Proposition 10.83.3. In particular, after replacing $A$ by $A_ f$ for some $f \in A$, $f \not\in \mathfrak p$ we may assume that $I/I^2$ is free of rank $r$. Pick $f_1, \ldots , f_ r \in I$ which give a basis of $I/I^2$. By Nakayama's lemma (see Algebra, Lemma 10.20.1) we see that, after another replacement $A \leadsto A_ f$ as above, $I$ is generated by $f_1, \ldots , f_ r$.

Proof of the “quasi-regular” case. Above we have seen that $I/I^2$ is free on the $r$-generators $f_1, \ldots , f_ r$. To finish the proof in this case we have to show that the maps $\text{Sym}^ d(I/I^2) \to I^ d/I^{d + 1}$ are isomorphisms for each $d \geq 2$. This is clear as the faithfully flat base changes $\text{Sym}^ d(IB/(IB)^2) \to (IB)^ d/(IB)^{d + 1}$ are isomorphisms locally on $B$ by assumption. Details omitted.

Proof of the “$H_1$-regular” and “Koszul-regular” case. Consider the sequence of elements $f_1, \ldots , f_ r$ generating $I$ we constructed above. By Lemma 15.30.15 we see that $f_1, \ldots , f_ r$ map to a $H_1$-regular or Koszul-regular sequence in $B_ g$ for any $g \in B$ such that $IB$ is generated by an $H_1$-regular or Koszul-regular sequence. Hence $K_\bullet (A, f_1, \ldots , f_ r) \otimes _ A B_ g$ has vanishing $H_1$ or $H_ i$, $i > 0$. Since the homology of $K_\bullet (B, f_1, \ldots , f_ r) = K_\bullet (A, f_1, \ldots , f_ r) \otimes _ A B$ is annihilated by $IB$ (see Lemma 15.28.6) and since $V(IB) \subset \bigcup _{g\text{ as above}} D(g)$ we conclude that $K_\bullet (A, f_1, \ldots , f_ r) \otimes _ A B$ has vanishing homology in degree $1$ or all positive degrees. Using that $A \to B$ is faithfully flat we conclude that the same is true for $K_\bullet (A, f_1, \ldots , f_ r)$. $\square$