The Stacks project

Lemma 15.32.4. Let $A \to B$ be a faithfully flat ring map. Let $I \subset A$ be an ideal. If $IB$ is a Koszul-regular (resp. $H_1$-regular, resp. quasi-regular) ideal in $B$, then $I$ is a Koszul-regular (resp. $H_1$-regular, resp. quasi-regular) ideal in $A$.

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

\[ I/I^2 \otimes _{A/I} B/IB = I/I^2 \otimes _ A B = IB/(IB)^2. \]

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 $K_\bullet (A, f_1, \ldots , f_ r)$. $\square$

Comments (0)

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 068N. Beware of the difference between the letter 'O' and the digit '0'.