The Stacks project

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15.31 Regular ideals

We will discuss the notion of a regular ideal sheaf in great generality in Divisors, Section 30.20. Here we define the corresponding notion in the affine case, i.e., in the case of an ideal in a ring.

Definition 15.31.1. Let $R$ be a ring and let $I \subset R$ be an ideal.

  1. We say $I$ is a regular ideal if for every $\mathfrak p \in V(I)$ there exists a $g \in R$, $g \not\in \mathfrak p$ and a regular sequence $f_1, \ldots , f_ r \in R_ g$ such that $I_ g$ is generated by $f_1, \ldots , f_ r$.

  2. We say $I$ is a Koszul-regular ideal if for every $\mathfrak p \in V(I)$ there exists a $g \in R$, $g \not\in \mathfrak p$ and a Koszul-regular sequence $f_1, \ldots , f_ r \in R_ g$ such that $I_ g$ is generated by $f_1, \ldots , f_ r$.

  3. We say $I$ is a $H_1$-regular ideal if for every $\mathfrak p \in V(I)$ there exists a $g \in R$, $g \not\in \mathfrak p$ and an $H_1$-regular sequence $f_1, \ldots , f_ r \in R_ g$ such that $I_ g$ is generated by $f_1, \ldots , f_ r$.

  4. We say $I$ is a quasi-regular ideal if for every $\mathfrak p \in V(I)$ there exists a $g \in R$, $g \not\in \mathfrak p$ and a quasi-regular sequence $f_1, \ldots , f_ r \in R_ g$ such that $I_ g$ is generated by $f_1, \ldots , f_ r$.

It is clear that given $I \subset R$ we have the implications

\begin{align*} I\text{ is a regular ideal} & \Rightarrow I\text{ is a Koszul-regular ideal} \\ & \Rightarrow I\text{ is a }H_1\text{-regular ideal} \\ & \Rightarrow I\text{ is a quasi-regular ideal} \end{align*}

see Lemmas 15.29.2, 15.29.3, and 15.29.6. Such an ideal is always finitely generated.

Proof. Let $I \subset R$ be a quasi-regular ideal. Since $V(I)$ is quasi-compact, there exist $g_1, \ldots , g_ m \in R$ such that $V(I) \subset D(g_1) \cup \ldots \cup D(g_ m)$ and such that $I_{g_ j}$ is generated by a quasi-regular sequence $g_{j1}, \ldots , g_{jr_ j} \in R_{g_ j}$. Write $g_{ji} = g'_{ji}/g_ j^{e_{ij}}$ for some $g'_{ij} \in I$. Write $1 + x = \sum g_ j h_ j$ for some $x \in I$ which is possible as $V(I) \subset D(g_1) \cup \ldots \cup D(g_ m)$. Note that $\mathop{\mathrm{Spec}}(R) = D(g_1) \cup \ldots \cup D(g_ m) \bigcup D(x)$ Then $I$ is generated by the elements $g'_{ij}$ and $x$ as these generate on each of the pieces of the cover, see Algebra, Lemma 10.22.2. $\square$

Lemma 15.31.3. Let $I \subset R$ be a quasi-regular ideal of a ring. Then $I/I^2$ is a finite projective $R/I$-module.

Proof. This follows from Algebra, Lemma 10.77.2 and the definitions. $\square$

We prove flat descent for Koszul-regular, $H_1$-regular, quasi-regular ideals.

Lemma 15.31.4. 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 15.31.2 $IB$ is a finite module, hence $I$ is a finite $A$-module by Algebra, Lemma 10.82.2. 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 10.82.3. 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 10.19.1) 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 15.29.15 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 $K_\bullet (A, f_1, \ldots , f_ r)$. $\square$

Lemma 15.31.5. Let $A$ be a ring. Let $I \subset J \subset A$ be ideals. Assume that $J/I \subset A/I$ is a $H_1$-regular ideal. Then $I \cap J^2 = IJ$.

Proof. Follows immediately from Lemma 15.29.9 by localizing. $\square$


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