The Stacks project

Proof. For any ideal $J$ of $R$ we have the short exact sequence $0 \to J \to R \to R/J \to 0$. Tensoring with $R/I$ we get an exact sequence $J \otimes _ R R/I \to R/I \to R/I + J \to 0$ and $J \otimes _ R R/I = J/JI$. Thus the equivalence of (1), (2), and (3) follows from Lemma 10.39.5. Moreover, these imply (4).

The implication (4) $\Rightarrow $ (5) is trivial. Assume (5) and let $x_1, \ldots , x_ n \in I$. Choose $y_ i \in I$ such that $x_ i = y_ ix_ i$. Let $y \in I$ be the element such that $1 - y = \prod _{i = 1, \ldots , n} (1 - y_ i)$. Then $x_ i = yx_ i$ for all $i = 1, \ldots , n$. Hence (6) holds, and it follows that (5) $\Leftrightarrow $ (6).

Assume (5). Let $x \in I$. Then $x = yx$ for some $y \in I$. Hence $x(1 - y) = 0$, which shows that $x$ maps to zero in $(1 + I)^{-1}R$. Of course the kernel of the map $R \to (1 + I)^{-1}R$ is always contained in $I$. Hence we see that (5) implies (9). Assume (9). Then for any $x \in I$ we see that $x(1 - y) = 0$ for some $y \in I$. In other words, $x = yx$. We conclude that (5) is equivalent to (9).

Assume (5). Let $\mathfrak p$ be a prime of $R$. If $\mathfrak p \not\in V(I)$, then $IR_{\mathfrak p} = R_{\mathfrak p}$. If $\mathfrak p \in V(I)$, in other words, if $I \subset \mathfrak p$, then $x \in I$ implies $x(1 - y) = 0$ for some $y \in I$, implies $x$ maps to zero in $R_{\mathfrak p}$, i.e., $IR_{\mathfrak p} = 0$. Thus we see that (7) holds.

Assume (7). Then $(R/I)_{\mathfrak p}$ is either $0$ or $R_{\mathfrak p}$ for any prime $\mathfrak p$ of $R$. Hence by Lemma 10.39.18 we see that (1) holds. At this point we see that all of (1) – (7) and (9) are equivalent.

As $IR_{\mathfrak p} = I_{\mathfrak p}$ we see that (7) implies (8). Finally, if (8) holds, then this means exactly that $I_{\mathfrak p}$ is the zero module if and only if $\mathfrak p \in V(I)$, which is clearly saying that (7) holds. Now (1) – (9) are equivalent.

Assume (1) – (9) hold. Then $R/I \subset (1 + I)^{-1}R$ by (9) and the map $R/I \to (1 + I)^{-1}R$ is also surjective by the description of localizations at primes afforded by (7). Hence (11) holds.

The implication (11) $\Rightarrow $ (10) is trivial. And (10) implies that (1) holds because a localization of $R$ is flat over $R$, see Lemma 10.39.18. $\square$

Proof. For example, by property (7) of Lemma 10.108.2 we see that $I = \mathop{\mathrm{Ker}}(R \to \prod _{\mathfrak p \in V(I)} R_{\mathfrak p})$ can be recovered from the closed subset associated to it. $\square$

Proof. Let $I$ be a pure ideal. Then since $R \to R/I$ is flat, by going down generalizations lift along the map $\mathop{\mathrm{Spec}}(R/I) \to \mathop{\mathrm{Spec}}(R)$. Hence $V(I)$ is closed under generalizations. This shows that the map is well defined. By Lemma 10.108.3 the map is injective. Suppose that $Z \subset \mathop{\mathrm{Spec}}(R)$ is closed and closed under generalizations. Let $J \subset R$ be the radical ideal such that $Z = V(J)$. Let $I = \{ x \in R : x \in xJ\} $. Note that $I$ is an ideal: if $x, y \in I$ then there exist $f, g \in J$ such that $x = xf$ and $y = yg$. Then

Verification left to the reader. We claim that $I$ is pure and that $V(I) = V(J)$. If the claim is true then the map of the lemma is surjective and the lemma holds.

Note that $I \subset J$, so that $V(J) \subset V(I)$. Let $I \subset \mathfrak p$ be a prime. Consider the multiplicative subset $S = (R \setminus \mathfrak p)(1 + J)$. By definition of $I$ and $I \subset \mathfrak p$ we see that $0 \not\in S$. Hence we can find a prime $\mathfrak q$ of $R$ which is disjoint from $S$, see Lemmas 10.9.4 and 10.17.5. Hence $\mathfrak q \subset \mathfrak p$ and $\mathfrak q \cap (1 + J) = \emptyset $. This implies that $\mathfrak q + J$ is a proper ideal of $R$. Let $\mathfrak m$ be a maximal ideal containing $\mathfrak q + J$. Then we get $\mathfrak m \in V(J)$ and hence $\mathfrak q \in V(J) = Z$ as $Z$ was assumed to be closed under generalization. This in turn implies $\mathfrak p \in V(J)$ as $\mathfrak q \subset \mathfrak p$. Thus we see that $V(I) = V(J)$.

Finally, since $V(I) = V(J)$ (and $J$ radical) we see that $J = \sqrt{I}$. Pick $x \in I$, so that $x = xy$ for some $y \in J$ by definition. Then $x = xy = xy^2 = \ldots = xy^ n$. Since $y^ n \in I$ for some $n > 0$ we conclude that property (5) of Lemma 10.108.2 holds and we see that $I$ is indeed pure. $\square$

Proof. If (1) holds, then $I = I \cap I = I^2$ by Lemma 10.108.2. Hence $I$ is generated by an idempotent by Lemma 10.21.5. Thus (1) $\Rightarrow $ (2). If (2) holds, then $I = (e)$ and $R = (1 - e) \oplus (e)$ as an $R$-module hence $R/I$ is flat and $I$ is pure and $V(I) = D(1 - e)$ is open. Thus (2) $\Rightarrow $ (1) $+$ (3). Finally, assume (3). Then $V(I)$ is open and closed, hence $V(I) = D(1 - e)$ for some idempotent $e$ of $R$, see Lemma 10.21.3. The ideal $J = (e)$ is a pure ideal such that $V(J) = V(I)$ hence $I = J$ by Lemma 10.108.3. In this way we see that (3) $\Rightarrow $ (2). By Lemma 10.78.2 we see that (4) is equivalent to the assertion that $I$ is pure and $R/I$ finitely presented. Moreover, $R/I$ is finitely presented if and only if $I$ is finitely generated, see Lemma 10.5.3. Hence (4) is equivalent to (1). $\square$

Proof. If any finite flat $R$-module is finite locally free then the support of $R/I$ where $I$ is a pure ideal is open. Hence the implication (2) $\Rightarrow $ (1) follows from Lemma 10.108.3.

For the converse assume that $R$ satisfies (1). Let $M$ be a finite flat $R$-module. The support $Z = \text{Supp}(M)$ of $M$ is closed, see Lemma 10.40.5. On the other hand, if $\mathfrak p \subset \mathfrak p'$, then by Lemma 10.78.5 the module $M_{\mathfrak p'}$ is free, and $M_{\mathfrak p} = M_{\mathfrak p'} \otimes _{R_{\mathfrak p'}} R_{\mathfrak p}$ Hence $\mathfrak p' \in \text{Supp}(M) \Rightarrow \mathfrak p \in \text{Supp}(M)$, in other words, the support is closed under generalization. As $R$ satisfies (1) we see that the support of $M$ is open and closed. Suppose that $M$ is generated by $r$ elements $m_1, \ldots , m_ r$. The modules $\wedge ^ i(M)$, $i = 1, \ldots , r$ are finite flat $R$-modules also, because $\wedge ^ i(M)_{\mathfrak p} = \wedge ^ i(M_{\mathfrak p})$ is free over $R_{\mathfrak p}$. Note that $\text{Supp}(\wedge ^{i + 1}(M)) \subset \text{Supp}(\wedge ^ i(M))$. Thus we see that there exists a decomposition

by open and closed subsets such that the support of $\wedge ^ i(M)$ is $U_ r \cup \ldots \cup U_ i$ for all $i = 0, \ldots , r$. Let $\mathfrak p$ be a prime of $R$, and say $\mathfrak p \in U_ i$. Note that $\wedge ^ i(M) \otimes _ R \kappa (\mathfrak p) = \wedge ^ i(M \otimes _ R \kappa (\mathfrak p))$. Hence, after possibly renumbering $m_1, \ldots , m_ r$ we may assume that $m_1, \ldots , m_ i$ generate $M \otimes _ R \kappa (\mathfrak p)$. By Nakayama's Lemma 10.20.1 we get a surjection

for some $f \in R$, $f \not\in \mathfrak p$. We may also assume that $D(f) \subset U_ i$. This means that $\wedge ^ i(M_ f) = \wedge ^ i(M)_ f$ is a flat $R_ f$ module whose support is all of $\mathop{\mathrm{Spec}}(R_ f)$. By the above it is generated by a single element, namely $m_1 \wedge \ldots \wedge m_ i$. Hence $\wedge ^ i(M)_ f \cong R_ f/J$ for some pure ideal $J \subset R_ f$ with $V(J) = \mathop{\mathrm{Spec}}(R_ f)$. Clearly this means that $J = (0)$, see Lemma 10.108.3. Thus $m_1 \wedge \ldots \wedge m_ i$ is a basis for $\wedge ^ i(M_ f)$ and it follows that the displayed map is injective as well as surjective. This proves that $M$ is finite locally free as desired. $\square$