Lemma 10.108.2. Let $R$ be a ring. Let $I \subset R$ be an ideal. The following are equivalent:

1. $I$ is pure,

2. for every ideal $J \subset R$ we have $J \cap I = IJ$,

3. for every finitely generated ideal $J \subset R$ we have $J \cap I = JI$,

4. for every $x \in R$ we have $(x) \cap I = xI$,

5. for every $x \in I$ we have $x = yx$ for some $y \in I$,

6. for every $x_1, \ldots , x_ n \in I$ there exists a $y \in I$ such that $x_ i = yx_ i$ for all $i = 1, \ldots , n$,

7. for every prime $\mathfrak p$ of $R$ we have $IR_{\mathfrak p} = 0$ or $IR_{\mathfrak p} = R_{\mathfrak p}$,

8. $\text{Supp}(I) = \mathop{\mathrm{Spec}}(R) \setminus V(I)$,

9. $I$ is the kernel of the map $R \to (1 + I)^{-1}R$,

10. $R/I \cong S^{-1}R$ as $R$-algebras for some multiplicative subset $S$ of $R$, and

11. $R/I \cong (1 + I)^{-1}R$ as $R$-algebras.

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. $\square$

Comment #3573 by shanbei on

In the beginning of the third line, instead of "J \otimes_R R/I=R/JI", it should be "J \otimes_R R/I=J/JI".

Comment #6648 by 七海辣辣米 on

In lemma 04PS, the first paragraph, there is a missed 0 in the second exact sequence.

Comment #6871 by on

OK, no, that is because the second exact sequence isn't a short exact sequence in general. So then we do not write the $0$ at the beginning. OK?

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