## 38.14 Examples of relatively pure modules

In the short section we discuss some examples of results that will serve as motivation for the notion of a *relatively pure module* and the concept of an *impurity* which we will introduce later. Each of the examples is stated as a lemma. Note the similarity with the condition on associated primes to the conditions appearing in Lemmas 38.7.4, 38.8.3, 38.8.4, and 38.9.1. See also Algebra, Lemma 10.65.1 for a discussion.

Lemma 38.14.1. Let $R$ be a local ring with maximal ideal $\mathfrak m$. Let $R \to S$ be a ring map. Let $N$ be an $S$-module. Assume

$N$ is projective as an $R$-module, and

$S/\mathfrak mS$ is Noetherian and $N/\mathfrak mN$ is a finite $S/\mathfrak mS$-module.

Then for any prime $\mathfrak q \subset S$ which is an associated prime of $N \otimes _ R \kappa (\mathfrak p)$ where $\mathfrak p = R \cap \mathfrak q$ we have $\mathfrak q + \mathfrak m S \not= S$.

**Proof.**
Note that the hypotheses of Lemmas 38.7.1 and 38.7.6 are satisfied. We will use the conclusions of these lemmas without further mention. Let $\Sigma \subset S$ be the multiplicative set of elements which are not zerodivisors on $N/\mathfrak mN$. The map $N \to \Sigma ^{-1}N$ is $R$-universally injective. Hence we see that any $\mathfrak q \subset S$ which is an associated prime of $N \otimes _ R \kappa (\mathfrak p)$ is also an associated prime of $\Sigma ^{-1}N \otimes _ R \kappa (\mathfrak p)$. Clearly this implies that $\mathfrak q$ corresponds to a prime of $\Sigma ^{-1}S$. Thus $\mathfrak q \subset \mathfrak q'$ where $\mathfrak q'$ corresponds to an associated prime of $N/\mathfrak mN$ and we win.
$\square$

The following lemma gives another (slightly silly) example of this phenomenon.

Lemma 38.14.2. Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $R \to S$ be a ring map. Let $N$ be an $S$-module. If $N$ is $I$-adically complete, then for any $R$-module $M$ and for any prime $\mathfrak q \subset S$ which is an associated prime of $N \otimes _ R M$ we have $\mathfrak q + I S \not= S$.

**Proof.**
Let $S^\wedge $ denote the $I$-adic completion of $S$. Note that $N$ is an $S^\wedge $-module, hence also $N \otimes _ R M$ is an $S^\wedge $-module. Let $z \in N \otimes _ R M$ be an element such that $\mathfrak q = \text{Ann}_ S(z)$. Since $z \not= 0$ we see that $\text{Ann}_{S^\wedge }(z) \not= S^\wedge $. Hence $\mathfrak q S^\wedge \not= S^\wedge $. Hence there exists a maximal ideal $\mathfrak m \subset S^\wedge $ with $\mathfrak q S^\wedge \subset \mathfrak m$. Since $IS^\wedge \subset \mathfrak m$ by Algebra, Lemma 10.96.6 we win.
$\square$

Note that the following lemma gives an alternative proof of Lemma 38.14.1 as a projective module over a local ring is free, see Algebra, Theorem 10.85.4.

Lemma 38.14.3. Let $R$ be a local ring with maximal ideal $\mathfrak m$. Let $R \to S$ be a ring map. Let $N$ be an $S$-module. Assume $N$ is isomorphic as an $R$-module to a direct sum of finite $R$-modules. Then for any $R$-module $M$ and for any prime $\mathfrak q \subset S$ which is an associated prime of $N \otimes _ R M$ we have $\mathfrak q + \mathfrak m S \not= S$.

**Proof.**
Write $N = \bigoplus _{i \in I} M_ i$ with each $M_ i$ a finite $R$-module. Let $M$ be an $R$-module and let $\mathfrak q \subset S$ be an associated prime of $N \otimes _ R M$ such that $\mathfrak q + \mathfrak m S = S$. Let $z \in N \otimes _ R M$ be an element with $\mathfrak q = \text{Ann}_ S(z)$. After modifying the direct sum decomposition a little bit we may assume that $z \in M_1 \otimes _ R M$ for some element $1 \in I$. Write $1 = f + \sum x_ j g_ j$ for some $f \in \mathfrak q$, $x_ j \in \mathfrak m$, and $g_ j \in S$. For any $g \in S$ denote $g'$ the $R$-linear map

\[ M_1 \to N \xrightarrow {g} N \to M_1 \]

where the first arrow is the inclusion map, the second arrow is multiplication by $g$ and the third arrow is the projection map. Because each $x_ j \in R$ we obtain the equality

\[ f' + \sum x_ j g'_ j = \text{id}_{M_1} \in \text{End}_ R(M_1) \]

By Nakayama's lemma (Algebra, Lemma 10.20.1) we see that $f'$ is surjective, hence by Algebra, Lemma 10.16.4 we see that $f'$ is an isomorphism. In particular the map

\[ M_1 \otimes _ R M \to N \otimes _ R M \xrightarrow {f} N \otimes _ R M \to M_1 \otimes _ R M \]

is an isomorphism. This contradicts the assumption that $fz = 0$.
$\square$

Lemma 38.14.4. Let $R$ be a henselian local ring with maximal ideal $\mathfrak m$. Let $R \to S$ be a ring map. Let $N$ be an $S$-module. Assume $N$ is countably generated and Mittag-Leffler as an $R$-module. Then for any $R$-module $M$ and for any prime $\mathfrak q \subset S$ which is an associated prime of $N \otimes _ R M$ we have $\mathfrak q + \mathfrak m S \not= S$.

**Proof.**
This lemma reduces to Lemma 38.14.3 by Algebra, Lemma 10.153.13.
$\square$

Suppose $f : X \to S$ is a morphism of schemes and $\mathcal{F}$ is a quasi-coherent module on $X$. Let $\xi \in \text{Ass}_{X/S}(\mathcal{F})$ and let $Z = \overline{\{ \xi \} }$. Picture

\[ \xymatrix{ \xi \ar@{|->}[d] & Z \ar[r] \ar[rd] & X \ar[d]^ f \\ f(\xi ) & & S } \]

Note that $f(Z) \subset \overline{\{ f(\xi )\} }$ and that $f(Z)$ is closed if and only if equality holds, i.e., $f(Z) = \overline{\{ f(\xi )\} }$. It follows from Lemma 38.14.1 that if $S$, $X$ are affine, the fibres $X_ s$ are Noetherian, $\mathcal{F}$ is of finite type, and $\Gamma (X, \mathcal{F})$ is a projective $\Gamma (S, \mathcal{O}_ S)$-module, then $f(Z) = \overline{\{ f(\xi )\} }$ is a closed subset. Slightly different analogous statements holds for the cases described in Lemmas 38.14.2, 38.14.3, and 38.14.4.

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