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
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
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
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
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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