38.7 Localization and universally injective maps
Lemma 38.7.1. Let $R \to S$ be a ring map. Let $N$ be a $S$-module. Assume
$R$ is a local ring with maximal ideal $\mathfrak m$,
$\overline{S} = S/\mathfrak m S$ is Noetherian, and
$\overline{N} = N/\mathfrak m_ R N$ is a finite $\overline{S}$-module.
Let $\Sigma \subset S$ be the multiplicative subset of elements which are not a zerodivisor on $\overline{N}$. Then $\Sigma ^{-1}S$ is a semi-local ring whose spectrum consists of primes $\mathfrak q \subset S$ contained in an element of $\text{Ass}_ S(\overline{N})$. Moreover, any maximal ideal of $\Sigma ^{-1}S$ corresponds to an associated prime of $\overline{N}$ over $\overline{S}$.
Proof.
Note that $\text{Ass}_ S(\overline{N}) = \text{Ass}_{\overline{S}}(\overline{N})$, see Algebra, Lemma 10.63.14. This is a finite set by Algebra, Lemma 10.63.5. Say $\{ \mathfrak q_1, \ldots , \mathfrak q_ r\} = \text{Ass}_ S(\overline{N})$. We have $\Sigma = S \setminus (\bigcup \mathfrak q_ i)$ by Algebra, Lemma 10.63.9. By the description of $\mathop{\mathrm{Spec}}(\Sigma ^{-1}S)$ in Algebra, Lemma 10.17.5 and by Algebra, Lemma 10.15.2 we see that the primes of $\Sigma ^{-1}S$ correspond to the primes of $S$ contained in one of the $\mathfrak q_ i$. Hence the maximal ideals of $\Sigma ^{-1}S$ correspond one-to-one with the maximal (w.r.t. inclusion) elements of the set $\{ \mathfrak q_1, \ldots , \mathfrak q_ r\} $. This proves the lemma.
$\square$
Lemma 38.7.2. Assumption and notation as in Lemma 38.7.1. Assume moreover that
$S$ is local and $R \to S$ is a local homomorphism,
$S$ is essentially of finite presentation over $R$,
$N$ is finitely presented over $S$, and
$N$ is flat over $R$.
Then each $s \in \Sigma $ defines a universally injective $R$-module map $s : N \to N$, and the map $N \to \Sigma ^{-1}N$ is $R$-universally injective.
Proof.
By Algebra, Lemma 10.128.4 the sequence $0 \to N \to N \to N/sN \to 0$ is exact and $N/sN$ is flat over $R$. This implies that $s : N \to N$ is universally injective, see Algebra, Lemma 10.39.12. The map $N \to \Sigma ^{-1}N$ is universally injective as the directed colimit of the maps $s : N \to N$.
$\square$
Lemma 38.7.3. Let $R \to S$ be a ring map. Let $N$ be an $S$-module. Let $S \to S'$ be a ring map. Assume
$R \to S$ is a local homomorphism of local rings
$S$ is essentially of finite presentation over $R$,
$N$ is of finite presentation over $S$,
$N$ is flat over $R$,
$S \to S'$ is flat, and
the image of $\mathop{\mathrm{Spec}}(S') \to \mathop{\mathrm{Spec}}(S)$ contains all primes $\mathfrak q$ of $S$ lying over $\mathfrak m_ R$ such that $\mathfrak q$ is an associated prime of $N/\mathfrak m_ R N$.
Then $N \to N \otimes _ S S'$ is $R$-universally injective.
Proof.
Set $N' = N \otimes _ R S'$. Consider the commutative diagram
\[ \xymatrix{ N \ar[d] \ar[r] & N' \ar[d] \\ \Sigma ^{-1}N \ar[r] & \Sigma ^{-1}N' } \]
where $\Sigma \subset S$ is the set of elements which are not a zerodivisor on $N/\mathfrak m_ R N$. If we can show that the map $N \to \Sigma ^{-1}N'$ is universally injective, then $N \to N'$ is too (see Algebra, Lemma 10.82.10).
By Lemma 38.7.1 the ring $\Sigma ^{-1}S$ is a semi-local ring whose maximal ideals correspond to associated primes of $N/\mathfrak m_ R N$. Hence the image of $\mathop{\mathrm{Spec}}(\Sigma ^{-1}S') \to \mathop{\mathrm{Spec}}(\Sigma ^{-1}S)$ contains all these maximal ideals by assumption. By Algebra, Lemma 10.39.16 the ring map $\Sigma ^{-1}S \to \Sigma ^{-1}S'$ is faithfully flat. Hence $\Sigma ^{-1}N \to \Sigma ^{-1}N'$, which is the map
\[ N \otimes _ S \Sigma ^{-1}S \longrightarrow N \otimes _ S \Sigma ^{-1}S' \]
is universally injective, see Algebra, Lemmas 10.82.11 and 10.82.8. Finally, we apply Lemma 38.7.2 to see that $N \to \Sigma ^{-1}N$ is universally injective. As the composition of universally injective module maps is universally injective (see Algebra, Lemma 10.82.9) we conclude that $N \to \Sigma ^{-1}N'$ is universally injective and we win.
$\square$
Lemma 38.7.4. Let $R \to S$ be a ring map. Let $N$ be an $S$-module. Let $S \to S'$ be a ring map. Assume
$R \to S$ is of finite presentation and $N$ is of finite presentation over $S$,
$N$ is flat over $R$,
$S \to S'$ is flat, and
the image of $\mathop{\mathrm{Spec}}(S') \to \mathop{\mathrm{Spec}}(S)$ contains all primes $\mathfrak q$ such that $\mathfrak q$ is an associated prime of $N \otimes _ R \kappa (\mathfrak p)$ where $\mathfrak p$ is the inverse image of $\mathfrak q$ in $R$.
Then $N \to N \otimes _ S S'$ is $R$-universally injective.
Proof.
By Algebra, Lemma 10.82.12 it suffices to show that $N_{\mathfrak q} \to (N \otimes _ R S')_{\mathfrak q}$ is a $R_{\mathfrak p}$-universally injective for any prime $\mathfrak q$ of $S$ lying over $\mathfrak p$ in $R$. Thus we may apply Lemma 38.7.3 to the ring maps $R_{\mathfrak p} \to S_{\mathfrak q} \to S'_{\mathfrak q}$ and the module $N_{\mathfrak q}$.
$\square$
The reader may want to compare the following lemma to Algebra, Lemmas 10.99.1 and 10.128.4 and the results of Section 38.25. In each case the conclusion is that the map $u : M \to N$ is universally injective with flat cokernel.
Lemma 38.7.5. Let $(R, \mathfrak m)$ be a local ring. Let $u : M \to N$ be an $R$-module map. If $M$ is a projective $R$-module, $N$ is a flat $R$-module, and $\overline{u} : M/\mathfrak mM \to N/\mathfrak mN$ is injective then $u$ is universally injective.
Proof.
By Algebra, Theorem 10.85.4 the module $M$ is free. If we show the result holds for every finitely generated direct summand of $M$, then the lemma follows. Hence we may assume that $M$ is finite free. Write $N = \mathop{\mathrm{colim}}\nolimits _ i N_ i$ as a directed colimit of finite free modules, see Algebra, Theorem 10.81.4. Note that $u : M \to N$ factors through $N_ i$ for some $i$ (as $M$ is finite free). Denote $u_ i : M \to N_ i$ the corresponding $R$-module map. As $\overline{u}$ is injective we see that $\overline{u_ i} : M/\mathfrak mM \to N_ i/\mathfrak mN_ i$ is injective and remains injective on composing with the maps $N_ i/\mathfrak mN_ i \to N_{i'}/\mathfrak mN_{i'}$ for all $i' \geq i$. As $M$ and $N_{i'}$ are finite free over the local ring $R$ this implies that $M \to N_{i'}$ is a split injection for all $i' \geq i$. Hence for any $R$-module $Q$ we see that $M \otimes _ R Q \to N_{i'} \otimes _ R Q$ is injective for all $i' \geq i$. As $- \otimes _ R Q$ commutes with colimits we conclude that $M \otimes _ R Q \to N_{i'} \otimes _ R Q$ is injective as desired.
$\square$
Lemma 38.7.6. Assumption and notation as in Lemma 38.7.1. Assume moreover that $N$ is projective as an $R$-module. Then each $s \in \Sigma $ defines a universally injective $R$-module map $s : N \to N$, and the map $N \to \Sigma ^{-1}N$ is $R$-universally injective.
Proof.
Pick $s \in \Sigma $. By Lemma 38.7.5 the map $s : N \to N$ is universally injective. The map $N \to \Sigma ^{-1}N$ is universally injective as the directed colimit of the maps $s : N \to N$.
$\square$
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