## 38.8 Completion and Mittag-Leffler modules

Lemma 38.8.1. Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $A$ be a set. Assume $R$ is Noetherian and complete with respect to $I$. The completion $(\bigoplus \nolimits _{\alpha \in A} R)^\wedge$ is flat and Mittag-Leffler.

Proof. By More on Algebra, Lemma 15.27.1 the map $(\bigoplus \nolimits _{\alpha \in A} R)^\wedge \to \prod _{\alpha \in A} R$ is universally injective. Thus, by Algebra, Lemmas 10.82.7 and 10.89.7 it suffices to show that $\prod _{\alpha \in A} R$ is flat and Mittag-Leffler. By Algebra, Proposition 10.90.6 (and Algebra, Lemma 10.90.5) we see that $\prod _{\alpha \in A} R$ is flat. Thus we conclude because a product of copies of $R$ is Mittag-Leffler, see Algebra, Lemma 10.91.3. $\square$

Lemma 38.8.2. Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $M$ be an $R$-module. Assume

1. $R$ is Noetherian and $I$-adically complete,

2. $M$ is flat over $R$, and

3. $M/IM$ is a projective $R/I$-module.

Then the $I$-adic completion $M^\wedge$ is a flat Mittag-Leffler $R$-module.

Proof. Choose a surjection $F \to M$ where $F$ is a free $R$-module. By Algebra, Lemma 10.97.9 the module $M^\wedge$ is a direct summand of the module $F^\wedge$. Hence it suffices to prove the lemma for $F$. In this case the lemma follows from Lemma 38.8.1. $\square$

In Lemmas 38.8.3 and 38.8.4 the assumption that $S$ be Noetherian holds if $R \to S$ is of finite type, see Algebra, Lemma 10.31.1.

Lemma 38.8.3. Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $R \to S$ be a ring map, and $N$ an $S$-module. Assume

1. $R$ is a Noetherian ring,

2. $S$ is a Noetherian ring,

3. $N$ is a finite $S$-module, and

4. for any finite $R$-module $Q$, any $\mathfrak q \in \text{Ass}_ S(Q \otimes _ R N)$ satisfies $IS + \mathfrak q \not= S$.

Then the map $N \to N^\wedge$ of $N$ into the $I$-adic completion of $N$ is universally injective as a map of $R$-modules.

Proof. We have to show that for any finite $R$-module $Q$ the map $Q \otimes _ R N \to Q \otimes _ R N^\wedge$ is injective, see Algebra, Theorem 10.82.3. As there is a canonical map $Q \otimes _ R N^\wedge \to (Q \otimes _ R N)^\wedge$ it suffices to prove that the canonical map $Q \otimes _ R N \to (Q \otimes _ R N)^\wedge$ is injective. Hence we may replace $N$ by $Q \otimes _ R N$ and it suffices to prove the injectivity for the map $N \to N^\wedge$.

Let $K = \mathop{\mathrm{Ker}}(N \to N^\wedge )$. It suffices to show that $K_{\mathfrak q} = 0$ for $\mathfrak q \in \text{Ass}(N)$ as $N$ is a submodule of $\prod _{\mathfrak q \in \text{Ass}(N)} N_{\mathfrak q}$, see Algebra, Lemma 10.63.19. Pick $\mathfrak q \in \text{Ass}(N)$. By the last assumption we see that there exists a prime $\mathfrak q' \supset IS + \mathfrak q$. Since $K_{\mathfrak q}$ is a localization of $K_{\mathfrak q'}$ it suffices to prove the vanishing of $K_{\mathfrak q'}$. Note that $K = \bigcap I^ nN$, hence $K_{\mathfrak q'} \subset \bigcap I^ nN_{\mathfrak q'}$. Hence $K_{\mathfrak q'} = 0$ by Algebra, Lemma 10.51.4. $\square$

Lemma 38.8.4. Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $R \to S$ be a ring map, and $N$ an $S$-module. Assume

1. $R$ is a Noetherian ring,

2. $S$ is a Noetherian ring,

3. $N$ is a finite $S$-module,

4. $N$ is flat over $R$, and

5. 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 $IS + \mathfrak q \not= S$.

Then the map $N \to N^\wedge$ of $N$ into the $I$-adic completion of $N$ is universally injective as a map of $R$-modules.

Proof. This follows from Lemma 38.8.3 because Algebra, Lemma 10.65.5 and Remark 10.65.6 guarantee that the set of associated primes of tensor products $N \otimes _ R Q$ are contained in the set of associated primes of the modules $N \otimes _ R \kappa (\mathfrak p)$. $\square$

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