The Stacks project

Lemma 10.96.1. Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $\varphi : M \to N$ be a map of $R$-modules.

  1. If $M/IM \to N/IN$ is surjective, then $M^\wedge \to N^\wedge $ is surjective.

  2. If $M \to N$ is surjective, then $M^\wedge \to N^\wedge $ is surjective.

  3. If $0 \to K \to M \to N \to 0$ is a short exact sequence of $R$-modules and $N$ is flat, then $0 \to K^\wedge \to M^\wedge \to N^\wedge \to 0$ is a short exact sequence.

  4. The map $M \otimes _ R R^\wedge \to M^\wedge $ is surjective for any finite $R$-module $M$.

Proof. Assume $M/IM \to N/IN$ is surjective. Then the map $M/I^ nM \to N/I^ nN$ is surjective for each $n \geq 1$ by Nakayama's lemma. More precisely, apply Lemma 10.20.1 part (11) to the map $M/I^ nM \to N/I^ nN$ over the ring $R/I^ n$ and the nilpotent ideal $I/I^ n$ to see this. Set $K_ n = \{ x \in M \mid \varphi (x) \in I^ nN\} $. Thus we get short exact sequences

\[ 0 \to K_ n/I^ nM \to M/I^ nM \to N/I^ nN \to 0 \]

We claim that the canonical map $K_{n + 1}/I^{n + 1}M \to K_ n/I^ nM$ is surjective. Namely, if $x \in K_ n$ write $\varphi (x) = \sum z_ j n_ j$ with $z_ j \in I^ n$, $n_ j \in N$. By assumption we can write $n_ j = \varphi (m_ j) + \sum z_{jk}n_{jk}$ with $m_ j \in M$, $z_{jk} \in I$ and $n_{jk} \in N$. Hence

\[ \varphi (x - \sum z_ j m_ j) = \sum z_ jz_{jk} n_{jk}. \]

This means that $x' = x - \sum z_ j m_ j \in K_{n + 1}$ maps to $x \bmod I^ nM$ which proves the claim. Now we may apply Lemma 10.87.1 to the inverse system of short exact sequences above to see (1). Part (2) is a special case of (1). If the assumptions of (3) hold, then for each $n$ the sequence

\[ 0 \to K/I^ nK \to M/I^ nM \to N/I^ nN \to 0 \]

is short exact by Lemma 10.39.12. Hence we can directly apply Lemma 10.87.1 to conclude (3) is true. To see (4) choose generators $x_ i \in M$, $i = 1, \ldots , n$. Then the map $R^{\oplus n} \to M$, $(a_1, \ldots , a_ n) \mapsto \sum a_ ix_ i$ is surjective. Hence by (2) we see $(R^\wedge )^{\oplus n} \to M^\wedge $, $(a_1, \ldots , a_ n) \mapsto \sum a_ ix_ i$ is surjective. Assertion (4) follows from this. $\square$

Comments (2)

Comment #4201 by Qijun Yan on

For Lemma 0315 (3), it seems to me that the operation of taking completion always preserves short exact sequences by Lemma 0AS0. Maybe I got something wrong.

Comment #4203 by on

In Lemma 87.4.5 in Section 87.4 the topology on the submodule is the topology inherited from (it is given by the submodules ). But in the current Section 10.96 there is no mention whatsoever of topologies or completion with respect to any topology. We are just considering -adic completion straight up. Maybe we should be a little bi more careful in the statement of Lemma 87.4.5. Thanks for the comment.

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