
Lemma 10.95.1. 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 10.19.1 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 $x$ which proves the claim. Now we may apply Lemma 10.86.1 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 10.38.12. Hence we can directly apply Lemma 10.86.1 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. $\square$

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