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
Comment #4203 by Johan on