The Stacks project

Proof. As $M$ is flat, each of the sequences

is short exact, see Lemma 10.39.12 and the sequence $0 \to K^\wedge \to P^\wedge \to M^\wedge \to 0$ is a short exact sequence, see Lemma 10.96.1. It suffices to show that we can find splittings $s_ n : M/I^ nM \to P/I^ nP$ such that $s_{n + 1} \bmod I^ n = s_ n$. We will construct these $s_ n$ by induction on $n$. Pick any splitting $s_1$, which exists as $M/IM$ is a projective $R/I$-module. Assume given $s_ n$ for some $n > 0$. Set $P_{n + 1} = \{ x \in P \mid x \bmod I^ nP \in \mathop{\mathrm{Im}}(s_ n)\}$. The map $\pi : P_{n + 1}/I^{n + 1}P_{n + 1} \to M/I^{n + 1}M$ is surjective (details omitted). As $M/I^{n + 1}M$ is projective as a $R/I^{n + 1}$-module by Lemma 10.77.7 we may choose a section $t : M/I^{n + 1}M \to P_{n + 1}/I^{n + 1}P_{n + 1}$ of $\pi$. Setting $s_{n + 1}$ equal to the composition of $t$ with the canonical map $P_{n + 1}/I^{n + 1}P_{n + 1} \to P/I^{n + 1}P$ works. $\square$