Lemma 10.101.3. Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $M$ be an $R$-module. Let $A$ be a set and let $x_\alpha \in M$, $\alpha \in A$ be a collection of elements of $M$. Assume

$I$ is nilpotent,

$\{ \overline{x}_\alpha \} _{\alpha \in A}$ forms a basis for $M/IM$ over $R/I$, and

$\text{Tor}_1^ R(R/I, M) = 0$.

Then $M$ is free on $\{ x_\alpha \} _{\alpha \in A}$ over $R$.

**Proof.**
Let $R$, $I$, $M$, $\{ x_\alpha \} _{\alpha \in A}$ be as in the lemma and satisfy assumptions (1), (2), and (3). By Nakayama's Lemma 10.20.1 the elements $x_\alpha $ generate $M$ over $R$. The assumption $\text{Tor}_1^ R(R/I, M) = 0$ implies that we have a short exact sequence

\[ 0 \to I \otimes _ R M \to M \to M/IM \to 0. \]

Let $\sum f_\alpha x_\alpha = 0$ be a relation in $M$. By choice of $x_\alpha $ we see that $f_\alpha \in I$. Hence we conclude that $\sum f_\alpha \otimes x_\alpha = 0$ in $I \otimes _ R M$. The map $I \otimes _ R M \to I/I^2 \otimes _{R/I} M/IM$ and the fact that $\{ x_\alpha \} _{\alpha \in A}$ forms a basis for $M/IM$ implies that $f_\alpha \in I^2$! Hence we conclude that there are no relations among the images of the $x_\alpha $ in $M/I^2M$. In other words, we see that $M/I^2M$ is free with basis the images of the $x_\alpha $. Using the map $I \otimes _ R M \to I/I^3 \otimes _{R/I^2} M/I^2M$ we then conclude that $f_\alpha \in I^3$! And so on. Since $I^ n = 0$ for some $n$ by assumption (1) we win.
$\square$

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