Lemma 10.100.1. Let $(R, \mathfrak m)$ be a local ring with nilpotent maximal ideal $\mathfrak m$. Let $M$ be a flat $R$-module. If $A$ is a set and $x_\alpha \in M$, $\alpha \in A$ is a collection of elements of $M$, then the following are equivalent:

$\{ \overline{x}_\alpha \} _{\alpha \in A}$ forms a basis for the vector space $M/\mathfrak mM$ over $R/\mathfrak m$, and

$\{ x_\alpha \} _{\alpha \in A}$ forms a basis for $M$ over $R$.

**Proof.**
The implication (2) $\Rightarrow $ (1) is immediate. We will prove the other implication by using induction on $n$ to show that $\{ x_\alpha \} _{\alpha \in A}$ forms a basis for $M/\mathfrak m^ nM$ over $R/\mathfrak m^ n$. The case $n = 1$ holds by assumption (1). Assume the statement holds for some $n \geq 1$. By Nakayama's Lemma 10.19.1 the elements $x_\alpha $ generate $M$, in particular $M/\mathfrak m^{n + 1}M$. The exact sequence $0 \to \mathfrak m^ n/\mathfrak m^{n + 1} \to R/\mathfrak m^{n + 1} \to R/\mathfrak m^ n \to 0$ gives on tensoring with $M$ the exact sequence

\[ 0 \to \mathfrak m^ nM/\mathfrak m^{n + 1}M \to M/\mathfrak m^{n + 1}M \to M/\mathfrak m^ nM \to 0 \]

Here we are using that $M$ is flat. Moreover, we have $\mathfrak m^ nM/\mathfrak m^{n + 1}M = M/\mathfrak mM \otimes _{R/\mathfrak m} \mathfrak m^ n/\mathfrak m^{n + 1}$ by flatness of $M$ again. Now suppose that $\sum f_\alpha x_\alpha = 0$ in $M/\mathfrak m^{n + 1}M$. Then by induction hypothesis $f_\alpha \in \mathfrak m^ n$ for each $\alpha $. By the short exact sequence above we then conclude that $\sum \overline{f}_\alpha \otimes \overline{x}_\alpha $ is zero in $\mathfrak m^ n/\mathfrak m^{n + 1} \otimes _{R/\mathfrak m} M/\mathfrak mM$. Since $\overline{x}_\alpha $ forms a basis we conclude that each of the congruence classes $\overline{f}_\alpha \in \mathfrak m^ n/\mathfrak m^{n + 1}$ is zero and we win.
$\square$

## Comments (1)

Comment #3621 by JuanPablo on

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