Lemma 10.78.8. Let $R$ be a local ring with maximal ideal $\mathfrak m$ and infinite residue field. Let $R \to S$ be a ring map. Let $M$ be an $S$-module and let $N \subset M$ be an $R$-submodule. Assume

1. $S$ is semi-local and $\mathfrak mS$ is contained in the Jacobson radical of $S$,

2. $M$ is a finite free $S$-module, and

3. $N$ generates $M$ as an $S$-module.

Then $N$ contains an $S$-basis of $M$.

Proof. Assume $M$ is free of rank $n$. Let $I \subset S$ be the Jacobson radical. By Nakayama's Lemma 10.20.1 a sequence of elements $m_1, \ldots , m_ n$ is a basis for $M$ if and only if $\overline{m}_ i \in M/IM$ generate $M/IM$. Hence we may replace $M$ by $M/IM$, $N$ by $N/(N \cap IM)$, $R$ by $R/\mathfrak m$, and $S$ by $S/IS$. In this case we see that $S$ is a finite product of fields $S = k_1 \times \ldots \times k_ r$ and $M = k_1^{\oplus n} \times \ldots \times k_ r^{\oplus n}$. The fact that $N \subset M$ generates $M$ as an $S$-module means that there exist $x_ j \in N$ such that a linear combination $\sum a_ j x_ j$ with $a_ j \in S$ has a nonzero component in each factor $k_ i^{\oplus n}$. Because $R = k$ is an infinite field, this means that also some linear combination $y = \sum c_ j x_ j$ with $c_ j \in k$ has a nonzero component in each factor. Hence $y \in N$ generates a free direct summand $Sy \subset M$. By induction on $n$ the result holds for $M/Sy$ and the submodule $\overline{N} = N/(N \cap Sy)$. In other words there exist $\overline{y}_2, \ldots , \overline{y}_ n$ in $\overline{N}$ which (freely) generate $M/Sy$. Then $y, y_2, \ldots , y_ n$ (freely) generate $M$ and we win. $\square$

There are also:

• 4 comment(s) on Section 10.78: Finite projective modules

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).