# Application of Nakayama's lemma

Let $R$ be a local ring with maximal ideal $\mathscr M$. Let, $M$ be a finitely generated $R$-module. Let, $x_1,x_2,\cdots,x_n\in M$ such that $\{\bar{x_1},\bar{x_2},\cdots ,\bar{x_n}\}$ is a basis of $M/M\mathscr{M}$ over $R/\mathscr{M}$. Then show that $\{x_1,x_2,\cdots , x_n\}$ generates $M$.

Let, $N$ be a submodule of $M$ which is spanned by $\{x_1,x_2,\cdots,x_n\}$. Then if I can show that $N+\mathscr{M}M=M$ then using Nakayama's lemma I have done. But I'm stuck here to show $N+\mathscr{M}M=M$.

How I can show it? Any hint. ?

• Let $x\in M$ be arbitrary. Write $\overline{x}$ using the given basis. And... Sep 8, 2017 at 5:42

Let $N$ be the $R$-submodule of $M$ spanned by the $x_i$, as you suggest. We want to show that $N + \mathscr{M}M = M$. Clearly $N + \mathscr{M}M \subseteq M,$ so suppose that $m\in M$. Because $\{x_1 + \mathscr{M}M,\dots, x_n + \mathscr{M}M\}$ is a $R/\mathscr{M}$-basis of $M/\mathscr{M}M$, it follows that $$m + \mathscr{M}M = \sum_{i = 1}^n\alpha_i x_i + \mathscr{M}M,$$ for some elements $\alpha_i\in R$. Thus, it follows that $$n = m - \sum_{i = 1}^n\alpha_i x_i\in\mathscr{M}M.$$ However, this proves the result: we have expressed an arbitrary element of $M$ as an $R$-linear combination of the $x_i$ and elements of $\mathscr{M}M$.