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I have read that Nakayama's Lemma has a nice consequence for local rings. If $R$ is a local ring with a finitely generated module $M$ and maximal ideal $\mathfrak{m}$, then we see $M/\mathfrak{m}M$ is a vector space over the field $R/\mathfrak{m}$. If a set $\{b_1,\dots,b_n\}$ is a basis of $M/\mathfrak{m}M$, then pulling the $b_i$ back to $M$ gives a set of generators of $M$.

What's the reason for this?

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    Read Theorem 2.3 in Matsumura's *Commutative ring theory*.2011-11-28

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(Mariano's comment suffices but let me write the idea as an answer for the sake of completeness. This post is community wiki so I will not gain reputation for upvotes (or lose reputation for downvotes!).)

If $b_1,\dots,b_n\in M$ and if $\overline{b_1},\dots,\overline{b_n}$ is an $R/m$-basis for $M/mM$, then let $N=b_1A+\cdots+b_nA$ (be the submodule of $M$ generated by $b_1,\dots,b_n$).

Exercise 1: Prove that $N+mM=M$.

Exercise 2: Prove that $N=M$ by remembering that $R$ is a local ring and applying Nakayama's lemma.

I hope this helps!