# proof of Nakayama’s lemma

(This proof was taken from [1].)

If $M$ were not zero, it would have a simple quotient, isomorphic to $R/\mathfrak{m}$ for some maximal ideal $\mathfrak{m}$ of $R$. Then we would have $\mathfrak{m}M\neq M$, so that $\mathfrak{a}M\neq M$ as $\mathfrak{a}\subseteq\mathfrak{m}$.

## References

• 1 Serre, J.-P. Local Algebra. Springer-Verlag, 2000.
Title proof of Nakayama’s lemma ProofOfNakayamasLemma 2013-03-22 13:16:50 2013-03-22 13:16:50 nerdy2 (62) nerdy2 (62) 6 nerdy2 (62) Proof msc 13C99