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
Canonical name	ProofOfNakayamasLemma
Date of creation	2013-03-22 13:16:50
Last modified on	2013-03-22 13:16:50
Owner	nerdy2 (62)
Last modified by	nerdy2 (62)
Numerical id	6
Author	nerdy2 (62)
Entry type	Proof
Classification	msc 13C99