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[parent] proof of Nakayama's lemma (Proof)

(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}$.

Bibliography

1
Serre, J.-P. Local Algebra. Springer-Verlag, 2000.



"proof of Nakayama's lemma" is owned by nerdy2.
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Cross-references: maximal ideal, isomorphic, quotient, simple
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This is version 3 of proof of Nakayama's lemma, born on 2002-12-15, modified 2002-12-15.
Object id is 3764, canonical name is ProofOfNakayamasLemma2.
Accessed 2330 times total.

Classification:
AMS MSC: 13C99 (Commutative rings and algebras :: Theory of modules and ideals :: Miscellaneous)
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nakayama's lemma proof by nerdy by sjm on 2008-05-16 05:20:41
Where is the hypothesis that M is finitely generated used in Serre's
proof?
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