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proof of Nakayama's lemma
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(Proof)
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(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 $\mathfrak{a}\subseteq \mathfrak{m}$
- 1
- Serre, J.-P. Local Algebra. Springer-Verlag, 2000.
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"proof of Nakayama's lemma" is owned by nerdy2.
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Cross-references: maximal ideal, isomorphic, quotient, simple, proof
There is 1 reference to this entry.
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 2895 times total.
Classification:
| AMS MSC: | 13C99 (Commutative rings and algebras :: Theory of modules and ideals :: Miscellaneous) |
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Pending Errata and Addenda
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