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

Let $ X = \{x_1, x_2, \dots, x_n\}$ be a minimal set of generators for $ M$, in the sense that $ M$ is not generated by any proper subset of $ X$.

Elements of $ \mathfrak{a}M$ can be written as linear combinations $ \sum a_i x_i$, where $ a_i \in \mathfrak{a}$.

Suppose that $ \vert X\vert > 0$. Since $ M = \mathfrak{a}M$, we can express $ x_1$ as a such a linear combination:

$\displaystyle x_1 = \sum a_i x_i.$
Moving the term involving $ a_1$ to the left, we have
$\displaystyle (1 - a_1)x_1 = \sum_{i > 1} a_i x_i.$
But $ a_1 \in J(R)$, so $ 1-a_1$ is invertible, say with inverse $ b$. Therefore,
$\displaystyle x_1 = \sum_{i > 1} b a_i x_i.$
But this means that $ x_1$ is redundant as a generator of $ M$, and so $ M$ is generated by the subset $ \{x_2, x_3, \dots, x_n\}$. This contradicts the minimality of $ X$.

We conclude that $ \vert X\vert = 0$ and therefore $ M = 0$.



"proof of Nakayama's lemma" is owned by mclase.
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Cross-references: subset, inverse, invertible, term, linear combinations, proper subset, generated by, generators, minimal

This is version 2 of proof of Nakayama's lemma, born on 2002-11-03, modified 2002-11-03.
Object id is 3564, canonical name is ProofOfNakayamasLemma.
Accessed 3110 times total.

Classification:
AMS MSC: 13C99 (Commutative rings and algebras :: Theory of modules and ideals :: Miscellaneous)
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