proof of Nakayama’s lemma

Let X={x1,x2,,xn} be a minimal set of generators for M, in the sense that M is not generated by any proper subsetMathworldPlanetmathPlanetmath of X.

Elements of 𝔞M can be written as linear combinationsMathworldPlanetmath aixi, where ai𝔞.

Suppose that |X|>0. Since M=𝔞M, we can express x1 as a such a linear combination:


Moving the term involving a1 to the left, we have


But a1J(R), so 1-a1 is invertiblePlanetmathPlanetmathPlanetmathPlanetmath, say with inversePlanetmathPlanetmathPlanetmath b. Therefore,


But this means that x1 is redundant as a generator of M, and so M is generated by the subset {x2,x3,,xn}. This contradicts the minimality of X.

We conclude that |X|=0 and therefore M=0.

Title proof of Nakayama’s lemma
Canonical name ProofOfNakayamasLemma
Date of creation 2013-03-22 13:07:46
Last modified on 2013-03-22 13:07:46
Owner mclase (549)
Last modified by mclase (549)
Numerical id 5
Author mclase (549)
Entry type Proof
Classification msc 13C99