equivalent formulation of Nakayama’s lemma

The following is equivalent to Nakayama’s lemma.

Let $A$ be a ring, $M$ be a finitely-generated $A$ -module, $N$ a submodule of $M$ , and $\mathfrak{a}$ an ideal of $A$ contained in its Jacobson radical. Then $M=\mathfrak{a}M+N\Rightarrow M=N$ .

Clearly this statement implies Nakayama’s Lemma, by setting $N$ to $0$ . To see that it follows from Nakayama’s Lemma, note first that by the second isomorphism theorem for modules,

$\frac{\mathfrak{a}M+N}{N}=\frac{\mathfrak{a}M}{\mathfrak{a}M\cap N}$

and the obvious map

$\mathfrak{a}M\to\mathfrak{a}\frac{M}{N}:am\mapsto a(m+N)$

is surjective; the kernel is clearly $\mathfrak{a}M\cap N$ . Thus

$\frac{\mathfrak{a}M+N}{N}\cong\mathfrak{a}\frac{M}{N}$

So from $M=\mathfrak{a}M+N$ we get $M/N=\mathfrak{a}(M/N)$ . Since $\mathfrak{a}$ is contained in the Jacobson radical of $M$ , it is contained in the Jacobson radical of $M/N$ , so by Nakayama, $M/N=0$ , i.e. $M=N$ .

Title	equivalent formulation of Nakayama’s lemma
Canonical name	EquivalentFormulationOfNakayamasLemma
Date of creation	2013-03-22 19:11:47
Last modified on	2013-03-22 19:11:47
Owner	rm50 (10146)
Last modified by	rm50 (10146)
Numerical id	4
Author	rm50 (10146)
Entry type	Theorem
Classification	msc 13C99