# 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 EquivalentFormulationOfNakayamasLemma 2013-03-22 19:11:47 2013-03-22 19:11:47 rm50 (10146) rm50 (10146) 4 rm50 (10146) Theorem msc 13C99