Theorem 2 (“Nakayama’s lemma”)

Let M be an A-module and $\mathfrak{a}$ a two-sided ideal of A. Suppose that one of the following conditions is satisfied:

(i) The A-module M is finitely generated, and $\mathfrak{a}$ is contained in the radical of A.

(ii) The ideal $\mathfrak{a}$ is nilpotent.

If N is a submodule of M such that $M = N +\mathfrak{a}M$, then we have N = M. In particular, if the module M is nonzero, then we have $M\not=\mathfrak{a}M$.

Suppose that M is finitely generated and that we have $\mathfrak{a}\subset \mathfrak{R}(A)$. Let N be a submodule of M such that $M = N +\mathfrak{a}M$. By Proposition 6, we have $M = N +\mathfrak{R}$(M), hence M = N by Proposition 2, b) of VIII, p. 153.

Now suppose that $\mathfrak{a}$ is nilpotent, and let N be a submodule of M such that $M = N +\mathfrak{a}M$. By induction on the integer $n\geqslant 0$, we establish the relation $M = N +\mathfrak{a}^nM$. By assumption, there exists an integer $n\geqslant 0$ such that $\mathfrak{a}^n= 0$; hence M = N.

The last assertion of the theorem follows from the above by taking N to be 0.