Lemma 10.19.1 (Nakayama's lemma). Let $R$ be a ring with Jacobson radical $\text{rad}(R)$. Let $M$ be an $R$-module. Let $I \subset R$ be an ideal.

If $IM = M$ and $M$ is finite, then there exists a $f \in 1 + I$ such that $fM = 0$.

If $IM = M$, $M$ is finite, and $I \subset \text{rad}(R)$, then $M = 0$.

If $N, N' \subset M$, $M = N + IN'$, and $N'$ is finite, then there exists a $f \in 1 + I$ such that $fM \subset N$ and $M_ f = N_ f$.

If $N, N' \subset M$, $M = N + IN'$, $N'$ is finite, and $I \subset \text{rad}(R)$, then $M = N$.

If $N \to M$ is a module map, $N/IN \to M/IM$ is surjective, and $M$ is finite, then there exists a $f \in 1 + I$ such that $N_ f \to M_ f$ is surjective.

If $N \to M$ is a module map, $N/IN \to M/IM$ is surjective, $M$ is finite, and $I \subset \text{rad}(R)$, then $N \to M$ is surjective.

If $x_1, \ldots , x_ n \in M$ generate $M/IM$ and $M$ is finite, then there exists an $f \in 1 + I$ such that $x_1, \ldots , x_ n$ generate $M_ f$ over $R_ f$.

If $x_1, \ldots , x_ n \in M$ generate $M/IM$, $M$ is finite, and $I \subset \text{rad}(R)$, then $M$ is generated by $x_1, \ldots , x_ n$.

If $IM = M$, $I$ is nilpotent, then $M = 0$.

If $N, N' \subset M$, $M = N + IN'$, and $I$ is nilpotent then $M = N$.

If $N \to M$ is a module map, $I$ is nilpotent, and $N/IN \to M/IM$ is surjective, then $N \to M$ is surjective.

If $\{ x_\alpha \} _{\alpha \in A}$ is a set of elements of $M$ which generate $M/IM$ and $I$ is nilpotent, then $M$ is generated by the $x_\alpha $.

## Comments (6)

Comment #3850 by Lucy on

Comment #3936 by Johan on

Comment #4236 by Aolong on

Comment #4415 by Johan on

Comment #4580 by Xavier on

Comment #4760 by Johan on