Nakayama Lemma

There are three “Nakayama lemmas”. Here we call them “module version”, “ideal version” and “general version”.

Let $U$ be Noetherian(or finitely generated). Then $U\to U/\mathrm{Rad}U$ is essential. Equivalently, if $V$ is a submodule of $U$ with $V+\mathrm{Rad}U=U$, then $V=U$.

Link to original

Let $M$ be a finitely generated $A$-module, and let $\alpha \subseteq A$ be an ideal with $\alpha \subseteq \mathrm{Jac}(A)$. If $\alpha M=M$, then $M=0$.

Link to original

Let $M$ be a finitely generated $R$-module, and let $I\subseteq R$ be an ideal of $R$. If $M=IM$, then there exists $f=1+r$ with $r\in I$ such that $f\cdot M=0$.

Link to original

Connections between them

Proposition

The “ideal version” can be proved by the “module version”.

\begin{proof} Assume that $M$ is a finitely generated $A$-module, $\alpha \subseteq \mathrm{Jac}A$ and $\alpha M=M$.

Since $\mathrm{Jac}(A)$ kills all simple modules $M/W$, we know $\mathrm{Jac}(A)M\subseteq W$ for each maximal submodule $W$ and so $J(A)M\subseteq \mathrm{Rad}(M)$. Note $\alpha \subseteq \mathrm{Jac}(A)$, then

$$M=\alpha M\subseteq \mathrm{Jac}(A)M\subseteq \mathrm{Rad}(M).$$

By the module version, because $(0)+\mathrm{Rad}(M)=M$, there is $(0)=M$ and we finish the proof. \end{proof}

Proposition

The “general version” can be proved by the “module version”.

\begin{proof} Assume that $M$ is a finitely generated $A$-module, $\alpha \subseteq \mathrm{Jac}A$ and $\alpha M=M$. Take $I=\alpha$ and apply the general version, then we have $f=1+r$ with $r\in I$ such that $f\cdot M=0$. Then $M=0$ because $1+r$ is a unit. \end{proof}