Nakayama's lemma. Let $M$ be a finitely generated $A$-module, $I\subseteq A$ an ideal such that $I\subseteq \mathfrak{R}(A)$ Jacobson's radical of $A$. If $IM=M$, then $M=0$.
I'd like to show that both the finitely generated hypothesis on $M$ and the inclusion hypothesis in the Jacobson's radical $\mathfrak{R}(A)$ are necessary. Explicitly, I'm looking for:
- A non finitely generated $A$-module $M\neq 0$ with an ideal $I\subseteq \mathfrak{R}(A)$ such that $IM=M$.
- A finitely generated $A$-module $M\neq 0$ with an ideal $I\nsubseteq \mathfrak{R}(A)$ such that $IM=M$.
I found that $\mathbb{Q}$ as $\mathbb{Z}_{(2)}$-module with the maximal ideal $I=(2)\cdot\mathbb{Z}_{(2)}$ works for the first request.
I couldn't find any example for the second request, apart from the trivial $\mathbb{Z}$ as $\mathbb{Z}$-module with $I=\mathbb{Z}$. Is there any example with an ideal $I\nsubseteq \mathfrak{R}(A)$ but still a proper ideal of $A$?
Best Answer
For the second request, you could take $A = \mathbb Z$, $M = \mathbb Z / 2\mathbb Z$ and $I = 3\mathbb Z$. Clearly, $IM = M$, but $M \neq 0$. The reason why Nakayama's lemma is not applicable here is that $I \nsubseteq \mathfrak{R}(A)$, since $I$ is not contained in the maximal ideal $2\mathbb Z$, for example.