Example needed for $\alpha M=M$, where $M$ is a module, and $\alpha$ an ideal of the ring

Question

Let $\alpha\lhd R$ be a non-trivial ideal of ring $R$, and let $M$ be an $R$-module. Could someone give me an example where $\alpha M=M$?

Motivation: Nakayama's lemma has this as an assumption.

Answer 1

If $\alpha = m\mathbb{Z}$ for some integer $m$, and $M$ is a finite abelian group of the form $$\prod\limits_i \mathbb{Z}/p_i^{e_i} \mathbb{Z}$$ for various, not necessarily distinct primes $p_i$, then $\alpha M = M$ provided $m$ is not divisible by any of the $p_i$. To see this, consider the abelian group homomorphism $f: M \rightarrow M$ given by $a \mapsto ma$. This map is injective, and its image is contained in $\alpha M$. So you have $$f(M) \subseteq \alpha M \subseteq M$$ By a cardinality argument, all these things are equal.

Answer 2

Let $R_1$, $R_2$ be any two rings with identity, $R=R_1\times R_2$, and $M=\alpha=R_1\times \{0\}$. Then $\alpha M=M^2=M$.

A perhaps more useful example we're talking about Nakayama's Lemma is a local ring with a nil idempotent Jacobson radical.

You can use $F[x_1, x_1,\ldots]/(\{x_i^2\mid i\in \Bbb N\}\cup\{x_i-x_{i+1}x_{i+2}\mid i\in \Bbb N\})$

Answer 3

A simple but nontrivial example is to take $R=\mathbb{Z}$, $M=\mathbb{Q}$, and $\alpha=n\mathbb{Z}$ for any $n>1$. We have $n\mathbb{Z}\cdot \mathbb{Q}=\mathbb{Q}$, since for any $x\in \mathbb{Q}$ we can write $x=n\cdot x/n$.