Let $R$ be a commutative ring with unit 1. Prove that an $R$-module is finitely generated if and only if it is a quotient of a free module $R^{n}$, for some $n>0$.

Let $M$ be a finitely generated $R$-module. Suppose now $I$ is an ideal of $R$, and $\phi$ is an $R$-homomorphism from $M$ to $M$ with the property that

$$\phi(M) \subset I \cdot M=\left\{m \in M \mid m=r m^{\prime} \quad \text { with } \quad r \in I, m^{\prime} \in M\right\}$$

Prove that $\phi$ satisfies an equation

$$\phi^{n}+a_{n-1} \phi^{n-1}+\cdots+a_{1} \phi+a_{0}=0$$

where each $a_{j} \in I$. [You may assume that if $T$ is a matrix over $R$, then $\operatorname{adj}(T) T=$ $\operatorname{det} T$ (id), with id the identity matrix.]

Deduce that if $M$ satisfies $I \cdot M=M$, then there is some $a \in R$ satisfying

$$a-1 \in I \quad \text { and } \quad a M=0 .$$

Give an example of a finitely generated $\mathbb{Z}$-module $M$ and a proper ideal $I$ of $\mathbb{Z}$ satisfying the hypothesis $I \cdot M=M$, and for your example, give an explicit such element $a$.