Suppose $M$ is a Noetherian $R$-module and $\phi: M \rightarrow M$ is an $R$-module endomorphism of $M$. So I did an exercise in my algebra textbook which was to show that $\ker(\phi^n) \cap \operatorname{im}(\phi^n) = 0$ and show that if $\phi$ is surjective, then $\phi$ is an isomorphism.
I thought to myself and wondered if $\phi$ was injective, would $\phi$ be an isomorphism? I couldn't really think of an example or a way to prove this, so I was wondering if $\phi$ was an isomorphism if $\phi$ was injective.
 A: Consider $\mathbb{Z} \rightarrow \mathbb{Z}$ multiplication by 2.
A: Claim: Let $M$ be a Artinian $R$-module and let $\phi: M \to M$ be an injective $R$-module homomorphism. Then $\phi$ is an isomorphism.
Proof: First note that $\operatorname{Im}\phi \supseteq \operatorname{Im}\phi^2 \supseteq\operatorname{Im}\phi^3\supseteq \dots$ is a chain of submodules of $M.$ Since $M$ is Artinian, there exists $n \in \mathbb N$ such that $\operatorname{Im}\phi^n = \operatorname{Im}\phi^{n+1}.$ Choose a least such $n.$ Let $m \in M.$ Choose $m' \in M$ such that $\phi^n(m)=\phi^{n+1}(m') = \phi^n(\phi(m')).$ Since $\phi$ is injective, $m = \phi(m').$ Hence $\phi$ is surjective.
Examples: (1). Let $\phi:k[x] \to k[x], x \mapsto x^2$ be a $k$-algebra homomorphism. Then $\phi$ is injective, but not surjective. Notice that in this example the module is not Artinian.
(2). Let $\phi: k[x_1, x_2, \dots] \to k[x_1, x_2, \dots], x_i \mapsto x_{i+1}$ be a $k$-algebra homomorphism. This is injective but not surjective. Also note that the map $\psi:k[x_1, x_2, \dots] \to k[x_1, x_2, \dots], x_1 \mapsto 0, x_i \mapsto x_{i-1}, i \geq 2$ is a $k$-algebra homomorphism. This is surjective, but not injective.
