Proposition

Let $M$ be a Noetherian $A$-module and $u: M \rightarrow M$ a module homomorphism. If $u$ is surjective, then $u$ is an isomorphism.
If $M$ is Artinian and $u$ is injective, then again $u$ is an isomorphism.

Solution

1

$\ker(u) \subset \ker(u^1) \subset \ker(u^2) \subset \cdots$ is an ascending chain of submodules of $M$. Since $M$ is Noetherian, there must exist $n \in \mathbb{N}$ such that $\ker(u^n) = \ker(u^{n + 1}) = \cdots$. We want to show that $\ker(u) = 0$ by using the following two properties repeatedly:

$\ker(u^n) = \ker(u^{n + 1})$, so if $u^{n + 1}(x) = 0$, then $u^n(x) = 0$.
$u$ is surjective, so for any $x \in M$, there exists $x’ \in M$ such that $u(x’) = x$.

Let $x_0 \in \ker(u)$. Since $\ker(u) \subset \ker(u^{n + 1})$, $u^{n + 1}(x_0) = 0$.

By Property 1, $u^n(x_0) = 0$.
By Property 2, $x_0 = u(x_1)$ for some $x_1 \in M$. Therefore, $0 = u^n(x_0) = u^n(u(x_1)) = u^{n + 1}(x_1)$.
By Property 1, $u^n(x_1) = 0$.
By Property 2, $x_1 = u(x_2)$ for some $x_2 \in M$. Therefore, $0 = u^n(x_1) = u^n(u(x_2)) = u^{n + 1}(x_2)$.
By repeating this process, we eventually obtain $x_0, x_1, x_2, \cdots, x_n \in M$ such that
- $0 = u^{n}(x_i)$ for each $i$,
- $x_i = u(x_{i + 1})$ for each $i = 0, \cdots, n - 1$.

Since $x_i = u(x_{i + 1})$ for each $i$, $x_0 = u^n(x_n)$. Therefore, $x_0 = 0$, so $\ker(u) = 0$. Thus $u$ is injective, so it is an isomorphism.

2

Let $M$ be Artinian and $u: M \rightarrow M$ be injective. Then $u(M) \subset M$, and thus $u^2(M) \subset U(M) \subset M$, and so on. Thus $\im(u) \supset \im(u^2) \supset \im(u^3) \supset \cdots$ is a descending chain of submodules of $M$. Since $M$ is Artinian, $\im(u^n) = \im(u^{n + 1}) = \im(u^{n + 2}) = \cdots$ for some $n \in \mathbb{N}$. Let $x \in M$. Then $u^n(x) = u^{n + 1}(y)$ for some $y \in M$ because $\im(u^n) = \im(u^{n + 1})$. Since $u^{n + 1}(y) = u^n(u(y))$, and $u^n$ is injective, $x = u(y)$. Thus $u$ is surjective, so $u$ is an isomorphism..