Fix a ring \(R\), an \(R\)-module \(M\), and an \(R\)-module homomorphism \(f:M\rightarrow M\).
If \(M\) satisfies the descending chain condition on submodules, show that if \(f\) is injective, then \(f\) is surjective.
Hint: note that if \(f\) is injective, so are \(f\circ f\), \(f\circ f\circ f\), etc.
Give an example of a ring \(R\), an \(R\)-module \(M\), and an injective \(R\)-module homomorphism \(f:M\rightarrow M\) which is not surjective.
If \(M\) satisfies the ascending chain condition on submodules, show that if \(f\) is surjective, then \(f\) is injective.
Give an example of a ring \(R\), an \(R\)-module \(M\), and a surjective \(R\)-module homomorphism \(f:M\rightarrow M\) which is not injective.
Solution
1. Suppose \(M\) satisfies the descending chain condition and \(f:M\to M\) is injective. Then \(f\) is surjective.
The images form a descending chain \[ M\supseteq f(M)\supseteq f^2(M)\supseteq\cdots. \] By the descending chain condition, there is some \(n\) such that \[ f^n(M)=f^{n+1}(M). \] Let \(x\in M\). Then \(f^n(x)\in f^n(M)=f^{n+1}(M)\), so there exists \(y\in M\) with \[ f^n(x)=f^{n+1}(y)=f^n(f(y)). \] Since \(f\) is injective, so is \(f^n\), hence \(x=f(y)\). Thus every \(x\in M\) lies in \(f(M)\) and \(f\) is surjective.
2. The descending chain condition is necessary: there are injective non-surjective endomorphisms without it.
Take \(R=\mathbb Z\), \(M=\mathbb Z\), and \[ f(n)=2n. \] This is a \(\mathbb Z\)-module homomorphism and is injective, but it is not surjective because \(1\notin f(\mathbb Z)\).
3. Suppose \(M\) satisfies the ascending chain condition and \(f:M\to M\) is surjective. Then \(f\) is injective.
The kernels form an ascending chain \[ \ker f\subseteq\ker f^2\subseteq\ker f^3\subseteq\cdots. \] By the ascending chain condition, there exists \(n\) such that \[ \ker f^n=\ker f^{n+1}. \] Let \(x\in\ker f\). Since \(f\) is surjective, so is \(f^n\), so choose \(y\in M\) with \(f^n(y)=x\). Then \[ f^{n+1}(y)=f(x)=0, \] so \(y\in\ker f^{n+1}=\ker f^n\). Therefore \[ x=f^n(y)=0. \] Thus \(\ker f=0\) and \(f\) is injective.
4. The ascending chain condition is necessary: there are surjective non-injective endomorphisms without it.
Let \(k\) be any field and let \[ M=\bigoplus_{j\ge1}ke_j. \] Define the \(k\)-linear map \[ f(e_1)=0, \qquad f(e_{j+1})=e_j\quad(j\ge1). \] Every basis vector \(e_j\) is \(f(e_{j+1})\), so \(f\) is surjective. But \(e_1\ne0\) and \(f(e_1)=0\), so \(f\) is not injective.