Let \(R\) be a commutative ring.
Suppose that \(R\) is noetherian. Prove that if \(\phi \colon R \to R\) is a surjective ring homomorphism, then \(\phi\) is injective.
If \(R\) is not noetherian, must a surjective ring homomorphism \(\phi \colon R \to R\) be injective? Prove or give a counterexample.
Solution
1. Suppose \(R\) is Noetherian and \(\phi:R\to R\) is surjective.Then the ascending chain\[ \ker\phi\subseteq\ker\phi^2\subseteq\ker\phi^3\subseteq\cdots \]stabilizes.
Each \(\ker\phi^n\) is an ideal of \(R\). Since \(R\) is Noetherian, every ascending chain of ideals stabilizes. Thus there exists \(N\ge1\) such that \[ \ker\phi^N=\ker\phi^{N+1}. \]
2. Under the hypotheses of 1, \(\ker\phi=0\).
Let \(a\in\ker\phi\). Since \(\phi\) is surjective, so is \(\phi^N\). Hence there exists \(b\in R\) with \[ \phi^N(b)=a. \] Then \[ \phi^{N+1}(b)=\phi(a)=0, \] so \(b\in\ker\phi^{N+1}\). By stabilization, \[ b\in\ker\phi^N. \] Therefore \[ a=\phi^N(b)=0. \] Thus \(\ker\phi=0\).
3. Therefore every surjective endomorphism of a Noetherian commutative ring is injective.
A ring homomorphism is injective exactly when its kernel is zero, and 2 gives \(\ker\phi=0\).
4. The conclusion fails without the Noetherian hypothesis.
Let \(k\) be any field and set \[ R=k[x_1,x_2,x_3,\ldots]. \] This ring is not Noetherian, because the ideals \[ (x_1)\subsetneq(x_1,x_2)\subsetneq(x_1,x_2,x_3)\subsetneq\cdots \] form a strictly increasing chain.
Define a \(k\)-algebra endomorphism \(\phi:R\to R\) by \[ \phi(x_1)=0, \qquad \phi(x_{i+1})=x_i\quad(i\ge1). \] The map is surjective because every generator \(x_i\) lies in its image: \[ x_i=\phi(x_{i+1}). \] But \(x_1\neq0\) and \[ \phi(x_1)=0, \] so \(\phi\) is not injective. Hence a surjective endomorphism of a non-Noetherian ring need not be injective.