Is an injective homomorphism necessarily surjective in commutative Artinian rings?

Question

I saw this on a problem set:

Assume that $R$ is a commutative ring with identity. Prove that if R is Artinian then the injective homomorphism $f:R\to R$ is surjective.

I know nothing about "modules", and the definition I know of Artinian rings is: "A ring is called Artinian if every descending chain of ideals stops" but I saw the following post when I was searching for this problem which looks similar: Injective homomorphism of Artinian modules is surjective

But the chain they used is $f(M) \supset f^2(M) \supset f^3(M) \supset \cdots$, and I'm pretty sure if I want to do the same thing here, $f(R)$ is not an ideal.

Also one of my friends mentioned that this problem is wrong, so now I know nothing. Please either give me a solution or a counter example. Thanks!

Answer 1

I agree that the question is ambiguously phrased; if $f$ is a morphism of $R$-modules, then the desired fact indeed holds, by the proof in the post that you link to. But the desired fact does not hold in general if $f$ is "only" a ring morphism. Indeed, let $F$ be your favorite field, and let $R=F(x_n:n\in\mathbb{N})$ be the function field over $F$ in infinitely many variables. Then the $F$-algebra morphism $R\to R$ induced by $x_n\mapsto x_{n+1}$ is injective, but not surjective, since $x_0$ is not in its image. But $R$ is a field, hence Artinian, since its only proper ideal is $(0)$.