A surjective ring homomorphism which is not an isomorphism

Question

Give an example of a ring $R$ and a surjective homomorphism $R \to R$ that is not an isomorphism.

Here is my approach. I know a lemma that says that if $R$ is a Noetherian ring, and $f: R \to R$ is a surjective homomorphism, then $f$ is an isomorphism. So the ring $R$ above must be non-Noetherian, which is hard(?) to find, as 'most' rings are Noetherian (or so I've been told in lectures). One example of a non-Noetherian ring that I know is $K[X,Y]/\langle X,XY, XY^2, XY^3, \dots \rangle$, and from this, I was trying to construct such a homomorphism. But I'm stuck; I can see no intuition, and no systematic way to do this except for guessing. Does anyone have a nice solution to this?

Note: I've done some searching, and it seems that upon learning higher mathematics, this is a standard problem. For example, When is a ring homomorphism $A \to A$ surjective but not injective. has a solution which uses a lot of mathematics that I am unfamiliar with. This problem is from a basic undergraduate algebra course, and I'm trying to work out the answer with only what I know.

Edit: The example of a 'non-noetherian' ring I give above is wrong.

Answer 1

Let $R=\mathbb{Z}[x_1,x_2,\ldots,x_n,\ldots]$. Consider the map induced by sending $x_1$ to $0\in R$, and sending $x_{k+1}$ to $x_k$ for $k=2,3,4,\ldots$. Verify this gives a surjective ring homomorphism, but that it is not injective (hence not an isomorphism).

Answer 2

Analogous to @Arturo's answer, consider a "left-shift" on the infinite product of $\Bbb Z_n$'s say. That's $(x_1,x_2,\dots)\mapsto(x_2,x_3,\dots)$.

As a corollary, invoking the theorem you mentioned, we have that $\prod_{k=1}^\infty\Bbb Z_n$ is not Noetherian.