Since $k[X]$ is PID, we know that any ideal generated by an irreducible polynomial $f$ is prime and thus maximal, so $A=k[X]/(f)$ is a field and thus $\operatorname{Spec}A$ is just a point. So we see that for any irreducible polynomial $f$, all the spectra $V(f)$ are homeomorphic topological spaces.
I wonder if this is true for $k[x,y]/(f(x,y))$, and more generally for $k[x_1,…,x_n]/(f(x_1,…,x_n))$.
That is:
1) Is it true that $\operatorname V((h(x_1,…,x_n)))\cong \operatorname V((f(x_1,…,x_n)))$ for any pair of irreducible polynomials?
(Where by $\cong$ I mean homeomorphism of topological spaces, because I think that if I ask for an isomorphism of affine schemes the claim is false already for $n=2$.)
2) I think it is true for $n=2$, can you point me to a proof or counterexample?
Edit.
Following @ThorWittich comment, I'm interested also in answers that assume something about the cardinality or characteristic of the field.
Notation. By $\operatorname V((f))$ I mean the space of all primes containing the ideal generated by $f$, with the Zariski topology. In other words, I just mean $\operatorname{Spec}k[x_1,…,x_n]/(f)$.
Best Answer
Both $y$ and $y^2-x^3-x$ are irreducible in say $\mathbb{C}[x,y]$, but the corresponding specs are not homeomorphic. One of them is simply connected and the other not.