Can $\operatorname{Spec}(R)$ be not homeomorphic to $\operatorname{Spec}(S)$ but be isomorphic as a poset?

Question

Looking at the definition of the spectrum of a ring made me wonder when/whether is its topology determined by the poset of prime ideals of the ring. Let $R$ and $S$ be commutative unital rings. Can the posets of prime ideals of $R$ and $S$ for inclusion be isomorphic but their spectrums not be homeomorphic?

I tried seeing what happens with $\operatorname{Spec}(\mathbb Z)$ and $\operatorname{Spec}(\mathbb K[t])$ for countable field $\mathbb K$ since the posets of primes of these rings are isomorphic. But their spectrums are also homeomorphic. Trying to see what happens with more complicated rings seems too difficult.

Answer 1

Things are good if the topological spaces are Noetherian, or slightly more generally, every closed set has only finitely many irreducible components: let $f : \operatorname{Spec}(R) \to \operatorname{Spec}(S)$ be a poset isomorphism, with inverse $g$. If $V(J) \subseteq \operatorname{Spec}(S)$ is Zariski-closed, then there exist finitely many $q_i \in \operatorname{Spec}(S)$ with $V(J) = \bigcup_i^n V(q_i)$, so $f^{-1}(V(J)) = g(\bigcup_i^n V(q_i)) = \bigcup_i^n g(V(q_i)) = \bigcup_i^n V(g(q_i))$ which is Zariski-closed, so $f$ is Zariski-continuous.

But in general things can go wrong: the paper "The Ordering of Spec $R$" by Lewis-Ohm constructs an explicit example (2.2). I won't recreate the example here (because it's kind of complicated, although the paper gives a really good explanation), but the main points are:

1. For a ring $R$, let $A(0)$ denote the property "All finitely generated flat modules are projective".

2. A theorem of Lazard asserts that whether $R$ has $A(0)$ depends only on $\operatorname{Spec}(R)$ up to homeomorphism.

3. For a poset $X$, a topology on $X$ is called spectral if it makes $X$ into a topological space homeomorphic to $\operatorname{Spec}(R)$ for some ring $R$. A celebrated result of Hochster gives equivalent criteria for a topology to be spectral.

4. Lewis and Ohm construct a poset $X$, and $2$ spectral topologies on $X$, such that one has $A(0)$ but the other does not. This proves the existence of $2$ rings whose spectra are order-isomorphic but not homeomorphic.

In fact, the poset is $1$-dimensional, with countably many minimal and maximal primes. Since a poset isomorphism on spectra is a homeomorphism in dimension $0$, and any ring with finitely many minimal or maximal primes is known to have $A(0)$, this example is as small as possible.