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

algebraic-geometrycommutative-algebraorder-theoryring-theoryzariski-topology

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.

Best Answer

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.

Related Question