1
$\begingroup$

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.

$\endgroup$
4
  • 1
    $\begingroup$ Unless I'm mistaken, schemes are, topologically, sober spaces so they are determined up to homeomorphism by their lattice of open subsets (by Stone duality), and therefore by their lattice of closed subsets, which means that $Spec(R)$ is determined up to homeomorphism by the lattice of radical ideals of $R$. I would guess that the lattice of prime ideals (ie the lattice of irreducible closed subsets) should also determine the structure, but that needs an extra argument. $\endgroup$ Apr 23, 2023 at 8:30
  • $\begingroup$ @CaptainLama Why invoke sobriety or Stone duality? Isn't any topology determined by the set of open/closed sets? $\endgroup$
    – math54321
    Apr 24, 2023 at 17:57
  • 1
    $\begingroup$ @math54321 By the set of open/closed subset as a sublattice of a given subset lattice. Not as an abstract lattice. For instance all indiscrete spaces have the same lattice of open subsets (two elements, one smaller than the other), but they are homeomorphic iff they have the same cardinality. $\endgroup$ Apr 24, 2023 at 21:49
  • $\begingroup$ @CaptainLama Notice that I said topology, not topological space. Here we already have a bijection on underlying sets, so the issue is purely about distinguishing topologies. In other words we don't need any characterizations as abstract posets/lattices $\endgroup$
    – math54321
    Apr 24, 2023 at 22:00

1 Answer 1

2
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .