1
$\begingroup$

Given a commutative ring $R$, it seems to me that there are two obvious topologies one can put on the set $\operatorname{Spec}R$ of its prime ideals. First is the Zariski topology, with closed sets like \begin{equation} V(I) = \{ \mathfrak{p} \in \operatorname{Spec}R \, | \, I \subseteq \mathfrak{p} \}, \end{equation} where $I \subseteq R$ is any ideal. Second is the poset topology given by inclusion of primes, which has a basis given by sets like $V(\mathfrak{q})$ for $\mathfrak{q} \in \operatorname{Spec}R$. These topologies produce compact spaces for any $R$, and moreover both are contractible when we take $R$ to be an integral domain; the generic point $(0)$ even makes $\operatorname{Spec}(R)$ into a non-Hausdorff cone in the second case.

But it seems to me that these topologies need not be the same in general—the poset topology might have fewer open sets. However, in light of the formula $V(IJ) = V(I) \cup V(J)$, I suspect the topologies coincide for Dedekind domains. I'm also not certain in general what ideals can be constructed by summing over primes (that is, what we might get by allowing arbitrary intersections).

Does anyone have any light they could shed on the matter?

$\endgroup$
  • $\begingroup$ Do you have a reference for "both are contractible"? $\endgroup$ – Watson Feb 6 '17 at 20:39
  • 1
    $\begingroup$ Both spaces have a point whose closure is everything, namely the ideal $(0)$, and hence can be deformation retracted to that point. The explicit homotopy is given by $h(0,\mathfrak{p}) = \mathfrak{p}$, and $h(t,\mathfrak{p}) = (0)$ if $t > 0$. To see this is continuous, note that any proper closed set $C$ in the target pulls back to $C \times \{0\}$. $\endgroup$ – claudio Feb 6 '17 at 20:50
  • $\begingroup$ The Zariski topology is the topology whose closed sets are given by the upper sets of ideals in the ideal lattice intersected with the prime sublattice. You're asking what do we get if we change to only looking at upper sets in the prime sublattice -- is that right? $\endgroup$ – walkar Feb 6 '17 at 20:53
  • $\begingroup$ That's a much more eloquent way of putting it, yes. $\endgroup$ – claudio Feb 6 '17 at 20:55
0
$\begingroup$

I had the same idea half an hour ago. The problem with that specific topology you want to give on the $Spec$ is that intersection of prime ideals is not in general a prime ideal, so it may be not exist a $q$ in $Spec$ such that $Spec(q)=Spec(a)$$\cap$$Spec(b)$ for $a,b$ in $Spec$. So the "poset topology" you want to give it isnt topology at all. You want the property "arbitrary intersection of ideals is ideal" of the ring. Besides that, it is indeed quite natural to generalize the zariski topology on more general structures than rings, (like lattices i think).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.