# What kind of subset is Spec(R_P) in Spec(R)?

While trying to prove some properties of a subset of a prime spectrum I arrived at the following question: Let $R$ be a commutative ring and let $P \in \mathrm{Spec}(R)$. We can consider $\mathrm{Spec}(R_P)$ as a subset of $\mathrm{Spec}(R)$. My question is: What kind of subset (subspace) is this? I guess it is not true that $\mathrm{Spec}(R_P)$ is open, right? Is it true that a neighborhood of the generic point in $\mathrm{Spec}(R_P)$ is a neighborhood of $P$ in $\mathrm{Spec}(R)$?

• Isn't Spec$(R_P)$ (considered as a subset of Spec$(R)$) the complement of $V(P)$? That would make it open. (This question seems a bit elementary for MO.) – Joe Silverman Feb 9 '13 at 13:08
• Well, V(P) contains all $Q \supset P$, so its complement is the set of all $Q \not\supset P$. But $Spec(R_P)$ consists of all $Q$ with $Q \cap (R \setminus P) = \emptyset$, so $Q \subset P$. This is not the complement of $V(P)$. Am I wrong? This would be great. :-) – Georg S. Feb 9 '13 at 13:13
• @Georg: You are of course right, since spectra are not necessarily totally ordered by inclusion. – Fred Rohrer Feb 9 '13 at 13:14
• $\mathrm{Spec}(R_f)$ is the complement of $V(f)$ (for $f \in R$), but $\mathrm{Spec}(R_P)$ is not the complement of $V(P)$ (for $P \in \mathrm{Spec}(R)$). – Martin Brandenburg Feb 9 '13 at 18:17

As already said, in general the image is not open and even not constructible. But it is pro-constructible (that is, locally an intersection of locally constructible subsets, see EGA IV.1.9.4, and 1.9.5(ix) because $X:=\mathrm{Spec}(R)$ is affine hence quasi-compact).

Denote this image by $S$. This is a subset of $X$ having exactly one closed point and which is stable by generization as pointed out by Martin. I claim that these two properties characterize all possibles images $S$ when $X$ is noetherian (this hypothesis could be weakened if we notice that $S$ is always quasi-compact, but I don't know how exactly). Here $X$ needs not be affine.

Indeed, let $s$ be the unique closed point of $S$. Let us show $S$ is the intersection of all open neighborhoods of $s$ in $X$ (then it is easy to show $S$ is the image of $\mathrm{Spec}(O_{X,s})$). Let $U$ be any open neighborhood of $s$ in $X$. Then $S\cap (X\setminus U)$ is noetherian hence admits a closed point if non-empty. But this would be a closed point of $S$ different from $s$. So $S\cap (X\setminus U)=\emptyset$ and $S\subseteq U$. Conversely, any point $x\in X$ in the intersecion of all $U\ni s$ is a generization of $s$ (otherwise the complementary of $\overline{\{ x\}}$ is an open neighborhood of $s$ not containing $x$).

Take $R=\mathbb{Z}$ and $P=0$. Then, ${\rm Spec}(R_P)$, considered canonically as a subset of ${\rm Spec}(R)$, consists precisely of the point $0$. In particular, it is not open in ${\rm Spec}(R)$. Hence, it is not a neighbourhood in ${\rm Spec}(R)$ of $0$, despite $0$ being the generic point of ${\rm Spec}(R_P)$ and ${\rm Spec}(R_P)$ being an open neighbourhood of its generic point in itself.

You might want to have a look at Section I.2.5, especially Proposition I.2.5.2, of EGA [1970 edition] (or Section I.2.4 in the first edition) to get some more general information about your situation.

• This is about the fact that $Spec(R_P)$ is homeomorphic to the "generization" of $P$ in $Spec(R)$? Does this perhaps imply an answer about my neighborhood question? – Georg S. Feb 9 '13 at 13:19

If $X$ is a scheme and $x \in X$, then the canonical morphism $j_x : \mathrm{Spec}(\mathcal{O}_{X,x}) \to X$ is a homeomorphism onto its image. The image consists of those $y \in X$ which generalize $x$, i.e. satisfy $x \in \overline{\{y\}}$. Locally, this is just the classification of prime ideals in the localization.

It follows that that the image is closed under generalizations, and contains $x$. In general, I doubt that anything more can be said.

Every open subset is stable under generalizations. The converse is true for constructible subsets of noetherian sober topological spaces (see here). Chevalley's theorem implies that the image of a morphism of finite presentation is constructible. However, $j_x$ is almost never of finite presentation.

For example, when $x$ is a generic point of $X$, this means that the image of $j_x$ equals $\{x\}$, which is usually neither open nor closed.

$\text{Spec } R_P$ is by the usual correspondences the set of all primes of $R$ which are contained in $P$. In particular, it is somehow dual to $V(P) =$ the set of primes which are contained in $P$. It is not generally equal to the complement of $V(P)$ plus the point $P$ itself (this is very rare although does occur in all valuation rings as Fred pointed out in the comments). It is not open, $\text{Spec } R_P$ as a subset of $R$ is closed under generization.

For example: I like to think of say $\text{Spec }$ of the local ring of the origin in $\mathbb{A}^2$ as the origin itself, some germ of every curve passing through the origin, and the generic point.

• "...does occur in some valuation rings": This is in fact the case in every valuation ring. – Fred Rohrer Feb 9 '13 at 14:48

It is a limit of open subsets (the subsets containing $P$.) Thus, though it is not set-theoretically an open subset, it behaves scheme-theoretically like an open subset.

• What do you mean by "behaves scheme-theoretically like an open subset."? – Martin Brandenburg Feb 9 '13 at 21:39
• e.g. if a property holds on every sufficiently small neighborhood of a point, it usually holds on the local ring. – Will Sawin Feb 9 '13 at 21:57