Definition 28.6.1. A scheme $S$ is said to be Jacobson if its underlying topological space is Jacobson.
28.6 Jacobson schemes
Recall that a space is said to be Jacobson if the closed points are dense in every closed subset, see Topology, Section 5.18.
Recall that a ring $R$ is Jacobson if every radical ideal of $R$ is the intersection of maximal ideals, see Algebra, Definition 10.35.1.
Lemma 28.6.2. An affine scheme $\mathop{\mathrm{Spec}}(R)$ is Jacobson if and only if the ring $R$ is Jacobson.
Proof. This is Algebra, Lemma 10.35.4. $\square$
Here is the standard result characterizing Jacobson schemes. Intuitively it claims that Jacobson $\Leftrightarrow $ locally Jacobson.
Lemma 28.6.3. Let $X$ be a scheme. The following are equivalent:
The scheme $X$ is Jacobson.
The scheme $X$ is “locally Jacobson” in the sense of Definition 28.4.2.
For every affine open $U \subset X$ the ring $\mathcal{O}_ X(U)$ is Jacobson.
There exists an affine open covering $X = \bigcup U_ i$ such that each $\mathcal{O}_ X(U_ i)$ is Jacobson.
There exists an open covering $X = \bigcup X_ j$ such that each open subscheme $X_ j$ is Jacobson.
Moreover, if $X$ is Jacobson then every open subscheme is Jacobson.
Proof. The final assertion of the lemma holds by Topology, Lemma 5.18.5. The equivalence of (5) and (1) is Topology, Lemma 5.18.4. Hence, using Lemma 28.6.2, we see that (1) $\Leftrightarrow $ (2). To finish proving the lemma it suffices to show that “Jacobson” is a local property of rings, see Lemma 28.4.3. Any localization of a Jacobson ring at an element is Jacobson, see Algebra, Lemma 10.35.14. Suppose $R$ is a ring, $f_1, \ldots , f_ n \in R$ generate the unit ideal and each $R_{f_ i}$ is Jacobson. Then we see that $\mathop{\mathrm{Spec}}(R) = \bigcup D(f_ i)$ is a union of open subsets which are all Jacobson, and hence $\mathop{\mathrm{Spec}}(R)$ is Jacobson by Topology, Lemma 5.18.4 again. This proves the second property of Definition 28.4.1. $\square$
Many schemes used commonly in algebraic geometry are Jacobson, see Morphisms, Lemma 29.16.10. We mention here the following interesting case.
Lemma 28.6.4. Examples of Noetherian Jacobson schemes.
If $(R, \mathfrak m)$ is a Noetherian local ring, then the punctured spectrum $\mathop{\mathrm{Spec}}(R) \setminus \{ \mathfrak m\} $ is a Jacobson scheme.
If $R$ is a Noetherian ring with Jacobson radical $\text{rad}(R)$ then $\mathop{\mathrm{Spec}}(R) \setminus V(\text{rad}(R))$ is a Jacobson scheme.
If $(R, I)$ is a Zariski pair (More on Algebra, Definition 15.10.1) with $R$ Noetherian, then $\mathop{\mathrm{Spec}}(R) \setminus V(I)$ is a Jacobson scheme.
Proof. Proof of (3). Observe that $\mathop{\mathrm{Spec}}(R) - V(I)$ has a covering by the affine opens $\mathop{\mathrm{Spec}}(R_ f)$ for $f \in I$. The rings $R_ f$ are Jacobson by More on Algebra, Lemma 15.10.5. Hence $\mathop{\mathrm{Spec}}(R) \setminus V(I)$ is Jacobson by Lemma 28.6.3. Parts (1) and (2) are special cases of (3).
Direct proof of case (1). Since $\mathop{\mathrm{Spec}}(R)$ is a Noetherian scheme, $S$ is a Noetherian scheme (Lemma 28.5.6). Hence $S$ is a sober, Noetherian topological space (use Schemes, Lemma 26.11.1). Assume $S$ is not Jacobson to get a contradiction. By Topology, Lemma 5.18.3 there exists some non-closed point $\xi \in S$ such that $\{ \xi \} $ is locally closed. This corresponds to a prime $\mathfrak p \subset R$ such that (1) there exists a prime $\mathfrak q$, $\mathfrak p \subset \mathfrak q \subset \mathfrak m$ with both inclusions strict, and (2) $\{ \mathfrak p\} $ is open in $\mathop{\mathrm{Spec}}(R/\mathfrak p)$. This is impossible by Algebra, Lemma 10.61.1. $\square$
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)