MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Question: Let $X$ be a noetherian integral scheme. Is there a dense open subscheme $U\subset X$ such that $U$ is Jacobson?

I am happy to allow $X$ to be excellent and then the question of course immediately reduces to $X$ excellent and regular. Equivalently, we could also ask whether every noetherian/excellent scheme admits a stratification into Jacobson schemes.

Note that if $X=\mathrm{Spec}(A)$ is affine, then the Jacobson radical may be $0$ even though $X$ is not Jacobson. An example is $X=\mathrm{Spec}(D[x])$ where $D$ is a DVR (there are many closed points of $X$ sitting in the generic fiber over $\mathrm{Spec}(D)$).

What if $X$ is essentially of finite type over a field?

share|cite|improve this question
up vote 9 down vote accepted

Here is a counterexample. Let $k$ be a field, $R=k[x,y]$. Choose a set $\Sigma$ of closed points of $\mathbb{A}^2_k=\mathrm{Spec}(R)$ such that:

(1) $\Sigma$ is Zariski-dense,
(2) for every $s\in\Sigma$ there is a curve $C$ containing $s$ such that $C\cap\Sigma$ is finite.

EDIT: (For instance, if $\mathrm{char}\,k=0$, take $\Sigma=\mathbb{Q}^2$, and for $s=(a,b)\in\Sigma$ take $C$ defined by $(x-a)^2+(y-b)^2=0$.)

Now let $R_1$ be the localization of $R$ at $\Sigma$, and $X=\mathrm{Spec}\,R_1$. Let $\emptyset\neq U\subset X_1$ be open. By (1), $U$ contains some $s\in \Sigma$, and by (2), $U$ has a one-dimensional closed subscheme with only finitely many closed points. Hence $U$ is not Jacobson.

share|cite|improve this answer
Thanks, Laurent. Nevertheless, schemes admitting stratifications into Jacobson schemes are quite common in applications so perhaps merit a name like "constructibly Jacobson". – David Rydh Feb 6 '13 at 9:33

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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