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?


1 Answer 1


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.

  • $\begingroup$ Thanks, Laurent. Nevertheless, schemes admitting stratifications into Jacobson schemes are quite common in applications so perhaps merit a name like "constructibly Jacobson". $\endgroup$
    – David Rydh
    Feb 6, 2013 at 9:33

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