# Jacobson radical = nilradical iff every open set of $\text{Spec}A$ contains a closed point.

I believe there is a typo in this problem statement. It is exercise 2.3 in Algebraic Geometry I: Schemes, with examples and exercises Ulrich Görtz and Torsten Wedhorn. In what follows below $A$ is a commutative ring with unit.

Show that the nilradical of $A$ is equal to the Jacobson radical of $A$ if and only if every open subset of $\text{Spec}A$ contains a closed point of $\text{Spec} A.$

My attempted proof of the forward direction:

Proof. We first parse the definitions in the problem statement. We recall that the nilradical of $A,$ denoted $\mathfrak N,$ is the ideal containing all nilpotent elements of $A,$ or equivalently the intersection of all the prime ideals of $A.$ The Jacobson radical of $A,$ denoted $\mathfrak R$ is defined as the intersection of all maximal ideals of $A.$ By definition 2.2. in GW, open sets of $\text{Spec} A$ are of the form $$\text{Spec} A \smallsetminus V(\mathfrak a) := \left\{\mathfrak p \in \text{Spec} A; \mathfrak a \not\subset \mathfrak p\right\}$$ for some ideal $\mathfrak a$ of $A.$ Finally we recall that by example 2.9 in GW, a closed point $\mathfrak m$ of $\text{Spec} A$ is precisely a maximal ideal of $A.$

With these definitions in mind, we need to show that $\mathfrak N = \mathfrak R$ if and only if for every ideal $\mathfrak a$ of $A$ there is a maximal ideal $\mathfrak m$ of $A$ so that $\mathfrak a \not\subset \mathfrak m.$ For the forward direction, we proceed by contrapositive. Let $\mathfrak a \subset A$ be an ideal and suppose it is contained in every maximal ideal of $A.$ In particular this implies that $\mathfrak a \neq (1).$ But then $V(\mathfrak a) \neq \emptyset$ which in turn implies that $\text{Spec} A \smallsetminus V(\mathfrak a) \neq \text{Spec} A.$ This can only be true if there exists a prime ideal $\mathfrak p$ of $A$ that does not contain $\mathfrak a.$ In other words, $\mathfrak a \subseteq \mathfrak R$ but $\mathfrak a \not\subset \mathfrak N,$ i.e., $\mathfrak R \neq \mathfrak N.\\$

I believe that one of the conditions in the problem statement cannot be met. Indeed, for any commutative ring $A,$ the open set $U:=\text{Spec}A \smallsetminus V(\mathfrak R),$ where $\mathfrak R$ is the Jacobson radical of $A,$ by definition could never include a closed point. Is there a typo in the problem statement? It seems like the problem would make sense if all open sets except $U$ were considered. On the other hand, I am quite new to Algebraic geometry and commutative algebra, so there is a good chance I have bungled the meaning of the problem statement.

I would also appreciate any hints for the reverse direction of the proof (please do not write out a whole proof, I am just looking for hints).

Thank you!

• $V(\mathfrak{R})$ is all of $\mathrm{Spec}(A)$, so your open set $U$ is empty. Perhaps the statement should be amended to read every nonempty open set? Jan 13, 2018 at 18:00
• @AlexWertheim Isn't $V(\mathfrak R) = \text{Spec}A$ iff $\mathfrak R = \mathfrak N?$ Jan 13, 2018 at 18:04
• Shoot, my bad. I typically denote the nilradical by $\mathfrak{R}$, and didn't read things carefully. This does get at your point though, doesn't it? If the statement is modified to read "nonempty open set", then the situation you describe is no longer problematic. Jan 13, 2018 at 18:10

Your reasoning does not work: you say that if $\operatorname{Spec}(A) \setminus V(\mathfrak{a}) \neq\operatorname{Spec}(A)$, then there is some prime ideal that does not contain $\mathfrak{a}$. But what it really says is that there is some prime ideal that does contain $\mathfrak{a}$, which is trivial, since it is contained in all maximal ideals.
There is no mistake: if $\mathfrak{N}=\mathfrak{R}$, then it is clear that $\operatorname{Spec}(A)\setminus V(\mathfrak{R}) = \emptyset$ contains no point (and hence no closed point). It is empty, so it's not a problem. But this example actually throws light on the other direction: if $\mathfrak{N} \neq \mathfrak{R}$, then the above is a nonempty set that contains no closed point.
For the direction you tried: if $\mathfrak{N}=\mathfrak{R}$, then an ideal that is contained in all maximal ideals is also contained in all prime ideals, hence its corresponding open set is the empty one.