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

Question

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.

The problem statement reads:

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., $\mathfrak R \neq \mathfrak N.\\$

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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!

Answer 1

There is a condition missing: every non-empty open set has to contain a closed point. This is important for both directions, as we'll see.

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.