Example of an affine scheme where closed points aren't dense.

Question

I'm looking for an example of an affine scheme where closed points aren't dense. It's easy to show (using Hilbert's Nullstellensatz) that if $A$ is a finitely generated algebra over a field, then the closed points of $\operatorname {Spec} A$ are dense. Therefore, I suppose that I can find an example where $A$ is not finitely generated. Specifically, I'm looking for an open set in $k[x_1,x_2,...]$ that doesn't contain any closed point. But then, how can it contain a prime ideal at all?

Answer 1

If $R$ is a discrete valuation ring (for example the localization of a PID at a maximal ideal), then $\mathrm{Spec}(R)$ has two points: One generic point, one closed point. Hence the closed points are not dense.

The polynomial ring in infinitely many variables $k[x_1,x_2,\dotsc]$ doesn't work: If $ f \in k[x_1,x_2,\dotsc]$ and $k[x_1,x_2,\dotsc,]_f$ has no maximal ideals, it follows that for all $a_1,a_2,\dotsc \in \overline{k}$ we have $f(a_1,a_2,\dotsc)=0$ (otherwise consider the maximal ideal $\ker(x_i \mapsto a_i)$), hence $f=0$ (reduce to the finite case, then use induction on the number of variables). Hence the closed points are dense in $\mathbb{A}^{\infty}_k$.

Answer 2

In general, if $A$ is not Jacobson (that is, if the Jacobson radical of $A$ is not equal to the nilradical) then the closed points of $\mbox{Spec}(A)$ are not dense. The converse is also true!