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

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?

• You ask for an affine scheme whose closed points aren't dense but you then ask about a specific affine scheme over a field. What about $\mathrm{Spec}(k[x]_{(x)})=\{(x),(0)\}$? – Keenan Kidwell Jan 19 '14 at 21:24
• @KeenanKidwell, it was my bad wording, I'm sorry. I meant I thought I could find an example if $A=k[x_1,...]$. Thank you for your example. I forgot that $x$ could also not vanish at $(0)$! – Rodrigo Jan 19 '14 at 22:06

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 $\mathbb{A}^{\infty}_k$.
• Dear Martin, I suppose you meant "transfinite induction on the number of variable". But how do you do it? A simple induction, which in this case is vacuous because we already know that $f=0$ in $\Bbb A^n_k \forall n$, will look like "$f=0$ in $\Bbb A^n_k \Rightarrow f=o$ in $\Bbb A^{n+1}_k$". – Rodrigo Jan 19 '14 at 23:37
• I mean induction. Remember how $k[x_1,x_2,\dotsc]$ is defined! Every polynomial is finite. – Martin Brandenburg Jan 19 '14 at 23:57
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!