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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?

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  • $\begingroup$ 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 $\mathrm{Spec}(k[x]_{(x)})=\{(x),(0)\}$? $\endgroup$ – Keenan Kidwell Jan 19 '14 at 21:24
  • $\begingroup$ @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 $(0)$! $\endgroup$ – Rodrigo Jan 19 '14 at 22:06
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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 $\mathbb{A}^{\infty}_k$.

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  • $\begingroup$ 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 $\Bbb A^{n+1}_k$". $\endgroup$ – Rodrigo Jan 19 '14 at 23:37
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    $\begingroup$ I mean induction. Remember how $k[x_1,x_2,\dotsc]$ is defined! Every polynomial is finite. $\endgroup$ – Martin Brandenburg Jan 19 '14 at 23:57
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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!

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