PID with infinitely many maximal ideals, irreducible, generic points.

I am trying to do this question and will appreciate if anyone gives comment on my attempt. I am sure there are mistakes somewhere, so I will be glad if someone points them out to me:

Let $A$ be a principal ideal domain with infinitely many maximal ideals. Show that any subset of $\mbox{Spec} A$ that consists of infinitely many closed points is irreducible, but does not contain a generic point.

Attempt

Suppose we have $S$ such a set. This set is irreducible in $\mbox{Spec} A$ if and only if its closure $\overline{S}$ is irreducible.

So I will work with $\overline{S}$. Next I claim that $\overline{S}=\mbox{Spec}A$. Since it is closed, and $A$ is PID, so $\overline{S}=V(f)$ for some element $f\in A$. Suppose $f\neq 0$. Since $A$ is also UFD, so there can only be finitely many maximal ideals belonging to $V(f)$. [Here I use the result that a point $\{\frak{p}\}$ in $\mbox{Spec}A$ is closed if and only if $\frak{p}$ is maximal ideals, and since it is PID, every prime ideals are maximal ideals anyway.] Hence $\overline{S}=\mbox{Spec}A$. Now $\mbox{Spec}A$ is irreducible because if we have $\mbox{Spec}A\subseteq V(f)\cup V(g)$ for some $f\in A$, $g\in A$, then $\mbox{Spec}A$ can only have finitely many closed points because $A$ is PID (and therefore UFD), which contradicts that $A$ has infinitely many maximal ideals.

Hence $\overline{S}$ is irreducible, and so is $S$.

For the second part, I guess it is because if there is a generic point $\{\frak{p}\}$ in $S$, then $\{\frak{p}\}$ itself is closed in $S$ and so it can't be generic because $S$ has infinitely many closed points.

In fact, what I tried for the second part was that $S$ is irreducible and closed, so it has a generic point, $\{\frak{p}\}$. So $S\subseteq \{\frak{p}\}$ which is absurd. But then this would only mean that $\overline{S}$ has no generic points, yet an irreducible closed set has a generic point.

Confusion

I a bit confused here because does that mean that from my argument, $\mbox{Spec}A$ does not have a generic point?? But a closed irreducible subspace has a generic point. Where have I gone wrong?

Thanks for pointing my mistakes out!

• In the statement, do you mean the points of $S$ are all closed in $\mathrm{Spec}(A)$ ? – Cantlog Sep 29 '13 at 15:45
• Yes. I think so. – enoughsaid05 Sep 29 '13 at 15:50
• $A$ being Noetherian implies that your closed set has a finite descomposition into irreducible components , say $V(\frak p_1)\cup\dots\cup$ $V(\frak p_n)$. But all of these sets are finite, if the prime ideal is not $(0)$; so the unique posibility for your closed set is being Spec($A$). And this has a generic point, namely $(0)$; because the only closed set that contains it is Spec($A$). – Daniel Mejia Sep 29 '13 at 17:12
• It didn't occur to me that $\{(0)\}$ is a generic point! – enoughsaid05 Oct 1 '13 at 8:06

However your reasonning can't be applied to $S=\mathrm{Spec}(A)$ because not all points of $\mathrm{Spec}(A)$ are closed.
• But if $A$ is a PID, then every point $\{\frak p\}$ in $\mbox{Spec} (A)$ is closed. This is because every prime ideal is maximal, and so it is closed. Am I right in the reasoning? – enoughsaid05 Sep 29 '13 at 16:05
• No. $A$ is a domain, hence $0$ is a prime ideal which is not maximal unless $A$ is a field. – Hagen Knaf Sep 29 '13 at 16:15