TITLE: PID with infinitely many maximal ideals, irreducible, generic points.
QUESTION [6 upvotes]: 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!

REPLY [6 votes]: It is not true in general that an irreducible topological space has a generic point. So there is no contradiction in your conclusion. 
However your reasonning can't be applied to $S=\mathrm{Spec}(A)$ because not all points of $\mathrm{Spec}(A)$ are closed.