If $R$ is a PID then $\mathrm{Spec}(R)\!=\!\mathrm{Max}(R)\!\cup\!\{0\}$?

Question

Is the following statement true?

If $R$ is a PID then $\mathrm{Spec}(R)\!=\!\mathrm{Max}(R)\!\cup\!\{0\}$

I know the reverse is false.

Answer 1

This is true. Since $R$ is a PID, we can write our prime $P$ as $(a)$, and we'll assume that $a \neq 0$. Suppose that we have $ P \subset I \subset R$, and write $I = (b)$. If $a \in (a) \subset (b)$, then there is an element $c$ so that $a = bc$. Now $a$ belongs to a prime, so either $b \in P$ or $c\in P$, If $b \in P$, then $I = (b)$, and this proves $P = I$. On the other hand, of $c \in P$, then there is a $d \in R$ so that $c = ad$. But this gives $a = bc = bad \implies 1 = bd$, since PIDs are integral domains, and $a \neq 0$. But this means that $b$ is a unit, and therefore that $I = (b) = R$.

All together we have proven that when $P \subset I \subset R$ either $I = P$ or $I = R$, and this is exactly what it means to be maximal.