Let $R$ be a PID and $I$ a prime ideal of $R$ s.t. $0 \subset I \subset 1_R$

Question

Let $R$ be a PID and $I$ a prime ideal of $R$ s.t. $0 \subset I \subset 1_R$ and let $I = \langle a \rangle$, where $a$ is a prime element of $R$.

My question is: is there any other prime ideal $J$ s.t. $0\subset J \subset I$?

Here is all I know: $a$ is prime and irreducible element in $R$ and $I$ is a maximal ideal of $R$. Also, I think the answer would be no(since $I$ is a prime ideal). But, then again I am not sure. Could anyone please help me with this? Thanks.

Answer 1

Suppose $0\subset J\subseteq I\subseteq R$ with $J$ prime. Since all ideals are principal, we have $J=(a)$ and $I=(b)$ for a prime $a$ and some $b\in R$. Since $I\supseteq J$ we have $a=bd$ for some $d\in R$. Since $a$ is prime we have $a\mid d$ or $a\mid b$.

If $a\mid d$ then $a=abe$ for some $e\in R$. Since every PID is an integral domain, we have $1=be$, and $I=R$. If $a\mid b$, then $b=ae$ for some $e\in R$. Let $x\in I$. Then $x=cb$ for some $c\in R$ and, by substituting the above, $x=cea$ so $J\subseteq I \subseteq J$ and $I=J$.

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Note this shows every prime ideal, not $0$, is maximal. It follows that no nonzero prime ideal contains another nonzero prime ideal.

Answer 2

Hint $\ $ For principal ideals: $\,\ \rm\color{#0a0}{contains} = \color{#c00}{divides},\,$ i.e. $\,(a)\supset (b)\iff a\mid b,\,$ hence

$$\begin{align} (p)\text{ is prime}\ \Rightarrow&\,\ \ p\ \text{ is prime}\\ \Rightarrow&\,\ \ p\ \text{ is irreducible}\\ \Rightarrow&\ \ \,p\ \ \text{ has no proper}\,\ {\rm\color{#c00}{divisor}}\,\ a\\ \Rightarrow&\ (p)\, \text{ has no proper } \,{\rm\color{#0a0}{container}}\,\ (a)\\ \Rightarrow&\ (p)\, \text{ is maximal} \end{align}\qquad$$