Let $R$ be an integral domain. Let $I$ and $J$ be nonzero prime ideals such that $I\subsetneq J$. If $I$ is principal, show that $J$ is not principal.

Question

I'm a little stuck on this question from my Abstract Algebra homework and I would like a hint (read: please don't solve this for me, I just want a little hint to get me on track). Here is the question in full:

Let $R$ be an integral domain. Let $I$ and $J$ be nonzero prime ideals in $R$ such that $I$ is properly contained in $J$, i.e., $I\subset J$ and $I\neq J$. If $I$ is principal, show that $J$ is not principal.

So far I've tried proving this by contradiction. So then suppose $I$ and $J$ are both principal. Then $I=\langle a\rangle$ and $J=\langle b\rangle$ for some nonzero $a,b\in R$. Since $\langle a\rangle\subset\langle b\rangle,$ we have that $b\,|\, a$ (we proved this in class), so $by=a$ for some $y\in R.$ Now since $J\setminus I$ is nonempty, let $x\in J\setminus I$. Then $x=bz$ for some $z\in R$ but $a$ does not divide $x.$ Now I get that $byz=az=xy$, and I feel like since $R$ is an integral domain I could potentially use this to show some kind of contradiction. Maybe, for instance, this could be used to show one of $a,b$ is zero, or this could lead to $x\in I$. Both of these would be contradictions and I'd be done. I also notice I haven't used the assumption the ideals are prime.

If it helps, I also proved as a lemma that if $\langle a\rangle$ is prime, then $a$ is prime: suppose $a\,|\,xy$ for some $x,y\in R$. Then $xy\in \langle a\rangle$. Since $\langle a\rangle$ is prime, $x\in\langle a\rangle$ or $y\in\langle a \rangle$. This shows $a\,|\,x$ or $a\,|\,y$. Thus, $a$ is prime.

So maybe since $a$ is prime, there's something about the $az=xy$ equation that could show $a$ divides $x$, giving a contradiction?

Does anyone have any hints for how I can proceed?

EDIT: This question has been flagged as a duplicate of In a PID every nonzero prime ideal is maximal, but here there is no assumption that $R$ is a PID. We are assuming $R$ is any integral domain.

You have shown $a=by$ and that $a,b$ are prime. Also, $I\subset J$ is strict. Try from here… — Andrew Hubery, Feb 17 '23 at 06:29
@AndrewHubery So then $a,|,b$ or $a,|,y$ since $a$ is prime, right? If $a,|,b$ then $a$ and $b$ are associates since $b,|,a$, which would mean $I=J$. That contradicts the proper containment. But now what if $a,|,y$? — blakedylanmusic, Feb 17 '23 at 06:45
@blakedylanmusic you haven't used that $R$ is an integral domain yet. If $a$ divides $y$, then $y = az$ for some $z \in R$. Hence, $by = bza = a$. What can you conclude? — Alex Wertheim, Feb 17 '23 at 06:46
@AlexWertheim One thing that pops into mind is that $bza-a=a(bz-1)=0,$ so since $a\neq 0$ we would have to have that $bz=1$. So then $b$ is a unit. That means $\langle b\rangle = R$. — blakedylanmusic, Feb 17 '23 at 06:50
How does that give a contradiction though? Can't $R$ still be a nonzero prime ideal, or no? — blakedylanmusic, Feb 17 '23 at 06:53
@blakedylanmusic no, prime ideals are necessarily proper ideals by definition, so your solution works out perfectly. — Alex Wertheim, Feb 17 '23 at 06:54
Oh right! I forgot that part of the definition lol. Thank you!! — blakedylanmusic, Feb 17 '23 at 06:55
By here in the linked dupe, prime ideals are maximal among prinicpal ideals. — Bill Dubuque, Feb 17 '23 at 08:55

Let $R$ be an integral domain. Let $I$ and $J$ be nonzero prime ideals such that $I\subsetneq J$. If $I$ is principal, show that $J$ is not principal.

0 Answers0