My question relates to this question, which is exercise 3 in Section 8.2 of Dummit and Foote. They ask to prove that a quotient of a PID by a prime ideal is again a PID. The answers to the previous question point out that in a PID $R$ a prime ideal is maximal, the quotient is a field, and any field is a PID since its only ideals are $0=(0)$ and $R=(1)$.

Fine, but before I turned to SE I undertook to answer this is on my own, and I came up with the following argument:

Let $P$ be an ideal of PID $R$, $\bar R=R/P$ the quotient, $\bar I$ any ideal of $\bar R$ and $\varphi$ the natural homomorphism from $R\to\bar R$. By the lattice isomorphism theorem $I=\varphi^{-1}(\bar I)$ is an ideal of $R$ that contains $P$. Since $R$ is a PID, there is an $a\in R$ such that $I=(a)$. Now let $\bar b$ be any element of $\bar I$, and $b$ a representative of $\bar b$ in $R$. Then there is some $r\in R$ such that $b=ra$. Since $\varphi$ is a homomorphism, $\varphi(b)=\bar b=\varphi(r)\varphi(a)=\bar r\bar a$. Therefore $\bar I=(\bar a)$, and $\bar I$ is principal. Since this argument placed no restrictions on $\bar I$, every ideal of $\bar R$ is principal. $\square$

I was bothered by this because I have always learned that if you don't use everything you're given when solving a problem, you're probably missing something, and I made no use of the information that $P$ is a prime ideal.

So here's my question: do you need the fact that $P$ is prime to prove that $R/P$ is a PID? If so, where did I go wrong? Can someone give me a counterexample so I can see where the argument fails?

Muchas gracias!


Yes, because $R/P$ is an integral domain iff $P$ is prime.

If you don't have the fact that PIDs are 1 dimensional at your disposal, you can still continue with "and any quotient of a principal ideal ring is also a principal ideal ring, therefore $R/P$ is a PID."

There is nothing wrong with what you wrote: it shows that quotient of a PIR is a PIR. The problem is that you didn't do anything to show the quotent is an integral domain.

  • $\begingroup$ @LeonAvery Yes, I use that abbreviation. $\endgroup$ – rschwieb Jan 23 '15 at 0:07
  • $\begingroup$ Well, that was simple! :-) $\endgroup$ – Leon Avery Jan 23 '15 at 0:09
  • $\begingroup$ Oops, it just got complicated again. I don't understand the sentence you just added. You seem to be saying that I can use the fact that any quotient of a principal ideal ring is a principal ideal ring to prove that any quotient of a principal ideal ring is a principal ideal ring... $\endgroup$ – Leon Avery Jan 23 '15 at 0:14
  • $\begingroup$ @LeonAvery no, I want to use that fact to show a quotient of a PID is a PIR. The proof that quotients of PIRs are PIRs is extremely simple, but let me know if you get stuck on it. $\endgroup$ – rschwieb Jan 23 '15 at 3:33

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