# Proving the quotient of a principal ideal domain by a prime ideal is again a principal ideal domain [closed]

the quotient of a principal ideal domain by a prime ideal is again a principal ideal domain.

This was from Abstract Algebra

Suppose $R$ is a PID and $P$ is a prime ideal of $R$.

If $P = 0$ then $R/P = \{r + 0 : r \in R\} = R$ which is a PID.

If $P \neq 0$ then $P$ is maximal since every nonzero prime ideal in a PID is a maximal ideal. Now $R/P$ is a field since an ideal $P$ is maximal if and only if $R/P$ is a field. A field is trivially a PID since it has only two ideals $(0)$ and $(1)$ both of which are principal.

HINT $\:$ Nonzero prime ideals are maximal in a PID, since "contains = divides" for principal ideals.

For example, $\rm\:\mathbb Z/p\:$ is the field $\rm\:\mathbb F_p\:,\:$ for prime $\rm\:p\in \mathbb Z\:,\:$ and $\rm\mathbb\: Q[x]/f(x)\:$ is an algebraic number field for prime $\rm\:f(x)\in \mathbb Q[x]\:.\:$ A field is trivially a PID since it has only two ideals $(0)$ and $(1)\:,\:$ both principal.

• What I know about the quotient ring R/P is that P is prime, and it is of a integral domain. Can you please help with the proof? – mary Apr 10 '11 at 20:27
• But since a non0 prime is max the quotient R/P is not only a domain but a ..., which is vacuously a PID. – Bill Dubuque Apr 10 '11 at 20:30
• @user8917, did you mean to comment on my answer? If so: I recommend you look in your textbook and see what statements are proven about prime ideals there. – Zev Chonoles Apr 10 '11 at 20:31

If the ideal is nontrivial, then the quotient will be a field, a PID; otherwise the quotient will be the original PID.

To prove the first assertion, prove that every nonzero prime ideal in a PID is necessarily maximal.

• In a PID every nonzero prime is maximal, as I note above. – H.Durham Dec 15 '13 at 13:45

Hint: A ring $R$ is a PID, by definition, when

1. $R$ is an integral domain, and
2. every ideal of $R$ is principal.

Given a ring $R$ and a prime ideal $P\subset R$, what do we know about the quotient ring $R/P$?

Given an ideal $I\subset R/P$, what do we know about the ideal $q^{-1}(I)\subset R$, where $q:R\rightarrow R/P$ is the quotient homomorphism?

Let $R$ be a PID, and suppose you're taking the quotient by $P=(p)$ a prime ideal of $R$. Let $P'$ be a non-zero ideal of $R/P$. Consider a collection of elements in $R$, $S=\{r\in R : \overline{r}\in P'\}$ where by $\overline{r}$ I mean the image of $r$ under the canonical homomorphism onto the quotient. If $r,s\in S$ then $\overline{r+s}=\overline{r}+\overline{s}\in S$ so $r+s\in S$. If $t\in R$ and $r\in S$ then $\overline{rt}\in S$ since $P'$ is an ideal of $R/P'$ and $\overline{r}\in P'$. Hence, $S$ is an ideal of $R$, so $S=(s)$. So, $P'=(\overline{s})$ since if $a\in P'$ there is some $r\in S$ such that $\overline{r}=a$ but $r=st$ so $a=\overline{r}=\overline{st}$. I think.....

• Also... yeah... prime ideals in PID's are special, so this should really be a one line proof! – Jonathan Beardsley Apr 10 '11 at 20:30

Hint: It's enough to show that the reduction of an ideal $I=(a)$ modulo the prime ideal $P$ is generated by $a$ modulo $P$.