Proving the quotient of a principal ideal domain by a prime ideal is again a principal ideal domain Please help me prove that 

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

This was from Abstract Algebra
 A: 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.
A: 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.
A: 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.
A: Hint: A ring $R$ is a PID, by definition, when


*

*$R$ is an integral domain, and

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