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

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

Solution 1:

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.

Solution 2:

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.