Let $R$ be a principal ideal domain and let $ I\neq \{0\} $ be a proper ideal. Prove that $R/I$ is a field iff it is a integral domain. if $R/I$ is a field then it is an integral domain by Def'$N$.
If $R/I$ is an integral domain then $I$ is a prime ideal. by theorem.
$I=\langle b \rangle $ for some $ b\in R $ since R is a principal ideal domain.
let $ a\in R $
WTS $ \exists r\in R $ such that $(a+I)(r+I)=(ar+I)=(1+I)$ or $(ar-1) \in I$
I believe i know that $R$ is an integral domain from the Def'$N$ that $R$ is a principal ideal domain. not really sure where to go from here but i think i have most of the pieces? i need something more before i can use the fact that $R$ is a principal ideal domain...
EDIT: We show that $I$ is maximal in the following way.
$I \subset J$ where $J= \langle c \rangle $ thus $ b \in \langle c \rangle $, so b=ck for some $ k \in R $.
But ⟨b⟩ is a prime ideal.
So either c or k are in I if c is in I we are done as $I=J$ in this case so assume that c isn't in I so thus $k \in I$. thus k=bl for some l so b=cbl or $b(cl−1)=0$ but b≠0 by Def'n so $cl−1=0$ or $cl=1$ but since $ \exists l \in R $ s.t $cl=1 $ c is a unit so $J=R $ and thus I is maximal ideal.
Thus by theorem $<b> $ is maximal is equivalnt to $ R/<b> $ is a field hence if R/I is a integral domain and $I\neq \{0\} $ R/I is a field as desired.
Your question basically boils down to show that in a PID $R$, every non-zero prime ideal is maximal.
So let's prove it. Take $P$ a non-zero prime ideal of $R$ and suppose for sake of contradiction that $P$ isn't maximal, then by Krull's theorem there is a maximal ideal $M$ of $R$ such that $P\subset M$. By hypothesis there are $p,m\in R$ such that $P=(p)$ and $M=(m)$. Then $p\in (p)\subset (m)$ implies that $p=mr$ for some $r\in R$. So $mr\in (p)=P$, which is a prime ideal, so by definition $m\in P$ or $r\in P$.
If $m\in P$, then $(m)=M\subset P$, so $P=M$, contradiction. Therefore it must be $r\in P$. This means that $r\in (p)$ which implies that there is some $s\in R$ such that $r=sp$. Hence we have $p=mr=msp$, so $p(1-ms)=0$. Now, as $p\neq 0$, we deduce that $ms=1$, so $1\in M$, which is a contradiction since $M$ is a maximal ideal. We then conclude that $P$ is a maximal ideal and we're done.