$R=\mathbb{Q}[x]$ and $J=\langle x^2\rangle$, Is $R/J$ a principal ideal ring? Let $R=\mathbb{Q}[x]$ and $J=\langle x^2\rangle$
Now I want to answer some question regarding $R/J$.
First of all it is not a field. Also $ax, bx \in R/J$, both are non zero but their product $abx^2$ is zero so it is not an integral domain. 
But Is it true that $R/J$ is an principal ideal ring? How should I show that every non zero ideal is principle.
 A: It is worth identifying what $R/J$ looks like, so I'll avoid using the theorem identifying ideals of $R/J$ with ideals of $R$ containing $J$. 
Indeed, for any polynomial $p \in R$, say $p = \sum_{k=0}^n a_kx^k$, we have that $p - (a_1x+a_0) \in J$. Furthermore, the difference of any two distinct polynomials of degree at most $1$ doesn't belong in $J$, so these are in different equivalence classes.
Therefore, the elements of $R/J$ are in $1-1$ correspondence with polynomials $ax+b$, for $a,b \in \mathbb Q$. That is, $R/J = \{(ax+b) + \langle x^2\rangle\}$.
Now, given an ideal $I \subset R/J$, each element of $I$ contains precisely one element of the form $ax+b$,with $a,b \in \mathbb Q$ (for the reason I mentioned earlier). 
Can you use this information? Try to use it first to show that $I$ must be principal(gcd?), then try to see if you can understand better what $I$ must be. In particular, I claim that either $I$ is empty, the whole space, or the ideal generated by $x + \langle x^2 \rangle$. Try to manipulate $ax+b + \langle x^2 \rangle$ using the ideal property to find other elements which an ideal containing $ax+b$ must necessarily contain.
