I have no idea how to prove this if it is true or to give a counter example if it is not true.
Is the homomorphic image of a G-domain is a G-domain?
A G-domain is an integral domain $R$ with quotient field $F$ such that there exists $0\neq{a}\in{R}$ such that $F=R[\frac{1_{R}}{a}]$.
Thanks!
Still seems a little off-topic, however...
Claim: Let $R$ be a commutative ring which is not a zero ring or a field. Then $R$ contains a proper ideal $I$ which is not prime.
Proof: Assume the contrary, i.e. every proper ideal of $R$ is prime.
1) Suppose $I,J$ are arbitrary ideals of $R$. Then $IJ \subseteq I \cap J \subseteq I,J$. If $IJ \subsetneq I, IJ \subsetneq J,$ the ideal $IJ$ is not prime (just take $x \in I \setminus IJ, y \in J \setminus IJ$. Then $xy \in IJ, x,y \notin IJ$). Therefore $IJ=I\cap J=I$ or $IJ=I\cap J=J$. Thus, $I\subseteq J$ or $J \subseteq I$, hence $R$ is unserial.
2) In particular, the Jacobson radical of $R$ is its only maximal ideal, say $M$. Consider any finitely generated ideal $I$ of $R$. Then, by the same argumentation as above, we have $IM=I$ (the case $IM=M$ can occur only if $I=M$, since we have $I \subseteq M$). By Nakayama's lemma, this implies $I=0$. This is a contradiction (every nonzero ring has a nonzero finitely generated ideal). QED
(EDIT: A much simpler argument is given by Karl Kronenfeld in comments.)
Now, consider a G-domain $R$ and its ideal $I$ which is not prime. Then the image of the canonical projection $\pi_I:R \rightarrow R/I$ is not an integral domain, in particular, it is not a G-domain.