The ring $$\dfrac{\mathbb{R}[x,y]}{\langle x^3-y^2\rangle}$$
a) is unique factorization domain and principal ideal domain
b) is unique factorization domain.
c) unique factorization domain but not principal ideal domain
d) not integral domain
$\textbf{My Try}$
I take homomorphism. $$\phi:\mathbb{R}[x,y]\to \mathbb{R}[x]$$ such that $$\phi(f(x,y))=f(x^2,x^3)$$
Here, $\dfrac{\mathbb{R}[x,y]}{\langle x^3-y^2\rangle}$ is not isomorphic to $\mathbb{R}[x]$ with kernel $\langle x^3-y^2\rangle$ because $\phi$ doesn't has image $x \in \mathbb{R}[x]$.
But it is isomorphic to $A=\{\sum a_i x^i \in \mathbb{R}[x]~|~a_1=0\}$, which is not P.I.D because $A$ generated by $\langle x^2,1\rangle$. So option (b) and (c) are true.
But in answer key option a and b are true.
They showed it is isomorphic to $R[x]$ and $\phi$ is surjective. But clearly it is not surjective. Please help me, i am stuck.