Let $R$ be an integral domain and let $p(x), q(x) \in R[x]$ be polynomials of different degrees.
It is clear that $R[x]/(p(x))$ and $R[x]/(q(x))$ are not isomorphic as $R$-modules because the isomorphism class of a free module over a commutative ring is determined by rank.
Can I use this to show that the quotient rings are not isomorphic as rings? I know that in general, module and ring homomorphisms are different and one need not imply the other, but it felt like information about the module "size" should be relevant to their structure as rings?
EDIT: Can I impose any conditions on $R$ to make this true? For example, if $R$ is a field, or if $R$ is a Euclidean domain like $\mathbb{Z}$, does this change anything?
Let $R=k[y]$ for some domain $k$ and let $p(x)=y$ and $q(x)=x$. Then $R[x]/(p(x))\cong k[x]\cong k[y]\cong R[x]/(q(x))$ as rings.