Let R and R' be two rings and we already know that they are isomorphic if there is a one to one and onto map between them (of course this map is a ring homomorphism). I have a question at this point.
If we find a surjective but not injective ring morphism between R and R', can we directly say they are not isomorphic?
For example, let $k$ be a field and $k[x,y]$ be polynomial ring with two variables. Let $I$ be the ideal generated by $xy-1$. I want to show $\frac{k[x,y]}{I}$ is not isomorphic to $k[x]$.
Let $\phi:\frac{k[x,y]}{I} \rightarrow k[x]$ map $f(x)+I$ to $f(x)$. Easily one can show $\phi$ is surjective. However, for example, $x+y+I$ has image $h(x)$, so $\phi(x+y+I)=\phi(h(x)+I)$, by showing $x+y-h(x) \notin I$, we can show $\phi$ is not injective. Can we say $\frac{k[x,y]}{I}$ is not isomorphic to $k[x]$.
In general, the existence of a surjective but non-injective ring homomorphism does not imply that two rings are not isomorphic. Consider e.g. the ring $\mathbb{Z}(x_1,x_2,\dots)$ of integer polynomials on countably many variables and the map that sends $x_1$ to itself and any other variable $x_j$ to $x_{j-1}$. This results in a surjective but non-injective ring homomorphism, yet any ring is isomorphic with itself.