This question is related to my previous question Existence of surjective but not injective map between rings imply being non isomorphic?
Here my question is, how can i show $\frac{k[x,y]}{I}$ and $k[x]$ are not isomorphic. I showed there exist a ring hom. which is surjective but not injective. However this is not enough to prove the being non isomorphic.
If you're considering your rings as $k$-algebras, the argument is easy.
Suppose $\phi \colon k[x,y]/(xy-1) \to k[x]$ is a $k$-algebra isomorphism. Because $\bar x$ is invertible, its image $\phi(\bar x)$ is invertible as well. But $k[x]^* = k^*$, so $\phi(\bar x) \in k^*$. This contradicts the fact the $\phi$ is an isomorphism of $k$-algebras.
If you're considering your rings as general rings, it takes a bit more work to derive a contradication from $\phi(\bar x) \in k^*$.
After you conclude that $\phi(\bar x) \in k^*$, look at $\phi(1 + \bar x)$. Note that $\phi(1 + \bar x) = 1 +\phi(\bar x) \in k$. Now there are two cases: either $1 + \phi(\bar x) \in k^*$ or $1 + \phi(\bar x) = 0$. The first case gives a contradiction because $1 + \bar x$ is not invertible in $k[x,y]/(xy-1)$, so its image under $\phi$ also cannot be invertible. The second case gives a contradiction because it means that $\phi(\bar x) = -1$ and therefore $\bar x = -1$, which is not the case.
As an aside, to see that $1 + \bar x$ is not invertible, consider the $k$-algebra homomorphism from $k[x,y]/(xy - 1)$ to $k$ induced by $x \mapsto -1$ and $y \mapsto -1$. (Note that this is a homomorphism because $xy - 1 \mapsto 0$). Under this homomorphism $1 + \bar x$ maps to $0$, which is not invertible, so $1 + \bar x$ is not invertible either.