Show $\ker (\phi)$ is a principal ideal Let $\phi : \mathbb{C}[x,y] \rightarrow \mathbb{C}[t]$ be the ring homomorphism which satisfies:
$\phi(x)=t^2,\ \phi(y)=t^2-t$ and $\phi(c)=c$
Show that the kernel of $\phi$ is a principal ideal.
What I found out: $\phi$ is surjective. Then, $\mathbb{C}[x,y]/\ker(\phi) \cong\mathbb{C}[t]$, but I couldn't arrive anywhere else.

You can show that $(x-y)^2-x$ is in the kernel so $((x-y)^2-x) \subseteq \text{ker}(\phi).$ To prove the equality note that the height of the kernel is one and show that $(x-y)^2-x=y^2-(2x)y+(x^2-x)$ is irreducible which can be done using the (generalization of) Eisenstein's criterion for the $\mathbb{C}[x][y]$ and the prime ideal $(x)$ of $\mathbb{C}[x].$ Therefore $\text{ker}(\phi)=((x-y)^2-x).$
