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.

Observe $\phi(x)=t^2,~\phi(y)=t^2-t~\Rightarrow \phi(x-y)=t~\Rightarrow \phi((x-y)^2)=\phi(x)$ and so we have a nontrivial element of $\ker\phi$, namely $(x-y)^2-x$. Writing $z=x-y$ and $t=\sqrt{x}$ we have
$$\frac{\Bbb C[x,y]}{((x-y)^2-x)}\cong\frac{\Bbb C[x][z]}{z^2-x}\cong\Bbb C[\sqrt{x}]=\Bbb C[t]$$
with the resulting isomorphism given by $x\mapsto t^2$ and $y\mapsto t^2-t$ (check) which is exactly $\phi$.
