I have been having trouble with an exercise in my abstract algebra course. It is as follows:
Let $f: \mathbb{C}[x,y] \rightarrow \mathbb{C}[t]$ be a homomorphism that is the identity on $\mathbb{C}$ and sends $x$ to $x(t)$ and $y$ to $y(t)$ such that $x(t)$ and $y(t)$ are not both constant. Prove that the kernel of $f$ is a principal ideal.
Best Answer
It is enough to suppose that $x(t)$ or $y(t)$ is not constant.
As the codomain of $f$ is an integral domain, the kernel $\mathfrak p$ of $f$ is a prime ideal. We have $\mathbb C[x,y]/\mathfrak p\simeq \mathbb C[x(t), y(t)]\subseteq \mathbb C[t]$.
$\mathfrak p$ is not a maximal ideal because otherwise $\mathbb C[x(t), y(t)]=\mathbb C$ and $x(t), y(t)$ would both be constant.
$\mathfrak p\ne 0$ because otherwise $f$ would induce a field extension $\mathbb C(x,y)\to \mathbb C(t)$, which is impossible by comparing the transcendence degrees. One can also say that, if $x(t)$ is not constant, then $\mathbb C[t]$ is integral over $\mathbb C[x(t)]$, hence $y(t)$ is integral over $\mathbb C[x(t)]$. An integral dependence relation gives rise to a non-zero element of $\mathfrak p$.
So $\mathfrak p$ is principal (I hope you know the classification of prime ideals in $\mathbb C[x,y]$).