TITLE: Prove that the kernel of a homomorphism is a principal ideal. (Artin, Exercise 9.13)
QUESTION [11 upvotes]: 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.

REPLY [5 votes]: A less elementary proof and valid over any field. 
Set $\mathfrak p=\ker f$ and observe that $K[X,Y]/\mathfrak p\simeq K[x(T), y(T)]$. Since at least one of the polynomials $x$ and $y$ is non-constant the ring extension $K[x(T), y(T)]\subseteq K[T]$ is integral, so $\dim K[X,Y]/\mathfrak p=\dim K[T]=1$. This shows that $\text{ht}\ \mathfrak p=1$ and height one prime ideals in UFDs are principal.