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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.

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Why did you edit out the definition of $f$? The question doesn't make sense like that. Also, I don't think it has much to do with complex analysis, not directly, anyway. – tomasz Nov 26 '12 at 1:25

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 $\mathbb C[x,y]/\mathfrak p\simeq \mathbb C[x(t), y(t)]\subseteq \mathbb C[t]$.

  1. $\mathfrak p$ is not a maximal ideal because otherwise $\mathbb C[x(t), y(t)]=\mathbb C$ and $x(t), y(t)$ would both be constant.

  2. $\mathfrak p\ne 0$ 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 $\mathfrak p$.

  3. So $\mathfrak p$ is principal (I hope you know the classification of prime ideals in $\mathbb C[x,y]$).

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In algebraic geometry langage, the zero-set of $\mathfrak p$ is the Zariski closure of the image of $\pi : \mathbb C\to \mathbb C^2$, $t\mapsto (x(t), y(t))$. As $\pi$ is not constant, its image is an algebraic curve. But in the affine plane, algebraic curves are defined by a single equation (which is then a generator of $\mathfrak p$). – user18119 Nov 25 '12 at 20:21

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.

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