Artin's Algebra, Exercise 11.9.13

Question

Let $\varphi: \mathbb{C}[x, y] \rightarrow \mathbb{C}[t]$ be a homomorphism that is the identity on $\mathbb{C}$ and sends $x \sim x(t)$. $y \leadsto y(t)$, and such that $x(t)$ and $y(t)$ are not both constant. Prove that the kernel of $\varphi$ is a principal ideal.

Prove that the kernel of a homomorphism is a principal ideal. (Artin, Exercise 9.13)

Here is the similar question posted 9 years ago. And if I comment my approach in the comments box on this post I think it is absurd. That's why I'm posting this question again with my approach please give me hints how to approach from here.

Here is my approach:-

Consider $\varphi: \mathbb{C}[x, y] \rightarrow \mathbb{C}[t]$ be a homomorphism that is the identity on $\mathrm{C}$ and sends $$ \begin{array}{l} x \rightarrow x(t) \\ y \rightarrow y(t) \end{array} $$ and such that $x(t)$ and $y(t)$ are not both constant.
To prove: The kernel of $\varphi$ is a principal ideal.
Claim: $\ker\varphi$ is principal. If not, then $\ker\varphi$ contains two elements $f, g$ that do not have a common factor. It is enough to show that they do not have a common factor in $\mathrm{C}(x)[y]$. For the proof suppose that; $$ h \in \mathbb{C}(x)[y] $$ It is a common factor, then: $h=a^{-1} h_{0}$ for some $a \in \mathbb{C}[x], h_{0} \in \mathbb{C}[x, y]$ by clearing denominator.

And i cannot approach from here please help me,, and if my argument is wrong then please give another hints or solution. Thank you

Answer 1

The ideal $(x^2,xy)$ shows that your start is wrong: it is not principal and every element of this ideal is divisible by $x$.

For an elementary approach to this problem see here. The main idea is to choose a polynomial $p_0\in\ker\phi$ such that $d_0=\deg_yp_0$ is minimal among the non-zero polynomials in $\ker\phi$ and the coefficient of $y^{d_0}$ has minimal degree in $x$ among the polynomials in $\ker\phi$ whose $y$-degree is $d_0$. Then show that $\ker\phi=(p_0)$ by using the long division in $\mathbb C(x)[y]$.