Show $\ker (\phi)$ is a principal ideal

Question

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.

Answer 1

You can show that $(x-y)^2-x$ is in the kernel so $((x-y)^2-x) \subseteq \text{ker}(\phi).$ To prove the equality note that the height of the kernel is one and show that $(x-y)^2-x=y^2-(2x)y+(x^2-x)$ is irreducible which can be done using the (generalization of) Eisenstein's criterion for the $\mathbb{C}[x][y]$ and the prime ideal $(x)$ of $\mathbb{C}[x].$ Therefore $\text{ker}(\phi)=((x-y)^2-x).$

Answer 2

Since $\phi(x)=t^2$ and $\phi(y)=t^2-t$, $\phi(x-y)=t$ and $\phi((x-y)^2-x)=0$. Thus $\langle (x-y)^2-x\rangle\subseteq\ker(\phi)$, and we want to show the reverse inclusion. So suppose $\phi(a)=0$ for some $a\in\Bbb C[x,y]$. Factor out every term divisible by $y^2$, and subtract off a multiple of $(x-y)^2-x$ to cancel these terms. Thus we are left with $$a=\sum_{k=0}^n(a_k+b_ky)x^k,$$

so \begin{align} \phi(a)&=\phi\left(\sum_{k=0}^n(a_k+b_ky)x^k\right)=\sum_{k=0}^n(a_k+b_k\phi(y))\phi(x^k)\\ &=\sum_{k=0}^n(a_k+b_k(t^2-t))t^{2k}=a_0+\sum_{k=1}^n(a_k+b_{k-1})t^{2k}-\sum_{k=0}^nb_kt^{2k+1}=0 \end{align}

But the last expression is simply a power series expression for the polynomial, with odd coefficients $-b_k$, even coefficients $a_k+b_{k-1}$, and constant term $a_0$. Since this is equal to the zero polynomial, all these must be zero, and hence $a=0$. Thus $\ker(\phi)\subseteq\langle (x-y)^2-x\rangle$.

Answer 3

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 $\phi$.