2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ One quick way: Any height one prime ideal in a Noetherian UFD is principal. $\endgroup$ – Ehsan M. Kermani Apr 8 '14 at 0:38
3
$\begingroup$

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 $\text{ker}(\phi)=((x-y)^2-x).$

$\endgroup$
  • $\begingroup$ I didn't know what "height" was, so I looked it up. Why does the kernel has height one? $\endgroup$ – Aloizio Macedo Apr 8 '14 at 13:31
  • $\begingroup$ Because this is the relation involving Krull's dimension $1=\text{dim}(\mathbb{C}[t])=\text{dim}(\mathbb{C}[x,y]/\text{ker})=\text{dim}(\mathbb{C}[x,y])-\text{height}(\text{ker})$ and $\text{dim}(\mathbb{C}[x,y])=2,$ so $\text{height}(\text{ker})=1.$ $\endgroup$ – Ehsan M. Kermani Apr 8 '14 at 16:07
2
$\begingroup$

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 $\phi$.

$\endgroup$
0
$\begingroup$

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 $\ker(\phi)\subseteq\langle (x-y)^2-x\rangle$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.