7
$\begingroup$

If I have a map between rings like $f\colon k[x_1,x_2]\to k[t],x_1\mapsto t^2-1,x_2\mapsto t^3-t$, how can I prove that the kernel is $\mathfrak{a}=(x_2^2-x_1^2(x_1+1))$?

I see that $\mathfrak{a}\subseteq \ker(f)$ as $x_2^2-x_1^2(x_1+1)$ is clearly mapped to $0$, but I don't see how to do the other direction.

My idea was to assume $p\in\ker(f)$ which would imply $p(t^2-1,t^3-t)=0$, but I am missing the connection to $\mathfrak{a}$.

$\endgroup$
4
$\begingroup$

Here's a proof that uses Krull dimension.

Observations: $\ker f$ and $\mathfrak a$ are prime.

$\ker f$ is prime because $k[x_1,x_2]/\ker f$ is an integral domain, $\mathfrak a$ is prime because is generated by an irreducible element of $k[x_1,x_2]$ which is an UFD, hence it is generated by a prime element.

In $\text{Im} f=k[t^2-1,t^3-t]$ the element $t^3-t$ is integral over $k[t^2-1]$, is a root of the polynomial $X^2-(t^2-1)^2((t^2-1)+1)$, hence $\dim k[t^2-1,t^3-1]=\dim k[t^2-1]=1$.

From this it follows that $\ker f$ has height $1$ but $k[x_1,x_2]$ has dimension $2$ henceforth $\ker f$ can contain only $(0)$ as a proper prime subideal, so $\mathfrak a=\ker f$.

$\endgroup$
  • $\begingroup$ It's more striking that $k[t^2-1,t^3-t]\subset k[t]$ is integral. Anyway, a lot of machinery to prove such a small result. $\endgroup$ – user26857 Jul 27 '15 at 22:50
  • 1
    $\begingroup$ @user26857 I agree that is a lot like shooting a little bird with a cannon, nonetheless it is a way to take familiarity with the instrument of dimension theory..... at least in my personal opinion. $\endgroup$ – Giorgio Mossa Jul 28 '15 at 7:06
5
$\begingroup$

Let $p\in\ker(f)$, that is, $p(t^2-1,t^3-t)=0$. Write $$p(x_1,x_2)=(x_2^2-x_1^2(x_1+1))q(x_1,x_2)+r(x_1,x_2)$$ with $\deg_{x_2}r\le1$. Then $r(x_1,x_2)=a(x_1)+b(x_1)x_2$ and from $p(t^2-1,t^3-t)=0$ we get $a(t^2-1)+b(t^2-1)(t^3-t)=0$. Now conclude that $a=b=0$. (In order to do this look at the degree of polynomials involved in the last equation.)

$\endgroup$

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.