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 $\mathfrak{a}$.


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 $\mathfrak a=\ker f$.

  • $\begingroup$ 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
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    $\begingroup$ @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

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.)


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