1
$\begingroup$

In this post there is a counterexample to the going down theorem.

I am pretty sure that the reason why it fails is because $R$ is not integrally closed in $A$, but I don't have any nice argument to show that this is true.

I would appreciate any help!

$\endgroup$
1
$\begingroup$

If you want to show that $R = \lbrace f \in K[X,Y] \colon f(0, 0) = f (1, 1) \rbrace$ is not integrally closed, then notice that $X$ is integral over $R$, and $X\notin R$.

$\endgroup$
  • $\begingroup$ An equation of integral dependeance would be $T^2-(x+y-y^2)T-(xy^2-xy)$ right? Thank you for your help! $\endgroup$ – math635 Oct 2 '15 at 16:36
  • $\begingroup$ @math635 It seems that $f(x,y)=x+y-y^2\notin R$: $f(0,0)=0$ and $f(1,1)=1$. $\endgroup$ – user26857 Oct 2 '15 at 16:40
  • $\begingroup$ @math635 A simple one: $T^2-T-X(X-1)$. $\endgroup$ – user26857 Oct 2 '15 at 16:42

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.