Going down fails when $R$ is not integrally closed.

Question

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!

score 1 · Accepted Answer · answered Oct 02 '15 at 15:53

1

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 $X\notin R$.

answered Oct 02 '15 at 15:53

user26857

52,094

An equation of integral dependeance would be $T^2-(x+y-y^2)T-(xy^2-xy)$ right? Thank you for your help! – math635 Oct 02 '15 at 16:36
@math635 It seems that $f(x,y)=x+y-y^2\notin R$: $f(0,0)=0$ and $f(1,1)=1$. – user26857 Oct 02 '15 at 16:40
@math635 A simple one: $T^2-T-X(X-1)$. – user26857 Oct 02 '15 at 16:42

Going down fails when $R$ is not integrally closed.

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