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Maybe this exercise comes from some textbook, but I do not know.

It said that this ring extension $k[x(x-1),x^2(x-1),z]\subset k[x,z]$ does not have the Going-Down property.

I observe that $k[x(x-1),x^2(x-1)]=\{f(x)\in k[x]|f(0)=f(1)\}$, and $k[x(x-1),x^2(x-1),z]$ is isomorphic to $k[x,y,z]/(y^2-xy-x^3)$. We have a morphism from $\mathrm{Spec}\ k[x,y,z]/(y^2-xy-x^3)$ to $\mathbb{A}^2$.

But I still have not solved the exercise. And I do not know how one found this counterexample. Why did he consider this?


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1 Answer 1

up vote 9 down vote accepted

HINT $\ $ Put $T = k[x],\ x_1 = x(x-1),\ R = k[x_1,xx_1].$ Note that $T[z]$ is the integral closure of $R[z]$ since $T$ is the integral closure of $R.$ To show that GD fails for $R[z]\subset T[z]$ consider the prime ideals $P = x_1\:(z-x)\ \subset R[z]$ and $Q = (x-1,z)\subset T[z].$ Note that $Q$ lies over the prime $Q\cap R[z] = (x_1,xx_1,z)\supseteq P$ but $T[z]$ has no prime in $Q$ lying over $P.$

Note that the GD theorem doesn't apply since $R$ is not integrally closed. Below is further discussion from Matsumura's Commutative Algebra, p. 32. enter image description here enter image description here

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Matsumura is too powerful, maybe before asking questions about commutative algebra we should first check his book. :) – wxu Sep 16 '11 at 8:44
I don't understand how you define $P$. – user26857 Nov 14 at 22:49

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