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In these notes by Gathmann the question is posed

Which of the following extensions of $\mathbb{C}[x]$ are integral?

  1. $\mathbb{C}[x,y,z]/(z^2 - xy)$
  2. $\mathbb{C}[x,y,z]/(z^2 - xy, y^3 - x^2)$
  3. $\mathbb{C}[x,y,z]/(z^2 - xy, x^3 - yz)$

So #2 is a composition of integral extensions (pass through $\mathbb{C}[x,y]/(y^3-x^2)$). Extension #1 I believe is not integral because it does not satisfy incomparability: lying over $(x) \triangleleft \mathbb{C}[x]$ is both $(x, z^2-xy)$ and $(x, z^2 - xy, y)$. (Edit: Whoops! Forgot to use prime ideals.)

How am I supposed to see the third one? Heuristics and "how to think about it" answers are much appreciated.

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    $\begingroup$ Item 3 can be also done by the incomparability. You can use the prime ideals $(x,z)$ and $(x,y,z)$. I think for your argument for item 1, you may want to say $(x,z)$ and $(x,y,z)$ since $(x, z^2-xy)$ is not a prime ideal. For item 1, you may use the dimension argument as well. $\endgroup$ – Youngsu Jul 4 '17 at 2:02
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    $\begingroup$ A probably silly question about #1: Set $K:=\mathbb C$, $B:=K[X,Y,Z]/(Z^2-XY)=K[x,y,z]$. It seems to me: first, the ideal $(x)$ is not prime because $z^2\in(x)$ but $z\notin(x)$, and, second, you're implicitly claiming that it is prime. What am I missing? $\endgroup$ – Pierre-Yves Gaillard Jul 4 '17 at 11:49
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    $\begingroup$ @Pierre-YvesGaillard You're right! Whoops. $\endgroup$ – Eric Auld Jul 4 '17 at 15:11
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    $\begingroup$ @Pierre-YvesGaillard Don't have time to think about it just now, so "I don't know". $\endgroup$ – Eric Auld Jul 4 '17 at 15:26
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    $\begingroup$ @Pierre-YvesGaillard What about this: $(0)\subsetneq (x,z)\subsetneq (x,y,z)$? $\endgroup$ – user26857 Jul 4 '17 at 22:03
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If $S=\mathbb{C}[x,y,z]/(z^2-xy,x^3-yz)$ were integral over $\mathbb{C}[x]$, then $S/(x)$ would be integral over $\mathbb{C}[x]/(x)=\mathbb{C}$. But $S/(x)=\mathbb{C}[y,z]/(z^2,yz)$ contains $\mathbb{C}[y]$ as a subring, and so is not integral over $\mathbb{C}$.

Here's a more geometric way to think about essentially the same argument. If $S$ were integral over $\mathbb{C}[x]$, then the induced morphism $\operatorname{Spec} S\to\mathbb{A}^1$ would be finite, and in particular would have finite fibers. In very concrete terms, this means that if you plug in any particular number for $x$, then the system of equations $z^2-xy=0$ and $x^3-yz=0$ should have only finitely many solutions. But if you plug in $x=0$, these equations just say $z^2=0$ and $yz=0$, and so $y$ can be anything as long as $z=0$.

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  • $\begingroup$ Probably silly question but: why the fact that $C[y] $ is a subring implies $S$ is not integral? $\endgroup$ – mattecapu Feb 16 at 10:25
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    $\begingroup$ $y$ is transcendental over $\mathbb{C}$, so it is not integral over $\mathbb{C}$. $\endgroup$ – Eric Wofsey Feb 16 at 15:54
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This solution is really pedestrian, using only the definition, there are probably a lot more elegant methods.

If $ S =\mathbb{C}[x,y,z]/(z^2 - xy, x^3 - yz)$ was integral over $R =\mathbb C[x]$, then for $I=(z) \triangleleft S $, $R/(R\cap I) \subset S/I$ would be integral.

Claim: $R \cap I = (x^3)$

Proof: Clearly, $(x^3) \subset R \cap I$. As $R$ is a PID, $R \cap I$ is generated by some divisor of $x^3$, so it suffices to show that $x^2 \notin R\cap I$. If this was false, we would have in $\mathbb{C}[x,y,z]$ $$x^2-zf(x,y,z)=(z^2-xy)g(x,y,z)+(x^3-yz)h(x,y,z)$$ for some $f,g,h \in \mathbb{C}[x,y,z]$. Pluggin in $z=0$ and $y=0$ gives $x^2=x^3h(x,0,0) \Rightarrow 1=xh(x,0,0)$ in $\mathbb C [x]$ which is absurd.

Now suppose that $\mathbb{C}[x]/(x^3) \subset \mathbb{C}[x,y,z]/(z^2 - xy, x^3 - yz,z) \cong \mathbb{C}[x,y]/(xy,x^3)$ was integral. Then there is some monic non-constant polynomial $f(T,x) \in \mathbb{C}[x]/(x^3)[T]$ such that $f(y,x) = 0$ in $\mathbb C[x,y]/(xy,x^3)$. Since $xy=0$, a lot of terms in this equation cancel out and we may assume that $f$ is of the form $f(T) = g(T)+h(x)$, where $g(T) \in \mathbb C[T]$ is monic and non-constant. Now lifting this to $\mathbb{C}[x,y]$ we get that that $f(y,x)=g(y)+h(x)$ is in $(xy,x^3)$. But every element of $(xy,x^3)$ is divisible by $x$. Now if $x \mid f(y,x)$, then $f(y,0) \equiv 0$, but clearly $f(y,0)=g(y)+h(0)$ is non-constant, as $g$ is non-constant. This finishes the proof.

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    $\begingroup$ You tried harder than necessary: first of all, it's clear that $z$ is integral over $\mathbb C[x]$ since $z^3=x^4$ in the bigger ring, so the only try is on $y$. If suppose $y$ is integral over $\mathbb C[x]$, then one has $$a_0(x)+a_1(x)y+\cdots+a_{n-1}(x)y^{n-1}+y^n\in(z^2-xy,x^3-yz).$$ Now send $x$ and $z$ to $0$ and get that $y$ is integral over $\mathbb C$, a contradiction. $\endgroup$ – user26857 Jul 4 '17 at 22:35

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