TITLE: Determine whether these extensions of $\mathbb{C}[x]$ are integral
QUESTION [10 upvotes]: In these notes by Gathmann the question is posed

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

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


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.

REPLY [11 votes]: 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 $z=0$.