TITLE: Counterexample to Noether normalization over $\mathbb{Z}$
QUESTION [5 upvotes]: The Noether normalization lemma states that if $k$ is a field and $A$ is a finitely generated $k$-algebra, then $A$ is module-finite over a polynomial ring over $k$ (or over $k$ itself). This generalization shows that if $k$ is replaced with a domain $D$, a finitely generated $D$-algebra $A$ is still module-finite over a polynomial ring over a localization $D_f$ of $D$ (or possibly module-finite over $D_f$ itself). One does need to localize, though: for example, $\mathbb{Z}[1/2]$ is not module-finite over any polynomial ring over $\mathbb{Z}$ (or $\mathbb{Z}$ itself).
I would like an example that shows the need of localization when the transcendence degree is at least 1 though. It seems to me that $\mathbb{Z}[x,y]/(2xy-1)$ is probably an example, but I'm not sure how to prove this:

Is it true that $A = \mathbb{Z}[x,y]/(2xy-1)$ is not module-finite over any polynomial ring over $\mathbb{Z}$? How would you prove it?

The proof of the generalized Noether normalization lemma given in the link, when applied to this case, leads to the conclusion that $A$ is module-finite over the polynomial ring $\mathbb{Z}[1/2][x]$ over the localization $\mathbb{Z}[1/2]$, suggesting that inverting $2$ is necessary. But how could we prove that there is no element $\alpha\in A$ such that $A$ is a finite $\mathbb{Z}[\alpha]$-module?

REPLY [2 votes]: If $A \subset B$ and $B$ is a finitely generated module over $A$, then given a prime ideal of $A$, there is a prime ideal of $B$ lying over the prime in $A$. In your example, if $\mathbb{Z}[t] \subset \mathbb{Z}[x,y]/(2xy−1)$ with the larger ring a finite module over the smaller one, there is not prime ideal lying over $2 \mathbb{Z}[t]$.