Can a finitely generated $\mathbb{Z}$-algebra contain $\mathbb{Q}$?

Question

Is there a ring between $\mathbb{Q}$ and $\mathbb{R}$ that is finitely generated as an algebra over $\mathbb{Z}$? My guess is there isn't.

I can see that it would have to be finitely generated over $\mathbb{Q}$ as well, and I think I can deal with algebraic generators. But if there are algebraically dependent transcendentals, I don't see how to exclude some rational. Why couldn't there be $\alpha$, $\beta$ transcendental, such for every prime $p$, $1/p$ is given by some integer polynomial in $\alpha,\beta$?

Martin Brandenburg · Accepted Answer · 2024-10-19T11:43:53.493

5

The residue fields of a finitely generated $\mathbb{Z}$-algebra (i.e. the quotients by maximal ideals) are finite fields (see here). There is no homomorphism from $\mathbb{Q}$ to a finite field.

PS: This works for every infinite field. See also Can a finitely generated ring contain an infinite field?

edited Oct 19 '24 at 11:43

answered May 23 '13 at 16:06

Martin Brandenburg

181,922

PS: We need no embedding to $\mathbb{R}$. So this has nothing to do with algebraic/transcendental numbers. – Martin Brandenburg May 23 '13 at 16:13
Well, that's the direction I was thinking, so feel free to remove the tag. – ronno May 23 '13 at 16:17

Can a finitely generated $\mathbb{Z}$-algebra contain $\mathbb{Q}$?

1 Answers1

Linked