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

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 $\alpha,\beta$?

• @Jared That's why I have $\alpha$ and $\beta$ – ronno May 23 '13 at 16:07

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: 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