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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 $\alpha,\beta$?

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

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

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