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 $\alpha,\beta$?

  • $\begingroup$ @Jared That's why I have $\alpha$ and $\beta$ $\endgroup$ – 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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.