Let $R$ be a subring of the field $K$ which is a finite extension of $\mathbb{Q}$. Is $R$ a finitely generated $\mathbb{Z}$-module? I want to say this is true, because we can think of $K$ as a finitely generated abelian group and hence is a finitely generated $\mathbb{Z}$-module, then since $R$ is a subring of $K$, $R$ is also a finitely generated $\mathbb{Z}$-module, but I'm not sure if this works.

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    $\begingroup$ $R=K=\mathbb{Q}$ is a counterexample. Perhaps you want to ask something different? $\endgroup$ – Martin Brandenburg Nov 1 '11 at 12:41
  • $\begingroup$ What if $R \subsetneq K$? $\endgroup$ – IOG Nov 1 '11 at 12:49
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    $\begingroup$ Why can you think of $K$ as a finitely generated abelian group? Fields of characteristic 0 are never finitely generated $\mathbb{Z}$-modules. $K$ is finite dimensional over $\mathbb{Q}$. $\endgroup$ – Bill Cook Nov 1 '11 at 13:08
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    $\begingroup$ Take $K=\mathbb{Q}$ and let $R$ be the ring of rational numbers with odd denominator, for an example where $R \neq K$ and $R$ is not finitely generated. There are plenty of other examples. $\endgroup$ – David E Speyer Nov 1 '11 at 13:11
  • $\begingroup$ You might be interested in orders, see en.wikipedia.org/wiki/Order_%28ring_theory%29 $\endgroup$ – Sebastian Nov 1 '11 at 15:04

If you have an inclusion of rings $\mathbb Z \subset K$ such that $K$ is a finitely generated algebra over $\mathbb Z$, then $K$ is guaranteed not to be a field.
So none of your $K$'s is finitely generated over $\mathbb Z$ as an algebra, much less as a module.

That said, if you take a subring $R\subset K$ of one of these fields $K$ of finite $\mathbb Q$-dimension, it may or may not be a finitely generated $\mathbb Z$-module .
For example, if you take $K= \mathbb Q[\sqrt [4]{7}]$ then the rings $\mathbb Z[\sqrt {7}],\mathbb Z[\sqrt [4]{7}]$ are finitely generated $\mathbb Z$- modules , but the fields $\mathbb Q[\sqrt {7}],\;\mathbb Q [\sqrt [4]{7}]$ or the ring $\mathbb Z_{(2)}[\sqrt [2]{7}]$ are not even finitely generated algebras .

Bourbaki, Algèbre Commutative, Ch.5, §3, Corollaire 3 contains the result stated in my first sentence. (I think an English translation exists)

  • $\begingroup$ Update: In the English translation Commutative Algebra of Bourbaki, the result I quote is on page 354. $\endgroup$ – Georges Elencwajg Nov 1 '11 at 14:18

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