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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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    $R=K=\mathbb{Q}$ is a counterexample. Perhaps you want to ask something different?2011-11-01
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    What if $R \subsetneq K$?2011-11-01
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    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 $\mathbb{Q}$.2011-11-01
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    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.2011-11-01
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    You might be interested in orders, see http://en.wikipedia.org/wiki/Order_%28ring_theory%292011-11-01

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