Let $A$ be a ring which is also a finitely generated $\mathbb{Z}$-module. If $A$ is an integral domain and $K$ is its field of fractions and $K$ has characteristic zero, then why is $K$ a finite dimensional vector space over $\mathbb{Q}$?

I see that $K$ contains $\mathbb{Q}$ but why is the extension finite?

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    $\begingroup$ More general: if $B\subset A$ is a finite ring extension of integral domains, then $Q(B)\subset Q(A)$ is a finite field extension. $\endgroup$ – user26857 May 14 '13 at 23:23

Let $a_1,\ldots,a_n$ generate $A$ as a $\mathbf{Z}$-module, and let $F=\mathbf{Q}(a_1,\ldots,a_n)\subseteq K$. Note that $\mathbf{Q}(a_1,\ldots,a_n)$ contains $A$, so in fact we must have $\mathbf{Q}(a_1,\ldots,a_n)=K$. Now, because $A$ is finite over $\mathbf{Z}$, it is in particular integral over $\mathbf{Z}$, which means that every element of $A$ is the root of a monic polynomial with coefficients in $\mathbf{Z}$. In particular, this is true of the $a_i$, and so each $a_i$ is algebraic over $\mathbf{Q}$ (integrality over a field is the same as algebraicity). So $K$ is generated over $\mathbf{Q}$ by finitely many algebraic elements, and therefore is a finite extension of $\mathbf{Q}$.

  • $\begingroup$ Great thanks for your answer. This is probably a stupid question, but can you describe to me how you are defining the field $\mathbb{Q}(a_1,\ldots,a_n)$? For example what is $a_1/2$? $\endgroup$ – Gazerun May 14 '13 at 23:35
  • $\begingroup$ The fact that $K$ has characteristic zero implies that $\mathbf{Q}$ is naturally a subfield of $K$ (it also means that $\mathbf{Z}$ is a subring of $A$). So $a_1/2$ means $a_1$ multiplied in $K$ by $1/2\in\mathbf{Q}\subseteq K$. In general, if $K$ is a field of characteristic zero (also known as an extension of $\mathbf{Q}$), then for elements $a_i\in K$, $i$ in some index set (finite or infinite), $\mathbf{Q}(\{a_i:i\in I\})$ is defined to be the smallest subfield of $K$ containing $\mathbf{Q}$ and the $a_i$ (equivalently the smallest subfield containing the $a_i$, as $\mathbf{Q}$ is in $\endgroup$ – Keenan Kidwell May 15 '13 at 0:40
  • $\begingroup$ there for free). You can see an explicit description of this field, the nature of which depends on whether or not the $a_i$ are algebraic over $\mathbf{Q}$ or not, in any text which treats basic field theory. For example, I know that Hungerford's Algebra has a proposition devoted to this very thing. $\endgroup$ – Keenan Kidwell May 15 '13 at 0:41

Show first that $K$ is an algebraic extension by considering the minimal polynomial over $\mathbb{Z}$ satisfied by any element of $A$. Then show that any finite generating set for $A$ as a $\mathbb{Z}$-module generates $K$ as a field. A finitely generated algebraic field extension is a finite field extension.


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