$R$ is subring of $\mathbb{C}$, $R$ is not field, $Q$ is field of fractions, then is $[Q:R]$ infinite?

Question

Let $R$ be a subring of $\mathbb{C}$ (so $\mathbb{Z}\subseteq R$) and assume that $R$ is not a field.

Let $Q$ be the quotient field (field of fractions) of $R$.

Is it true that $Q$ is not a finitely generated $R$-module?

Answer 1

Let $K$ be a field and $R$ be a subring of $K$ such that $K$ is a finitely generated $R$-module.

Let $x \in R$ be nonzero. Then, as $K$ is a finitely generated $R$-module, $1/x \in K$ is integral over $R$, ie there are $a_0,\ldots,a_{n-1} \in R$ with $\frac{1}{x^n}=\sum_{k=0}^{n-1}{\frac{a_k}{x^k}}$. It follows that $\frac{1}{x}=\sum_{k=0}^{n-1}{a_kx^{n-1-k}} \in R$.

So $R$ is a field.

Answer 2

It is true that for an integral domain $R$ that is not a field its field of fractions cannot be a finitely generated $R$-module. Indeed, let $Q$ be a field of fractions of $R$ and suppose that it is a finitely generated $R$-module. Then

$$R\subseteq Q$$

is a finite (and hence integral) extension of rings. Now it is a fact from commutative algebra, that then $R$ would be a field. This is a contradiction.

It is true however, that there exists a ring $R$ such that its field of fractions is a finitely generated as an $R$-algebra. Pick

$$R = \big\{\frac{a}{b}\in \mathbb{Q}\,\big|\,b\mbox{ is odd}\big\}$$

Clearly $R$ is not a field, because $2\in R$ is not invertible. Moreover, $R[\frac{1}{2}] = \mathbb{Q}$ and of course $\mathbb{Q}$ is a field of fractions of $R$. So $\mathbb{Q}$ is generated over $R$ by single element.

Remark. In general this holds for every DVR and there are many DVRs inside $\mathbb{C}$.