Finitely generated quotient field

Question

If $K$, the quotient field of a commutative integral domain $R$, is finitely generated as an $R$-module, is $R$ necessarily a field?

Thanks for any help.

Answer 1

Assume that the fractions $a_1/b_1,a_2/b_2,\ldots, a_t/b_t$ generate $K$ as an $R$-module. Let $s\in R$ be an arbitrary non-zero element. As the product $r=sb_1b_2\cdots b_t$ is non-zero, the fraction $1/r$ exists in $K.$ Therefore there exist elements $r_i\in R$, $1\le i\le t$, such that $$ \frac1r=\sum_{i=1}^tr_i\cdot\frac{a_i}{b_i}. $$ Using this it should be easy to see that $1/s\in R$.

Answer 2

Since $K$ is a field and is integral ($\Leftarrow$ module finite) over $R$ , $R$ is a field too : Proposition 5.7 here .