# Finitely generated quotient field

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.

## 2 Answers

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 $1/s\in R$.

• Another question: when P is a prime ideal of R, could one say that R_P/PR_P is the quotient field of R/P? – karparvar May 2 '14 at 7:21
• @karparvar, if you have another question, ask it as such, citing this one if relevant – vonbrand May 2 '14 at 9:13

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

• Jyrki's answer is more didactic because self-contained, but I could never resist the temptation of a one-liner :-) – Georges Elencwajg May 2 '14 at 12:10