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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.

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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 $1/s\in R$.

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  • $\begingroup$ 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? $\endgroup$ – karparvar May 2 '14 at 7:21
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    $\begingroup$ @karparvar, if you have another question, ask it as such, citing this one if relevant $\endgroup$ – vonbrand May 2 '14 at 9:13
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Since $K$ is a field and is integral ($\Leftarrow$ module finite) over $R$ , $R$ is a field too : Proposition 5.7 here .

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

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