Quotient field of a domain Let $A$ be a commutative domain and $K=Quot(A)$, its field of fractions (quotient field).
Prove that $K$ is a f.g. $A$-module if and only if $A=K$.
 A: Let $\{\frac{1}{a_1},\ldots,\frac{1}{a_n}\}$ generate $K$ as an $A$ module, with each $a_j\in A$, and let $x$ be an element of $K$.  Then there exist $b_1,\ldots,b_n$ in $A$ such that $\frac{x}{a_1a_2\cdots a_n}=\sum\limits_{j=1}^n\frac{b_j}{a_j}$.  Multiplying both sides of the equation by $a_1a_2\cdots a_n$ shows that $x$ is in $A$.
A: Such modules $\rm\:\!\supsetneq A\:$ are not closed under multiplication so are not rings. Below I show that this holds much more generally, for reasons ubiquitous in commutative algebra. First, notice that
$\rm\displaystyle\qquad\frac{1\:}{a_1}\not\in A\:\Rightarrow\:A$-module $\rm\displaystyle\:A\left<\frac{\:\!1\,}{a_1}\right>\:$ is not a ring, else $\rm\displaystyle\ \frac{1\:}{a_1^2} = \frac{a\:}{a_1}\:\Rightarrow\: \frac{1\:}{a_1} = a\in A\:$
Ditto for $\rm\displaystyle\:B = A\left<\frac{1\:}{a_1},\ldots,\frac{1\:}{a_n}\right> $ since, if $\rm\:B\:$ is a ring, then $\rm\displaystyle\:B = A\left<\frac{1\:}a\right>\:$ for $\rm\:a = a_1\cdots a_n.$
Hence, $\rm\ B\:$ is a ring (e.g. a field) $\rm \iff B = A \iff 1/a_i \in A\quad$ QED
More generally, this is simply a special case of: $ $ integral extensions cannot introduce inverses. Indeed, if $\rm\:A[1/a]\:$ lies in some subring of $\rm\:K\:$ that is a finitely generated $\rm\:A$-module, then by the determinant trick, $\rm\:1/a\:$ is integral over $\rm\:A,\:$ i.e. $\rm\:1/a\:$ is a root of a monic polynomial $\rm\:f\in A[x].\:$ Now, as in the Rational Root Test, clearing denominators shows that $\rm\:a\:|\:1 = $ lead coefficient of $\rm\:\! f,\:$ hence $\rm\:1/a\in A.\:$ Note that, unlikes inverses, it is possible for more general proper fractions to be roots of monic polynomials,  e.g. over $\rm\:A = \mathbb Z[\sqrt{-3}]\:$ we have $\rm\:w^3 = 1\:$ for $\rm\:w = (-1\!+\!\sqrt{-3})/2.$
