Can a quotient field ever be finitely generated as an algebra? If A is a commutative integral domain that's not a field, and let $K$ be the quotient field of A. We know that $K$ is not finitely generated as an A-module. But can $K$ ever be finitely generated as an A-algebra?
 A: In fact, the class of such $A$ is not only nonempty, but important enough to have a standard name: Goldman domains.  See page 117 of Pete L. Clark's notes on commutative algebra here.
A: Sure. Let $A$ be the integers localized at $(2)$; that is,
$$A = \left\{ \frac{a}{b}\in\mathbb{Q}\;\Bigm|\; a,b\in\mathbb{Z}, b\gt 0, \gcd(a,b)=\gcd(2,b)=1\right\}.$$
The field of quotients of $A$ is $\mathbb{Q}$, and is equal to $A[\frac{1}{2}]$, so it is generated as an $A$-algebra by $1$ and $\frac{1}{2}$. 
More generally, any UFD $R$ with only finitely many pairwise non-associated irreducibles will have a field of fractions that is finitely generated as an $R$-algebra: just take $m_1,\ldots,m_k$ to be a maximal list of pairwise non-associates irreducibles in $R$, and the field of fractions will be equal to the subalgebra $R[\frac{1}{m_1},\ldots,\frac{1}{m_k}]$.
Added. In fact:
Theorem. Let $R$ be a UFD. The following are equivalent:


*

*The field of fractions $K$ of $R$ is finitely generated as an $R$-algebra.

*$R$ has only finitely many pairwise non-associate irreducible elements.
Proof. If $R$ has only finitely many non-associate irreducible elements (possibly $0$), $m_1,\ldots,m_k$, then the subalgebra $R[\frac{1}{m_1},\ldots,\frac{1}{m_k}]$ in $K$ equals all of $K$: every element of $K$ can be written as $\frac{a}{b}$ with $a,b\in R$, $b\neq 0$, and $b$ can be factored into irreducibles $b=um_1^{\alpha_1}\cdots m_k^{\alpha^k}$, where $\alpha_1,\ldots,\alpha_k$ are nonnegative integers and $u$ is a unit of $R$. Then
$$\frac{a}{b} = au^{-1}\left(\frac{1}{m_1}\right)^{\alpha_1}\cdots\left(\frac{1}{m_k}\right)^{\alpha_k}\in R\left[\frac{1}{m_1},\ldots,\frac{1}{m_k}\right].$$
Conversely, suppose that $K$ is finitely generated as an $R$-algebra. We may assume that the set that generated $K$ is made up entirely of fractions of the form $\frac{1}{b}$, because any element $\frac{a}{b}$ can be replaced with $\frac{1}{b}$ and we still get the same $R$-subalgebra. Moreover, we can assume that $b$ is irreducible, because if $b=m_1^{\alpha_1}\cdots m_r^{\alpha_r}$, then we can replace $\frac{1}{b}$ with $\frac{1}{m_1},\ldots,\frac{1}{m_r}$. Thus, we may assume that $K$ is generated as an $R$-algebra by the multiplicative inverses of a finite set of pairwise non-associated irreducible elements of $R$, $m_1,\ldots,m_k$. Now let $m\in R$ be any irreducible. We can express $\frac{1}{m}$ as a sum of multiples of powers of the $m_i^{-1}$, so we have
$$\frac{1}{m} = \frac{a_1}{m_1^{\alpha_1}} + \cdots + \frac{a_k}{m_k^{\alpha_k}} = \frac{b_1a_1+\cdots+b_ka_k}{m_1^{\alpha_1}\cdots m_k^{\alpha_k}}$$
where 
$$b_j = \frac{m_1^{\alpha_1}\cdots m_k^{\alpha_k}}{m_i^{\alpha_i}}.$$
Then we must have that $(b_1a_1+\cdots+b_km_k)m = m_1^{\alpha_1}\cdots m_k^{\alpha_k}$, so $m$ divides $m_1^{\alpha_1}\cdots m_k^{\alpha_k}$, and hence $m$ is an associate of one of $m_1,\ldots,m_k$. Thus, $R$ has only finitely many pairwise non-associate irreducibles, as claimed. $\Box$
