# 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?

-

## 2 Answers

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:

1. The field of fractions $K$ of $R$ is finitely generated as an $R$-algebra.
2. $R$ 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. $\Box$

-
@Pete: Take a finite set $\frac{a_1}{b_1},\ldots,\frac{a_r}{b_r}$ of elements of $K$ that generate $K$ as an $A$-algebra. If these elements generate, then so do $\frac{1}{b_1},\ldots,\frac{1}{b_r}$, because we may obtain the original list from these by suitable multiplication by elements of $A$. If these elements generate, then we can replace, say, $\frac{1}{b_1}$ by $\frac{1}{m_1},\ldots,\frac{1}{m_r}$ as in the argument, because from these elements we can obtain $\frac{1}{b_1}$ by suitable multiplication of powers of elements of the new set. Am I making a mistake somewhere? –  Arturo Magidin Feb 2 '11 at 15:58
@Pete L. Clark: I'm unclear on something about your comment: what does "every overring is a localization" mean? $\mathbb{Z}$ is Dedekind with torsion Picard group, so this would imply that the overring $\mathbb{Z}[x]$ is "a localization"... not of $\mathbb{Z}$, surely, so of what? Or do you mean "every overring inside the field of quotients"? In any case, I'm not dealing with every overring, just the full ring of quotients, which can always be obtained by localizing at the nonzero elements, $S^{-1}D$ with $S=D-\{0\}$. –  Arturo Magidin Feb 2 '11 at 20:02
sorry, the term "overring" has a technical meaning in this context which is not what one would otherwise guess: an overring of a domain is ring intermediate between that domain and its field of fractions. But about your argument --- oops! You're right. I somehow forgot that we were looking at the full fraction field. I apologize for the (i.e., my) confusion. –  Pete L. Clark Feb 2 '11 at 21:12
@Pete: Phew! Thanks for the clarification. –  Arturo Magidin Feb 2 '11 at 21:18

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.

-
Funny, I was going to say something very similar. Oh, well: +1. –  Pete L. Clark Feb 2 '11 at 5:00