# When is a field a nontrivial field of fractions?

If we take any integral domain, then we can define a field of fractions by taking equivalence classes of ordered pairs of elements, the same way that the rational numbers are constructed from the integers. My question is:

What fields (of characteristic $0$) are isomorphic to the field of fractions of some integral domain (that's not a field)?

For instance, is the field of constructible real numbers a nontrivial field of fractions? What about the algebraic real numbers? What about arbitrarily real closed fields? And what about if we restrict ourselves to integral domains which are models of Peano arithmetic, or models of Robinson arithmetic? (EDIT: for those less acquainted with logic and model theory, let me ask this: what if we restricted the integral domains to ones that are discretely ordered rings?) I should mention that my motivation for asking these sorts of questions is my MathOverflow question.

Any help would be greatly appreciated.

• What about the field of rational functions, that is, $x \mapsto \frac{P(x)}{Q(x)}$ with $P$ and $Q$ being some polynomials? – dtldarek Sep 18 '13 at 6:51
• I think any. Choose a transcendence basis $(X_i)$ for your field $k$. Let $R$ be the integral closure of $\mathbb{Z}[\{X_i\}]$ in $k$. Evidently $R\ne k$, but $\text{Frac}(R)=k$. – Alex Youcis Sep 18 '13 at 6:52
• @AlexisYoucis That's not quite the sort of thing I was looking for. What if we restrict the integral domains to models of Peano arithmetic or Robinson arithmetic? If you don't know what I'm talking about, for starters what if we restrict the integral domains to discretely ordered rings? – Keshav Srinivasan Sep 18 '13 at 7:20
• @KeshavSrinivasan I'm not sure, but it does seem like a big part of your question was just "which fields are fraction fields of an integral domain which is not a field". The above answers that. – Alex Youcis Sep 18 '13 at 7:22
• mathoverflow.net/questions/47103/… – user26857 Sep 18 '13 at 8:02

Every field of characteristic zero is the fraction field of some integral domain which is not a field. Indeed, let $k$ be your field and let $(X_i)$ be a transcendence basis for $k$ over $\mathbb{Q}$. Consider the ring $R$ which is the integral closure of $\mathbb{Z}[\{X_i\}]$ in $k$. Note then that $R\ne k$ (since integral extensions preserve dimension), but it's a common fact that $k=\text{Frac}(R)$.
• A trivial argument: $R\neq k$ because $\mathbb{Q}-\mathbb Z$ is not contained in $R$. – user26857 Sep 18 '13 at 8:18
• @YACP Sure, that works too. Or if $R/S$ is integral, both domains, then $R$ is a field if and only if $S$ is a field (although this is just a special fact of what I said for dimension $0$). – Alex Youcis Sep 18 '13 at 8:20