According to my lecture notes:
If $K$ is a field, we say $A \subset K$ is a valuation subring iff for every $0 \neq x\in K$, we have $x \in A$ or $x^{-1} \in A$.
Then, if $A \subset K$ is a valuation subring, we have $K=Frac(A)$.
I’m not sure if I have the right intuition for this… this means that for every $a, b \in A$ with $b \neq 0$, we’d have that either $ab^{-1} \in A$ or $a^{-1}b \in A$, right? Could someone please give an example of how this works?
Also, I’m trying to find a simple proof of this fact. So far, I can see that $Frac(A) \subset K$ since $Frac(A)$ is the smallest field containing $A$. But why are they equal?
If a subring contains $x \neq 0$ then the field must contain $x^{-1}$ as well. Since inverses are unique every nonzero element of $F$ can partitioned into pairs $\{x,x^{-1}\}$, not necessarily distinct, and we know that at least one element in $A$ from each of these pairs comes from $F$. So if we enrich $A$ by constructing inverses for the nonzero elements of $A$ we're just including every element from every partition, meaning all of $F$ and so we're done.
The key idea here is that the elements of $A$ must have at least one element from each of these partitions which is how we can assure that all of $F$ is included in the field of fractions. It ensures you're traversing all of $F$.