# Fields of quotients

If R is an integral domain, show that the field of quotients Q is the smallest field containing R in the following sense: If R is a subset of F, where F is a field, show that F has subfield K such that R is a subset of K and K is isomorphic to Q.

I have trouble interpreting this question. My understanding is that we assume R is a subset of F, we want to prove that there exists K which is a subfield of F such that R is a subset of K and K is isomorphic to Q. That means we want to prove K is a subfield of F. Am I right? If I am right, then how to prove K is a subfield of F. Do I have to prove K is a subring first, then prove K is a field?

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Yes, $K$ needs to be a subfield of $F$, but that will follow immediately from the fact that it is a field (since it is isomorphic to $Q$, which is a field) and that it is contained in $F$. The real meat of the problem is finding $K$. – Arturo Magidin Mar 25 '12 at 4:38
The way to go is to find $Q$ inside of $F$, and since $Q$ is a field, its automatically a subfield of $F$. To understand why the problem is true, just think about it, what does any field containing $R$ need to have? Well of course it needs all the elements of $R$, but since it needs to be a field, it also needs to contain all the inverses of elements in $R$. But of course we can find these inverses in $F$ since it's a field. Now, theres no guarantee this is all of $F$, $F$ may have some extra stuff, but if we just take these out of $F$, we have $Q$! $R$'s field of fractions. – Deven Ware Mar 25 '12 at 4:45
View $R$ as a subring of $Q$. Let $F$ be a field and $f:R\to F$ a monomorphism. Then there is a unique morphism $g:Q\to F$ which extends $f$. – Pierre-Yves Gaillard Mar 25 '12 at 4:59

Hint: Suppose that $R\subseteq F$. Then, for all nonzero $x\in R$, $x\in U(F)=F\setminus\{0\}$. So, think about the field generated by the following subset of $K$: $\{xy^{-1}\,\vert\,x\in R, y\in R\setminus\{0\}\}$. Can you find an isomorphism from this field to $Q$?
Well, technically the quotient field is a set of equivalence classes of $R\times (R\setminus\{0\})$. – user5137 Mar 25 '12 at 6:23