# Field of Fractions for Commutative Ring with Identity

I'm trying to complete a problem that is walking me through creating a field of fractions $F$ from a commutative ring with identity (and no non-zero divisors) $R$. The construction first asks me to define the equivalence relation on $R$ such that $(r,s)~\sim(r',s') \iff rs'=sr'$. Denoting each equivalence class suggestively by $r/s$, I'm to let the field $F$ equal the set of all such equivalence classes such that s is non-zero. I then have to show that adding and multiplying as though they're fractions in $\mathbb{Q}$ is well-defined. I check that these operations indeed make $F$ a field. So here's where I begin to have some trouble...

I'm to define an injective ring homomorphism $i:R\rightarrow F$, essentially an inclusion like one might have from $\mathbb{Z}$ to $\mathbb{Q}$. My proposal here is $i(r)=r/1$, and methinks this is kosher. And the last part completely stumps me. I'm supposed to show that I've constructed the smallest field containing $R$. Namely, I need to prove that if $h:R\rightarrow K$ is any injective homomorphism into a field $K$, then there exists a unique injective homomorphism $H:F\rightarrow K$ such that $H(i) = h$. I guess I will then have demonstrated an example of a universal mapping property. I'm really confused as to how to prove the existence of such an H and how to show its uniqueness, and I also can't quite figure out how showing both these things demonstrates that $F$ is the smallest field containing $R$.

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There's only one possible way to define $H$: $H(r/s) = h(r)/h(s)$. Check this works. Now suppose you have another field with this universal property – show it is isomorphic to $F$. – Zhen Lin Sep 14 '12 at 11:51