Wikipedia states that
The field of fractions of R is characterised by the following universal property:
If $h : R \rightarrow F$ is an injective ring homomorphism from R into a field F, then there exists a unique ring homomorphism $g : \text{Frac}(R) \rightarrow F$ which extends $h$.
I don't understand how we can show this. I get that $g$ will be equal to $h \circ \iota$, where $\iota(x, 1) = x$ on all $(x, 1)$, but how can we uniquely that extend to the rest of $\text{Frac}(R)$?
Best Answer
Consider $g : \operatorname{Frac}(R) \to F$ given by $g(a,b) = h(a)h(b)^{-1}$. Since $h(1) = 1$, $g(x,1) = h(x)$ for all $x \in R$, which means that $h = g \circ \iota$, where $\iota : R \to \operatorname{Frac}(R)$ is given by $\iota(x) = (x,1)$.
Added: Suppose that $g' : \operatorname{Frac}(R) \to F$ is another ring homomorphism such that $h = g' \circ \iota$. Then note that $g'(x,1) = h(x)$ for all $x \in R$, and then, for $b \in R \setminus \{0\}$ we have $$1 = h(1) = g'(1,1) = g'(b,b) = g'((b,1)(1,b)) = g'(b,1)g'(1,b) = h(b)g'(1,b)$$ that is, $g'(1,b) = h(b)^{-1}$. Hence, for each $(a,b) \in \operatorname{Frac}(R)$ we have $$g'(a,b) = g'((a,1)(1,b)) = g'(a,1)g'(1,b) = h(a)h(b)^{-1} = g(a,b).$$