# unique ringhomomorphism from the Field of fractions to another field

$R$ is a ring, $L$ a field and $K$ the fraction field constructed from $R$.

For any injective ring homomorphism $f=R \rightarrow L$, there is a unique ring homomorphism $\tilde{f}:K \rightarrow L$ such that $\tilde{f}(\frac{r}{1}) = f(r)$ for all $r \in R$.

I think the map $\phi=R \rightarrow K$ will help to prove the existence, but I don't see why $\tilde{f}$ is unique, because we don't know wat $\tilde{f}(\dfrac{a}{b})$ is (with b different from $0$ and $1$).

• Well, you do know what $f(b)\tilde f(\frac{a}{b})$ is, don't you? – Johannes Huisman May 19 '16 at 7:29