Why do we require uniqueness in the universal property for a fraction field? I'm a university student taking a course in abstract algebra. My professor recently introduced fraction fields, giving this definition:

Let $R$ be an integral domain. There exists a field $F$, called the field of fractions of $R$, with the following properties:
(i) There is an injective ring homomorphism $\alpha: R \rightarrow F$.
(ii) If $\beta$ is an injective homomorphism from $R$ to some field $K$, then there is a $\textit{unique}$ homomorphism $\phi: F \rightarrow K$ such that $\phi\circ\alpha = \beta$.

$\phi$, being a field homomorphism, is also injective.
The rationale given for this definition was that it constrains $F$ to be the smallest field, up to isomorphism, containing a subring isomorphic to $R$. However, it is not clear to me how the uniqueness constraint serves that purpose. The fact that $\phi$ is injective already forces $F$ to be at least as small as $K$, in terms of cardinality. What do we gain by requiring $\phi$ to be unique?
 A: For example, if you take $R = \mathbb{R}$ and $F = \mathbb{C}$ with injective homomorphism $\alpha=\beta :R\to F=K$ given by $x \mapsto x$ (i.e. the inclusion). Then, you have two choices of $\phi$, namely, the idenitiy and the complex conjugate. Thus, $\mathbb{C}$ is not the fraction field of $\mathbb{R}$. This is expectable since the fraction field of a field should be itself. More generally, a choice of such a $\phi:F \to K$ means a choice of a homomorphism $F\to K$ that fixes $R$ (note the injections $\alpha,\beta$ allow us to see $R$ as subrings of $F$ and $K$). Thus, the uniquenss means that if we have a field $K$ that contains $R$, then there is only one possible homomorphism $\phi:F\to K$ that fixes $R$. Now, to answer your question, the point is that this property uniquely charaterizes $F$ together with injection $\alpha:R\subset F$ (i.e. how we regard $R$ as a subring of $F$) up to (unique) isomorphism fixing $R$ since if we have another $\alpha':R\subset F'$ satisfying the universal property, then we have (unique) homomorphism $\phi:F\to F'$ and $\psi:F' \to F$ fixing $R$, but then $\phi\circ \psi:F\to F$ and $\psi\circ \phi:F'\to F'$ are homomorphisms fixing $R$, so by the uniqueness of such maps, they must be the identity maps, which shows $\phi$ and $\psi$ are the inverses to each other. There are many equivalent ways to state this using category theory (e.g. using the language of initial object/adjoint functors (or equivalently representable functors)), so maybe you can ask your professor if you are interested. (By the way, having the same cardinality does not imply two fields are isomorphic (as rings). Again, if we think about $\mathbb R$ and $\mathbb C$, then they have the same cardninality, but they are not isomorphic: one is algebraically closed and the other is not.)
