Is it possible to describe $Q(x)$ as the extension field of $R$ freely generated by $\{x\}$? Given an integral domain $R$, the polynomial ring $R[x]$ can be defined as the commutative $R$-algebra freely generated by $\{x\}$. Also, let $Q$ denote the field of fractions associated to $R$. Then we can describe $Q(x)$ as the field of fractions of $R[x]$. But suppose we want to describe $Q(x)$ more directly. By an extension field of $R$, let us mean an $R$-algebra that just happens to be a field.
Question. Is it possible to describe $Q(x)$ as the extension field of $R$ freely generated by $\{x\}$?
What I mean is that okay, we've got a forgetful functor "extension fields of $R$" $\rightarrow \mathbf{Set}$. This should have a left-adjoint, call it $L$. Then I'm thinking that $Q(x)$ can be described as $L(\{x\})$. Is this correct? If not, what is the correct way to describe $Q(x)$?

In the category of extension fields of $R$, the morphisms $Q(x) \rightarrow S$ correspond bijectively to the elements of $S$ that are transcendental (over $Q$). So if you define $\mathcal{R}$ to be the functor that sends an extension field of $R$ to the set of elements that are transcendental (over $Q$), then its left adjoint $\mathcal{L}$ has the property that $\mathcal{L}(\{x\}) = Q(x)$.
