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 case of rings and polynomials, the universal property of $R[x]$ is that for any ring $R'$ with a homomorphism $\phi:R\to R'$ and any $y\in R'$ there is a unique homomorphism $R[x]\to R'$ with $x\mapsto y$.
But the elements of $\mathbb Q(x)$ are rational functions of $x$, which means they can't be evaluated everywhere - there is no "evaluation at $y$" morphism from $Q(x)\to Q$ with $y=1$, for example, since $\frac{1}{x-1}\in\mathbb Q(x)$ can't be evaluated at $1$.
So if you define "freeness" with "universal properties," there isn't a sense in which $\mathbb Q(x)$ is free.
It is certainly analogous to freeness in some ways, just not in that way.
