Is it possible to describe $Q(x)$ as the extension field of $R$ freely generated by $\{x\}$?

Question

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 $Q(x)$?

Answer 1

The category of fields is not well-behaved. There is no free extension field on given generators. You have to specify the relations between the generators, as well as the minimal polynomials of these generators. But even if you do that, it's not always possible to find a universal solution.

If $K$ is a field, then $K(x)$ may be described as a field equipped with a homomorphism $K \to K(x)$ and with a transcendental element $x$ such that for every field homomorphism $K \to L$ and some trancendental element $\alpha \in L$ over $K$ there is a unique homomorphism $K(x) \to L$ which extends $K \to L$ and maps $x$ to $\alpha$. [This is basically the definition of transcendence.] In other words, $K(x)$ represents the functor $K / \mathsf{Fld} \to \mathsf{Set}$ mapping $K \to L$ to the set of transcendent elements.

But this functor has no left adjoint. This would mean that for every set $S$ we have a universal extension $K(S)$ of $K$ equipped with a map from $S$ to the set of trancendent elements. It is not hard to show that this must be the usual function field, if it exists. Now if $S$ has two elements $x,y$, then $x,y$ are algebraically independent in $K \to K(S)$, which by universality would imply that every two trancendent elements are algebraically independent, which is absurd (consider $x$ and $x-1$ in $K(x)$).

Answer 2

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.

Answer 3

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 $\mathcal{L}(\{x\}) = Q(x)$.