Unitary rings : free A-algebra (polynomial ring) = fields :?

Question

I was studying field theory when a question came to my mind.

The essence of the question is in the title, what I'm looking for is the type of algebraic structure matching the ?. I'll elaborate on that a bit.

Given a commutative unitary ring A and a subset S of its underlying set, let's call P the unitary ring of polynomial with coefficients in A having a variable for each element of S. The unitary subring of A generated by S turns out to be the image of the evaluation morphism from P to A mapping each variable to the corresponding element. This is linked to the fact that P is the (or 'a', to your taste) free commutative unitary A-algebra generated by (a set equipotent to) S (namely, the set of variables).

My question is the following: how can one similarly interpret the circumstance of considering fields instead of commutative unitary rings, and generated subfields instead of generated unitary subrings? I know that a subfield generated by S may be viewed as the set of evaluations of algebraic fractions with coefficients in K evaluated in elements of S, which is, in turn, (isomorphic to) the quotient field of P, but is that a sort of free structure somewhat generated by the variables? Or has it any universal property (additional to the universal property of quotient fields)? Or is there any other kind of structure underneath, better explaining the situation?

Answer 1

The category of fields is poorly behaved categorically, and there is no free field on a set of variables, for multiple reasons. To understand generated subfields in this style it is necessary, as far as I know, to work temporarily with a ring which is not a field, as follows:

Let $K$ be a field and $S \subseteq K$ be a subset of it. The subring generated by $S$ is the image of the homomorphism $\mathbb{Z}[S] \to K$ which extends the inclusion $S \to K$. To get the subfield generated by $S$ we need to add in all inverses, but only of those elements of $\mathbb{Z}[S]$ whose images in $K$ are nonzero, and not any others. This is a localization of $\mathbb{Z}[S]$, but which localization it is depends on the details of how $S$ sits inside $K$, or equivalently on the kernel of the map $\mathbb{Z}[S] \to K$. For example:

• If $K$ has characteristic $0$ then we invert all nonzero elements of $\mathbb{Z}$, but if $K$ has characteristic $p$ then we do not invert $p$ (because its image is $0$!).
• If $s \in S$ is transcendental then we invert all monic polynomials $p(s) \in \mathbb{Z}[s]$, but if $s$ is algebraic and satisfies a polynomial $m_s(s) \in \mathbb{Z}[s]$ then we do not invert $m_s$.

The most uniform behavior occurs if the elements of $S$ are algebraically independent; in that case you invert everything (except possibly the characteristic) and get the field $\mathbb{F}_p(S)$ where $p = \text{char}(K)$ (with the convention that $\mathbb{F}_0 = \mathbb{Q}$). So $\mathbb{F}_p(S)$ is the "free field of characteristic $p$ on a set of algebraically independent transcendentals."

This can also be done, funnily enough, in the language of meadows, or equivalently commutative von Neumann regular rings. This is sort of an "equational envelope" of the theory of fields where we weaken the inverse axiom from $xx^{-1} = 1$ (for $x$ nonzero) to $xx^{-1} x = x$ and $x^{-1} x x^{-1} = x^{-1}$ (for all $x$!). The category of meadows is much better behaved than the category of fields, and in particular there exists a free meadow on a set $S$ (which is the "meadowization" of $\mathbb{Z}[S]$, obtained by adjoining all weak inverses). The free meadow does, in fact, have the property that its image in a field $K$ induced by an inclusion $S \to K$ is the subfield generated by $S$. But this is pretty obscure stuff and I don't know if it's useful for anything.