Can we axiomatize a field starting with the binary operations and only “equational” axioms?

Question

The usual field axioms include the existence of (additive and multiplicative) identities and inverses. Is there a set of field axioms where all axioms are purely equational (see below for what I mean)?

The Wikipedia article on fields contained (and still contains, slightly rewritten) an intriguing section on “Alternative axiomatizations”:

Because of the relations between the operations, one can alternatively axiomatize a field by explicitly assuming that there are four binary operations (add, subtract, multiply, divide) with axioms relating these, …

This is something I'm interested in, and I wonder whether it's true: can I see an example of such a set of axioms? Or prove that one does not exist?

Specifically (because whatever Wikipedia is talking about may turn out not to be the thing I want), I'm thinking of a definition something like the following: a field is a set $F$ along with four operations $(+, -, \times, \div)$ satisfying the following axioms (here $a, b, c, d$ denote any elements of $F$):

$$\begin{align} a + b &= b + a \\ a + (b + c) &= (a + b) + c \\ a + (b - c) &= (a + b) - c \\ a - (b - c) &= (a - b) + c \\ a + (b - b) &= a \quad \rlap{\text{(maybe we need something like this?)}} \\ a \times b &= b \times a \\ &\dots \end{align}$$ where each axiom is simply an equation (or a term-rewriting rule: if we have an expression of the form on the left, then we can transform it to the one on the right, maybe do these transformations until we get a canonical form), with no axioms of the form “there exist…” (like assuming $0$ or $1$ or additive or multiplicative inverses). If such a system does not result in a field, what's missing?

(I'm trying to see whether, by starting with four arbitrary operations defined on a set $S$ and introducing equational constraints on the operations—such as commutativity, associativity, etc.—whether we can finally reach a state where we know these are all the constraints. I know this axiomatization may seem weird, but there do exist weird ones like Tarski's axiomatization of the reals.)

Answer 1

No, and more generally there is no axiomatization of fields using any number of operations and only equational axioms. If such an axiomatization existed, then any product of two fields would have a field structure (just use the operations on each coordinate separately). But, for instance, there is no field structure on the underlying set of the product $\mathbb{F}_2\times\mathbb{F}_3$.

Answer 2

This is a great question! The answer is no, fields cannot be so axiomatized.

The key observation is that equations are preserved by products: if $A, B$ are structures satisfying some set of equations, then $A\times B$ also satisfies those equations. But the product of two fields is never a field.

By contrast, it's easy to check that rings are axiomatizable by equations (so are groups, monoids, and many other interesting classes of structures).$^*$

Classes of structures which can be axiomatized by equations are called (perhaps confusingly) varieties, and their study is part of universal algebra. Let me end by mentioning one of the fundamental theorems of universal algebra:

HSP theorem: Let $\mathcal{V}$ be a (nonempty) collection of (nonempty) algebras (that is, structures in some language containing only function symbols) which is closed under isomorphism (that is, $A\cong B, A\in\mathcal{V}\implies B\in\mathcal{V}$). Then $\mathcal{V}$ is a variety if and only if $\mathcal{V}$ is closed under substructures, homomorphic images, and arbitrary Cartesian products.

One direction of the theorem is relatively easy: show that products, substructures, and homomorphic images all preserve equations. The other direction is more interesting. Roughly, supposing $\mathcal{V}$ is closed under H, S, and P (hence the name of the theorem) and $A$ is some algebra satisfying every equation which is true of every element of $\mathcal{V}$, we want to show $A\in\mathcal{V}$. We use closure under products to construct a very large "free" algebra (analogous to a free group) in $\mathcal{V}$, and then show that $A$ is the homomorphic image of a substructure of this algebra.

(Note that there is a broader theme going on here: what kinds of first-order sentences are preserved by what kinds of algebraic operations? You can find some details on this in Hodges' model theory book(s).)

---

$^*$Actually, there is an important subtlety here: the language matters! Think about groups. If we have a symbol for the identity element and a symbol for the inverse operation as well as a symbol for the group operation, then the usual group axioms are all equational. If, however, we don't have a symbol for these, then we need more complicated axioms (in particular, we need to say "there exists some element such that ..." which is not equational).

Indeed, the class of groups in the language containing only the group operation is not a variety! This follows from the fact that substructures preserve equations: every equation true in $\mathcal{Z}=(\mathbb{Z}; +)$ is also true in $\mathcal{N}=(\mathbb{N}; +)$, so any variety containing the former also contains the latter. The point is that a more expressive language lets equations say more.