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.)

  • $\begingroup$ Note that division is not a binary operation on a field, since it is not defined if the second input is $0$. You might still try to axiomatize fields using a generalized version of division (which, say, always outputs $0$ when you divide by $0$). $\endgroup$ Feb 27, 2018 at 3:41
  • $\begingroup$ Not what you are asking, but I never liked the standard clause that $1$ needs to not equal $0$. It turns out that instead of that, you can just require that $\mathbb{F}$ have at least two elements and that implies (with the other axioms) that $1\neq0$. Or you can require that $+$ and $\times$ be distinct operations (meaning that there is at least one pair such that $a+b\neq a\times b$). That again implies (with the other axioms) that $1\neq0$. $\endgroup$
    – 2'5 9'2
    Feb 27, 2018 at 4:01
  • $\begingroup$ @alex.jordan Nice, good to know! And that reminds me: another cool thing is that we don't need to require commutativity of addition: from $(1+x)(1+y)=(1+y)(1+x)$ and the distributive law, commutativity of multiplication, etc., we can work out $x+y = y+x$. (Maybe you need additive inverses to be both left- and right-; don't remember.) $\endgroup$ Feb 27, 2018 at 19:19

2 Answers 2


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 $\mathbb{F}_2\times\mathbb{F}_3$.

  • $\begingroup$ Thanks, maybe I should have thought about the question more: as division by $0$ is what's causing the problem here, what about a ring? Can we get one starting with just $(+, -, \times)$ and some equations? $\endgroup$ Feb 27, 2018 at 3:45
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    $\begingroup$ You can axiomatize rings with just equations. You just need to include $0$, $1$ as nullary operations and $-$ as a unary operation, so you can state the axioms about units and inverses without saying "there exists"... $\endgroup$ Feb 27, 2018 at 3:48
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    $\begingroup$ You can't do it without symbols for $0$ and $1$, though. For instance, if your only operations are $+$, $-$, and $\times$, then the empty set (with the unique binary operation on it) will automatically satisfy all your axioms, but the empty set is not a ring. $\endgroup$ Feb 27, 2018 at 3:49
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    $\begingroup$ Even if you rule out the empty algebra, you're still in trouble (take the even numbers), at least if by "ring" you mean "unital ring". $\endgroup$ Feb 27, 2018 at 3:53
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    $\begingroup$ @YuvalFilmus Not for a ring : in a ring it is possible that $0=1$ (of course that only happens when the ring is trivial). $\endgroup$
    – Arnaud D.
    Feb 27, 2018 at 16:36

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.

  • $\begingroup$ Thank you for the reference to an entire topic of study! Universal algebra and varieties look very relevant and interesting. $\endgroup$ Feb 27, 2018 at 3:56
  • $\begingroup$ Actually what I was trying to do was something like this: if I start with a finite set $\{a,b,c,d,\dots}$ (standing for a subset of the real numbers, say), and consider all possible “expressions” formed out of them like $a+b(c-d)$, and reduce these expressions to equivalence classes according to known rules (e.g. $a+b(c-d)=(c-d)b+a$), how many (and what) equations do I need, to know I've “covered everything”, and inequivalent expressions truly are so (say in the sense that they can take distinct values for at least some values of $a,b,c,d$)? Does universal algebra help with this sort of thing? $\endgroup$ Feb 27, 2018 at 4:02
  • $\begingroup$ @ShreevatsaR Yes, that is one of the things which universal algebra studies: how many equations are needed to completely describe a given situation (I'm phrasing this vaguely since what you've described is only one instance of this sort of question). $\endgroup$ Feb 27, 2018 at 4:16
  • $\begingroup$ I think it is worth pointing out that classes of structures axiomatized by equations are first defined as "equational classes," while classes closed under homomorphic images, substructures, and products are called "varieties." Birkhoff's theorem -- which says that equational classes and varieties are one and the same -- is, in my mind, the first non-trivial result in universal algebra. $\endgroup$
    – Bey
    Mar 29, 2018 at 1:13

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