## In what sense are fields an algebraic theory?

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Since there is no "free field generated by a set", it would seem that

1) there is no monad on Set whose algebras are exactly the fields

and

2) there is no Lawvere theory whose models in Set are exactly the fields

(Are 1) and 2) correct?)

Fields don't form a variety of algebras in the sense of universal algebra since the field axioms can´t be written as identities (since the axiom for multiplicative inverses has the restriction that the element be non-zero).

I guess fields are an algebraic theory in a more general universal algebra sense of being defined by operations on a single set with a set of first order axioms.

Is there any better sense in which they are algebraic or are fields just not really algebraic in nature?

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## 5 Answers

1 and 2 are correct, for a simple reason. If C is a category satisfying either 1 or 2 then C has a terminal object. But there is no terminal object in the category of fields (and ring homomorphisms), because there are no maps between fields of different characteristic.

For the same reason, the category of fields is not an essentially algebraic theory (mentioned in Andrew's answer). An essentially algebraic theory can be defined as, simply, a small category with finite limits. A model or algebra for an essentially algebraic theory T is a finite-limit-preserving functor T --> Set. (Of course, you can consider models in other finite limit categories too.) And the category of models always has a terminal object.

This embodies the idea that Andrew was describing, of a theory where some operations are only partially defined, but (and this is crucial!) the domain of definition is itself defined by equations. You can see some rough connection between finite limits and this intuitive idea if you consider pullbacks in Set. A pullback in Set is, after all, the set of pairs satisfying some equation.

I don't know in what sense the theory of fields is algebraic. It's partly because of its failure to be algebraic in any of the usual senses that one often chooses to work with commutative rings rather than fields, in algebraic geometry and in topos theory, for instance.

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Thanks for the confirmation! I doubted that fields would be even essentially algebraic but wasn't sure that there wasn't a sneaky way to do it. – Andrew Stacey Oct 29 2009 at 7:59

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Combining your answer with Andrew Stacey's, might this be a reason to think that there would be an interesting notion of algebraic geometry over meadows? – Steven Gubkin Feb 3 2010 at 17:58

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Fields are not algebraic. An algebraic theory, for example, has free objects: there are free rings, free groups, a free monoids. The free functor is left adjoint to the forgetful functor to sets (okay, I'm talking about models in Set). There are, though, no free fields.

One can extend one's idea of an "algebraic theory" to an "essentially algebraic theory" in which partially defined operations are allowed (it's not clear to me that fields satisfy those since you need to specify the domain in terms of other operations whereas it seems that one can only specify the domain of the inverse as the complement of such a subset). Or, (maybe, but I doubt it), one could define a field as a Z2-graded algebraic theory where 0 is in degree 0 and everything else is in degree 1. Here, a grading should be regarded simply as a labelling system.

Alternatively, one can talk about meadows. Meadows are algebraic theories which are modified versions of fields. Instead of multiplicative inverses, there is a unary operation ι:M → M which satisfies the identity xι(x)x = x. Defining ι(x) = x-1 for non-zero x, and ι(0) = 0 turns any field into a meadow. The relationship between meadows and fields is quite strong.

An arXiv search throws up 68 references (at time of writing; for some reason google doesn't turn up anything particularly relevant, even when combined with the word "field"). One prominent name is that of Jan Bergstra.

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I didn't know about meadows, thanks for telling me about them! – Omar Antolín-Camarena Oct 28 2009 at 16:00
Is there a way to talk about a "field object" in a category? – Dinakar Muthiah Oct 29 2009 at 0:08
There's probably some way to make sense of that ("Anyone saying something is impossible is more than likely to be interrupted by some idiot doing it."), but I think it would be so contrived that none of the intuition from group objects or ring objects would carry over.  The problem is that the inverse is defined on the complement of a subset and that's not obvious how to generalise to an arbitrary category. – Andrew Stacey Oct 29 2009 at 8:02

As previous answers have said, fields are not algebraic. They are also not essentially algebraic, because categories of models of essentially algebraic theories have an initial object, and the category of fields instead has a set of initial objects -- Z and Z_p for each prime p. (There is no map from a field of one characteristic to a field of a different characteristic, so there can't be a single initial object.)

Fields are models of a a theory which is essentially algebraic plus allows specification of disjunctions. In the language of sketches, fields are the models of a "finite sum sketch." This was proved in Diers' thesis and is spelled out in the paper "The formal description of data types using sketches" by Charles Wells and Michael Barr, in volume 298 of the Springer Lecture Notes in Computer Science, 1988. For a general overview of sketches, see "Sketches: Outline with References" at http://www.cwru.edu/artsci/math/wells/pub/pdf/sketch.pdf .

ADDED 17 November 2009. The category of models of a finite-sum theory is not as nice as the models of an algebraic theory or even an essentially algebraic theory. Generally, the more different kinds of things you can specify in a sketch, the more awkward the category of models is. The category of fields is pretty awkward!
It does have filtered colimits and is closed under ultraproducts. Field theorists have made considerable use of the closure under ultraproducts.

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Is it possible that the category of field of fixed characteristic p are essentially algebraic? – Charles Rezk Oct 29 2009 at 3:01

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Categories of models of essentially algebraic theories are categories of models of finite limit sketches, and thus locally presentable, which the category of fields of a given characterstic is not (for instance it lacks a terminal object or coproducts). – Reid Barton Oct 29 2009 at 3:14
It also follows from this answer that the category of fields is accessible, which can be regarded as a sort of "algebraicity". – Mike Shulman Oct 29 2009 at 3:37

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Yes, I've always thought that locally presentable and accesible categories (see Adamek and Rosicky's book of the same name) have really bad PR--sounds like the most boring and technical thing in the world but actually the theory is quite beautiful and illuminating. – Reid Barton Oct 29 2009 at 3:45

A small remark that might be helpful as well is motivated by the Birkhoff theorem - in most usual senses, algebras for the given algebraic theory are closed under products, which fields are not.

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Algebraic theory are those definited in Universal ALgebras ang formalizated by B.Lawvere in categorical formalism. But this is only the first elementary level of what is called "Categorical Logic", in "right" categories is possible formalizing the "language and terms" used in the further logics, for example $\exists$, $\Rightarrow$, ecc.. For define a Field structure (beyond the machinary used for Rings) you need use the existential $\exists$, or the "not", similar thing for define the notion of "Local commutative ring". See P. Johnstone "Topos Theory" or (much) better "Sketches of an Elephant: A Topos Theory Compendium 2"

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