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Compared to the categories of other “common” algebraic objects like groups and rings, it seems that fields as a whole are missing some important properties:

  • There are no initial or terminal objects
  • There are no free fields
  • No products or coproducts
  • Every arrow is a mono (maybe not a bad thing, but still indicates how restrictive the category is)

A logician once told me in passing that part of the reason is that the properties for fields contain a decidedly “weird” property, namely that every element in a field except zero has a multiplicative inverse. If I understood him correctly, this property is sufficiently different from the others that the category of all such objects loses some features. But I have no idea if this was a heuristic or a proven theorem.

Ducky
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  • If a zero had an inverse it would not be a zero. – markvs Jul 14 '20 at 01:40
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    We're sorry. Really. – Will Jagy Jul 14 '20 at 01:40
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    That basically is why, because it implies there can be no nontrivial maps between different characteristics. – pancini Jul 14 '20 at 01:46
  • I can think of a way to test this in some extremely weak sense. What if you add in the arrows for the ring homomorphisms but don't add any additional objects to FIELD? If the category is still non-nice, maybe zero is not to blame. – Greg Nisbet Jul 14 '20 at 01:47
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    @GregoryNisbet You basically don't gain anything: since the kernel of a commutative ring homomorphism is always an ideal, every homomorphism from a ring which happens to be a field is either injective or sends everything to zero. So all that happens is that you add a new "zero morphism" between every pair of fields. – Noah Schweber Jul 14 '20 at 02:08
  • @NoahSchweber That is, if you are considering non-neccesarily-unital morphisms. – Jackozee Hakkiuz Jul 14 '20 at 02:30
  • @JackozeeHakkiuz Good point - if you're looking at morphisms of unital rings, then of course you don't gain any new ones at all. – Noah Schweber Jul 14 '20 at 02:35
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    Things look a bit better if you look at the category of fields with a given characteristic: you then get initial objects and free objects. – Rob Arthan Jul 14 '20 at 02:41
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    You don't get free objects (other than the free object on $\varnothing$) – every rational function has a pole somewhere... – Zhen Lin Jul 14 '20 at 03:09
  • @ZhenLin, setting aside the constants as you obviously meant, it depends on the field, if I understand you correctly. For example, $1/(x^2 + 1) \in \mathbb R(x)$ has no pole …. – LSpice Jul 14 '20 at 10:54
  • It doesn't have a pole when evaluated in $\mathbb{R}$, of course, but it does in $\mathbb{C}$. – Zhen Lin Jul 14 '20 at 11:37
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    Fields don't form a variety in the universal algebra sense. By the HSP theorem, this means that fields are not closed under either homomorphic images, or subalgebras, or products (where these are defined relative to total operations of multiplication and addition but not the partial operation of division). This lack of algebraic niceness leads to a lack of categorical niceness. – John Coleman Jul 14 '20 at 15:24
  • Was that logician by any chance Kim Bruce? He told me something very similar back in about 1975 -- that the existence of a negation in the field axioms led to problems of a specific kind (perhaps something like "the homomorphic image of a field is not a field," although I'd have to know what definition of homomorphism was being used for that to make sense...) – John Hughes Jul 19 '20 at 15:37

2 Answers2

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There is a precise sense in which the concept of field is not algebraic like, say, the concept of ring or group or vector space etc.: it is a theorem that any kind of mathematical structure that is defined as having a set of elements and some fixed list of total operations of constant finite arity obeying some fixed list of unconditional equations gives rise to a category with certain nice properties (which I omit for the moment). The usual definition of field has a partially defined operation – inversion – as well as an inequality ($0 \ne 1$), which means the theorem is not applicable; the fact that the category of fields does not have the nice properties of algebraic categories tells us there is actually no way of defining fields so that the theorem applies.

So what does being algebraic buy us, and how do we recognise an algebraic category without thinking about the logical form of the definition? Well, a category is equivalent to a category of algebraic structures if and only if it has all of the following properties:

  • It has limits for all small diagrams and colimits for small filtered diagrams.
  • There is an object $A$ such that the functor $\mathrm{Hom} (A, -)$ has a left adjoint, is monadic, and preserves colimits for small filtered diagrams.

In fact, it follows that such a category has colimits for small diagrams in general, but this fact is not needed in the theorem. Note that the object $A$ is not unique up to isomorphism; this is essentially the phenomenon of Morita equivalence.

Zhen Lin
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    Fancy seeing you here! – Asaf Karagila Jul 14 '20 at 09:48
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    @Asaf For some reason I seem to have too much free time these last few months... – Zhen Lin Jul 14 '20 at 11:33
  • Could you also add a source for the complete theorem that you mention at the start? Thanks! –  Jul 14 '20 at 12:22
  • I no longer remember where it is written in the literature. The "only if" direction is easy, at any rate; the "if" direction requires more work but is easy if you know about Lawvere theories, locally presentable categories, etc. – Zhen Lin Jul 14 '20 at 13:00
  • Thanks for the helpful answer! So then the failure of products, free fields, etc. can all be seen as a failure for certain diagrams to have limits? – Ducky Jul 14 '20 at 13:15
  • Products are a particular kind of limit. Free objects are what you get by applying the left adjoint of the forgetful functor. It is difficult to reason about counterfactuals but since the category of fields is a finitely accessible category and the forgetful functor is finitely accessible, you could say that if the category had limits then everything would have worked. – Zhen Lin Jul 14 '20 at 13:30
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    Does the inequality $0\neq 1$ matter so much? Adding the zero ring to the category of fields does not seem to alter the landscape much, since it would be an isolated object anyway. – tomasz Jul 14 '20 at 13:59
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    There are many (interrelated) reasons for excluding the zero ring. From an algebraic geometry perspective: fields are points, but the zero ring is just empty; from the algebra perspective: the quotient of a commutative ring by an ideal is a field if and only if the ideal is maximal, etc. – Zhen Lin Jul 15 '20 at 01:42
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    @Brahadeesh You can find this theorem, and a number of related ones, in Adamek-Rosicky-Vitale's book on algebraic theories. – Kevin Carlson Jul 15 '20 at 04:44
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This is a very minor comment on this statement:

The usual definition of field has a partially defined operation – inversion – as well as an inequality ($0 \ne 1$), which means the theorem is not applicable

Actually the issue is not that inversion is partially defined. You can of course extend inversion to a total function $(-)^{-1} : F \to F$ by just declaring that $0^{-1} = 0$. This offends our mathematical sensibilities but it can be done. The issue is that if you do this you still need to state a non-equational axiom, namely that $a \cdot a^{-1} = 1$ for all $\color{red}{a \neq 0}$, and that "$a \neq 0$" part is the real problem.

This $a \neq 0$ requirement prevents fields from being closed under products (by which I don't mean categorical products but cartesian products with inherited operations): as you can see from the fact that $(1, 0)$ and $(0, 1)$ are zero divisors in a product $F_1 \times F_2$ of two fields, the precise issue is that a nonzero element in $F_1 \times F_2$ is not a pair of nonzero elements in $F_1$ and $F_2$. So even if we define $(-)^{-1}$ on $F_1 \times F_2$ pointwise (by either taking the inverse or sending $0$ to $0$), the non-equational axiom above is not satisfied. So, by the easy direction of the HSP theorem fields are not a variety in the sense of universal algebra, as John Coleman says in the comments.

From here we could ask something like: what is the best "equational approximation" to the non-equational field axioms? The simplest one I can think of is that $a \cdot a^{-1} \cdot a = a$, meaning that $a^{-1}$ is a kind of "weak inverse." We could also require that $(a^{-1})^{-1} = a$, which implies $a^{-1} \cdot a \cdot a^{-1} = a^{-1}$, and we could also require $(ab)^{-1} = a^{-1} b^{-1}$. (And we just drop $0 \neq 1$ entirely.) It would be interesting to know whether these, together with the commutative ring axioms, imply all equational axioms satisfied by fields. (Edit: I've asked a question about this here.)

These are perfectly sensible equational axioms and they now hold not only for fields but for all products of fields, if $a^{-1}$ is taken to be the operation which componentwise either inverts or sends $0$ to $0$, as above. They also holds for subrings of products of fields which are closed under weak inverse. The resulting category must in fact have all limits, colimits, etc. now, although the coproduct seems to be strange. I guess there must even be a free such thing on $n$ generators but I don't know what it looks like.

Every ring in this category is von Neumann regular.

Interestingly this category includes all rings satisfying $a^n = a$ for all $a$ and a fixed $n$, where we take $a^{-1} = a^{n-2}$ to be the weak inverse operation; this condition is the subject of a classical theorem of Jacobson, who showed that they are commutative.

Qiaochu Yuan
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