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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.

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    $\begingroup$ We're sorry. Really. $\endgroup$ – Will Jagy Jul 14 at 1:40
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    $\begingroup$ That basically is why, because it implies there can be no nontrivial maps between different characteristics. $\endgroup$ – Elliot G Jul 14 at 1:46
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    $\begingroup$ 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. $\endgroup$ – Rob Arthan Jul 14 at 2:41
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    $\begingroup$ You don't get free objects (other than the free object on $\varnothing$) – every rational function has a pole somewhere... $\endgroup$ – Zhen Lin Jul 14 at 3:09
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    $\begingroup$ 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. $\endgroup$ – John Coleman Jul 14 at 15:24
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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.

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    $\begingroup$ Fancy seeing you here! $\endgroup$ – Asaf Karagila Jul 14 at 9:48
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    $\begingroup$ @Asaf For some reason I seem to have too much free time these last few months... $\endgroup$ – Zhen Lin Jul 14 at 11:33
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    $\begingroup$ 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. $\endgroup$ – tomasz Jul 14 at 13:59
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    $\begingroup$ 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. $\endgroup$ – Zhen Lin Jul 15 at 1:42
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    $\begingroup$ @Brahadeesh You can find this theorem, and a number of related ones, in Adamek-Rosicky-Vitale's book on algebraic theories. $\endgroup$ – Kevin Arlin Jul 15 at 4:44

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