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Quotient field vs Functor

  1. Oct 17, 2005 #1

    Hurkyl

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    I'm embarassed because I'm surely missing something obvious... the very first exercise in Categories for the Working Mathematician is:

    Show how each of the following constructions can be regarded as a functor: The field of quotients of an integral domain; the Lie algebra of a Lie group.

    The latter one is easy, since any map f of manifolds gives rise to a map f* of the tangent spaces, and as I recall, it all works out properly.

    But the former is bothering me. Assuming commutative domains, taking quotients cannot be a map Domain-->Field, and considering the map [itex]\mathbb{Z} \rightarrow \mathbb{Z}_5[/itex]:

    Taking the quotient field cannot be covariant because that would yield a map of fields [itex]\mathbb{Q} \rightarrow \mathbb{Z}_5[/itex], and it cannot be contravariant because that would yield a map of fields [itex]\mathbb{Z}_5 \rightarrow \mathbb{Q}[/itex].

    So in what way have I confused myself? Or is this actually a bad exercise?
     
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  3. Oct 17, 2005 #2

    matt grime

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    What about the map sending everything to zero? I know it doesn't send 1 to 1 but sometimes people define things so that 1 isn't that distinguished in fields.
     
  4. Oct 18, 2005 #3

    Hurkyl

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    CWM defines that the morphisms of Rng (objects rings with 1) preserve the unit. I'd imagine it expects fields to be the same.
     
  5. Oct 19, 2005 #4
    the obvious map from Q to Z5 which takes a/b to f(a)/f(b) is not actually a map, since it would take 1/5 to 1/0. But if we could eliminate that danger, this would be a nice functorial field homomorphism.

    the problem is that any field homomorphism is a monomorphism, and functors preserve monicness, so if you want a functor from that takes integral domains to fields, you have to consider only monomorphisms. The category of integral domains with monomorphisms as morphisms is Dom<sub>m</sub>, as opposed to the category of integral domains with all ring homomorphisms. The map from Z to Z5 is of course not a monomorphism. Look instead at something like Z->Z or Z2->F<sub>4</sub> and it should work fine, with the functor taking the ring homomorphism to the field homomorphism a/b |--> f(a)/f(b). The injectivity of f will guarantee that f(b) is not zero.

    This is explained in MacLane on page 56.

    By the way, the functor taking Dom_m to Fld is the adjoint to the forgetful functor.
     
  6. Oct 19, 2005 #5
    I guess you can see that this won't be functorial when you consider Z --> Z2 --> F_4
     
  7. Oct 19, 2005 #6

    Hurkyl

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    Ah, that would do it, wouldn't it? I didn't think about taking a non-full subcategory of Rng!
     
  8. Oct 20, 2005 #7
    yes, well as you can see, it's not a very obvious solution, and so probably not so great as the very first problem in the textbook. at any rate, nothing to be embarrassed about, not seeing the answer.

    A long time ago, me and a buddy spent a long time looking for the answer to number 5 in that same problem set: find two functors on Grp whose object functions are the identity. Obviously one is the identity functor. I eventually came up with an answer for another functor (actually, an infinity of unconstructable answers by invoking AC), but that seems far too ugly an answer. So if you have any thoughts you wanna share on that one, I'd like to hear them.
     
  9. Oct 20, 2005 #8

    Hurkyl

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    I thought about it on the way into work today: we could "replace" a single group in Grp with a conjugate.

    That is, let h be an element of the group G.

    Any morphism f whose target is G gets replaced with x -> h f(x) h^-1.
    Any morphism f whose source is G gets replaced with x -> f(h^-1 x h)

    So the map F given by:
    If f does not have G as source or target, F(f)(x) = f(x)
    If f does has G as target, but not as source, F(f)(x) = h f(x) h^-1
    If f does has G as source, but not as target, F(f)(x) = f(h^-1 x h)
    If f has G as both source and target, F(f)(x) = h f(h^-1 x h) h^-1
    should be a functor.


    This could be generalized: for any choice "function" on the objects of Grp, we can play the above game. We don't need the axiom of choice, because all groups have a distinguished object: their identity. (The axiom of choice might not even be applicable! e.g. if you posit that all sets are small, this can't be a set-theoretic choice function)
     
  10. Oct 20, 2005 #9
    yeah, that seems to do it. Though your construction, like mine, is unnatural. It requires a lot of arbitrariness.

    Well, you probably shouldn't choose the identity element to conjugate by, or else your functor will be the identity functor. However, you don't need AC to choose one group from the objects of Grp, nor do you need it to choose one element from that one group. You only need AC to choose infinitely many things.

    I don't understand what you mean here. As far as I know, not only can there be a set-theoretic choice function on small sets of sets, AC guarantees that there is one, for any small set (and big sets too).

    By the way, my answer to this question was the following: define an equivalence relation between homomorphisms G-->H if there there exist automorphisms a on G and b on H so that f~g if f=agb. Then take as your functor one which assigns to each homomorphism its equivalence class. Or if you prefer, using AC, takes each homomorphism to any representative of its equivalence class.

    This construction is similar to yours, I think, in that it takes advantage of the fact that you can throw in an automorphism in front of or behind any morphism, and not change too much about it. Your automorphism was a single conjugation. I left mine unspecified, both in form and in number. But both constructions required arbitrary choices, I was hoping there was some natural functor, but I guess there isn't.
     
  11. Oct 21, 2005 #10

    Hurkyl

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    I didn't think that construction would be appropriate, since it's not really a functor from Grp -> Grp, but from Grp -> Grp/~. Your invocation of the AC might make it Grp -> Grp, but I'm not 100% convinced it's a functor. (How do you know that you can find a compatable collection of representatives, in the sense that if, for the equivalence classes f g = h, that the representatives also satisfy that equation?)

    My comments on the AC were for my generalization, where I conjugate every group. I don't need the AC because I can start with the distinguished choice "function" that sends each group to its identity, and can construct perturbations of that.

    My comments that it might not be applicable is if I decide to work with a model of ZFC in which all sets are small, which doesn't have a set of all small groups, and thus I could not invoke the AC, if I wanted to.
     
  12. Nov 1, 2005 #11

    mathwonk

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    elementary comment on the domain to field thing: this construction is a special case of "localization", or formation of rings of quotients in classical commutative algebra.

    This is a functor of pairs, (R,S) where S is a multiplicative set in (the commutative ring with identity) R. then the correct morphisms preserve these multiplicative sets. For a domain the canonical multiplicative set S is the set of all non zero elements.

    Usually S contains 1 and no pairs of zero divisors. Or simply S is closed under multiplication and does not contain zero. The canonical set in an R which may not be a domain, is the set of non zero divisors. Then any map to a ring which sends everything in S to a unit, extends uniquely to a morphism of the localization of R at S.

    or any map of pairs (R,S) to (R',S') taking S into S', induces a map of the localizations.

    see Zariski Samuel pages 221-222, especially theorem 14, p.222.
     
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