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Proving Hom(Q/Z, Q/Z) is isomorphic to \hat{Z}

  1. Sep 7, 2010 #1
    Hi folks, long time reader, first time poster.


    So I'm trying to prove that $\hat{Z} = \varprojlim Z/nZ$ is isomorphic to Hom(Q/Z, Q/Z).

    Problem is, I have a lot of trouble describing even a single endomorphism of that torsion group (except, of course, the trivial one, that sends everything to 0.)

    Q/Z can be considered as the set of all roots of unity. and I should be able to find an endomorphism on that group, just by permuting key roots of unity (simple example: {1,-1,i,-1}) But I haven't been able to generalize this for the whole thing.

    I also tried a pretty naive thing, taking a \in \hat{Z}, say a=(a_1, a_2, ...) and trying to define some \theta : Q/Z \to Q/Z, with \theta (m/n) = (m * a_n) / n. which I believe defines a map from \hat{Z} into Hom(Q/Z, Q/Z), but I don't have much in the way of an inverse, especially since I can't describe anything but trivial members of Hom(Q/Z, Q/Z)

    I'm aware that \hat{Z} is also the product of all the p-adic integers Z_p, but I'm having trouble proving that as well.

    In any case, I'm stuck, I don't want to spend much more time on this (Because its just one small bit of a much larger project), so I thought I'd try asking the kind and clever folks here for some help.


    so, Hom(Q/Z, Q/Z) is isomorphic to \hat{Z}. Any ideas on how to prove it?

    Thanks.
     
    Last edited: Sep 7, 2010
  2. jcsd
  3. Sep 7, 2010 #2

    Hurkyl

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    Q/Z is an Abelian group, a.k.a. a Z-module. Thus multiplication by an integer is an endomorphism.
     
  4. Sep 7, 2010 #3
    indeed, but that's hardly all endomorphisms!
     
  5. Sep 7, 2010 #4

    Hurkyl

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    Sure; but since you had overlooked it, I thought I'd point it out and see if you got somewhere.


    My first instinct on the problem is to write Q/Z as the union of all of its cyclic subgroups. I suppose that's just a convoluted way of saying you can write elements as m/n, though.
     
  6. Sep 7, 2010 #5
    Thanks! actually, I think my naive approach works. if you define a reverse map like so:

    given f \in End(Q/Z), define an a=(a_1, a_n, ... ) \in \prod_n Z/nZ by a_n = n f(1/n). if m|n, then we can write n=mr and mr f(1/n) - m f(1/m) is obviously divisible by m, so a meets the appropriate congruence relations to be in \hat{Z}. and its the inverse map of the one I defined above. So unless I'm missing something else, I think I have it...
     
  7. Sep 7, 2010 #6

    Hurkyl

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    The functions theta you defined before aren't homomorphisms -- sorry, I misread and thought you had noticed that already.
     
  8. Sep 7, 2010 #7
    (that plus sign was suppose to be multiplication. I corrected the error in the edit.)

    they aren't? I thought I proved that for a \in \hat{Z}, \theta_a by \theta_a ( m/n ) = a_n * m/n was. Like so:

    [tex]\theta_a (m_1 / n_1 + m_2 / n_2 ) = \theta_a ( (n_2 m_1 + n_1 m_2) / n_1 n_2) = a_{n_1 n_2} (n_2 m_1 + n_1 m_2) / n_1 n_2[/tex]

    [tex]\theta_a (m_1/n_1) + \theta_a(m_2 / n_2) = a_{n_1} m_1 / n_1 + a_{n_2} m_2 / n_2 = (a_{n_1} n_2 m_1 + a_{n_2} n_1 m_2) / n_1 n_2[/tex]

    now, by def of [tex]a \in \hat{Z}[/tex], we have [tex]a_{n_1 n_2} = q n_1 + a_{n_1} = r n_2 + a_{n_2}.[/tex] Or in other words,
    [tex]a_{n_1} = a_{n_1 n_2} - q n_1[/tex] and [tex]a_{n_2} = a_{n_1 n_2} - r n_2[/tex].

    so then [tex]a_{n_1} n_2 m_1 + a_{n_2} n_1 m_2 = a_{n_1 n_2} (n_2 m_1 + n_1 m_2) - n_1 n_2 (m_1 q + m_2 r)[/tex]

    divide that by [tex]n_1 n_2[/tex] and the negative term is 0 (since we're in Q/Z, the n_1*n_2 turns it into an integer and integers are modded out). And that proves its a homomorphism.

    Or am I confused and/or making mistakes?
     
    Last edited: Sep 7, 2010
  9. Sep 7, 2010 #8

    Hurkyl

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    Changing the + to * changes things. :wink:
     
  10. Sep 7, 2010 #9
    Ok. First, it shouldn't be hard to prove that hom(Q/Z, Q/Z) is isomorphic to hom(Q/Z, R/Z), which is the Pontryagin dual of the discrete group Q/Z. So it suffices to prove that Q/Z and [tex]\hat{\mathbb{Z}}[/tex] are dual to each other.

    (In case you're unfamiliar with this duality: If G is a locally compact abelian topological group, let Ch(G) = hom(G, R/Z). Then Ch is a contravariant functor, and Ch2 is naturally isomorphic to the identity functor on the category of locally compact abelian topological groups; hence Ch is a duality of categories and thus interchanges limits and colimits.)

    Now note:
    [tex]\mathbb{Q}/\mathbb{Z} = \lim_{\longrightarrow} N^{-1}\mathbb{Z}/\mathbb{Z}.[/tex]
    and
    [tex]\hat{\mathbb{Z}} = \lim_{\longleftarrow} \mathbb{Z}/N\mathbb{Z}.[/tex]
    (The direct and inverse limits are taken over N and divisibility.) It is elementary to show that Ch(N-1Z/Z) is naturally (in N) isomorphic to Z/NZ. Since Ch interchanges limits and colimits, we get the desired result.


    Also, you can prove that [tex]\hat{\mathbb{Z}}[/tex] is the product of the Zp, purely by using the universal properties of limits and products, and the Chinese remainder theorem. (Show that the former satisfies the universal property of the latter.) The key is that if N = p1n1...pknk is a factorization of N into powers of distinct primes, then Z/NZ is isomorphic to Z/p1n1Z × ... × Z/pknkZ.

    Let me know if anything needs clarification. I don't mind writing this stuff up in more detail. (Although I guess you need to know some terminology from category theory.)
     
    Last edited: Sep 7, 2010
  11. Sep 8, 2010 #10
    I've barely seen any category theory before, but its on my list of things to do. Thanks for that, you've given me another example to try to understand. :)
     
  12. Sep 8, 2010 #11
    This article has fewer prerequisites, but it describes Ch(Q). Then Ch(Q/Z) is isomorphic to the subgroup of Ch(Q) consisting of elements with kernel containing Z, which presumably you can show is isomorphic to [tex]\hat{\mathbb{Z}}[/tex].

    www.math.uconn.edu/~kconrad/blurbs/gradnumthy/characterQ.pdf
     
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