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Math Help - Basis of dual space involving free Z-module

  1. #1
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    Basis of dual space involving free Z-module

    Hello friends,

    I am truly stuck on this problem. Any guidance would be supremely appreciated.

    Let F=Sum over x in X of Zx be a free Z-module with an infinite basis X. Then {f_x : x in X} does not form a basis of F*.

    Relevant notes:

    f_x is a map F -> Z given by f_x(y) = d_xy, where d is the Kronecker delta

    Hints given: F* is isomorphic to the direct product over x in X Zx, but under this isomorphism, f_y -> {d_xy * x}, which is in the direct product over x in X of Zx.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Re: Basis of dual space involving free Z-module

    Quote Originally Posted by empyreandance View Post
    Hello friends,

    I am truly stuck on this problem. Any guidance would be supremely appreciated.

    Let F=Sum over x in X of Zx be a free Z-module with an infinite basis X. Then {f_x : x in X} does not form a basis of F*.

    Relevant notes:

    f_x is a map F -> Z given by f_x(y) = d_xy, where d is the Kronecker delta

    Hints given: F* is isomorphic to the direct product over x in X Zx, but under this isomorphism, f_y -> {d_xy * x}, which is in the direct product over x in X of Zx.
    Hey man, it's a little hard to decipher precisely what's going on here, but I'm guessing you're supposed to show that \left(\mathbb{Z}^{\oplus\lambda}\right)^\vee \not\cong \mathbb{Z}^{\oplus\lambda} (or what you're asking is basically equivalent to this) where \lambda is some infinite cardinal and ^\vee denotes the dual module. I think the easiest way to do this is to recall the way Hom reacts with coproducts so that



    \left(\mathbb{Z}^{\oplus\lambda}\right)^\vee=\text  {Hom}_\mathbb{Z}(\mathbb{Z}^{\oplus\lambda}, \mathbb{Z})\cong \left(\text{Hom}_\mathbb{Z}(\mathbb{Z},\mathbb{Z})  )^\lambda\cong \mathbb{Z}^{\lambda}


    So, what you really want to prove is that for an infinite cardinal one has that \mathbb{Z}^{\oplus\lambda} \not\cong \mathbb{Z}^\lambda. But, since \mathbb{Z} is countable this is trivial for a stupid reason-- \mathbb{Z}^\lambda and \mathbb{Z}^{\oplus\lambda} aren't equipotent. Indeed, it's trivial from Cantor's theorem that |\mathbb{Z}^\lambda|\geqslant |2^\lambda|>\lambda. That said, |\mathbb{Z}^{\oplus\lambda}|=\lambda since \lambda is an infinite cardinal--I leave this to you to work out.
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  3. #3
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    Re: Basis of dual space involving free Z-module

    That makes perfect sense and yes, that was what I was asking. My apologies for the exposition. Thank you so very much!
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