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Homomorphism confusion

  1. Jan 30, 2007 #1
    1. The problem statement, all variables and given/known data
    Here's the problem I've been trying to get my mind around:
    Prove that there exists an isomorphism of [tex]F^n[/tex] into [tex]Hom(Hom(F^n,F),F)[/tex].

    I'm missing something. Here's what I get to:

    [tex]F^n[/tex] is an n-tuple and [tex] F[/tex] is a field. So I can see that there is a set of homomorphisms from [tex]F^n[/tex] into [tex]F[/tex].
    It would be a finite n-tuple mapped into an infinite field, so there would just be a finite number of elements mapped infinitely.

    I used n=3 and the real numbers as an example for the sake of trying to understand this:
    Define a homomorphism [tex]T:F^n \rightarrow F[/tex] as follows:
    so [tex]x_3[/tex] is everything else besides 1 and 2.

    Then I get confused trying to define [tex]Hom(Hom(F^n,F),F)[/tex] Am I just mapping everything back into the origional [tex]F[/tex]? And isn't that just what I started with, which is [tex]F[/tex]?(or the reals in my attempted example)

    So the whole point is to show that there is an isomorphim from [tex]F^n[/tex] into the [tex]Hom((HomF^n,F),F)[/tex] But it looks to me like F^n is a finite n-tuple and I can't get my mind around how there can be an isomprphism between an infinite field and a finite n-tuple.

    What am I missing? Where have I gone wrong?

    any clarification will be greatly appreciated.
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
    Last edited: Jan 30, 2007
  2. jcsd
  3. Jan 30, 2007 #2


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    Homework Helper

    I don't really follow what you've tried so far.

    Remember, Hom(F^n,F) is the set of linear functions from F^n into F. That is, functions f where x is in F^n and f(x) is in F, and which satisfy f(ax+by)=af(x)+bf(y), where x,y are vectors in F^n and a,b are scalars (in F). Such linear functions from a vector space into the underlying field are usually called linear functionals.

    Note that the linearity of the function implies that its value at an arbitrary vector x is completely determined by its values on a basis of F^n. So, for example, if F=R and n=3, there is a bijection between 3-tuples (a,b,c), where a,b,c are real, and linear functionals, given by sending (a,b,c) to the functional f defined by f(1,0,0)=a, f(0,1,0)=b, f(0,0,1)=c, so that f(x,y,z)=ax+by+cz. You can show that the set of such linear functionals forms another vector space over F, and in this case the bijection just mentioned is in fact a vector space isomorphism.

    Does this help at all?
  4. Jan 30, 2007 #3
    That makes a whole bunch of sense...but this question is in the section before we study what a basis is. I just went through that section so I see what you are talking about. I don't think I can apply the basis to this problem BUT I'm going to rewrite my definition of the homomorphism[tex]F^n \rightarrow F[/tex] and try again.
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