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Algebriac Structures

  1. Jan 25, 2007 #1
    1. The problem statement, all variables and given/known data

    HI, I'm working on this:
    If n>m, prove that there is a homomorphism of F^(n) onto F^(m) with a kernel W which is isomorphic to F^(n-m).

    2. Relevant equations

    Def: If U and V are vector spaces over F (a field) then the mapping T of U onto V is said to be a homomorphism if:
    a) (u1+u2)T=u1T+u2T
    b) (a u1)T=a(u1)T

    If T in addition is one-to-one, we call it and isomorphism. The Kernel ot T is defined as {u in U|uT=0} where o is the identity element of the addition in V.

    3. The attempt at a solution
    These are my thoughts:
    It seems trivial to me that there is a homomorphism from F^(n) onto F^(m) since n>m. I just don't know how to formalize that argument. I can't get a picture in my mind to write it down.
    The part about the kernel of the homomorphism isomorphic to F^(n-m) also seems to be intuitively simple..the kernel will have n-m elements in it, so there's got to be an isomorphism between the kernel and F^(n-m).

    Please help me clarify and formalize this.
  2. jcsd
  3. Jan 25, 2007 #2


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    Homework Helper

    what is F^n explicitly? Write down a general element in F^n and it should be obvious where to send this in F^m to get a homomorphism (usually called a linear map in this case) with the desired properties. Think about the case F=R, the real numbers.
  4. Jan 25, 2007 #3
    Without working out the details explicitly, I would imagine the multivariate projection map from [tex] F^n \rightarrow F^m [/tex] should satisfy the homomorphism conditions.

    Furthermore, have you considered using the first isomorphism theorem for rings to help formally show that [tex] F^{n-m} [/tex] is isomorphic to the kernel?

    It shouldn't be too hard to show that the projection map [tex] \pi (x) [/tex] is surjective, and so that [tex]Image(\pi(x))=F^m[/tex] is a subring of [tex]F^n[/tex]. Then [tex]F^m[/tex] is isomorphic to [tex]F^n \setminus_{Ker(\pi)}[/tex]. Consider the natural map between these two sets and then see if you can get anything from there.

    This might be a bit over the top, but was the first thing that came to my mind.
  5. Jan 27, 2007 #4
    Thanks for the input. It's been a year since I studied rings and I'm all rusty. I'm pretty sure I understand this one and I think I got it.

    My next question is this:Prove that there exists an isomorphism from [tex]F^n[/tex] into [tex]Hom(Hom(F^n , F),F).[/tex]
    Again, I'm all rusty on this stuff, so any input will be helpful.
    My confusion here lies in that I just finished proving (by contradiction) that [tex]F^1[/tex] is not isomorphic to [tex]F^n[/tex] for n>1.
    So I'm not sure what [tex]F[/tex] is or what [tex]Hom(F^n,F)[/tex] looks like. It seems to me that [tex]F^n[/tex] is finite and [tex]F[/tex] is infinite and then my mind just starts going in circles. Please help me understand this.
    Last edited: Jan 27, 2007
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