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Group Homomorphisms

  1. Oct 28, 2007 #1
    [SOLVED] Group Homomorphisms

    Thanks in advance for any help on this problem
    I can't even pretend that I know how to go about this question. I'm quite lost. Though thus far studying modern algebra hasn't been too difficult (knock on wood) and I've been understanding I'm struggling with this weeks problem set. Anyway here's the question:

    1. The problem statement, all variables and given/known data

    Let G be a group of all polynomials with real coefficients under addition. For each [tex]f[/tex] in G let [tex]\int f[/tex] denote the antiderivative of [tex]f[/tex] that passes through the point (0,0) Show that the mapping [tex]f \rightarrow \int f[/tex] from G to G is a homomorphism. What is the kernel of this mapping? Is this mapping a homomorphism if [tex]\int f[/tex] denotes the antiderivative that passes through (0,1)

    2. Relevant equations

    To show that a mapping is homomorphic I must show :
    [tex]\Phi (ab) = \phi(a)\phi(b)[/tex]

    The kernel is all the elements in G that map to the identity in G

    3. The attempt at a solution

    What I of think is that if [tex]g[x], h[x] \in G[/tex] then [tex]\phi(h[x]+g[x] = \int(h[x]+g[x]) = \int h[x] + \int g[x] = \phi(h[x]) + \phi(g[x])[/tex] So to me it seems that would suffice to show that it is homomorphic. However, I have a nagging feeling that by not using the fact that in integral passes through the point (0,0) I have made an error.

    As far as for the part about the kernel, I'm not sure at all.

    Thanks again for any help. It's much appreciated.

    Last edited: Oct 28, 2007
  2. jcsd
  3. Oct 28, 2007 #2


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    Actually, you did "use the fact that the integral passes through (0,0)" (perhaps by mistake!). The "anti-derivative" of a polynomial is a polynomial of degree one higher with the constant of integration as constant term. That will "go through (0,0)" if and only if you take the constant of integration to be 0- which you, in effect, did by "ignoring" the constant of integration!
  4. Oct 28, 2007 #3
    It always seems so obvious when someone points it out. Thanks so much HallsofIvy. I should be able to finish the question from there.
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