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Math Help - Radical is an Ideal

  1. #1
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    Radical is an Ideal

    Let S be the set of all scalar homomorphisms of an algebra X. Then the radical rad{X} of X is defined to be rad{X}=intersection_{f in {S}} {Ker{f}}.

    Can anyone help me prove that a radical of an algbera is an ideal?

    Thanks
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  2. #2
    Super Member girdav's Avatar
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    Re: Radical is an Ideal

    Take r in the radical , x an element of the algebra and f a scalar homomorphism. Show that f(rx)=0=f(xr).
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  3. #3
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    Re: Radical is an Ideal

    Quote Originally Posted by girdav View Post
    Take r in the radical , x an element of the algebra and f a scalar homomorphism. Show that f(rx)=0=f(xr).
    Well f(xr)=f(x)f(r)=0.f(r)=0. A similar result holds for f(rx). But this surely can't be a proof.
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  4. #4
    Super Member girdav's Avatar
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    Re: Radical is an Ideal

    What's wrong with this?
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  5. #5
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    Re: Radical is an Ideal

    Do we not need to prove that it is a subspace of X too?

    I'm really confused and having be drowning in functional analysis problems all day. Are you able to offer a full proof?
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  6. #6
    Super Member girdav's Avatar
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    Re: Radical is an Ideal

    Yes, we also need to check that the radical is stable by sums: the sum of two elements of the ideal still is in the ideal. After that, I think we get all the condition that an ideal needs to satisfy (ideal in sense of ring theory, I mean).
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  7. #7
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    Re: Radical is an Ideal

    So if x, y in Ker(f) and lambda, mu in C, then

    f(lambda(x) + mu(y)) = f(lambda(x)) + f(mu(y))= lambda f(x) + mu f(y) = lambda.0 + mu.0 = 0, so that lambda(x)+mu(y) in Ker(f)?
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