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Math Help - Homomorphisms

  1. #1
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    Homomorphisms

    R and S are rings.
    Show that if f:R\to S is a homomorphism, and if a is a unit of S. Show, in fact, that f(a^{-1})=f(a)^{-1} for any unit of a of R.

    f(a^{-1})-f(a)^{-1}=f((a^{-1})+(-a)^{-1})=f(0)=0

    Is this how I show this?
    Last edited by dwsmith; July 12th 2011 at 12:01 PM. Reason: Forget a (
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  2. #2
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    Re: Homomorphisms

    Quote Originally Posted by dwsmith View Post
    R and S are rings.
    Show that if f:R\to S is a homomorphism, and if a is a unit of S. Show, in fact, that f(a^{-1})=f(a)^{-1} for any unit of a of R.

    f(a^{-1})-f(a)^{-1}=f(a^{-1})+(-a)^{-1})=f(0)=0

    Is this how I show this?
    No. When you use the expression f(a)^{-1} you have assumed it exists, but you have to first show that it exists. Also, in the first equality it looks like (trying to read around your type) you have assumed that -f(a)^{-1}=f((-a)^{-1}), which is neither given.

    Instead, just notice that f(a^{-1})f(a)=f(a^{-1}a)=f(1)=1. The conclusion follows immediately.
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    Re: Homomorphisms

    Quote Originally Posted by hatsoff View Post
    you have assumed that -f(a)^{-1}=f((-a)^{-1}), which is neither given.
    My book states the following:

    f is a homomorphism if f satisfies the following properties:
    (1)  f(r+r')=f(r)+f(r')
    (2) f(r\cdot r')=f(r)f(r')
    (3) f(1)=1

    If f satifies 1-3, then
    (4) f(0)=0
    (5) f(-r)=-f(r)

    So I didn't assume anything. I used what was given. It is stated that f is a homomorphism so it satisfies 1-3 which means 4 and 5 are also satisfied.
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    MHF Contributor Drexel28's Avatar
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    Re: Homomorphisms

    Quote Originally Posted by dwsmith View Post
    My book states the following:

    f is a homomorphism if f satisfies the following properties:
    (1)  f(r+r')=f(r)+f(r')
    (2) f(r\cdot r')=f(r)f(r')
    (3) f(1)=1

    If f satifies 1-3, then
    (4) f(0)=0
    (5) f(-r)=-f(r)

    So I didn't assume anything. I used what was given. It is stated that f is a homomorphism so it satisfies 1-3 which means 4 and 5 are also satisfied.
    Well the first mistake you made was assuming that f(a)\in S^\times.
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    Re: Homomorphisms

    Quote Originally Posted by Drexel28 View Post
    Well the first mistake you made was assuming that f(a)\in S^\times.
    I don't know what you mean by that so it doesn't make any sense to me.
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    MHF Contributor Drexel28's Avatar
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    Re: Homomorphisms

    Quote Originally Posted by dwsmith View Post
    I don't know what you mean by that so it doesn't make any sense to me.
    S^\times is the set of units of S--I was saying that in your proof you are assuming f(a) is a unit, which is part of what you are trying to prove.
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    Re: Homomorphisms

    since a is a unit, a^{-1} exists. since f is a homomorphism:

    1_S = f(1_R) = f(aa^{-1}) = f(a)f(a^{-1}), so....
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    MHF Contributor Drexel28's Avatar
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    Re: Homomorphisms

    Quote Originally Posted by Deveno View Post
    since a is a unit, a^{-1} exists. since f is a homomorphism:

    1_S = f(1_R) = f(aa^{-1}) = f(a)f(a^{-1}), so....
    A previous user had said the exact same thing.
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  9. #9
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    Re: Homomorphisms

    no, but it was isomorphic.
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