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Math Help - Homomorphism

  1. #1
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    Homomorphism

    Hi, new to the board.

    I need help proving that f: Z5 to Z7 has no ring homomorphisms.
    Thanks in anticipation.
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  2. #2
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    Quote Originally Posted by PauKelome View Post
    Hi, new to the board.

    I need help proving that f: Z5 to Z7 has no ring homomorphisms.
    Thanks in anticipation.
    That is never true. Everything always has a "trivial-homomorphism".

    Consider,
    f:\mathbb{Z}_5\to \mathbb{Z}_7
    f(x)=0
    Is indeed a homomorphism.

    Perhaps you meant an isomorphism.
    Last edited by ThePerfectHacker; October 28th 2006 at 03:27 PM.
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  3. #3
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    Yes that is what I mean.
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  4. #4
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    Quote Originally Posted by PauKelome View Post
    Yes that is what I mean.
    Then the answer is trivial.
    Since |\mathbb{Z}_5|<|\mathbb{Z}_7|
    There is no way to create an isomorphism.
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  5. #5
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    OK. I don't think I got my question answered correctly. By my definition, a ring homomorphism exists when R and S are rings, then the function f: R to S adheres to the following properties:
    f(a+b)=f(a)+f(b) for all a and b in R
    f(ab)=f(a)f(b) for all a and b in R
    f(1)=1.

    So I need to prove that f: Z5 to Z7 and f: Z6 to Z7 do not have ring homomorphisms. Which conditions fail?

    Thanks for any further help on this issue. Pau
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  6. #6
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    Quote Originally Posted by PauKelome View Post
    OK. I don't think I got my question answered correctly. By my definition, a ring homomorphism exists when R and S are rings, then the function f: R to S adheres to the following properties:
    f(a+b)=f(a)+f(b) for all a and b in R
    f(ab)=f(a)f(b) for all a and b in R
    f(1)=1.

    So I need to prove that f: Z5 to Z7 and f: Z6 to Z7 do not have ring homomorphisms. Which conditions fail?

    Thanks for any further help on this issue. Pau
    First, a ring homomorphism is,
    \phi:R\to R'
    Such that,
    \phi(a+b)=\phi(a)+'\phi(b)
    \phi(ab)=\phi(a)\cdot ' \phi(b)
    In no way, is it defined for,
    \phi(1)=1'.
    Yes, it happens to be true if a ring homomorphism preserves unity and zero's for the two rings but that can easily be proved from the first two statements, thus it is not necessarily.
    ---
    Now, returning to the question. Again, there does exist a ring homomorphism. The trivial-homomorphism can be made to exist between any two rings or groups.
    Define,
    \phi:R\to R' as,
    \phi(x)=0 for all x\in R.

    Maybe, you wish to show there does not exist no other homomorphism.
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  7. #7
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    This problem usually asks to show that there are no non-trivial homomorphism between these two rings. One can show that \phi (nx) = n\phi (nx)\quad \& \quad \phi (x^n ) = \left[ {\phi (x)} \right]^n is true for any ring homomorphism. Use that to complete your problem.
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  8. #8
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    Quote Originally Posted by Plato View Post
    This problem usually asks to show that there are no non-trivial homomorphism between these two rings. One can show that \phi (nx) = n\phi (nx)\quad \& \quad \phi (x^n ) = \left[ {\phi (x)} \right]^n is true for any ring homomorphism. Use that to complete your problem.
    Assume,
    \phi:\mathbb{Z}_5\to\mathbb{Z}_7
    Is a homomorphism.
    Then by the fundamental homomorphism theorem,
    \ker (\phi) is an ideal in \mathbb{Z}_7.
    Since,
    \mathbb{Z}_7 is a field it has no proper-nontrivial ideals. Thus, \ker (\phi)=\{0\} \mbox{ or }\mathbb{Z}_7
    Thus,
    \phi[\mathbb{Z}_5]\simeq \mathbb{Z}_7
    Or,
    \phi[\mathbb{Z}_5]\simeq \{0\}
    But the first isomorphism cannot exist because,
    |\mathbb{Z}_5|<|\mathbb{Z}_7|
    Thus,
    \phi[\mathbb{Z}_5]\simeq \{0\}
    Which means, the function, \phi maps everything into a single element. Which must be \phi(x)=0
    Which is exactly the trivial homomorphism.
    (The full details are to the reader to finish).
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