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Math Help - Ring Homomorphism

  1. #1
    Senior Member vincisonfire's Avatar
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    Ring Homomorphism

    Can you help me. Here is my work.
    (a) Prove that there is no ring homomorphism Z/5Z → Z.
    (b) Prove that there is no ring homomorphism Z/5Z→ Z/7Z.
    (a) Axiom of ring homomorphism are
    f(1) = 1
    f(a+b) = f(a) + f(b)
    f(ab) = f(a)*f(b)
    Consequences:
    f(0) = 0
    f(-r) = -f(r)
    Homomorphism must therefore be
    f(0) = 0
    f(1) = 1
    f(4 mod 5) = f(-1 mod 5) = -f(1) = -1
    f(4) = f(2) + f(2) = 2*f(2) = -1
    f(2) =  -1 \cdot 2^{-1}
    f(2) can't exist because there is no inverse with respect to multiplication in Z.
    (b)Homomorphism must therefore be
    f(0) = 0
    f(1) = 1
    f(4 mod 5) = f(-1 mod 5) = -f(1) = -1 mod 7 = 6 mod 7
    f(4) = f(2) + f(2) = 2*f(2) = 6
    f(2) =  6 \cdot 2^{-1} = 6 * 4 = 24 = 3 mod 7
    f(2+3) = f(5) = f(0) = 0 = f(2) + f(3)
    f(3) = -f(2) = -3 = 4 mod 7
    Recapitulate
    f(0) = 0
    f(1) = 1
    f(2) = 3
    f(3) = 4
    f(4) = 6
    f(1+3) = f(4) = 6 not equal to f(1) + f(3) = 1+4 = 5
    Contradiction and there is no homomorphism from Z/5Z→ Z/7Z
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  2. #2
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    Yes.
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  3. #3
    Senior Member vincisonfire's Avatar
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    Eh don't make fun of me we're not all best number theorist
    Just joking thanks.
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  4. #4
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    I don't really understand one thing. In b) we are assuming that f: Z/5Z -> Z/7Z is a homomorphism, so f(1) = 1 but also f(2) = f(1+1) = f(1)+f(1) = 1+1 = 2? And by the same argument f(a mod 5) = a mod7? I feel a bit bad about this because we have to show that NO function is a ring homom. and don't think it is okay to "assume" what it the specific rule of the mapping. I am a bit confused with all that as you can see...

    But could we say that f(2)=2 and f(2)= f(-3) = -f(3) = -3 = 4, but this is a contradiction because 2 is not equal to 4 in Z/7Z?
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    Quote Originally Posted by luzerne View Post
    and don't think it is okay to "assume" what it the specific rule of the mapping. I am a bit confused with all that as you can see...
    No it is okay to assume. You assume that if f is a ring homomorphism then we arrive at a contradiction. Which means that f cannot be a ring homorphism. And that shows there are no ring homomorphisms.
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