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Math Help - What kind of proof to use? Ring Homomorphisms

  1. #1
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    What kind of proof to use? Ring Homomorphisms

    Hi here is the question
    prove that every ring homomorphism phi from Z_n to itself has the form phi(x)=ax where a^2=a

    im just wondering if direct proof is the best option and if it is solid enough so far i have something along the lines
    we know that every group homomorphism from Z_n to itself has the form phi(1)=k where 0<k<n-1 (should be or equal too) so phi(x) = phi(1+1...+1)=phi(1)+phi(1)...=kx
    and phi(x)=phi(1*x)=phi(1)*phi(x)=k*k*x=k^2*x which equals kx if k^2=k so in order for phi to be a ring homomorphism in addition to being a group homomorphism then k^2=k so let k=a and there it is..? will this proof work?
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by fizzle45 View Post
    Hi here is the question
    prove that every ring homomorphism phi from Z_n to itself has the form phi(x)=ax where a^2=a

    im just wondering if direct proof is the best option and if it is solid enough so far i have something along the lines
    we know that every group homomorphism from Z_n to itself has the form phi(1)=k where 0<k<n-1 (should be or equal too) so phi(x) = phi(1+1...+1)=phi(1)+phi(1)...=kx
    and phi(x)=phi(1*x)=phi(1)*phi(x)=k*k*x=k^2*x which equals kx if k^2=k so in order for phi to be a ring homomorphism in addition to being a group homomorphism then k^2=k so let k=a and there it is..? will this proof work?
    It looks right by choosing x\in\mathbb{Z}_n^{\times} for the last part.
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    Quote Originally Posted by fizzle45 View Post
    Hi here is the question
    prove that every ring homomorphism phi from Z_n to itself has the form phi(x)=ax where a^2=a

    im just wondering if direct proof is the best option and if it is solid enough so far i have something along the lines
    we know that every group homomorphism from Z_n to itself has the form phi(1)=k where 0<k<n-1 (should be or equal too) so phi(x) = phi(1+1...+1)=phi(1)+phi(1)...=kx
    and phi(x)=phi(1*x)=phi(1)*phi(x)=k*k*x=k^2*x which equals kx if k^2=k so in order for phi to be a ring homomorphism in addition to being a group homomorphism then k^2=k so let k=a and there it is..? will this proof work?

    I can't see why not: it looks just fine to me.

    Tonio
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