# Thread: What kind of proof to use? Ring Homomorphisms

1. ## 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?

2. Originally Posted by fizzle45
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 $\displaystyle x\in\mathbb{Z}_n^{\times}$ for the last part.

3. Originally Posted by fizzle45
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