Proving Ring Homomorphism of \phi: Zp \rightarrow Zp

  • Thread starter Thread starter phyguy321
  • Start date Start date
  • Tags Tags
    Ring
phyguy321
Messages
45
Reaction score
0

Homework Statement


Prove that \phi : Zp \rightarrow Zp,
\phi (a) = a p is a ring homomorphism, find the ker \phi


Homework Equations





The Attempt at a Solution


So show that a _{p} + b _{p} = (a + b)p?
and (ab)p = (ap)(bp)?
 
Last edited:
Physics news on Phys.org
Yes, show those equalities mod p. The second is easy. For the first think about the binomial theorem. Is p supposed to be a prime?
 
p is prime.
so show a mod p + b mod p = (a+b) mod p
and (ab) mod p = (a mod p)*(b mod p)
how do i do that?
 
phyguy321 said:
p is prime.
so show a mod p + b mod p = (a+b) mod p
and (ab) mod p = (a mod p)*(b mod p)
how do i do that?

Those are always true whether p is prime or not. You must have already proved them. Your problem is to prove (a+b)^b mod p=(a+b) mod p when p is prime. I told you how to do that. Use the binomial theorem on (a+b)^p.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top