Prove that phi:Z_p mapped to Z_p, phi(a)=a^p is a ring homomorphism and find ker phi.
phi(xy)=(xy)^p=(x^p)*(y^p) ---> It is a commutative ring.
phi(x)phi(y)=(x^p)*(y^p).
Thus, phi(xy)=phi(x)*phi(y)
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phi(x+y)=(x+y)^p=x^p+px^{p-1}y+...+pxy^{-1}+y^p
Note that all term in middle of binomial expansion are divisible by p. Hence, this simplifies to x^p+y^p
Thus,
phi(x+y)=phi(x)+phi(y)
Note,
ker(phi) = {0}