Prove that phi:Z_p mapped to Z_p, phi(a)=a^p is a ring homomorphism and find ker phi.

Printable View

- April 17th 2007, 12:04 PMchadlyterMore ring homoorphism
Prove that phi:Z_p mapped to Z_p, phi(a)=a^p is a ring homomorphism and find ker phi.

- April 17th 2007, 12:42 PMThePerfectHacker
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)

---

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}