Finite Fields and ring homomorphisms HELP

[cheeee]

Homework Statement

Assuming the mapping Z --> F defined by n --> n * 1F = 1F + ... + 1F (n times) is a ring homomorphism, show that its kernel is of the form pZ, for some prime number p. Therefore infer that F contains a copy of the finite field Z/pZ.

Also prove now that F is a finite dim vector space over Z/pZ; if this dim. is denoted d, then show that F has exactly p^d elements.

I know that the kernel of a ring homomorphism is defined as ker(f) = {a in Z : f(a) = 0}
but I am still having trouble exactly where to go from this...it appears that the only element of Z s.t. f(a) = 0, is 0 which would map to 0 * 1F = 0. But how is this of the form pZ, for some prime p??

Any help or push in the right direction would be great...thanks.

 

[ystael]

If $$R$$ and $$S$$ are rings, the kernel of a ring homomorphism $$\phi: R \to S$$ is an ideal of $$R$$. What are the ideals of $$\mathbb{Z}$$?

Of those ideals, only some of them can be the kernel of a homomorphism $$\mu: \mathbb{Z} \to F$$ given by $$\mu(n) = n \cdot 1_F$$. The others are incompatible with one of your hypotheses.

 

[cheeee]

Okay I get that the all of the ideals of Z are of the form mZ for some integer m, but I am still not sure how n*1F implies that the kernel must be a prime ideal of Z?

 

[ystael]

The important thing here is that $$F$$ is a field (or at least an integral domain). If the kernel of $$\mu$$ is $$(m) = m\mathbb{Z}$$, and $$m = rs$$ is composite, think about the equation $$\mu(m) = \mu(r)\mu(s)$$.