Rational numbers with denoms not divisible by a prime p mod I is isomorphic to Z_p

1. Dec 3, 2011

1. The problem statement, all variables and given/known data
Show that the ring of rational numbers whose reduced form denominator is not divisble by a prime, p, mod an ideal the set of elements of the above set whose numerators are divisible by p is isomorphic to Z_p

2. Relevant equations

3. The attempt at a solution
It seems very trivial: Use 1st homomorphism theorem with phi(a/b) = a(modp), but I am having a hard time showing that such a mapping is actually a homomorphism additively. I.E., phi(a/b + c/d) = phi(ad+bc/bd) = ad+bc mod(p) =/= a modp + b modp = phi(a/b) + phi (c/d).

I am stuck here and any help would be appreciated.

Thanks,

2. Dec 3, 2011

micromass

Re: Rational numbers with denoms not divisible by a prime p mod I is isomorphic to Z_

Instead of constructing an isomorphism directly, maybe you can show that it's a field with p elements. This would imply it as well.

3. Dec 4, 2011

Re: Rational numbers with denoms not divisible by a prime p mod I is isomorphic to Z_

Worked well, I showed the following:
I is maximal in R, since if we have N an ideal with N=/= I then, there is a/b in N with p|\a -> a=/= 0, for if a=0, p|a. Then, there exists a^-1 in Q s.t. a*a^-1 = a^-1 * a = 1. b=/= 0 -> there exists b^-1 with the same property. Since N is an ideal, (a^-1/b^-1)*(a/b) is in N -> 1 is in N -> N=R -> I is maximal -> R/I is a field (R is commutative with 1).

Then, the number of distinct additive cosets r+I is precisely p QED

Thanks