# Property of a ring homomorphism

1. Nov 9, 2008

### Geekster

1. The problem statement, all variables and given/known data

Suppose that f:R->Q (reals to rationals) is a ring homomorphism. Prove that f(x)=0 for every x in the reals.

2. Relevant equations

Homomorphisms map the zero element to the zero element.
f(0) = 0
f(-a)=-f(a)

and finally,
f(a - b) = f(a) - f(b)

3. The attempt at a solution

My guess is go with contradiction and say, Suppose that f(a != 0) != 0 for some a in the reals.

But I don't see where to go from there. A hint or suggestion would be nice.

2. Nov 9, 2008

### Geekster

I think I have an answer, let me know if you think my proof is ok.

Suppose there exists some a in the reals such that f(a)!=0. Then f(1*a)=f(1)*f(a)=f(a).
So f(1) must be the identity in Q. But that allows us to get numbers like 2 since f(1+1)=f(1)+f(1)=2. But then we have 2=f(2)=f(sqrt(2)*sqrt(2))=f(sqrt(2))*f(sqrt(2)),
and we know that there is no rational who's square is 2, hence our assumption that there exist some a in the reals such that f(a)!=0 is false.

Therefore, f(x)=0 for all x in the reals.

Let me know if there is anything wrong with the proof.

Thanks