That shows that either [tex]f(1) = 1[/tex] or [tex]1 - f(1)[/tex] is a zero divisor, which is not what the problem asks for.You cannot "cancel" f(x) because in a ring, elements do not in general have multiplicative inverses.
But consider the following. Since f is nontrivial, there exists x in R such that f(x) [itex]\neq[/itex]0. Then f(x)=f(x)f(1) and substracting f(x), we get 0=f(x)(1-f(1)).
And now consider the two possible cases: 1-f(1)=0 and 1-f(1)[itex]\neq[/itex]0.
No, I didn't assume that. Let me fill in the steps:f(1R)= f(1R)f(1R)
f(1R)- 1s.f(1R) = 0
but when you write this are you not already assuming that f(1R) = 1s which is what we are trying to prove
Yes, we use it when we rule out the possibility f(1R)=0 (in post #4).yes but this proof does not use the given property that it is a nontrivial homomorphism
No, it DOES use the fact that it's a nontrivial homomorphism to eliminate the second of the three possibilities that I listed, and therefore either the first or the third must be true, which is exactly what the problem asks for! Go back and re-read the part that starts "There are three possibilities:"yes but this proof does not use the given property that it is a nontrivial homomorphism,
there is some x in R for f(x) /= 0s
i feel like this is how we should start
f(x)1s= f(x) = f (x1R)= f(x)f(1R)
f(x) - f(x)f(1R) =0
Yes, that's true by definition of the multiplicative identity: if you multiply something by it, you get back the same something. It never hurts to show as many details as possible in elementary proofs like this, because the goal is not only for every step to be correct and justified, but to show that you understand what the justification is. The more details you write, the more the grader will be convinced of this.ok, thanks !.
so it is ok to start this proof by saying
consider 1R=1R.1R then,
(the steps in post 6 are correct)