- #1
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hkhk
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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.
f(1R)= f(1R)f(1R)
so
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 but this proof does not use the given property that it is a nontrivial homomorphism
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)
f(x) - f(x)f(1R) =0
ok, thanks !.
so it is ok to start this proof by saying
consider 1R=1R.1R then,
(the steps in post 6 are correct)