# Ring homomorphism

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quasar987
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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) $\neq$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)$\neq$0.

ab = 0 where a and b are zero divisors

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jbunniii
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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) $\neq$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)$\neq$0.
That shows that either $$f(1) = 1$$ or $$1 - f(1)$$ is a zero divisor, which is not what the problem asks for.

Here is a slight modification that works:

$$f(1_R) = f(1_R) f(1_R)$$

so

$$f(1_R) (1_S - f(1_R)) = 0$$

There are three possibilities:

$$1_S - f(1_R) = 0$$ and therefore $$f(1_R) = 1_S$$

or

$$f(1_R) = 0$$

or

$$f(1_R)$$ is a zero divisor and so, incidentally, is $$1_S - f(1_R)$$

The second possibility is ruled out because if it were true, then

$$f(x) = f(1_R) f(x) = 0$$

for every x, i.e., f is the trivial homomorphism.

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

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jbunniii
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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
No, I didn't assume that. Let me fill in the steps:

$$1_R = 1_R \cdot 1_R$$

So

$$f(1_R) = f(1_R \cdot 1_R)$$

But f is a homomorphism, so

$$f(1_R \cdot 1_R) = f(1_R) f(1_R)$$

Therefore we have

$$f(1_R) = f(1_R) f(1_R)$$

Now subtract $$f(1_R)f(1_R)$$ from both sides:

$$f(1_R) - f(1_R)f(1_R) = 0$$

Now, all of the quantities in this equation are members of the ring S. Thus I can multiply any of them by $$1_S$$ without changing anything. I choose to multiply the first $$f(1_R)$$ by $$1_S$$:

$$f(1_R) 1_S - f(1_R)f(1_R) = 0$$

Now factor $$f(1_R)$$ from both terms using using the distributive property:

$$f(1_R) (1_S - f(1_R)) = 0$$

Now you can proceed as I did above.

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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

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quasar987
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yes but this proof does not use the given property that it is a nontrivial homomorphism
Yes, we use it when we rule out the possibility f(1R)=0 (in post #4).

Sorry for misguiding you, I though I saw how to get that f(1) is a zero divisor from the fact that 1-f(1) is but I was mistaken.

jbunniii
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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
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:"

ok, thanks !.
so it is ok to start this proof by saying
consider 1R=1R.1R then,

(the steps in post 6 are correct)

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jbunniii