# Ring homomorphism

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## Answers and Replies

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

(the steps in post 6 are correct)

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.