Q&A: can a homomorphism mapping the integers to the integers be the zero map?
Question by monika: can a homomorphism mapping the integers to the integers be the zero map?
The original question asks to show that if a homomorphism from the integers to the integers has a non-trivial kernel, then it must be the zero map. But I don’t understand how a homomorphism can be the zero map–which maps every element to zero–AND map the integers to themselves.
Best answer:
Answer by (Don’t fear) The Reaper
Hi, a map f : Z —> Z defined by f(a) = 0 for any a € Z is indeed an homomorphism, since you have that
- f(a + b) = 0 = 0 + 0 = f(a) + f(b),
- f(a*b) = 0 = 0*0 = f(a)*f(b)
(the first one alone if you are considering Z as an additive group, both of them if you are considering it as a ring). The point is that the question says “to”, not “onto”: the only request is for the image to be a subset (subgroup or subring) of Z, and {0} is.
Now, going to the original question. If f has a non-trivial kernel you have that f(a) = 0 for some a € Z. Since f is a homomorphism this leads to
f(a) = af(1) = 0,
which implies f(1) = 0 since a is not and Z is an integral domain. But if f(1) = 0 then for any b € Z you have
f(b) = bf(1) = b*0 = 0,
which shows that if f has a non-trivial kernel then it is the zero map.
I hope this helped you. Bye!
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yes, sending everything to 0 is a homomorphism. All you are doing is making the kernel equal to the whole group