OK. I don't think I got my question answered correctly. By my definition, a ring homomorphism exists when R and S are rings, then the function f: R to S adheres to the following properties:
f(a+b)=f(a)+f(b) for all a and b in R
f(ab)=f(a)f(b) for all a and b in R
f(1)=1.
So I need to prove that f: Z5 to Z7 and f: Z6 to Z7 do not have ring homomorphisms. Which conditions fail?
Thanks for any further help on this issue. Pau
First, a ring homomorphism is,
Such that,
In no way, is it defined for,
.
Yes, it happens to be true if a ring homomorphism preserves unity and zero's for the two rings but that can easily be proved from the first two statements, thus it is not necessarily.
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Now, returning to the question. Again, there does exist a ring homomorphism. The trivial-homomorphism can be made to exist between any two rings or groups.
Define,
as,
for all .
Maybe, you wish to show there does not exist no other homomorphism.
Assume,
Is a homomorphism.
Then by the fundamental homomorphism theorem,
is an ideal in .
Since,
is a field it has no proper-nontrivial ideals. Thus,
Thus,
Or,
But the first isomorphism cannot exist because,
Thus,
Which means, the function, maps everything into a single element. Which must be
Which is exactly the trivial homomorphism.
(The full details are to the reader to finish).