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
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
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.
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.
for all .
Maybe, you wish to show there does not exist no other homomorphism.
Is a homomorphism.
Then by the fundamental homomorphism theorem,
is an ideal in .
is a field it has no proper-nontrivial ideals. Thus,
But the first isomorphism cannot exist because,
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).