Hi, new to the board.

I need help proving that f: Z5 to Z7 has no ring homomorphisms.

Thanks in anticipation.

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- October 26th 2006, 10:38 AMPauKelomeHomomorphism
Hi, new to the board.

I need help proving that f: Z5 to Z7 has no ring homomorphisms.

Thanks in anticipation. - October 26th 2006, 04:26 PMThePerfectHacker
- October 26th 2006, 06:08 PMPauKelome
Yes that is what I mean.

- October 26th 2006, 06:12 PMThePerfectHacker
- October 27th 2006, 05:50 PMPauKelome
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 - October 28th 2006, 02:55 PMThePerfectHacker
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. - October 28th 2006, 03:13 PMPlato
This problem usually asks to show that there are no non-trivial homomorphism between these two rings. One can show that is true for any ring homomorphism. Use that to complete your problem.

- October 28th 2006, 03:25 PMThePerfectHacker
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).