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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- Oct 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. - Oct 26th 2006, 04:26 PMThePerfectHacker
- Oct 26th 2006, 06:08 PMPauKelome
Yes that is what I mean.

- Oct 26th 2006, 06:12 PMThePerfectHacker
- Oct 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 - Oct 28th 2006, 02:55 PMThePerfectHacker
First, a ring homomorphism is,

$\displaystyle \phi:R\to R'$

Such that,

$\displaystyle \phi(a+b)=\phi(a)+'\phi(b)$

$\displaystyle \phi(ab)=\phi(a)\cdot ' \phi(b)$

In no way, is it defined for,

$\displaystyle \phi(1)=1'$.

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.

Define,

$\displaystyle \phi:R\to R'$ as,

$\displaystyle \phi(x)=0$ for all $\displaystyle x\in R$.

Maybe, you wish to show there does not exist no other homomorphism. - Oct 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 $\displaystyle \phi (nx) = n\phi (nx)\quad \& \quad \phi (x^n ) = \left[ {\phi (x)} \right]^n$ is true for any ring homomorphism. Use that to complete your problem.

- Oct 28th 2006, 03:25 PMThePerfectHacker
Assume,

$\displaystyle \phi:\mathbb{Z}_5\to\mathbb{Z}_7$

Is a homomorphism.

Then by the fundamental homomorphism theorem,

$\displaystyle \ker (\phi)$ is an ideal in $\displaystyle \mathbb{Z}_7$.

Since,

$\displaystyle \mathbb{Z}_7$ is a field it has no proper-nontrivial ideals. Thus, $\displaystyle \ker (\phi)=\{0\} \mbox{ or }\mathbb{Z}_7$

Thus,

$\displaystyle \phi[\mathbb{Z}_5]\simeq \mathbb{Z}_7$

Or,

$\displaystyle \phi[\mathbb{Z}_5]\simeq \{0\}$

But the first isomorphism cannot exist because,

$\displaystyle |\mathbb{Z}_5|<|\mathbb{Z}_7|$

Thus,

$\displaystyle \phi[\mathbb{Z}_5]\simeq \{0\}$

Which means, the function, $\displaystyle \phi$ maps everything into a single element. Which must be $\displaystyle \phi(x)=0$

Which is exactly the trivial homomorphism.

(The full details are to the reader to finish).