# Homomorphism

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• October 26th 2006, 10:38 AM
PauKelome
Homomorphism
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 PM
ThePerfectHacker
Quote:

Originally Posted by PauKelome
Hi, new to the board.

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

That is never true. Everything always has a "trivial-homomorphism".

Consider,
$f:\mathbb{Z}_5\to \mathbb{Z}_7$
$f(x)=0$
Is indeed a homomorphism.

Perhaps you meant an isomorphism.
• October 26th 2006, 06:08 PM
PauKelome
Yes that is what I mean.
• October 26th 2006, 06:12 PM
ThePerfectHacker
Quote:

Originally Posted by PauKelome
Yes that is what I mean.

Then the answer is trivial.
Since $|\mathbb{Z}_5|<|\mathbb{Z}_7|$
There is no way to create an isomorphism.
• October 27th 2006, 05:50 PM
PauKelome
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 PM
ThePerfectHacker
Quote:

Originally Posted by PauKelome
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,
$\phi:R\to R'$
Such that,
$\phi(a+b)=\phi(a)+'\phi(b)$
$\phi(ab)=\phi(a)\cdot ' \phi(b)$
In no way, is it defined for,
$\phi(1)=1'$.
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,
$\phi:R\to R'$ as,
$\phi(x)=0$ for all $x\in R$.

Maybe, you wish to show there does not exist no other homomorphism.
• October 28th 2006, 03:13 PM
Plato
This problem usually asks to show that there are no non-trivial homomorphism between these two rings. One can show that $\phi (nx) = n\phi (nx)\quad \& \quad \phi (x^n ) = \left[ {\phi (x)} \right]^n$ is true for any ring homomorphism. Use that to complete your problem.
• October 28th 2006, 03:25 PM
ThePerfectHacker
Quote:

Originally Posted by Plato
This problem usually asks to show that there are no non-trivial homomorphism between these two rings. One can show that $\phi (nx) = n\phi (nx)\quad \& \quad \phi (x^n ) = \left[ {\phi (x)} \right]^n$ is true for any ring homomorphism. Use that to complete your problem.

Assume,
$\phi:\mathbb{Z}_5\to\mathbb{Z}_7$
Is a homomorphism.
Then by the fundamental homomorphism theorem,
$\ker (\phi)$ is an ideal in $\mathbb{Z}_7$.
Since,
$\mathbb{Z}_7$ is a field it has no proper-nontrivial ideals. Thus, $\ker (\phi)=\{0\} \mbox{ or }\mathbb{Z}_7$
Thus,
$\phi[\mathbb{Z}_5]\simeq \mathbb{Z}_7$
Or,
$\phi[\mathbb{Z}_5]\simeq \{0\}$
But the first isomorphism cannot exist because,
$|\mathbb{Z}_5|<|\mathbb{Z}_7|$
Thus,
$\phi[\mathbb{Z}_5]\simeq \{0\}$
Which means, the function, $\phi$ maps everything into a single element. Which must be $\phi(x)=0$
Which is exactly the trivial homomorphism.
(The full details are to the reader to finish).
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