Hi, new to the board.
I need help proving that f: Z5 to Z7 has no ring homomorphisms.
Thanks in anticipation.
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,
$\displaystyle \phi:R\to R'$
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.
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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,
$\displaystyle \phi:R\to R'$ as,
$\displaystyle \phi(x)=0$ for all $\displaystyle x\in R$.
Maybe, you wish to show there does not exist no other homomorphism.
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.
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).