Gcd(m, n)=1 implies Zmn isomph to Zm x Zn

1. Nov 11, 2011

ArcanaNoir

1. The problem statement, all variables and given/known data
show that if m, n are relatively prime, that is, greatest common divisor of m and n is 1, then $\mathbb{Z} _mn \approx \mathbb{Z} _m \times \mathbb{Z} _n$

2. Relevant equations

I need to show that $\theta$ is operation preserving, and I need to show that it is one to one and onto.

3. The attempt at a solution

For theta, $\theta ([a]_{mn} + _{mn}) = \theta ([a+b]_{mn})=([a+b]_m,[a+b]_n)=$
$([a]_m+_m,[a]_n+_n)=([a]_m,[a]_n)+([b)_m,_n)= \theta ([a]_{mn}) + \theta (_{mn})$
Did I assume anything I shouldn't have there?
I'm going to consult my notes about proving 1-1. going to try the kernel thing.
As for onto, how do I show that?
I'm concerned that I haven't used the fact that m, n are relatively prime.

Last edited: Nov 11, 2011
2. Nov 11, 2011

I like Serena

Hi Arcana!

You assumed your definition of theta, which I have to deduce from your work now.
Beyond that your proof of the preservation of the operation is correct.

So what's left is to proof that theta is 1-1, for which you need that m and n are relatively prime.

3. Nov 11, 2011

ArcanaNoir

The definition of theta was given in a previous problem. I'm just tired of typing latex. Guess I need more practice so I'm not so slow at it.
going to consult notes about kernel thingy now for 1-1, but what about onto?

4. Nov 11, 2011

I like Serena

How many elements does each set have?
Are they the same?

I'm not sure what your kernel thingy is, but if you can say each element is mapped onto a different element, you can deduce "onto" by the fact that each set has the same number of elements.

Last edited: Nov 11, 2011
5. Nov 11, 2011

ArcanaNoir

difficulty:
theta is 1-1 iff ker(theta)= {e_Z_mn}
$$\theta ([a]_{mn})=([0]_m,[0]_n) \Rightarrow$$
$$m \mid a \wedge n \mid a \Rightarrow$$
$$a=km \wedge a=jn$$

I know that being relatively prime is the reason a must =mn but I'm not sure why, so I don't know what to say next.

6. Nov 11, 2011

I like Serena

You have n|a.

So from a=km, you know that n|(km).
Since n does not have any common factors with m, it must be that n|k.

7. Nov 11, 2011

I like Serena

This means that k=i n.
And in turn that:
a=km=(in)m=i(mn)

8. Nov 11, 2011

ArcanaNoir

What is i? 1? And why does k=in? Sorry I'm being dense here.

9. Nov 11, 2011

I like Serena

Since n|k, there must be a number i such that k=in.

10. Nov 11, 2011

I like Serena

Not dense, I know how hard it is to follow other people's reasoning in algebra.
The trick is to set up your own reasoning.