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

  • Thread starter ArcanaNoir
  • Start date
  • #1
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Homework Statement


show that if m, n are relatively prime, that is, greatest common divisor of m and n is 1, then [itex] \mathbb{Z} _mn \approx \mathbb{Z} _m \times \mathbb{Z} _n [/itex]


Homework Equations



I need to show that [itex] \theta [/itex] is operation preserving, and I need to show that it is one to one and onto.

The Attempt at a Solution



For theta, [itex] \theta ([a]_{mn} + _{mn}) = \theta ([a+b]_{mn})=([a+b]_m,[a+b]_n)= [/itex]
[itex] ([a]_m+_m,[a]_n+_n)=([a]_m,[a]_n)+([b)_m,_n)= \theta ([a]_{mn}) + \theta (_{mn}) [/itex]
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:

Answers and Replies

  • #2
I like Serena
Homework Helper
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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
768
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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
I like Serena
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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.
 
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  • #5
768
4
difficulty:
theta is 1-1 iff ker(theta)= {e_Z_mn}
[tex] \theta ([a]_{mn})=([0]_m,[0]_n) \Rightarrow [/tex]
[tex] m \mid a \wedge n \mid a \Rightarrow [/tex]
[tex] a=km \wedge a=jn [/tex]

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
I like Serena
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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
I like Serena
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This means that k=i n.
And in turn that:
a=km=(in)m=i(mn)
 
  • #8
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What is i? 1? And why does k=in? Sorry I'm being dense here.
 
  • #9
I like Serena
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What is i? 1? And why does k=in? Sorry I'm being dense here.
Since n|k, there must be a number i such that k=in.
 
  • #10
I like Serena
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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.
I just hope my comments can help you in your reasoning.
 

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