1. ## equation

Show that $(m,n) = 1 \Rightarrow (m) \cap (n) = (mn)$. so $mx+ny = 1$. Now $(m) = \{cm: c \in R \}$ and $(n) = \{dm: d \in R \}$. So show that there is some common element which implies that they must be equal?

2. Originally Posted by Sampras
Show that $(m,n) = 1 \Rightarrow (m) \cap (n) = (mn)$. so $mx+ny = 1$. Now $(m) = \{cm: c \in R \}$ and $(n) = \{dm: d \in R \}$. So show that there is some common element which implies that they must be equal?

I think you meant $(n) = \{dn: d \in Z \}$

Let me prove the following for you

if $m|x$ and $n|x$ then $mn|[\gcd(m,n)x]$

we have $k_1m+k_2n = \gcd(m,n)$
= $xk_1m+xk_2n = x\gcd(m,n)$

Now mn|LHS of the equation above. Hence the result.

In this case $gcd(m,n)=1$, so $mn|x$ Hence the result

3. Originally Posted by aman_cc
I think you meant $(n) = \{dn: d \in Z \}$

Let me prove the following for you

if $m|x$ and $n|x$ then $mn|[\gcd(m,n)x]$

we have $k_1m+k_2n = \gcd(m,n)$
= $xk_1m+xk_2n = x\gcd(m,n)$

Now mn|LHS of the equation above. Hence the result.

In this case $gcd(m,n)=1$, so $mn|x$ Hence the result
Can you use fact that either $(m) \cap (n)$ is disjoint or equal? Thus show that if there is some element in $(m) \cap (n)$ then $(m) = (n)$? Thus $(m) \cap (n) = (mn)$?

4. Sorry - but unable to follow you question completely.

5. Originally Posted by aman_cc
I think you meant $(n) = \{dn: d \in Z \}$

Let me prove the following for you

if $m|x$ and $n|x$ then $mn|[\gcd(m,n)x]$

we have $k_1m+k_2n = \gcd(m,n)$
= $xk_1m+xk_2n = x\gcd(m,n)$

Now mn|LHS of the equation above. Hence the result.

In this case $gcd(m,n)=1$, so $mn|x$ Hence the result
How does this show the result?

6. Originally Posted by Sampras
How does this show the result?
Hi - Maybe I am not getting your question, but here is the logic I have used.

We have proved if m|x and n|x then => mn|x

So let
1. x in (m) and x in (n) => x in (mn)
2. x in (mn) => x in (m) and x in (n)

Hence
$
(m) \cap (n) = (mn)$

Essentially all I am saying is $(m) \cap (n) = (LCM(mn))

$