equation

**Sampras** · October 8th 2009, 01:29 AM

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?

**aman_cc** · October 8th 2009, 01:46 AM

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

**Sampras** · October 8th 2009, 01:58 AM

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

**aman_cc** · October 8th 2009, 02:20 AM

Sorry - but unable to follow you question completely.

**Sampras** · October 14th 2009, 05:01 AM

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?

**aman_cc** · October 14th 2009, 05:13 AM

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
$<br /> (m) \cap (n) = (mn)$

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

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