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Let $m,n$ be any two positive integers. Note $\widehat X$ the set of positive divisors of $x$.

$\widehat X = \{ d : d \mid x\}$

(do not confuse it with $\hat a = \{x : x \equiv a \mod m\}$)

Assume $(m,n)=1$. How could one prove that

$\widehat {MN}=\{d:d\mid mn\}$

is the product $\widehat M \cdot \widehat N$ in the sense that all divisors of $mn$ appear in the product

$\left(\sum d_i+\sum d_i d_j+\cdots+ d_1 d_2 \cdots d_r \right) \left(\sum e_i+\sum e_i e_j+\cdots+ e_1 e_2 \cdots e_r \right)$

(clearly $e_1 e_2 \cdots e_r=m$ and $ d_1 d_2 \cdots d_r=n$) where $d_i$ are the divors of $n$ and $e_i$ those of $m$?

This is essential in the proof that if $f$ is multiplicative, then $F(n)=\sum_{d \mid n}f(d)$ also is.

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    I think I understand the question, but the notation is confusing, what is $\sum d_i+ \sum d_i d_j +\cdots+d_1 d_2\cdots d_r$? e.g. if $n=12$ then are the $d_i=1,2,3,4,6,12$? Then $d_1 d_2\cdots d_r=1728$. Wouldn't it suffice to say all divisors of $mn$ appear in the product $(\sum d_i)(\sum e_i)$?2012-06-20

2 Answers 2

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Let's see if I correctly understood what you asked:

$\,1\,.-\,\,$ Let $\,d\mid MN\Longrightarrow\,$ every prime divisor of $\,d\,$ divides either $\,M\,$ or $\,N\,$, but not both as $\,(M,N)=1\,$ , so putting$d_M:=\{\,p\mid d\;\;;\;\;p\mid M\,\,,p\,\,\,\text{a prime}\}\,\,,\,d_N:=\{\,p\mid d\;\;;\;\;p\mid N\,\,,p\,\,\,\text{a prime}\}$we get that $d=a_1\cdot\ldots\cdot a_k\cdot b_1\cdot\ldots\cdot b_k\,\,,\,a_i\in d_M\,\,,\,b_i\in d_N\,$ so $\,d\,$ is of the required form

$\,2\,.-\,$ On the other hand, if $\,\,d=a_1\cdot\ldots\cdot a_k\cdot b_1\cdot\ldots\cdot b_k\,\,,\,a_i\in d_M\,\,,\,b_i\in d_N\,$ , then clearly $\,d\mid MN\,$

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Let $d\in \widehat{MN}$ and $(d,m)=g\in \widehat{M}$. Then $(\frac{d}{g},m)=1\Rightarrow \frac{d}{g}\mid N$ by Euclid's Lemma. So $\frac{d}{g}\in \widehat{N}$ and $d=g\frac{d}{g}\in \widehat{M}\cdot\widehat{N}$.