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I ran across this while reading through a matrix theory proof. I'm not sure how to show this. Any help would be much appreciated.

My attempt:

gcd($k$,$m$) = $r$ $\implies$ $kx + my = r$, for integers $x$ and $y$ (by B$\acute{e}$zout's identity). Hence $\dfrac{k}{r}x + \dfrac{m}{r}y =1$, and so it follows that gcd($\tfrac{k}{r}$,$\tfrac{m}{r}$) = 1; i.e. $k/r$ and $m/r$ are coprime.

Now, we also have $k \mid \ell m \implies \dfrac{k}{r} \mid \dfrac{\ell m}{r}$. I feel like the solution to this is right in front of me, but I don't see how I can make the jump that $k/r \mid \ell$.

Dan P.
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2 Answers2

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To finish: $\,(k/r,m/r)=1\,$ and $\,k/r\mid \ell(m/r)\,\Rightarrow\, k/r\mid \ell\, \,$ by Euclid's Lemma. More directly:

$$ k\mid m\ell,k\ell \,\Rightarrow\,k\mid (m\ell,k\ell) = (m,k)\ell\,\Rightarrow\, k/(m,k)\mid \ell$$

Bill Dubuque
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For any prime $p$ let $a,b,c,d,e$ be the non-negative integers such that $p^a,p^b,p^c,p^d,p^e $ are the largest powers of $p$ that divide, respectively, $k,l,m,r,k/r. $

We have $a\leq b+c $ and $ d=\min(a,c).$ And $r|k$ so $e=a-d. $

It suffices to show that $e\le b.$

Now $e=a-d=a-\min (a,c).$

If $a\le c$ then $\min(a,c)=a,$ so $e=a-\min(a,c)=a-a=0\le b.$

If $a>c$ then $\min(a,c)=c,$ so $e=a-\min (a,c)=a-c \le (b+c)-c=b.$

DanielWainfleet
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