is $\operatorname{lcm}(a,\gcd(b,c))=\gcd(\operatorname{lcm}(a,b),\operatorname{lcm}(a,c))$? it should be true for positive integers because the commutative algebra by Atiyah says addition, intersection on the ideals of $\mathbb{Z}$ is distributive. how should i solve it?

  • $\begingroup$ Could you define $a,b,c$? $\endgroup$ – mrnovice Feb 26 '17 at 0:54
  • $\begingroup$ Try it for $(a,b,c)=(2,3,5)$ - it fails as currently written. $\endgroup$ – Joffan Feb 26 '17 at 0:56
  • $\begingroup$ sorry i edited my question $\endgroup$ – Mathcho Feb 26 '17 at 1:02
  • $\begingroup$ i am self studying, belive or not. i tried to compute both sides: left side: a*gcd(b,c)/gcd(a,b,c) and right side: gcd(ab/gcd(a,b),ac/gcd(a,c)) using lcm(a,b)=ab/gcd(a,b) then stuck. $\endgroup$ – Mathcho Feb 26 '17 at 1:16

One method that works in any gcd domain is to eliminate lcms using $\,{\rm lcm}(x,y) = xy/(x,y).\ $ Doing so and clearing denominators, then cancelling $a$ reduces it to

$$(a,b)(b,c)(c,a)\, =\, (a,b,c)(ab,bc,ca)$$

which is true since both sides $= (aab,aac,abb,abc,acc,bbc,bcc)$ by gcd laws (distributive etc).

  • $\begingroup$ See this answer for the dual (and further detail). $\endgroup$ – Bill Dubuque Feb 26 '17 at 2:22
  • $\begingroup$ okay now i see i had to proceed few steps further...... thanks for the great answer. $\endgroup$ – Mathcho Feb 26 '17 at 11:19

Well its not elegant but let $a=\prod p_i^{a_i} $, $b=\prod p_i^ {b_i} $ and $c=\prod p_i^{ci} $ where $p_i $ are prime and the powers can be zero.

The power of the $p_i $ factor of $lcm (a,\gcd (b,c))$ is $T=\max (a_i,\min (b_i,c_i)) $

The power for $\gcd (lcm (a,b),lcm (a,c))$ is $S=\min (\max (a_i,b_i),\max (a_i,c_i)) $.

If $a_i$ is the least of the three then $T=\min (b_i,c_i) $ and $S=\min (b_i,c_i) $ so equality holds.

If $a_i $ is the most of the three $S=T=a_i $ and equality holds.

If $a_i $ is strictly in between, then $a_i > \min (b_i,c_i) $ so $T=a_i $. Likewise one of $b_i$ or $c_i $ is greater than $a_i $ and the other smaller. So $a_i=$ one of the max pairs, but max of all three is the other max pair. So $S $ is the min of the pairs is $a_i $. So equality holds.

  • $\begingroup$ Good. But it can be shortened by noting that if $b\leq c$ then $\max (a,b)\leq \max (a,c)$ so $\min (\max (a,b),\max (a,c))=\max (a,b)$, and also $\max (a,\min (b,c))=\max (a,b).$ Then interchange $ b $ with $ c.$ $\endgroup$ – DanielWainfleet Feb 26 '17 at 7:32
  • $\begingroup$ this is fairly neat. thanks. $\endgroup$ – Mathcho Feb 26 '17 at 11:16
  • $\begingroup$ I thought of that, but I only wanted to do three cases; not four with wolog. I think my wording got obscure at the third point. $\endgroup$ – fleablood Feb 26 '17 at 16:23

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