2
$\begingroup$

Let $a$, $b$, $c \in \mathbb{N}$. $[a, b]$ denotes $\mathrm{lcm}(a, b)$ and $(a,b)$ denotes $\gcd(a, b)$

Show that

  1. $(a,[b,c]) = [(a,b),(a,c)]$.
  2. $[a,(b,c)] = ([a,b],[a,c])$.
  • 1
    Certainly prime factorization works, but it would be nice to see a proof that avoids it.2011-09-14
  • 0
    These properties say that $\mathbb N$ with the divisor relation is a [distributive lattice](http://en.wikipedia.org/wiki/Distributive_lattice).2011-09-14
  • 0
    as @Srivatsan Narayanan said, it would be nice to see a proof by definition. I donot know if every such properties that is something (equations) about gcd and lcm hold in ufd always hold in a [GCD](http://en.wikipedia.org/wiki/Gcd_domain) domain?2011-09-14

1 Answers 1

6

HINT $\ $ Using the basic GCD laws (associative, commutative, distributive) and, furthermore, employing $\rm\:[x,y] = xy/(x,y)\:$ to eliminate LCMs, we obtain

$$\begin{array}{lrll} &\rm(a,[b,c]) &=&\rm [(a,b),(a,c)]&\qquad\qquad\qquad\qquad\quad \\ \iff &\rm(a,bc/(b,c)) &=&\rm (a,b)(c,a)/(a,b,c)& \\ \iff &\rm(a,b,c)(ab,ac,bc) &=&\rm (a,b)(a,c)(b,c)& \end{array} $$

true since both sides $\rm\: =\: (abc, baa,caa, abb,cbb, acc,bcc)\:$ (i.e. all trinomials except cubes), after expanding, by distributivity. The dual identity is proved similarly, yielding the same equality.

  • 0
    Ah of course, $(a,b)(c,d) = (a(c,d),b(c,d)) = ((ac,ad),(bc,bd)) = ((ac,ad),bc,bd) = (ac,ad,bc,bd)$ so you can just multiply them out. I should give up maths for today :(2011-09-14
  • 0
    @Thi That's why I always stress the GCD laws when teaching. These GCD proofs are all quite trivial once one intuitively grasps that GCD arithmetic is essentially the same as integer arithmetic (modulo a couple GCD-specific laws). Such axiom-based proofs are more general than proofs depending on primes (UFDs) or the Bezout identity (PIDs), e.g. the same proof often works for gcds and ideals (which is abstracted by *divisor theory*), e.g. the [Freshman's Dream](http://math.stackexchange.com/questions/10400/comaximal-ideals-in-a-commutative-ring/10416#10416) $\rm\ (A+B)^n\:=\:A^n + B^n$2011-09-14