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$D_{m}$ is a boolean algebra iff there is no prime $p$ such that $p^2 \mid m.$ How can I prove this? Could anyone help me please?

amrsa
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    Is $D_m$ the divisors of $m$ with $d_1\wedge d_2$ the least common multiple, $d_1\vee d_2$ the greatest common divisor and $\neg d=m/d$? In that case proving that if $m$ is square free, then $D_m$ is a Boolean algebra is just checking the definition. Or alternatively note that a divisor in that case is equivalent as having the set of primes dividing that divisor and $\wedge$ and $\vee$ correspond to union and intersection of sets. Conversely, if $D_m$ is a Boolean algebra, then for a prime $p|m$ if $p^2|m$ then $p\vee \neg p=\operatorname{gcd}(p,m/p)=p\neq 1$. – MoonLightSyzygy Dec 20 '19 at 22:01
  • @MoonLightSyzygy You could elaborate that comment to make a full answer (starting with the observation that $D_m$ is a distributive lattice, for every $m$). – amrsa Dec 21 '19 at 10:56

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A Boolean algebra is a bounded distributive lattice in which each element has a (necessarily unique) complement.

It is clear that $(D_m,\gcd,\mathrm{lcm},m,1)$ is a bounded lattice.
It is not difficult to show that it is distributive (otherwise check this answer, for example).

Now, if $m$ is square-free, given $a \in D_m$, let $b = m/a$ and let us check that $a$ and $b$ are complements of each other.
Let $d=\gcd(a,b)$, and suppose, for a contradiction, that $p \mid d$, for some prime $p$.
Then $p \mid a$ and $p \mid b$, whence $p^2 \mid ab=m$, a contradiction with $m$ being square-free.
Hence no such $p$ exists and therefore, $d = \gcd(a,b) = 1$.
Now given that $\gcd(a,b)=1$, it follows that $\mathrm{lcm}(a,b)=ab=m$.
Thus $a$ and $b$ are complements of each other and $D_m$ is a Boolean algebra.

For the converse, suppose $p^2\mid m$, for some prime $p$.
Can you see that there is no $b \in D_m$ such that $\gcd(p,b)=1$ and $\mathrm{lcm}(p,b)=m$?

amrsa
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