$D_{m}$ is a boolean algebra iff there is no prime $p$ such that $p^2 \mid m$

Question

$D_{m}$ 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?

Answer 1

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 $\mathrm{lcm}(p,b)=m$?