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For what values of n the set of divisors of n under partial order relation divides is a complemented lattice?

Complemented lattice: If every element of a lattice has at least one complement, it is called complemented lattice.

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Under which partial order relation? The relation $x \mid y$? – goblin Sep 30 '13 at 13:41
Yup. That is clearly stated. – user221458 Sep 30 '13 at 13:45
Right. Yes I missed that. – goblin Sep 30 '13 at 13:47
Is this homework? – Marc van Leeuwen Sep 30 '13 at 13:47
up vote 2 down vote accepted

If $n$ is a product of distinct primes, then this lattice is complemented. The top element is $n$, the bottom element is $1$. Any divisor $j$ has $n/j$ for a complement, since $n/j$ and $j$ are relatively prime (i.e., have meet $1$) and have join $n$.

On the other hand, if $\gcd(p, n) > p$ for some prime $p$, i.e., if $n$ has a squared-prime factor $p^2$, then $n/p$ has no complement in the lattice. So we have characterized all $n$ which yield a complemented lattice.

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If $n$ is a product of $k$ prime numbers, then every element (except $1$ and $n$) has $k-1$ complements. In particular, if $n = pq$ with $p$ and $q$ prime, then every element has exactly one complement. If $p^2$ divides $n$ for some prime $p$, then $p$ has no complement.

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Can u find p and q for 30 ? – user221458 Sep 30 '13 at 13:38

The lattice in question is a product over all primes$~p$ (or just those that divide$~n$) of partial orders 9lattices) that are chains of length the multiplicity of$~p$ as factor of$~n$. A product lattice is complemented if and only if all its factors are. So it boils down to finding which chains are complemented lattices; it is obvious that that chains of length $l\leq1$ are and those of length $l>1$ are not. So the necessary and sufficient condition is that every prime have multiplicity $0$ or $1$ in$~n$, in other words that $n$ be square-free.

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