1. **Problem statement:** We consider the lattice formed by the divisors of 105, denoted as 105D, with the partial order defined by divisibility. We need to (i) draw the Hasse diagram, (ii) find the complement of each element, and (iii) find the set of atoms. 2. **Step (i): Draw the Hasse diagram of 105D.** - First, find all divisors of 105. Since $105 = 3 \times 5 \times 7$, the divisors are all products of these primes in any combination: $$1, 3, 5, 7, 15, 21, 35, 105$$ - The Hasse diagram connects elements $a$ and $b$ if $a$ divides $b$ and there is no $c$ such that $a$ divides $c$ and $c$ divides $b$ (cover relation). - The diagram levels: - Bottom: $1$ - Next level: $3, 5, 7$ - Next level: $15 = 3 \times 5$, $21 = 3 \times 7$, $35 = 5 \times 7$ - Top: $105 = 3 \times 5 \times 7$ - Edges: - $1$ to $3, 5, 7$ - $3$ to $15, 21$ - $5$ to $15, 35$ - $7$ to $21, 35$ - $15, 21, 35$ to $105$ 3. **Step (ii): Find the complement of each element in 105D.** - In a lattice of divisors with order by divisibility, the complement of an element $a$ is an element $b$ such that $a \wedge b = 1$ (greatest common divisor) and $a \vee b = 105$ (least common multiple). - For each divisor $a$, find $b$ such that $\gcd(a,b) = 1$ and $\mathrm{lcm}(a,b) = 105$. - Since $105 = 3 \times 5 \times 7$, the complement of $a$ is the product of the primes in $105$ not in $a$. - Complements: - $1$: complement is $105$ - $3$: complement is $35$ (since $35 = 5 \times 7$) - $5$: complement is $21$ (since $21 = 3 \times 7$) - $7$: complement is $15$ (since $15 = 3 \times 5$) - $15$: complement is $7$ - $21$: complement is $5$ - $35$: complement is $3$ - $105$: complement is $1$ 4. **Step (iii): Find the set of atoms of 105D.** - Atoms in a lattice are elements that cover the minimum element (here, 1). - The elements that cover 1 are the prime divisors: $3, 5, 7$. - Therefore, the set of atoms is $\{3, 5, 7\}$. **Final answers:** - (i) Hasse diagram levels and edges as described. - (ii) Complements: $1 \leftrightarrow 105$, $3 \leftrightarrow 35$, $5 \leftrightarrow 21$, $7 \leftrightarrow 15$. - (iii) Atoms: $\{3, 5, 7\}$.