An exercise ask to find atoms and join-irreducible elements for the set of divisors of 360. I know how to find them by drawing the lattice but it seems difficult in this case.
Is there another way to find atoms? If not, is there a easy way to draw such a lattice?
The join-irreducible elements are precisely the prime powers dividing $360$:
Let $d=p^k$ be a divisor of $360$. Then $d=a\vee b$ implies $a=p^k$ or $b=p^k$, so $d$ is join-irreducible. Conversely, let $d$ be a join-irreducible divisor of $360$ and write $d=a\times b$ with $\gcd(a,b)=1$. Then $d=a\vee b$ and hence $d=a$ or $d=b$, meaning that $b\mid a$ or $a\mid b$, respectively. This holds for every factorization $d=a\times b$ and hence $d$ is a prime power.
The atoms are the numbers with no nontrivial proper divisors, i.e. the primes.