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?

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    $\begingroup$ I would be surprised if there isn't a nice characterization of what you're looking for. But to draw the lattice: It's essentially a $4 \times 3 \times 2$ box, due to the factorization of $360$ (three distinct primes, to various powers). To see what I mean, try to draw the lattice of divisors of $72 = 2^3 \cdot 3^2$ as a $3\times 2$ rectangle. $\endgroup$ – pjs36 Jun 24 '16 at 15:31

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.

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    $\begingroup$ It took me some time but now, when I'm solving lattice, problems is like I know kung fu. Thank You. $\endgroup$ – Lindow Jun 29 '16 at 17:52

How to draw a lattice; I think it would get tricky if you had more than three prime divisors, but $360$ is not too bad:

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