# How to draw a lattice for the divisors of big numbers?

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?

• 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. – 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.
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: