Prime ideals of a lattice

Question

Given the lattice generated by natural numbers (all except $0$) ordered by divisibility, how can I construct its prime ideals? How does a familiy of prime ideals without any given element $a$ look like?

Answer 1

Let $\mathbf D = (D,|) = (D,\gcd,\mathrm{lcm})$ and $\mathcal I_p(\mathbf D)$ denote the set of prime ideals of $\mathbf D$.

Let $I_{p^k}$ denote the set of natural numbers which are not multiples of the prime power $p^k$.
It is easy to check that $I_{p^k} \in \mathcal I_p(\mathbf D)$.

Now let $I$ be any other kind of ideal of $\mathbf D$, that is, there exist prime powers $p^{k}$ and $q^{k'}$ (with $p \neq q$) such that $p^k,q^{k'} \notin I$. Let $n$ be any member of $I$, and define $u=p^kn$ and $v=q^{k'}n$.
Clearly, $u,v \notin I$ (for otherwise $p^k,q^{k'} \in I$), but $\gcd(u,v)=n \in I$, and so $I$ is not a prime ideal.

Hence, an ideal $I$ of $\mathbf D$ is prime iff $I=I_{p^k}$, for some prime number $p$ and natural $k$.

It follows from that description that $$X_a = \{I \in \mathcal I_p(\mathbf D) : a \notin I\} = \{I_{p^k} : p^k \not\mid a\}.$$

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Update: I changed $I_p$, for some prime number $p$ to $I_{p^k}$, to some prime number $p$ and natural $k$.
Indeed, the set of numbers not multiples of any power of a prime is a prime ideal.