If you were to create a graph with ideals in $\mathbb{Z}$ as vertices and an edge between $(m)$ and $(n)$ if $(m) \subseteq (n)$ or $(m) \supseteq (n)$, then you would get something that looks like:
(for the purposes of the diagram $n$ is really denoting $(n)$ in particular $1 = (1) = \mathbb{Z}$)
Now $(m) \subseteq (n)$ if and only if $n \mid m$, and so you obtain the same graph if you take the set of vertices to simply be the integers and draw an edge between $m$ and $n$ if $m \mid n$ or $n \mid m$.
Now if you were to create a graph with subfields of $\overline{\mathbb{F}_{q}}$, the algebraic closure of the field with $q$ elements, and you draw an edge between $\mathbb{F}_{q^r}$ and $\mathbb{F}_{q^s}$ if $\mathbb{F}_{q^r}$ is a subfield of $\mathbb{F}_{q^s}$ or $\mathbb{F}_{q^s}$ is a subfield of $\mathbb{F}_{q^r}$ you obtain an isomorphic graph. The following map from $(t) \mapsto \mathbb{F}_{q^t}$ is clearly a bijection and it preserves operation because $\mathbb{F}_{q^r}$ is a subfield of $\mathbb{F}_{q^s}$ if and only if $r \mid s$.
My question is there a depper reason why you obtain isomorphic graphs, deeper that the referenced subfield criterion? Or does this perhaps highlight a relationship between $\overline{\mathbb{F}_{q}}$ and $\mathbb{Z}$?