Question about an inverse limit.

Question

Define a partial order on $\Bbb{N}$ to be $n \leq m$ iff $n = m $ or $n |m$ and there's a twin prime dividing $m$ and not $n$. It's easy to see that it's a poset. Define a system of abelian groups $A_n = \Bbb{Z}/n$ with $p_{nm} : A_m \to A_n$ the natural projections. Clearly $p_{ii} = \text{id}$. And if $i \leq j \leq k$, $p_{ik} = p_{ij} \circ p_{jk}$ So take the inverse limit and call it $A$. Consider the inverse system in which $n \leq m$ iff $n | m$ with the usual projections and call it $B$. Then is $A$ a just a subring of $B$ or is it an ideal?

Answer 1

Note that both systems in question consist of the same set of abelian groups. The only difference is that the system defining $A$ has less arrows in it. It follows that there is a natural inclusion $B\hookrightarrow A$ (and not vice versa). By definition, $B$ and $A$ are rings, and under this embedding we may say that $B$ is a subring of $A$.