Let $A \subset \Bbb{Z}$. $A$ is a called an anti-ideal when
- $x, y \in A \implies x - y \notin A$.
- $\Bbb{Z}\ni z \neq \pm 1 \neq a\in A \implies za \notin A$.
(Notice that from before only $z$ had the $\neq \pm \text{a unit}$ constraint. In particular, thanks @Christopher, $\{-1, 1\}$ is no longer a maximal anti-ideal.)
Conjecture. The collection $\mathcal{A}$ of anti-ideals of $\Bbb{Z}$ forms a matroid.
Proof. $\varnothing \in \mathcal{A}$ satisfies 1. & 2. vacuously. And clearly $\mathcal{A}$ is closed under taking subsets. Thus all we need to prove is that if $A, B \in \mathcal{A}$ are anti-ideals such that $|A| \gt |B|$, then there exists $x \in A \setminus B$ such that $B \cup \{x\} \in \mathcal{A}$. This is true since if no such $x$ exists then we can say that either $A - B = \{ a - b : a \in A, b \in B\} \subset B$ or $z A \subset B$ for all $z \in \Bbb{Z}$ each of which implies $|A| \leq |B|$ since $f(x) = x - b$ and $g(x) = zx : A \to B$ are always injective. $\square$
Can you please verify this argument for me? By the way, the set of of all odd primes is an anti-ideal.
It's still not true. Take $A = \{8, 10\}$, $B = \{2\}$. Neither $\{2, 8\}$ nor $\{2, 10\}$ is an anti-ideal.