Is the set of odd primes a maximal "Anti-ideal"? Let $A$ be a commutative ring with $1$.  Define an anti-ideal to be a subset $I \subset A$ such that:


*

*If $x \in A$ is not a unit and $y \in I$ then $xy \notin I$.

*If $x,y \in I$ then $x - y \notin I$.


Define a maximal anti-ideal in the same way as a maximal ideal.  That is $I$ is maximal if for any $A \supset S \supsetneq I, $ then either of the above two conditions is not met for $S$.
The anti-ideals are closed under $\cap$ and taking subsets $J \subset I$.
The set $P$ of odd primes in $\Bbb{Z}$ is an example of an anti-ideal.  Is it maximal?  
 A: I'll assume that in this context, $P$ is the set of odd primes of $\mathbb{Z}$, (i.e., positive or negative ordinary odd primes).

Suppose $Q$ is an anti-ideal which properly contains $P$.

Then $Q$ can't contain any odd non-prime non-units, since any such number would be an odd prime times a non-unit.

Also, $Q$ can't contain $0$ since $0=(0)(3)$.

Hence any element of $Q\setminus P$ must be a power of $2$, or the negative of a power of $2$.

Let $x\in Q\setminus P$.

Then if $k$ is a positive integer, we can't have $2^kx \in Q$, and we can't have $-2^kx$ in $Q$, hence $Q$ contains at most one positive power of $2$, and at most one negative power of $2$, and if both, then they must be equal in absolute value.

Let's explore . . .

Fix a positive integer $m$, and suppose at least one of $2^m,-2^m$ is in $Q$.

Then there are $3$ possibilities for $Q$, namely 
$$Q=P\cup\{2^m\}$$
$$Q=P\cup\{-2^m\}$$
$$Q=P\cup\{-2^m,2^m\}$$

In all $3$ cases, $Q$ is an anti-ideal if and only if both


*

*$2^m$ is not the sum of two positive odd primes.

*$2^m$ is not the difference of two positive odd primes.



If such an $m$ exists, then $P$ is not maximal, else $P$ is maximal.

As far as I'm aware, it's not known if such a positive integer $m$ exists.
