Define $S \subset \Bbb{N}$ to be the square-free integers $s$ i.e. such that no $p^2 \mid s$ for any $p \in \Bbb{P}$ a prime number.
Then it is easy to see that $(S, \oplus)$ forms an boolean group via $a \oplus b := \frac{ab}{\gcd(a,b)^2}$, and that this extends to a monoid operation $(\Bbb{N}, \oplus)$ via $n \oplus m := \frac{\text{rad}(n)\text{rad}(m)}{\gcd(\text{rad}(n), \text{rad}(m))^2} \in S$.
As a consequence we have that on ideals $\mathcal{S} = \{ (s) \leqslant \Bbb{Z} : s \in S\}$, we have the group law:
$$ (a)\oplus(b) = [(a)(b):(a)^2 + (b)^2] $$
where $[I:J]$ is the Ideal quotient - Wikipedia.
But while this $\oplus$ on square-free numbers acts like XOR (or symmetric difference) on the sets of prime divisors of an input $n \in \Bbb{N}$, we of course have a boolean ring with the operation corresponding to AND being $\gcd$. We'll let $\gcd \equiv \circ$. In $\mathcal{S}$ of course we can put $(\gcd(a,b)) \equiv (a) + (b)$, so that clearly the addition of two ideals is defined in every ring, and we don't really require $\gcd$ to be defined.
$$ \mathcal{S} = (\mathcal{S}, \oplus, \circ) $$
forms a ring, where I believe identity for either operation is just the ideal $(1)$.
Question. Since the ideals of $\Bbb{Z}$ generated by a square-free number $s \in S$ correspond abstractly to the Radical of an ideal - Wikipedia, i.e. $(s) \in \mathcal{S} \iff \sqrt{(s)} = (s)$ itself; a question naturally becomes does the set $\mathcal{S}(R)$ of radical ideals of a ring $R$ also form a boolean ring under $A \oplus B = [AB:A^2 + B^2]$ and $A\circ B:= A + B$, for all $A,B \in \mathcal{S}(R)$.
