The set $\mathcal{S}(R)$ of all radical ideals of a ring $R$ might form a boolean ring with $\circ=$ ideal addition and $\oplus=$ a certain quotient.

Question

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.

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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 $A,B \in \mathcal{S}(R)$.

Answer 1

First, some corrections:

• In the case of $\mathbb{Z}$, while the identity for $\oplus$ is $(1)$, the identity for $\circ$ (the sum operation on ideals) should be $(0)$. But this is a problem, because $0$ is not square-free! As a result, $S$ is not a Boolean ring, since the multiplication operation $\oplus$ has no identity element.
• Of course, $(0)$ is a radical ideal. But if you try to extend your definition of $\oplus$ to include the zero ideal, you'll run into trouble... The point is that the square-free integers correspond to finite sets of primes, which do not form a Boolean algebra, since the set of all primes (which would be the identity for the intersection operation) is not finite.
• In the general case, the sum of radical ideals need not be radical (for example, in $k[x,y]$, $(y)$ and $(y-x^2)$ are radical, but $(y) + (y-x^2) = (y,x^2)$, which is not radical). So your $\circ$ operation on radical ideals should be $A\circ B = \sqrt{A+B}$.
• Similarly, I don't see why $A\oplus B = [AB:A^2+B^2]$ should be radical when $A$ and $B$ are radical, though I don't have a counterexample in mind.

It is a fact that if $I$ is a radical ideal, then for any ideal $J$, $[I:J]$ is radical. (Proof: Suppose $r^n\in [I:J]$. I claim $r\in [I:J]$. Let $j\in J$. Since $r^n\in [I:J]$, $r^nj\in I$, so $r^nj^n = (rj)^n \in I$, and since $I$ is radical, $rj\in I$. Thus $r\in [I:J]$.)

So as a replacement for the operation $A\oplus B = [AB:A^2+B^2]$, I suggest $A\oplus B = [\sqrt{AB}:A+B]$. By the previous fact, this is a binary operation on radical ideals, and it gives the expected answer in the case of $\mathbb{Z}$.

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As for your main question: As I pointed out above, we don't even get a natural Boolean algebra/ring structure on the set of radical ideals in $\mathbb{Z}$, much less an arbitrary ring. What we do get, though, is a distributive lattice structure. (See here.)

Recall that for any ring $R$, $\DeclareMathOperator{\Spec}{Spec}\Spec(R)$ is the space of prime ideals in $R$, equipped with the topology generated by basic open sets of the form $D(f) = \{p\in \Spec(R)\mid f\notin p\}$ for $f\in R$.

Now there is an inclusion-reversing bijection between radical ideals in $R$ and closed sets in $\Spec(R)$. Since the closed sets (in any topology) form a distributive lattice, the radical ideals in $R$ have a natural distributive lattice structure with operations $I\vee J = \sqrt{I+J}$ and $I\wedge J = \sqrt{IJ} = \sqrt{I\cap J}$. This is the point of view in Proof 2 of Martin Brandenburg's answer to the question I linked above.

Thinking topologically, the closest thing we can get to symmetric difference as an operator on closed sets is $A\oplus B = \overline{(A\cup B)\setminus (A\cap B)}$, where the bar indicates topological closure. I think this operation should correspond to the radical ideal $[\sqrt{AB}:A+B]$ that I suggested above, but I have not carefully checked the details.