I was thinking about different rules for rings and fields and I decided to analise the following structure: $(\mathcal{P}(X), \cup, \cap)$, where "$\mathcal{P}(X)$" denotes the set of subsets of a given set $X$.
I noticed this followed all the axioms of commutative rings with identity, where the identity of "$\cap$" is "$X$", except for the inverse element of addition, so I decided to switch the "$\cup$" operation for something that could work.
My next try was with the following operation: $+ : \mathcal{P}(X) \times \mathcal{P}(X) \rightarrow \mathcal{P}(X)$
\begin{cases}
A + B = A\cup B, \text{ if $A\cap B = \emptyset$} \\
A + B = \emptyset, \text{ if $A\cap B \neq \emptyset$}
\end{cases}
In this case, I got the inverse element, and the operation is still commutative and has a neutral element (i.e. the empty set). But since there were many inverses, I assume this is not associative, though I haven't personally checked.
In short, I would really like an operation that, together with the intersection, turns $\mathcal{P}(X)$ into a ring. Does anyone know of any such operation, or know any reason why it should not exist?
Thanks in advance!
Best Answer
Note that $$\mathcal{P}(X) \to {}^X\mathbb{F}_2 \\ A \mapsto \mathbf{1}_A,$$ which maps sets to indicator functions, is a bijection from the set $\mathcal{P}(X)$ to the ring $${}^X\mathbb{F}_2 = \{f : X \to \mathbb{F}_2\}.$$ It therefore induces a ring structure on $\mathcal{P}(X)$.
In particular, this induces $$A\cdot B := A\cap B \\ A + B := A\Delta B$$
If you want $A\cdot B := A\cup B$, then instead consider the map $A \mapsto \mathbf{1}_{X\setminus A}$ which induces, once again, $$A+B:=A \Delta B$$ (it seems we can't get away from it so easily).
If you want, try other bijections to find other "natural" ring structures, but those two are probably the most natural.