Find zero and one element for given ring

Question

Given $ M$ and the powerset $\mathcal{P}(M) $,
let $A,B \subseteq M$ and define:
$A \bigoplus B := A \backslash B \cup B \backslash A$
$A * B := A \cap B$
Then is $R_M = \langle \mathcal{P}(M); \bigoplus, *\rangle$ a commutative ring.

What is the zero element and whats the one element of $R_M$? And what is $R^\times_M$?

I Know that for the one element (e.g. 1 being the element): $1*A = A*1 = A$
And for the zero element (e.g. 0 being the element): $0*A=0$

But how do I transfer this to my sets?

Thanks in advance :)

Answer 1

For the identity of the $\ast$ operation, think about what you're doing. You need to find a set such that when you intersect it with any other subset of $X$ of $M$, you still get $X$ back. That means that your candidate must be a superset of all subsets of $M$...

For the identity of the $\oplus$ operation, think along similar lines. You need a set such that for any subset $X$ of $M$, the symmetric difference of $X$ and the set is just $X$. In particular your candidate must be contained in the complement of $X$ for every subset $X$ of $M$...

These lines of thought lead to the answers. But seriously, the answers are the likely suspects. You really should have checked them first!

Let $1$ be the $\ast $ identity, and let's think about all the units of this ring. It turns out that that units are as sparse as they get in this ring. Verify that $A\ast A=A$ for every $A$ in the ring. Suppose $A\ast B=1$ (i.e. that $A$ is a unit!) and then combine that with the information about $A\ast A=A$ to draw a conclusion.

Answer 2

$ \newcommand{\calc}{\begin{align} \quad &} \newcommand{\calcop}[2]{\notag \\ #1 \quad & \quad \text{"#2"} \notag \\ \quad & } \newcommand{\endcalc}{\notag \end{align}} $To me, it seems the simplest approach is to calculate $\;0\;$ and $\;1\;$.

For $\;0\;$, we know that $\;0 \oplus A = A\;$ for all $\;A\;$. And it helps to know that a simpler definition of $\;\oplus\;$ is $$ \tag{0} x \in A \oplus B \;\equiv\; x \in A \not\equiv x \in B $$ Using this, and letting $\;x\;$ implicitly range over $\;M\;$, we calculate

$$\calc \langle \forall A :: 0 \oplus A = A \rangle \calcop{\equiv}{set extensionality; definition (0) for $\;\oplus\;$} \langle \forall A :: \langle \forall x :: x \in 0 \not\equiv x \in A \equiv x \in A \rangle \rangle \calcop{\equiv}{logic: simplify} \langle \forall A :: \langle \forall x :: x \not\in 0 \rangle \rangle \calcop{\equiv}{definition of $\;\emptyset\;$; logic: remove unused $\;\forall A\;$} 0 = \emptyset \endcalc$$

For $\;1\;$ we know that $\;1 * A = A\;$ for all $\;A\;$, so we calculate

$$\calc \langle \forall A :: 1 * A = A \rangle \calcop{\equiv}{definition of $\;*\;$} \langle \forall A :: 1 \cap A = A \rangle \calcop{\equiv}{set extentionality; definition of $\;\cap\;$} \langle \forall A :: \langle \forall x :: x \in 1 \land x \in A \;\equiv\; x \in A \rangle \rangle \calcop{\equiv}{logic: simplify $\;P \equiv P \land Q\;$ to $\;\lnot P \lor Q\;$; merge $\;\forall\;$ quantifications} \langle \forall A,x :: x \not\in A \;\lor\; x \in 1 \rangle \calcop{\equiv}{logic: simplify by moving parts not using $\;A\;$ out of $\;\forall A\;$} \langle \forall x :: \langle \forall A :: x \not\in A \rangle \;\lor\; x \in 1 \rangle\calcop{\equiv}{$\;\langle \forall A :: x \not\in A \rangle\;$ is false, witness $\;A := \{x\}\;$; simplify} \langle \forall x :: x \in 1 \rangle \calcop{\equiv}{$\;x\;$ ranges over $\;M\;$; definition of $\;\subseteq\;$; $\;1 \subseteq M\;$} 1 = M \endcalc$$

This completes the calculation of both identities.