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Taking $R$ to be the set of subsets of an arbitrary set $U$, we have $A, B \in R$. Letting $A+B = (A\setminus B)\cup(B\setminus A)$ and $AB = A\cap B$, show that $R$ is a Boolean Ring.

I know that to be a Boolean Ring, the ring $R$ must satisfy $x^2 = x \hspace{2mm} \forall x \in R$. I haven't the slightest clue where to start with this. Any help is appreciated.

tdashrom
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    Well, per definition, what is $A^2$ for some $A\in R$? – Mastrem Mar 29 '18 at 17:17
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    So far you have asked why $A^2=AA=A\cap A$ is the same thing as $A$. Are you really stuck there? The "hard"(?) part is going to be verifying the ring axioms with the operations. In that case even, Mastrem's advice of writing down what the definitions say first is the best advice you can get. – rschwieb Mar 29 '18 at 17:20
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    Actually, the part of whether or not it is a ring is already on the site in more than one place so I guess stick to your question about it being boolean. – rschwieb Mar 29 '18 at 17:23

1 Answers1

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First, you want to show that $R$ satisfies the ring axioms. That is

  1. $(R,+)$ is an abelian group.
  2. Multiplication is associative.
  3. Multiplication is distributive with respect to addition.

You can show 1. by showing that i) addition is associative, ii) addition is commutative, iii) there exists an additive identity, and iv) there exists additive inverses.

Let $A,B,C \in R$, and let $A^C$ denote the compliment of $A$.

i) The operation $+$ defined by $A+B = (A\setminus B)\cup(B\setminus A)$ is commonly referred to as the "symmetric difference" and the (lengthy) proof of associativity is shown here: https://proofwiki.org/wiki/Symmetric_Difference_is_Associative

ii) Addition is commutative.

Since $\cup$ is commutative, $A+B = (A\setminus B)\cup(B\setminus A) = (B\setminus A)\cup(A\setminus B) = B+A$.

iii) Additive identity.

Suppose $0_R = \emptyset$. Then for any $A \in R$, $A + 0_R = (A\setminus \emptyset)\cup(\emptyset \setminus A) = (A\cap \emptyset^C)\cup(\emptyset \cap A^C) = (A\cap U)\cup(\emptyset \cap A^C) = A\cup \emptyset=A$. Since $+$ is commutative, it follows that $0_R + A = A + 0_R = A.$

iv) Additive inverses.

Suppose $(-A) = A$. Then for any $A \in R$, $A + (-A) = A + A = (A\setminus A)\cup(A\setminus A) = (A\cap A^C)\cup(A\cap A^C) = \emptyset \cap\emptyset = \emptyset = 0_R$.

2) Multiplication is associative.

Since $\cap$ is associative, $A(BC) = A\cap(BC) = A\cap(B\cap C) = (A\cap B)\cap C = (AB)C$.

3) Multiplication is distributive with respect to addition. \begin{align} A(B+C) &= A\cap(B+C) \\ &= A\cap((B\setminus C)\cup(C\setminus B)) \\ &= (A\cap(B\setminus C))\cup(A\cap(C\setminus B)) \\ &= (A\cap B\cap C^C)\cup(A\cap C\cap B^C) \\ &= ((A\cap B)\cap (C^C\cup A^C))\cup((A\cap C)\cap (B^C\cup A^C)) \\ &= ((AB)\cap(AC)^C)\cup((AC)\cap(AB)^C) \\ &= AB + AC \end{align}

Now all you need to show is that the ring $R$ is also a Boolean ring, which is simple since

$A^2 = AA = A\cap A = A$.

Joshua Hanson
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