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I'm studying for an abstract algebra exam and one of the review questions was this:

Let $X$ be a set, and $\mathcal P(X)$ be the power set of $X$. Consider the operations $\Delta$ = symmetric difference (a.k.a. "XOR"), and $\bigcap$ = intersection.
a) Does $\Delta$ and $\bigcap$ make $P(X)$ into a ring?
b) If so, is it a ring with unity?
c) Is the ring commutative?
d) Is it a field?

For parts a) and b), I think it does form a ring with unity, but I'm not quite sure how to get started on proving it.

For part c), it is a commutative ring since $\mathcal P(X)$ is closed under symmetric difference and intersection, right?

Not even sure how to get started on d).

I'd really like to understand this question fully, so any kind of input would be tremendously helpful. Thank you!

Eugene Zhang
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Max
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  • Do you know what's a ring? A ring with unity, and so on? For instance, I can show you how to prove that $\cap$ is a commutative operation (which means $a \cap b = b \cap a$ for $a,b \in P(X)$).

    $\forall x, (x \in a \cap b \longleftrightarrow (x \in a$ and $x \in b) \longleftrightarrow (x \in b$ and $x \in a) \longleftrightarrow x \in b \cap a)$, so $a \cap b = b \cap a$.

     – nombre Oct 26 '16 at 19:47
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    Well, if you know what a ring is, then you can confirm that a set with operations is a ring or not. Do you understand how to confirm that the ring axiom hold? It is a commutative ring since it's closed under... Well no, there are more axioms than just having a binary operation defined. And none of that directly proves the multiplication is commutative, either way. How about you systematically start to verify the axioms? It is all very mechanical. – rschwieb Oct 26 '16 at 20:36
  • I agree with @rschwieb. For (a), look at the definition of "ring". It will be about a set with two operations on it, usually called $+$ and $\cdot$. Throughout the definition, where it refers to that set, take it as meaning $\mathcal P(X)$. Where it refers to $+$, take it as meaning $\Delta$. Where it refer to $\cdot$, take it as meaning $\cap$. Check each requirement in the definition. If they're all satisfied, then you have a ring; if even one requirement is violated, it's not a ring. Then proceed similarly with the other 3 parts of the problem. – Andreas Blass May 29 '20 at 00:29

2 Answers2

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I believe that the best way to study the properties of this structure is to note that $\mathcal P(X)$ with these operations is isomorphic to the set $\mathcal{F}$ of functions $X \to E$, where $E = \{ 0, 1\}$ is the field with two elements, and $\mathcal{F}$ is endowed with pointwise sum and product, that is, for $f, g \in \mathcal{F}$ we have $$ (f + g)(x) = f(x) + g(x),\qquad (f \cdot g)(x) = f(x) \cdot g(x). $$ The isomorphism is given by $$ X \supseteq A \mapsto \left(f : x \mapsto \begin{cases}1 & \text{if $x \in A$}\\0 &\text{if $x \notin A$}\end{cases}\right) $$

Andreas Caranti
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Let symmetric difference be correspond to addition and intersection correspond to multiplication. Also let $\varnothing$ be the empty set and correspond to $0$, along with $X$ be the universe and correspond to $1$. Then we can prove that symmetric difference and intersection of $\mathcal{P}(X)$ satisfy every ring axiom. Suppose $A,B,C\in \mathcal{P}(X)$.

  1. Commutativity of addition: $$ A\bigtriangleup B=(A-B)\cup (B-A)=(B-A)\cup (A-B)=B\bigtriangleup A $$
  2. Additive identity: $$ A\bigtriangleup \varnothing =(A-\varnothing )\cup (\varnothing -A)=A\cup \varnothing =A $$

  3. Additive inverse: $$ A\bigtriangleup A=(A-A)\cup (A-A)=\varnothing \cup \varnothing =\varnothing $$ So additive inverse of $A$ is $A$ itself.

  4. Associativity of addition: By this post, there is $$ (A\bigtriangleup B)\bigtriangleup C=A\bigtriangleup (B\bigtriangleup C) $$

  5. Commutativity and associativity of multiplication are due to the fact that intersection is commutative and associative.
  6. Multiplication identity is satisfied for $$ A\cap X= A $$
  7. Multiplication is distributive with respect to addition: \begin{align} (A\bigtriangleup B)\cap C&=((A-B)\cup (B-A))\cap C \\ &=(A\cap C-B\cap C)\cup (B\cap C-A\cap C) \\ &=(A\cap C)\bigtriangleup (B\cap C) \end{align}

Hence $\mathcal{P}(X)$ under symmetric difference and intersection is a commutative ring with the unity.

Obviously it is not true that for any set $A\in\mathcal{P}(X)$, there is a set $B\in \mathcal{P}(X)$ that $A\cap B=X$. Thus $\mathcal{P}(X)$ under symmetric difference and intersection is not a field.

Eugene Zhang
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