**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.3 Exercise 7.3.15**

Solution: Let $\Phi : \mathcal{P}(X) \rightarrow {}^X\mathbb{Z}/(2)$ be the mapping given by $A \mapsto \chi_A$, and let $A,B \in \mathcal{P}(X)$.

We first consider $\Phi(A+B)$. There are 4 cases.

(1) Suppose $x \in A$ and $x \in B$. Then $x \notin A \setminus B$ and $x \notin B \setminus A$, so that $$x \notin (A \setminus B) \cup (B \setminus A) = A+B. $$Thus $$\Phi(A+B)(x) = 0 = 1 + 1 = \Phi(A)(x) + \Phi(B)(x) (\Phi(A)+\Phi(B))(x)$$ Hence $\Phi(A+B) = \Phi(A) + \Phi(B)$.

(2) Suppose $x \in A$ and $x \notin B$. Then $x \in A \setminus B$, so that $$x \in (A \setminus B) \cup (B \setminus A) = A+B.$$ Then $$\Phi(A+B)(x) = 1 = 1+0 = \Phi(A)(x) + \Phi(B)(x) = (\Phi(A)+\Phi(B))(x),$$ and we have $\Phi(A+B) = \Phi(A) + \Phi(B)$.

(3) The case $x \notin A$ and $x \in B$ is similar to the previous one.

(4) Suppose $x \notin A$ and $x \notin B$. Then $x \notin A \setminus B$ and $x \notin B \setminus A$, so that $$x \notin (A \setminus B) \cup (B \setminus A) = A+B.$$ Thus $$\Phi(A+B)(x) = 0 = 0+0 = \Phi(A)(x) + \Phi(B)(x) = (\Phi(A)+\Phi(B))(x).$$ Hence $\Phi(A+B) = \Phi(A)+\Phi(B)$.

Thus for all $A,B \in \mathcal{P}(X)$, $\Phi(A+B) = \Phi(A)+\Phi(B)$.

Now we consider $\Phi(AB)$. Again there are four cases.

(1) Suppose $x \in A$ and $x \in B$. Then $x \in A \cap B = AB$, so that $$\Phi(AB)(x) = 1 = 1 \cdot 1 = \Phi(A)(x) \cdot \Phi(B)(x) = (\Phi(A) \cdot \Phi(B))(x). $$Thus $\Phi(AB) = \Phi(A)\Phi(B)$.

(2) Suppose $x \in A$ and $x \notin B$.Then $x \notin A \cap B = AB$, so that $$\Phi(AB)(x) = 0 = 1 \cdot 0 = \Phi(A)(x) \cdot \Phi(B)(x) = (\Phi(A) \cdot \Phi(B))(x).$$ Thus $\Phi(AB) = \Phi(A)\Phi(B)$.

(3) The case $x \notin A$ and $x \in B$ is similar to the previous one.

(4) Suppose $x \notin A$ and $x \notin B$. Then $x \notin A \cap B = AB$, and we have $$\Phi(AB)(x) = 0 = 0 \cdot 0 = \Phi(A)(x) \cdot \Phi(B)(x) = (\Phi(A)\Phi(B))(x).$$ Thus $\Phi(AB) = \Phi(A)\Phi(B)$.

Thus for all $A,B \in \mathcal{P}(X)$, $\Phi(AB) = \Phi(A)\Phi(B)$, and $\Phi$ is a ring homomorphism.

Next we show that $\Phi$ is surjective. Let $\omega \in {}^X \mathbb{Z}/(2)$, and let $A_\omega = \omega^\ast(1)$. ($\omega^\ast$ denotes preimage.) I claim that $\Phi(A_\omega) = \omega$. To that end, suppose $x \in A_\omega$. Then $\Phi(A_\omega)(x) = 1 = \omega(x)$. Likewise, if $x \notin A_\omega$, then $\Phi(A_\omega)(x) = 0 = \omega(x)$. Thus $\Phi$ is surjective.

Next we show that $\Phi$ is injective. To that end, suppose $\Phi(A) = \Phi(B)$. Suppose $x \in A$; then $\Phi(A)(x) = 1 = \Phi(B)(x)$, and we have $x \in B$. Thus $A \subseteq B$. Symmetrically, $B \subseteq A$, and we have $A = B$. Thus $\Phi$ is injective.

Thus $\Phi$C is a ring isomorphism.