In ACC on the top of page 34, I read:
The construct Boo of boolean algebras is isomorphic to the construct BooRng of boolean rings and ring homomorphisms.
This contradicts my intuition since all compositions in BooRng-objects are finite, contrary compositions in Boo-objects. How could an isomorphic functor between BooRng and Boo be constructed? Will it map Boo-morphisms on ring homomorphism which preserve 1?
(Throughout ring is meant to mean a unital ring, and ring homomorphism is meant to mean a unital ring homomorphism; I also make skip pointing out relatively-trivial uses of Boolean algebraic facts.)
We define a map $\mathscr{F}: \mathbf{BoolRing} \rightarrow \mathbf{BoolAlg}$ by sending a Boolean ring $(R,\oplus,\cdot, 0,1)$ to the Boolean Algebra $(R,\wedge, \vee, \neg,0,1)$ where $\wedge,\vee,\neg$ are defined as follows: \begin{equation*} x\wedge y = x\cdot y \quad \text{and} \quad x\vee y = x\oplus y \oplus x\cdot y \quad \text{and} \quad \neg x =1\oplus x. \end{equation*} $\mathscr{F}$ sends a ring homomorphism $\varphi: R\rightarrow S$ to the Boolean Algebra homomorphism $\varphi: R\rightarrow S$.
This functor has an inverse $\mathscr{G}: \mathbf{BoolAlg}\rightarrow \mathbf{BoolRing}$ given by sending a Boolean algebra $(R,\wedge, \vee, \neg,0,1)$ to the Boolean Ring $(R,\oplus,\cdot,0,1)$ where $\oplus,\cdot$ are defined as follows: \begin{equation*} x\cdot y = x\wedge y \quad \text{and} \quad x\oplus y = (x\vee y)\wedge \neg (x\wedge y). \end{equation*} Again, $\mathscr{G}$ sends a Boolean Algebra homomorphism $\varphi: R\rightarrow S$ to the ring homomorphism $\varphi: R\rightarrow S$.
Naturally, we must show that these notions are all well-defined: i.e. that $\mathscr{F} (R,\oplus,\cdot,0,1)$, as described above, actually is a Boolean Algebra, that $\mathscr{G}(R,\wedge,\vee,\neg,0,1)$, also as described above, is a Boolean ring, that $\varphi: (R,\oplus_R,\cdot_R,0,1)\rightarrow (S,\oplus_S,\cdot_S,0,1)$ is a morphism in $\mathbf{BoolRing}$ implies that $\varphi: \mathscr{F}(R,\oplus_R,\cdot_R,0,1)\rightarrow \mathscr{F}(S,\oplus_S,\cdot_S,0,1)$ is a morphism in $\mathbf{BoolAlg}$, and that $\varphi: (R,\wedge_R,\vee_R,\neg_R,0,1)\rightarrow (S,\wedge_S,\vee_S,\neg_S,0,1)$ is a morphism in $\mathbf{BoolAlg}$ implies that $\varphi: \mathscr{G}(R,\wedge_R,\vee_R,\neg_R,0,1)\rightarrow \mathscr{G}(S,\wedge_S,\vee_S,\neg_S,0,1)$ is a morphism in $\mathbf{BoolRing}$.
Well-Definedness of $\mathscr{F}$ on objects
Suppose you have a Boolean ring $(R,\oplus, \cdot,0,1)$. Then we define $\vee$, $\wedge$, and $\neg$ as follows: \begin{equation*} x\wedge y = x\cdot y \quad \text{and} \quad x\vee y = x\oplus y \oplus x\cdot y \quad \text{and} \quad \neg x =1\oplus x. \end{equation*} Then $(R,\wedge,\vee,\neg,0,1)$ can be checked to be a Boolean algebra; certainly $\wedge$ and $\vee$ are commutative since $\cdot$ and $\oplus$ are by definition commutative. Similarly, $\wedge$ is clearly associative because $\cdot$ is by definition associative. $\vee$ is associative, for given $x,y,z\in R$, we have \begin{align} x\vee (y\vee z) & =x\vee (y\oplus z \oplus (y\cdot z)) \\ & =x\oplus (y\oplus z\oplus (y\cdot z)) \oplus (x\cdot (y\oplus z\oplus (y\cdot z))) \\ & =x\oplus y \oplus z \oplus (y\cdot z)\oplus (x\cdot y)\oplus (x\cdot z)\oplus (x\cdot (y\cdot z)) \\ & = (x\oplus y \oplus (x\cdot y))\oplus z\oplus (x\oplus y \oplus (x\cdot y))\cdot z \\ & = (x\vee y)\vee z \end{align} For distributivity of $\vee$ over $\wedge$, we have \begin{align} (x\vee y) \wedge (x\vee z) & = (x\oplus y \oplus (x\cdot y))\cdot (x\oplus z \oplus (x\cdot z)) \\ & = x^2\oplus (x\cdot z)\oplus (x^2\cdot z)\oplus (x\cdot y)\oplus (y\cdot z)\oplus (x\cdot y\cdot z)\oplus (x^2\cdot y)\oplus \\ & \quad \quad \oplus (x\cdot y\cdot z) \oplus (x^2\cdot y \cdot z) \\ & = x\oplus (y\cdot z)\oplus (x\cdot (y\cdot z)) \\ & = x \vee (y\cdot z) \end{align} where I have used the fact that $x\oplus x=0$ in a Boolean ring. For distributivity of $\wedge$ over $\vee$, we have \begin{align} (x\wedge y)\vee (x\wedge z) & = (x\cdot y)\oplus (x\cdot z)\oplus (x\cdot y\cdot x\cdot z) \\ & = (x\cdot y)\oplus (x\cdot z) \oplus (x\cdot y\cdot z) \\ & = (x\cdot (y\oplus z))\oplus (x\cdot (y\cdot z)) \\ & = x\cdot (y\oplus z \oplus (y\cdot z)) \\ & = x\wedge (y\vee z) \end{align} Next, we prove the identity laws: \begin{equation*} x \vee 0 = x\oplus 0 \oplus (x\cdot 0)=x\oplus 0 \oplus 0 = x \quad \text{and} \quad x\wedge 1 = x\cdot 1= x \end{equation*} Finally, we have complements: \begin{equation*} x \vee (\neg x)=x\vee (1\oplus x)=x\oplus (1\oplus x)\oplus (x\cdot (1\oplus x))=1\oplus (x\oplus x)=1 \end{equation*} and \begin{equation*} x\wedge (\neg x) = x\cdot (1\oplus x)= x\oplus x^2=x\oplus x = 0 \end{equation*} thus showing that $(R,\wedge, \vee, \neg, 0,1)$ is a Boolean Algebra.
Well-definedness of $\mathscr{G}$ on objects
Conversely, given a Boolean Algebra $(R,\wedge, \vee, \neg, 0,1)$, we define \begin{equation*} x\cdot y = x\wedge y \quad \text{and} \quad x\oplus y = (x\vee y)\wedge \neg (x\wedge y). \end{equation*} Naturally, we must check that $(R,\oplus, \cdot, 0,1)$ is a Boolean ring. Firstly, $\cdot$ is clearly associative (and also commutative, which while not necessary to show that $(R,\oplus,\cdot,0,1)$ is a Boolean ring, is helpful for showing distributivity) and $\oplus$ is clearly commutative. By the idempotent law, we also see that $x\cdot x= x\vee x= x$. $1$ also automatically an identity of $\cdot$, since it is an identity of $\vee$. $0$ is an identity of $\oplus$ for $x\oplus 0 = (x \vee 0 )\wedge \neg (x\wedge 0)= x\wedge \neg 0=x\wedge 1 = x$. Thus, all that remains is to show that $\oplus$ is associative and that $\cdot$ distributes over $\oplus$. \begin{align} (x\cdot y)\oplus (x\cdot z) & = (x\wedge y) \oplus (x\wedge z) \\ & =((x\wedge y)\vee(x\wedge z))\wedge \neg ((x\wedge y)\wedge (x\wedge z)) \\ & =(x\wedge (y\vee z)) \wedge \neg (x\wedge (y\wedge z)) \\ & = (x\wedge (y\vee z)) \wedge (\neg x \vee \neg (y\wedge z)) \\ & = ((x\wedge (y\vee z)) \wedge (\neg x)) \vee ((x\wedge (y\vee z)) \wedge \neg (y\wedge z)) \\ & = (0\wedge (y\vee z)) \vee ((x\wedge (y\vee z)) \wedge \neg (y\wedge z)) \\ & = 0 \vee ((x\wedge (y\vee z))\wedge \neg (y\wedge z)) \\ & = (x\wedge (y\vee z))\wedge \neg (y\wedge z) \\ & = x\wedge ((y\vee z)\wedge \neg (y\wedge z)) \\ & = x\cdot (y\oplus z) \end{align} showing distributivity of $\cdot$ over $\oplus$ (this shows both right and left distributivity by the fact that $\cdot$ is commutative). Finally, to show that $\oplus$ is associative, we have \begin{align} x\oplus (y\oplus z) & = x\oplus ((y\vee z) \wedge \neg (y\wedge z)) \\ & = (x\vee ((y\vee z) \wedge \neg (y\wedge z)))\wedge \neg (x\wedge ((y\vee z) \wedge \neg (y\wedge z))) \\ & = (x\vee ((y\vee z) \wedge (\neg y\vee \neg z)))\wedge \neg (x\wedge ((y\vee z) \wedge \neg (y\wedge z))) \\ & = ((x\vee y\vee z) \wedge (x\vee \neg y \vee \neg z)) \wedge \neg ( x\wedge (y\vee z) \wedge \neg (y\wedge z)) \\ & = ((x \vee y \vee z) \wedge (x\vee \neg y \vee \neg z)) \wedge (\neg x \vee \neg (y\vee z) \vee (y\wedge z)) \\ & = (x\vee y \vee z) \wedge (((x \vee \neg y \vee \neg z) \wedge \neg x) \\ & \quad \quad \vee (( x \vee \neg y \vee \neg z) \wedge (\neg y \wedge \neg z)) \vee (( x\vee \neg y \vee \neg z) \wedge (y \wedge z))) \\ & = (x\vee y \vee z) \wedge (((x \wedge \neg x) \vee (\neg y \wedge \neg x) \vee ( \neg z \wedge \neg x)) \\ & \quad \quad \vee ((x\wedge \neg y \wedge \neg z) \vee (\neg y \wedge \neg y \wedge \neg z)\vee (\neg z \wedge \neg z \wedge \neg y)) \\ & \quad \quad \vee (( x\wedge y \wedge z) \vee (\neg y \wedge y \wedge z) \vee (\neg z \wedge y \wedge z))) \\ & = (x\vee y \vee z) \wedge (( \neg y \wedge \neg x) \vee (\neg z \wedge \neg x) \vee (x\wedge \neg y \wedge \neg z) \vee (\neg y \wedge \neg z) \vee (\neg y \wedge \neg z)\vee (x\wedge y \wedge z)) \\ & = (x\wedge y \wedge z) \vee (x\wedge \neg \wedge \neg z) \vee (\neg x \wedge y \wedge \neg z) \vee (\neg x \wedge \neg y \wedge z) \end{align} This final form is symmetric in $x,y,z$, so that because of the commutativity of $\oplus$, we find that $x\oplus (y\oplus z) =(y\oplus z) \oplus x = (\sigma(y) \oplus \sigma(z) ) \oplus \sigma(x)$, where $\sigma$ is any permutation of $x,y,z$. In particular, taking $\sigma(y)=x$, $\sigma(z)=y$, and $\sigma(x)=z$, we find that $x\oplus (y\oplus z)=(x\oplus y)\oplus z$. This completes the proof that $(R,\oplus, \cdot, 0,1)$ is a Boolean ring.
Well-Definedness of $\mathscr{F}$ and $\mathscr{G}$ on morphisms
Finally, we must describe how the proposed functor acts on morphisms: it sends a morphism $\varphi: R\rightarrow S$ in either category to $\varphi: R\rightarrow S$ in the other category. To make sure that this is well-defined, we must show that $\varphi: R\rightarrow S$ is a ring homomorphism if and only if it is also a Boolean algebra homomorphism, with the ring structure and Boolean algebra structures as above. To this effect, suppose that $\varphi: R\rightarrow S$ is a ring homomorphism between the Boolean rings $(R,\oplus_R, \cdot_R, 0,1)$ and $(S, \oplus_S, \cdot_S,0,1)$. Then in particular $\varphi(0)=0$ and $\varphi(1)=1$ (the former a fact about group homomorphisms, the latter by definition). Additionally, \begin{equation*} \varphi(x \wedge_R y) = \varphi (x \cdot_R y)=\varphi(x) \cdot_S \varphi(y) = \varphi(x) \wedge_S \varphi(y) \end{equation*} so $\varphi$ preserves meets. Finally \begin{equation*} \varphi (x \vee_R y) = \varphi(x \oplus_R y\oplus_R (x\cdot_R y)) = \varphi(x)\oplus_S \varphi(y) \oplus_S (\varphi(x) \cdot_S \varphi(y)) = \varphi(x) \vee_S \varphi(y) \end{equation*} so $\varphi$ preserves joins, and hence is a Boolean algebra homomorphism. All that remains is to show that the reverse holds true: suppose $\varphi: R\rightarrow S$ is a Boolean algebra homomorphism between Boolean algebras $(R,\wedge_R, \vee_R, \neg_R, 0,1)$ and $(S, \wedge_S, \vee_S, \neg_S, 0,1)$. Then by definition $\varphi(1)=1$. Next, \begin{equation*} \varphi(x \cdot_R y) = \varphi(x \wedge_R y)=\varphi(x)\wedge_S \varphi(y)= \varphi(x) \cdot_S \varphi(y) \end{equation*} so $\varphi$ preserves the product, and \begin{equation*} \varphi( x \oplus_R y) = \varphi((x\vee_R y) \wedge_R \neg_R (x\wedge_R y))=(\varphi(x) \vee_S \varphi(y))\wedge_S \neg_S (\varphi(x) \wedge_S \varphi(y))=\varphi(x) \oplus_S \varphi(y) \end{equation*} so $\varphi$ also preserves addition, and hence is a ring homomorphism.
$\mathscr{F}$ and $\mathscr{G}$ are inverses
To see that this is an inverse, suppose we have a Boolean ring $(R,\oplus, \cdot, 0,1)$, which gives rise via $\mathscr{F}$ to a Boolean algebra $(R, \wedge, \vee,\neg, 0,1)$ as above. Then denote by $(R, \boxplus,\ast,0,1)$ the Boolean ring that comes from this Boolean algebra via $\mathscr{G}$. Then we find that $x \cdot y = x\wedge y = x\ast y$ for every $x,y\in R$, so $\cdot=\ast$. Next, \begin{align} x \boxplus y & = (x \vee y) \wedge \neg (x \wedge y) \\ & = (x\vee y) \cdot (1\oplus (x\cdot y)) \\ & =(x\oplus y \oplus (x\cdot y))\cdot (1\oplus (x\cdot y)) \\ & =(x\cdot 1) \oplus (x^2\cdot y) \oplus (y\cdot 1) \oplus (x\cdot y^2)\oplus (x\cdot y \cdot 1) \oplus (x\cdot y)^2 \\ & = x\oplus (x^2\cdot y) \oplus y \oplus (x\cdot y^2)\oplus (x\cdot y) \oplus (x\cdot y)^2 \\ & = x\oplus (x\cdot y)\oplus y \oplus (x\cdot y) \oplus (x\cdot y)\oplus (x\cdot y) \\ & = x\oplus y \end{align} for every $x,y\in R$, so $\oplus = \boxplus$.
Likewise, suppose we have a Boolean algebra $(R,\wedge, \vee, \neg, 0,1)$, which gives rise via $\mathscr{G}$ to a Boolean ring $(R,\oplus, \cdot, 0,1)$, which itself gives rise via $\mathscr{F}$ to the Boolean algebra $(R, \&, |,\sim , 0,1)$. Then $x\wedge y = x\cdot y = x \& y$ for every $x,y\in R$, so $\wedge= \&$. Next, \begin{align} x | y & = x \oplus y \oplus (x\cdot y) \\ & = x\oplus (y\oplus (x\wedge y)) \\ & = x\oplus ((y\vee (x\wedge y))\wedge \neg (y \wedge (x\wedge y))) \\ & = x \oplus (y \wedge \neg (y \wedge x)) \\ & = x \oplus ((y \wedge \neg y) \vee (y\wedge \neg x)) \\ & = x \oplus (y\wedge \neg x) \\ & = (x \vee (y\wedge \neg x)) \wedge \neg (x\wedge (y\wedge \neg x)) \\ & = ((x\vee y) \wedge (x\vee \neg x)) \wedge 1 \\ & = x\vee y \end{align} for every $x,y\in R$, so $\vee=|$. Lastly, $\sim x = 1\oplus x=(1\vee x) \wedge \neg (1\wedge x)=1 \wedge \neg x=\neg x$, so $\neg = \sim $.
Functoriality of $\mathscr{F}$ and $\mathscr{G}$ is obvious, since they are identities on the underlying sets and functions.