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\newtheorem{prop}{Proposition} \newtheorem{cor}{Corollary} \newtheorem*{utheorem}{Theorem} \newtheorem*{ulemma}{Lemma} \newtheorem*{uprop}{Proposition} \newtheorem*{ucor}{Corollary} \theoremstyle{definition} \newtheorem{defn}{Definition} \newtheorem{example}{Example} \newtheorem*{udefn}{Definition} \newtheorem*{uexample}{Example} \theoremstyle{remark} \newtheorem{remark}{Remark} \newtheorem{note}{Note} \newtheorem*{uremark}{Remark} \newtheorem*{unote}{Note} %------------------------------------------------------------------- \begin{document} %------------------------------------------------------------------- \section*{Boolean ring} \hypertarget{boolean_rings}{}\section*{{Boolean rings}}\label{boolean_rings} \noindent\hyperlink{definitions}{Definitions}\dotfill \pageref*{definitions} \linebreak \noindent\hyperlink{properties}{Properties}\dotfill \pageref*{properties} \linebreak \noindent\hyperlink{examples}{Examples}\dotfill \pageref*{examples} \linebreak \noindent\hyperlink{terminology}{Terminology}\dotfill \pageref*{terminology} \linebreak \noindent\hyperlink{analogues}{Analogues}\dotfill \pageref*{analogues} \linebreak \hypertarget{definitions}{}\subsection*{{Definitions}}\label{definitions} A [[ring with unit]] $R$ is \textbf{Boolean} if the operation of multiplication is [[idempotent]]; that is, $x^2 = x$ for every element $x$. Although the terminology would make sense for rings without unit, the common usage assumes a unit. Boolean rings and the [[ring]] [[homomorphisms]] between them form a [[category]] $Boo Rng$. \hypertarget{properties}{}\subsection*{{Properties}}\label{properties} \begin{itemize}% \item $R$ has characteristic $2$ (meaning that $x + x = 0$ for all $x$):\begin{displaymath} 2 x = 4 x - 2 x = 4 x^2 - 2 x = (2 x)^2 - 2 x = 2 x - 2 x = 0 . \end{displaymath} \item $R$ is commutative (meaning that $x y = y x$ for all $x, y$):\begin{displaymath} y x = x + y - x - y + y x = (x + y)^2 - x^2 - y^2 + y x = x^2 + x y + y x + y^2 - x^2 - y^2 + y x = x y + 2 y x = x y . \end{displaymath} \end{itemize} Define $x \vee y$ to mean $x + x y + y$. Then: \begin{itemize}% \item $\vee$ is commutative (as it would be in any commutative ring) and idempotent:\begin{displaymath} x \vee x = x + x^2 + x = 3 x = x . \end{displaymath} \item The absorption law ($x \vee x y = x$) also holds:\begin{displaymath} x \vee x y = x + x^2 y + x y = x + 2 x y = x . \end{displaymath} We could now prove the other absoprtion law to conclude that $R$ is a [[lattice]] using multiplication as [[meet]] and $\vee$ as [[join]]. \item But in fact, we can skip that step since it follows the distributive law ($x (y \vee z) = x y \vee x z$):\begin{displaymath} x (y \vee z) = x (y + y z + z) = x y + x y z + x z = x y + x^2 y z + x z = x y + (x y) (x z) + x z = x y \vee x z . \end{displaymath} Thus $R$ is a [[distributive lattice]]. \end{itemize} Next define $\neg{x}$ to be $x + 1$. Then: \begin{itemize}% \item $\neg{x}$ is a pseudocomplement of $x$ (meaning that $x (\neg{x}) = 0$):\begin{displaymath} x (\neg{x}) = x (x + 1) = x^2 + x = 2 x = 0 . \end{displaymath} By relativising from $x + 1$ to $x y + x + 1$, we can show that $R$ is a [[Heyting algebra]]. \item But don't bother, because $\neg{x}$ is also an op-pseudocomplement of $x$:\begin{displaymath} x \vee \neg{x} = x + x (x + 1) + (x + 1) = x^2 + 3 x + 1 = x + x + 1 = 1 . \end{displaymath} Therefore, $\neg{x}$ is a [[complement]] of $x$, and $R$ is a [[Boolean algebra]]. \end{itemize} Conversely, starting with a Boolean algebra $R$ (with the meet written multiplicatively), let $x + y$ be $x (\neg{y}) \vee (\neg{x}) y$ (which is called [[exclusive disjunction]] in $\{\top,\bot\}$ and [[symmetric difference]] in $2^X$). Then $R$ is a Boolean ring. In fact, we have: Boolean rings and Boolean algebras are equivalent. This extends to an equivalence of [[concrete category|concrete categories]]; that is, given the underlying [[set]] $R$, the set of Boolean ring structures on $R$ is [[natural isomorphism|naturally]] (in $R$) [[bijection|bijective]] with the set of Boolean algebra structures on $R$. Here is a very convenient result: although a Boolean ring $R$ is a [[rig]] in two different ways (as a ring or as a distributive lattice), these have the same concept of [[ideal]]! \hypertarget{examples}{}\subsection*{{Examples}}\label{examples} The most common example is the [[power set]] $P(S)$ of any set $S$. It is a Boolean ring with [[symmetric difference]] as the addition and the intersection of sets as the multiplication. \hypertarget{terminology}{}\subsection*{{Terminology}}\label{terminology} Back in the day, the term `ring' meant (more often than now is the case) a possibly \emph{non}unital ring; that is a [[semigroup]], rather than a [[monoid]], in [[Ab]]. This terminology applied also to Boolean rings, and it changed even more slowly. Thus older books will make a distinction between `Boolean ring' (meaning an idempotent semigroup in $Ab$) and `Boolean algebra' (meaning an idempotent monoid in $Ab$), in addition to (or even instead of) the difference between $+$ and $\vee$ as fundamental operation. This distinction survives most in the terminology of $\sigma$-[[sigma-ring|rings]] and $\sigma$-[[sigma-algebra|algebras]]. \begin{remark} \label{}\hypertarget{}{} We pause to note that ``idempotent monoid'' doesn't make sense a priori in a general monoidal category: generally speaking the idempotency axiom would be expressed by an equation \begin{displaymath} 1_M = \left(M \stackrel{\delta_M}{\to} M \otimes M \stackrel{mult}{\to} M \right) \end{displaymath} where $\delta_M$ is an appropriate diagonal map, not generally available for monoidal categories. But in a \emph{[[concrete category|concrete]]} monoidal category $\mathbf{M}$ where the underlying-set functor $U: \mathbf{M} \to Set$ is [[lax monoidal functor|lax monoidal]], the meaning is that the evident equation holds: \begin{displaymath} 1_{U(M)} = \left(U(M) \stackrel{\delta}{\to} U(M) \times U(M) \stackrel{\lambda}{\to} U(M \otimes M) \stackrel{U(mult)}{\to} U(M) \right) \end{displaymath} where $\lambda$ is a lax monoidal constraint. \end{remark} \hypertarget{analogues}{}\subsection*{{Analogues}}\label{analogues} Inasmuch as a [[semilattice]] is a commutative idempotent monoid, a Boolean ring may be defined as a semilattice in $Ab$. However, with Boolean rings, we do not need to hypothesize commutativity; it follows. That is, any idempotent monoid in $Ab$ is commutative; indeed, any idempotent [[magma]] in $Ab$ is commutative. [[!redirects Boolean ring]] [[!redirects boolean ring]] [[!redirects Boolean rings]] [[!redirects boolean rings]] [[!redirects Boo Rng]] [[!redirects BooRng]] [[!redirects Boo Ring]] [[!redirects BooRing]] \end{document}