Exercise 1.24

Question

Let $L$ be a lattice, in which the $\sup$ and $\inf$ of two elements $a,b$ are denoted by $a \lor b$ and $a \land b$ respectively. $L$ is a \emph{Boolean lattice} (or \emph{Boolean algebra}) if
  \begin{enumerate}[label=\roman*)]
    \item $L$ has a least element and a greatest element (denoted by $0,1$ respectively).
    \item Each of $\lor$, $\land$ is distributive over the other.
    \item Each $a \in L$ has a unique ``complement'' $a' \in L$ such that $a \lor a' = 1$ and $a \land a' = 0$.
  \end{enumerate}
  (For example, the set of all subsets of a set, ordered by inclusion, is a Boolean lattice).
  \par Let $L$ be a Boolean lattice. Define addition and multiplication in $L$ by the rules
  \begin{equation*}
    a + b = (a \land b') \lor (a' \land b), \qquad ab = a \land b.
  \end{equation*}
  Verify in this way $L$ becomes a Boolean ring, say $A(L)$.
  \par Conversely, starting from a Boolean ring $A$, define an ordering on $A$ as follows: $a \le b$ means that $a = ab$. Show that, with respect to this ordering, $A$ is a Boolean lattice. [The $\sup$ and $\inf$ are given by $a \lor b = a + b + ab$ and $a \land b = ab$, and the complement by $a' = 1 - a$.] In this way we obtain a one-to-one correspondence between (isomorphism classes of) Boolean rings and (isomorphism classes of) Boolean lattices.

Amit Awasthi · Accepted Answer

\begin{proof}[Proof that $A(L)$ is a Boolean ring]
  $0 \land 0 = 0$ implies $0' = 0$. So, $L$ is an abelian group with respect to addition since it is abelian by definition of $\sup,\inf$, has inverses by $iii)$, and has identity $0$ since $a + 0 = (a \land 0') \lor (a' \land 0) = 0 \lor 0 = 0$ by definition of $\sup,\inf$. Multiplication is associative by definition of $\inf$, distributes over addition by $ii)$, is commutative by definition of $\inf$, and has identity $1$ since $1a = 1 \lor a = a$ by definition of $\sup$.
  \par $L$ is a Boolean ring since $x^2 = x \land x = x$ for all $x \in L$ by definition of $\inf$.
\end{proof}
\begin{proof}[Proof that $L(A)$ is a Boolean lattice]
  $A$ is a poset under $\le$ since $a \le a$ is the condition $a = a^2$, $a \le b$ and $b \le a$ imply both $a = ab$ and $b = ab$ so $a = b$, and since $a \le b$ and $b \le c$ imply both $a = ab$ and $b = bc$ so $a = ac$, i.e., $a \le c$. $\sup,\inf$ define upper and lower bounds for two elements in $A$, respectively, and so it remains to show they are least upper and greatest lowest bounds, respectively. So suppose $c < \sup\{a,b\}$ but both $a \le c$ and $b \le c$. Then, $a = ac$ and $b = bc$, so $\sup\{a,b\} = a + b + ab = ac + bc + abc = \sup\{a,b\}c$ implies $c = 1$, a contradiction. Suppose $1 
e \inf\{a,b\} < c$ but both $c \le a$ and $c \le a$; note the $\inf\{a,b\} = 1$ case is trivial since this implies $a=b=1$. Then, $c = ac$ and $c = bc$, so $\inf\{a,b\}c = abc = c$ implies $c = 0$, a contradiction.
  \par $i)$. $0$ is the least element since $a \land 0 = 0$ for any $a \in A$. $1$ is the greatest element since $a \lor 1 = a + 1 + a = 1$ for any $a \in A$ by Exercise $1.11i)$.
  \par $ii)$. Using Exercise $1.11i)$,
  \begin{align*}
    a \lor (a \land b) &= a + ab + a^2b = a + ab + ab = a,\
    a \land (a \lor b) &= a(a \lor b) = a(a + b + ab) = a^2 + ab + a^2b = a + ab + ab = a.
  \end{align*}
  \par $iii)$. We see $a \land a' = aa' = a(1-a) = 1$ and $a \lor a' = a + (1-a) + a(1-a) = 1 + 1 = 0$ by Exercise $1.11i)$, and $a'$ is unique since multiplicative inverses are unique.
\end{proof}
\begin{proof}[Proof of correspondence]
  The operations $L \mapsto A(L)$ and $A \mapsto L(A)$ are bijections on sets, and so to show we have a bijection on isomorphism classes, it suffices to show that the lattice structure on $L$ and $L(A(L))$ are the same, and similarly the ring structure on $A$ and $A(L(A))$ are the same.
  \par If $L$ is a Boolean lattice, then $A(L)$ has $ab = a \land b$, and so $L(A(L))$ has $a \leqslant b$ defined by $a = ab = a \land b$. Now $a \leqslant b$ if and only if $a = a \land b$ if and only if $a \le b$.
  \par If $A$ is a Boolean ring, then $L(A)$ has $a \lor b = a + b + ab$ and $a \land b = ab$. Letting $\oplus,\otimes$ be the operations on $A(L(A))$, $a \otimes b = a \land b = ab$, and using Exercise $1.11i)$,
  \begin{align*}
    a \oplus b &= (a \land b') \lor (a' \land b) = ab' \lor a'b = ab' + a'b + aa'bb'\
    &= a(1-b) + (1-a)b + a(1-a)b(1-b)\
    &= a - ab + b - ab + (a-a^2)(b-b^2) = a + b.\qedhere
  \end{align*}
\end{proof}

Exercise 1.24

Answers

Comments

Exercise 1.24

Answers

Comments

Add answer