# Definitionally equivalence between Boolean algebras and Boolean rings

On page 17, Introduction to Boolean Algebras,Steven Givant,Paul Halmos(2000):

Motivated by this set-theoretic example, we can introduce into every Boolean algebra $A$ operations of addition and multiplication very much like symmetric difference and intersection; just define: $$p+q=(p\land q')\lor (p' \land q)$$$$p\cdot q =p \land q$$ Under these operations, together with 0 and 1 (the zero and unit of the Boolean algebra), $A$ becomes a Boolean ring. Conversely, every Boolean ring can be turned into a Boolean algebra with the same zero and unit; just define operations of join, meet, and complement by$$p \lor q = p+q+p \cdot q$$ $$p \land q = p \cdot q$$ $$p'= p + 1$$

It seems to me with the help of the above two sets of equations, every propostion involving Boolean algebra can be translated into the language of Boolean ring without difficulty, and vice versa. But it's not the case, since:

The point of view of Boolean algebras makes it possible to give a simple and natural description of an example (due to Sheffer [54]) that would be quite awkward to treat from the point of view of Boolean rings. Let $m$ be a positive integer, and let $A$ be the set of all positive integral divisors of $m$. Define the Boolean structure of $A$ by the equations $$0=1$$$$1=m$$$$p \land q = \mathbb{gcd}\{p, q\}$$$$p \lor q = \mathbb{lcm}\{p, q\}$$$$p'=m/p$$ It turns out that, with the distinguished elements and operations so defined, A forms a Boolean algebra if and only if $m$ is square-free (that is, m is not divisible by the square of any prime).

I have no clue as to why in this case, it "would be quite awkward to treat from the point of view of Boolean rings".

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Therefore one expects that talking about Boolean algebras will make sense for approaching order-theoretic examples like the lattice of divisors of a squarefree positive integer, and talking about Boolean rings will make sense for approaching commutative algebraic examples like the space of continuous functions from a topological space $X$ to $\mathbb{F}_2$ (with the discrete topology).