# Does the variety of Boolean Algebras contain no proper nontrivial subvarieties/subquasivarieties?

Consider the variety, in the sense of universal algebra, of Boolean Algebras in the language $$\{\cup,\cap,',0,1\}$$, where $$'$$ represents complementation, and the other symbols are well known. I conjecture that the only proper subvariety of Boolean Algebras is the trivial variety of one-element algebras. In fact, I make the stronger conjecture that the only proper subquasivariety of Boolean Algebras is, again, the class of one-element algebras. Are either of these conjectures true?

This is true. By Stone's theorem, any Boolean algebra is a subdirect product of copies of the two element Boolean algebra. Now suppose that $$\mathbf{K}$$ is a nontrivial variety/quasi variety of Boolean algebras. These are closed under sub algebras, and under products, hence, under subdirect products; moreover, the two element Boolean algebra is a sub algebra of any Boolean algebra. Hence $$\mathbf{K}$$ is simply all Boolean algebras.
• (Your penultimate sentence needs "nontrivial.") More generally, if ${\bf V}$ is the variety generated by an algebra $A$ and every nontrivial $B\in{\bf V}$ has $A$ as a subalgebra, then ${\bf V}$ has no nontrivial proper subvarieties. Commented Jul 20 at 18:55