Galois field of order 2 constituting a Boolean algebra

Question

We know that the the set $\{0,1\}$ constitutes a Boolean Algebra over the usual $OR$ and $AND$ operations. However, because of the lack of an additive inverse for $1$ this does not produce a Galois field. Also, we know that the above set forms $GF(2)$ when the operators are $XOR$ and $AND$. However, this definition doesn't follow the De Morgan's Theorem and hence not a Boolean Algebra. Can anyone think of a definition of the operators which works both ways? That is, a Galois field of order 2 constituting a Boolean Algebra.

Answer 1

I don't really understand what you're asking; it sounds to me like you've already shown that the Boolean algebra and field structures on $\{ 0, 1 \}$ are different, and you want them to somehow be the same anyway?

In any case, you might be interested in the notion of a Boolean ring. This is an alternative axiomatization of Boolean algebras (using AND and XOR instead of AND and OR), and the Boolean ring structure on $\{ 0, 1 \}$ is also the usual field structure on $\mathbb{F}_2$. Boolean algebras and Boolean rings are "the same thing" in a very strong sense, even though they involve different operations satisfying different axioms.