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.


1 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.

  • $\begingroup$ Well, what I really want to find is two binary operators on $\{0,1\}$, that will turn it into a Boolean Algebra as well as a Galois field. AND,OR pair makes it a Boolean Algebra but not a Galois field. XOR,AND pair makes it a Galois field but not a Boolean Algebra. Which pair of binary operators would make $\{0,1\}$ a Boolean Algebra as well as a Galois field? $\endgroup$ May 21, 2016 at 3:08
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    $\begingroup$ @samur: this is impossible. There are exactly two Boolean algebra structures and exactly two field structures (the other one is given by switching $0$ and $1$ in both cases), and you can just check that none of them match up. $\endgroup$ May 21, 2016 at 7:25
  • $\begingroup$ Thanks @Qiaochu !So does that mean there is no field of order two on which a Boolean Algebra can be defined? $\endgroup$ May 21, 2016 at 14:20

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