After trying to make sense of first order logic from an algebraic point of view I started to read about boolean algebras (similar to the explanations given here: wikipedia on boolean algebras. I also read through this post: Are Elementary Algebra and Boolean Algebra Algebras over a Ring (or Field)? (not necessarily all linked pages in full).

As I understand it, taking the initial boolean algebra, call it $B$, as an example, it is an algebra over a ring, or more specifically over the field $\mathbb{Z}_2$ (which is the underlying set for the whole boolean algebra, so it is an algebra over itself in a sense). The scalar multiplication then coincides with the usual multiplication in $B$. As an $R$-module $M$ becomes an $R$-algebra when $M$ is a ring itself this means that $B$ would be a ring. My questions are these:

Q1.) Is everything I mentioned above correctly understood? If not, would you please explain what I got wrong and/or link to other resources that give a thorough explanation?

Q2.) In the ring of two elements the multiplication table corresponds to the AND operation but the addition corresponds to the XOR operation and not to OR (otherwise $\mathbb{Z}_2$ would obviously not become an additive group). Have I misunderstood something fundamental here or don't we use the OR operation in the initial boolean algebra? And if so, what algebraic structure could this be given?

My level is undergraduate with an introductory course in abstract algebra covering the most fundamental parts of "Abstract Algebra" by John A. Beachy and William D. Blair, not including Galois theory.


Have I misunderstood something fundamental here or don't we use the OR operation in the initial boolean algebra?

Consider the $\vee$-$\wedge$ definition of a Boolean algebra. It turns out that you can turn this into a real boolean ring by defining $a\cdot b=a\wedge b$ and $a+b=(a\vee b)\wedge\neg(a\wedge b)$. As a result, one of the boolean algebra laws says that $a^2=a$ for any $a$ in the algebra.

In the context I've given above, $\vee$ acts like $OR$ and $\wedge$ acts like AND. The "new" addition operation we made acts like XOR. If you want to examine boolean algebras as $F_2$-algebras, then I think you want to shift to the version using the $+$ mentioned above, so that $a+a=0$ for all $a$ in the ring (contrasting with the previous $a\vee a=a$). This is where the characteristic of the field $F_2$ comes into play.

it is an algebra over [...] the field $\Bbb Z_2$

Yes, that's necessary but not sufficient. For example, the ring of polynomials over the field of two elements $F_2[x]$ is an $F_2$ algebra, but $x^2\neq x$.

You can prove using the technique here, however, that every boolean algebra $(B, +,\cdot)$ is isomorphic to a subring of $\prod_{i\in I}F_2$ for some index set $I$. This is an $F_2$-algebra, yes, but it is only a small part of the class of $F_2$-algebras.


We have seen two fundamental ways to study boolean algebras: using $\vee/\wedge$ or using $+/\cdot$. The first one conforms to more of the way we think with AND and OR, and the second one creates a nice associative ring structure that is easy to analyze. You can freely get between the two with these conversions:

$$a\wedge b=a\cdot b \\ a+b=(a\vee b)\wedge\neg(a\wedge b)\\ a\vee b=a\cdot b+a+b $$

  • $\begingroup$ I think I'm starting to get a clearer picture. To conclude, do we have any algebraic structure ( preferably among the ones mentioned here: en.wikipedia.org/wiki/Algebraic_structure ) that we can put on the boolean algebra defined using OR instead of XOR, i.e using $\vee$ / $\wedge$? Or does this just create something "else" completely? $\endgroup$ – Christopher.L Jun 16 '16 at 18:08
  • $\begingroup$ @Christopher.L We've just discussed how $\cdot/+$ makes it into an $F_2$-algebra, and the structure using $\vee/\wedge$ is at least a lattice. $\endgroup$ – rschwieb Jun 16 '16 at 19:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.