# ra.rings and algebras – Birkhoff & # 39; s completeeness theorem made in practice Birkhoff's exhaustiveness theorem (see Theorem 14.19 here) states that an equation that is true in all models of an algebraic theory can be proven in equational logic.

Question. Proof of Birkhoff's completeness theorem produce for each specific equation a proof in equation logic? If so, can you demonstrate this with an instructive example?

In fact, I suspect the answer is "no", but I'm not entirely sure.

Consider the following well-known statement: If $$R$$ is a ring in which each element $$r in R$$ satisfied $$r ^ 2 = r$$ (that is to say. $$R$$ is boolean), then $$R$$ is commutative. There is the following exaggerated evidence: $$R$$ is reduced, so a sub-direct product of domains $$R_i$$. From the cards $$R to R_i$$ are surjective, each domain $$R_i$$ satisfies the same equation and must then be isomorphic to $$mathbb {F} _2$$. In particular, $$R_i$$ is commutative. Since $$R to prod_ {i in I} R_i$$ is injective, it follows that $$R$$ is commutative. So, according to Birkhoff's theorem, it must be a little equational proof of this. My question is not what an equation proof looks like – this is just a basic algebra exercise. I would like to see how (if possible) it can be extracted from the proof of overkill.

I think the proof of Birkhoff's theorem in this particular case works as follows: Let us consider the free ring on two generators $$mathbb {Z} langle X, Y rangle$$ and take the quotient from relationships $$r ^ 2 = r$$ for all elements $$r$$. This is the free boolean ring $$R$$ on two generators. By exaggerated evidence, $$XY = YX$$ is holding $$R$$. It means that $$XY = YX$$ can be derived from relationships $$r ^ 2 = r$$ in $$mathbb {Z} langle X, Y rangle$$. But we don't to have a derivation, right? How to produce, for example, the following equation?

begin {align *} XY – YX & = bigl ((X + Y) ^ 2 – (X + Y) bigr) – bigl (X ^ 2-X bigr) – bigl (Y ^ 2-Y bigr) \ & phantom {=} + bigl ((YX) ^ 2 – (YX) bigr) – bigl ((- YX) ^ 2 – (-YX) bigr), end {align *}
As far as I know, we are not even guaranteed a priori to have a proof that works without the axiom of choice, as this is used in the structure theorem for reduced rings in the proof of overkill? (At first, this can be confusing, because the axiom of choice is certainly not allowed in equational proof, but in reality the axiom of choice is used to show the existence of an equational proof.)

I am interested in the more complicated applications of Birkhoff's theorem. This is just an example to start. You can also choose other examples if they are more informative. Posted on