Equational axiomatization and first-order axiomatization of the class of fields united with the class of Boolean algebras

Question

This is really two questions in one, one about universal algebra and one about model theory. Let our signature be $\{+,*,-,0,1\}$. I have recently realized that both fields and Boolean algebras share that same signature. If we interpret $+$ as addition, $*$ as multiplication, $-$ as additive inverse, we get fields, and if we interpret $+$ as union, $*$ as intersection, and $-$ as complementation, we get Boolean algebras. So, now, consider the class $K$ which is the union of the class of fields along with the class of Boolean algebras. My first question is, is there a finite equational axiomatization of the variety generated by $K$, and if so, can someone give me an explicit finite basis? My second question is, is the first-order theory $Th(K)$ generated by $K$ finitely axiomatizable, and if so, can someone give me an explicit finite set of axioms?

Answer 1

My first question is, is there a finite equational axiomatization of the variety generated by $K$, and if so, can someone give me an explicit finite basis?

The answer to both of these questions is Yes, but I will explain the answer to the first part only and then hint about how to do the second part.

Let $L$ be the language of rings using symbols $(+,*,-,0,1)$ with arities $(2,2,1,0,0)$. Let $\mathcal F$ be the variety generated by all fields in this language and let $\mathcal C$ be the variety of commutative rings in this language.

Claim. $\mathcal C=\mathcal F$.

Reasoning. Since every field is a commutative ring, $\mathcal F\subseteq \mathcal C$. For the other inclusion, if $R\in \mathcal C$, then $R$ is a quotient of a free commutative ring $D$. Free commutative rings, like $D$, are integral domains, so $D$ is embeddable in its field of fractions $Q\in {\mathcal F}$. Thus, $R$ is a quotient of a subring $D$ of a member $Q\in \mathcal F$. Since $\mathcal F$ is closed under the formation of quotients and subrings, $R\in \mathcal F$. \\\

Now let $\mathcal B$ be the variety of Boolean algebras in the language. The question is whether the variety ${\mathcal V}(K) = {\mathcal B}\vee {\mathcal C}$ has a finite equational axiomatization. If we interpret the symbols of $L$ as suggested by Qiaochu Yuan in his comment on Sep 3, 2023, then $\mathcal B$ is a variety of commutative rings, so ${\mathcal B}\subseteq {\mathcal C}$, so ${\mathcal V}(K) = {\mathcal B}\vee {\mathcal C} = {\mathcal C}$. Everyone knows a finite axiomatization for the variety $\mathcal C$ of commutative rings, so there is nothing more to do.

But let's return to the original formulation of the problem. The original post asks that we interpret $+$ as union and $-$ as complementation in $\mathcal B$. With this interpretation, neither $\mathcal B$ nor $\mathcal C$ contains the other. In fact, ${\mathcal B}\cap \mathcal C$ is the variety of 1-element algebras in this signature. The problem of determining whether ${\mathcal V}(K) = {\mathcal B}\vee {\mathcal C}$ has a finite equational axiomatization is now harder. In general, the join of two finitely based varieties in the same language need not be finitely based.

Fortunately (for this problem), it is known that the join of two independent finitely based varieties of finite signature must be finitely based. I don't know any place where the proof of this statement has been written down explicitly in the form that is needed here, but it can be extracted from the material on pages 20-21 of

O. C. Garcia and W. Taylor
The lattice of interpretability types of varieties.
Mem. Amer. Math. Soc. 50, no. 305, 1984.

or from Section 10.5 of

Ralph S. Freese, Ralph N. McKenzie, George F. McNulty, Walter F. Taylor,
Algebras, Lattices, Varieties: Volume III
Mathematical Surveys and Monographs Volume: 269; 2022.

Let me explain why $\mathcal B$ (the variety of Boolean algebras in this language) and $\mathcal C$ (the variety of commutative rings in this language) are independent. One establishes independence of two varieties $\mathcal U$ and $\mathcal V$ by exhibiting a binary 'decomposition term' $d(x,y)$ in the common language. This is a term such that $\mathcal U\models d(x,y)\approx x$ and $\mathcal V\models d(x,y)\approx y$. For this problem, the following term works: $$ d(x,y) = ((1+(-1))*x) + ((-1)*(-1)*y). $$ One readily verifies that $\mathcal B\models d(x,y)\approx x$ and $\mathcal C\models d(x,y)\approx y$ using the interpretations of the symbols given in the problem statement. In the language of the book by Freese, McKenzie, McNulty, Taylor that I mentioned above, this represents ${\mathcal V}(K)$ as the 'inner product' of ${\mathcal B}$ and ${\mathcal C}$ (Definition 10.73). Then, using their 'standard equational product axiomatization' of the inner product (Definition 10.73), one finds that an inner product of two finitely based varieties of finite signature is again finitely based.

Notes. Where I say above that $\ldots$ it can be extracted from $\ldots$ I am glossing over some things. The sources cited discuss the 'inner product' of varieties, ${\mathcal U}\vee {\mathcal V}={\mathcal U}\times {\mathcal V}$, only in the setting where every fundamental operation $F(x_1,\ldots,x_n)$ has the property that either ${\mathcal U}\models F(x_1,\ldots,x_n)\approx x_1$ or ${\mathcal V}\models F(x_1,\ldots,x_n)\approx x_1$. Therefore, if you want to cite their results you must reduce to this setting. One way to do this is to convert the original signature $(+,*,-,0,1)$ to a different 'derived signature' $(+_{\mathcal B},+_{\mathcal C},*_{\mathcal B},*_{\mathcal C},-_{\mathcal B},-_{\mathcal C},0_{\mathcal B},0_{\mathcal C},1_{\mathcal B},1_{\mathcal C},d)$ where $d$ is defined in the original signature by the formula above, and the other symbols of the derived signature are defined in terms of the original signature as follows: $x+_{\mathcal B} y:=d(x+y,x)$, and $x+_{\mathcal C} y:=d(x,x+y)$, (and the same pattern of definitions for $*, -, 0, 1$). Then you have ${\mathcal B}\models x_1+_{\mathcal B} x_2\approx x_1+x_2$, ${\mathcal B}\models x_1+_{\mathcal C} x_2\approx x_1$, ${\mathcal C}\models x_1+_{\mathcal B} x_2\approx x_1$, and ${\mathcal C}\models x_1+_{\mathcal C} x_2\approx x_1+x_2$ as needed to apply their results. Another restriction on the results in these sources is that they avoid the use of $0$-ary operations. You can reduce to this case by replacing a $0$-ary symbol like $1$ with a $1$-ary symbol $1(x)$ and adding an axiom $1(x)\approx 1(y)$ which forces $1(x)$ to interpret as a $1$-ary constant function.

Answer 2

Since both the first-order theories of fields and of Boolean algebras are finitely axiomatizable, so is the theory of the structures which are either fields or Boolean algebras. More generally, for any theories $T,S$ the model class of $$\{\varphi\vee\psi:\varphi\in T,\psi\in S\}$$ is always the union of the model classes of $T$ and $S$, and is obviously finitely axiomatizable if both $S$ and $T$ are.

Of course this doesn't address the equational part of the question, since equational logic doesn't allow disjunction, but this does solve the first-order aspect. EDIT: Keith Kearnes has now answered this much harder question, and this answer should really be considered a footnote to his.