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Consider the lattice of equational theories of a single binary operation $*$. The meet of the theory axiomatized by commutativity and the theory axiomatized by associativity is simply the theory axiomatized by both of them. What about the join? Is there a finite axiomatization of the join? I conjecture that it is axiomatized by the equation, $(x*y)*x=x*(y*x)$. Is this true? If not, is there some other finite axiomatization, and if so, can someone exhibit such a finite axiom set.

user107952
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    In less lattice-theoretic language, you're just asking about equational axiomatizations of the set of equational consequences of the sentence $$[\forall x,y(xy=yx)]\vee[\forall x,y,z(x(yz)=(xy)z)],$$ right? – Noah Schweber Jul 12 '20 at 22:42
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    Algebras satisfying your proposed equation are called flexible. Flexible algebras include algebras which are neither commutative nor associative such as the octonions. – Rob Arthan Jul 12 '20 at 22:50
  • The join may be not finitely axiomatizable. – markvs Jul 12 '20 at 22:53
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    Basically, we want to finitely axiomatize $\mathrm{HSP}(A\cup C)$ where $A$ is the class of associative magmas (semigroups) and $C$ is the class of commutative magmas. – Berci Jul 12 '20 at 23:00
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    @Berci: it's clearly finitely axiomatisable by the axiom Noah gave. The question is whether it is axiomatisable by equations (which it isn't, see my answer). – Rob Arthan Jul 12 '20 at 23:11
  • What is clearly finitely axiomatizable? – markvs Jul 13 '20 at 01:01
  • I would suggest posting this question to MO. – markvs Jul 13 '20 at 01:02
  • What would an infinite axiomatization look like? – Joel Adler Jul 26 '20 at 07:34

1 Answers1

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Partial answer: I don't know about a finite equational basis, but the "flexible law" $$(x*y)*x=x*(y*x)\tag1$$ is insufficient: the identity $$((x*y)*z)*(x*(y*z))=(x*(y*z))*((x*y)*z)\tag2$$ follows from the associative law, and it follows from the commutative law, but it does not follow from $(1)$.

Let $\mathbb F$ be any field not of characteristic $2$ or $3$. If we define $$x*y=\frac23x+\frac13y$$ then the magma $(\mathbb F,*)$ satisfies $(1)$ for all $x$ and $y$, but does not satisfy $(2)$ if $z\ne x$.

Note that $$x*y=y*x\iff x=y.$$ Now $$(x*y)*z=\frac23\left(\frac23x+\frac13y\right)+\frac13z=\frac49x+\frac29y+\frac39z$$ and $$x*(y*z)=\frac23x+\frac13\left(\frac23y+\frac13z\right)=\frac69x+\frac29y+\frac19z,$$ so $$(x*y)*z=x*(y*z)\iff z=x,$$ so $$(x*y)*x=x*(y*x),$$ that is, $(1)$ holds. Also $$((x*y)*z)*(x*(y*z))=(x*(y*z))*((x*y)*z)$$$$\iff(x*y)*z=x*(y*z)\iff z=x,$$ so $(2)$ does not hold unless $z=x$.

bof
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