Does cartesian product satisfy any nontrivial equational identity?

Question

This is a question concerning the model theory of set theory. Let $M$ be a model of ZFC set theory, and consider the algebraic structure $(M;\times)$, where $\times$ is (the Kuratowski implementation of) Cartesian product. Does that structure satisfy any nontrivial equational identity? A nontrivial identity is any identity other than $t=t$ for some term $t$. I conjecture that it does not, but how to prove it?

Answer 1

Remember that if $\mathfrak{X}$ is a subalgebra (in the universal-algebra sense) of $\mathfrak{Y}$, then the equational theory of $\mathfrak{X}$ contains that of $\mathfrak{Y}$. So to show that the equational theory of a given algebra $\mathfrak{A}$ is trivial, it suffices to show that that algebra has a subalgebra isomorphic to the free algebra (in the appropriate language) $\mathfrak{F}_n$ on $n$ generators for each $n$; in particular, the existence of an embedding $\mathfrak{F}_n\hookrightarrow\mathfrak{A}$ means that the $n$-variable equational theory of $\mathfrak{A}$ is trivial.

For example, here's how this plays out with the Wiener ordered pair $\langle\cdot,\cdot\rangle_W$ and associated Cartesian product $\times_W$. This is a bit nicer for our purposes than Kuratowski's definition, since it satisfies the following: $$\vert A\vert=\vert B\vert=1\quad\implies\quad A\times_WB=\{\{\{\{a\},\emptyset\},\{\{b\}\}\}: a\in A,b\in B\}=\{ \{\{A,\emptyset\},\{B\}\}\}.$$

From this it's easy-but-tedious to prove that - working within a fixed model $V\models\mathsf{ZFC}$ - for any singletons $A_1,...,A_n\in V$ the subalgebra of $(V,\times_W)$ generated by the $A_i$s is free.

For the Kuratowski definition, a bit more care is needed but the same result still holds. In particular, I think if we take each $A_i$ to be a different infinite cardinal then the resulting subalgebra should be free.