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Consider an algebraic structure $(S,+,*)$, where $+$ and $*$ are simply two different binary operations, not necessarily addition or multiplication. We say that $+$ and $*$ do not enmesh if the equational identities of $(S,+)$ united with the equational identities of $(S,*)$ suffice to generate all the equational identities of $(S,+,*)$. We say they do enmesh if it is not the case that they do not enmesh. For example, over the reals, $+$ and $*$ do enmesh, because of the distributive law connecting the two. I am finding it hard to come up with two explicit binary operations that do not enmesh. Can someone give an example, preferably two operations on a finite set, preferably as small a finite set as possible?

user107952
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  • What exactly are equational identities? Equations in which $+$ and $*$ operate on variables such that the equation is fulfilled for all elements of $S$ being substituted for the variables? – joriki Mar 24 '20 at 17:23
  • @joriki Yes, that is correct. – user107952 Mar 24 '20 at 17:24
  • A trivial example: let $+$ and $\ast$ binary constant functions, with distinct values. – Berci Mar 24 '20 at 17:25
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    @Berci: That yields the equational identity $a+a=(a*a)+a$. – joriki Mar 24 '20 at 17:26
  • @user107952: I assume you're excluding tautological identities that equate identical expressions, e.g. $a(a+a)=a(a+a)$. – joriki Mar 24 '20 at 17:30
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    You can always consider the term algebra (aka, absolutely free algebra) on the signature $(2,2)$. The problem then, is that it is infinite (doesn't satisfy any non-trivial identity). – amrsa Mar 24 '20 at 17:30
  • @amsra The question then, is whether there is a finite set with two distinct operations with this property. Maybe the finite set can even be a set with two elements. – user107952 Mar 24 '20 at 17:45
  • I guess you can start with a few identities for each operation (but none involving both) and then compute the free algebra on that variety over some small sets; perhaps you can find one such combination that gives a finite (and even small) free algebra? But I really don't want to try that since I wouldn't know where to start from. My previous example works, and it is the laziest possible example, since there is nothing to check. But it's not finite; term algebras are only finite if trivial... – amrsa Mar 24 '20 at 18:00
  • And then there's the trivial example when $|S|=1$. – amrsa Mar 24 '20 at 18:07
  • @amsra The title question states that I want a set with at least two elements. I wonder if there is a non-trivial finite example, and if so, what is the smallest finite example. Maybe it is not two, maybe it is three or even higher. – user107952 Mar 24 '20 at 18:18
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    @Berci: The example of 2 constant functions with different constant values works. The objection involving the identity $a+a=(aa)+a$ is not correct. If $+$ and $$ are constant, then the equational theory of $(S,+)$ contains the identity $w+x=y+z$ and the equational theory of $(S,)$ contains $wx=yz$. The union of these theories entails the identity $a+a=(aa)+a$. – Keith Kearnes Mar 25 '20 at 03:54

1 Answers1

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I can think of an example which might be considered too trivial, in a way...

The variety of Left-zero semigroups (given by the identity $xy=x$) is such that the free algebra on $n$ elements has $n$ elements. The same applies to Right-zero semigroups.
So take an $n$-element set ($n=2,3,\ldots$ whatever) and make $x+y=x$ and $x*y=y$.
The free algebra on that set has the same number of elements (it's trivial to check that it has the universal mapping property).

So an equational base for the resulting variety is given by those two identities.

amrsa
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