Suppose we are working in the language of a binary operation symbol $\*$. Let $S$ be a set of equations which generate precisely the same equational theory generated by the set containing the commutative and associative equations: $\{ x\*y=y\*x, (x\*y)\*z=x\*(y\*z) \}$. Suppose further that $S$ has at least three elements. Must $S$ have at least one redundant equation? I asked this on MSE, but got no response, so I am asking it here.



A question regarding equational bases of the theory of the commutative and associative properties
Yes it's true.


Write $S=(s\_i=t\_i)\_{i\in I}$ with $s\_i,t\_i$ in some free magma on $k\_i$ generators, where each of the $k\_i$ variables appears in either $s\_i$ or $t\_i$.


First, since $\mathbf{N}$ with addition satisfies this theory, we see that the length of $s\_i$ and $t\_i$ (number of letters) are the same. Call it $n\_i$. We suppose no relation is redundant.


If $n\_i\ge 3$ for all $i$, then we have a non-commutative model of the theory ($\{x,y,u,v,0\}$ with $xy=u$, $yx=v$, other products equal $0$), contradiction.


So there exists $j$ with $n\_j\le 2$. If $k\_j=4$, we have the relation $x\_1x\_2=x\_3x\_4$ which obviously doesn't hold in every commutative semigroup. Same if $k\_j=3$ (relation $x\_1x\_2=x\_1x\_3$ or $x\_1x\_2=x\_3x\_1$). Also $k\_j=1$ would mean either the redundant relation $x\_1=x\_1$, or the relation $x\_1=x\_2$ not true in every commutative semigroup. So $k\_j=2$ and we have the commutativity relation (since $x\_1x\_2=x\_1x\_2$ would be redundant). For the same reason, $n\_i\ge 3$ for all $i\neq j$ since otherwise we would obtain a redundant relation.


If $n\_i\ge 4$ for all $i\neq j$ there is a non-associative model $\{x,y,z,u,0\}$ with $xy=yx=z$, $xz,zx=u$ all other products equal 0.


Hence $n\_i=3$ for some $i$. So $k\_i\le 6$. We claim that if $k\_i\ge 4$, the relation $s\_i=t\_i$ does not hold in every commutative semigroup. Indeed, among commutative semigroup, such a relation reduces to one of the following:
$$x\_1x\_2x\_3=x\_4x\_5x\_6,\;x\_1x\_2x\_3=x\_1x\_4x\_5,\; x\_1x\_2x\_3=x\_4x\_5x\_5,$$ $$x\_1x\_2x\_2=x\_3x\_4x\_4,\;x\_1x\_2x\_3=x\_1x\_4x\_4,\;x\_1x\_2x\_3=x\_4x\_4x\_4,\;x\_1x\_2x\_3=x\_1x\_1x\_4,$$
none of which holds in every commutative semigroup.


If $k\_i=3$, for commutative semigroup the relation $s\_i=t\_i$ yields either $x\_1x\_2x\_3=x\_1x\_2x\_3$, or a nontrivial equality which is not true in every commutative semigroup. So $s\_i=t\_i$ is either the associativity relation (modulo commutative changes), or a trivial (hence redundant) equality. So we have the associativity relation, and all other ones are redundant.


It remains to deal with $k\_i\le 2$. If $k\_i=1$, since $xxx$ has only one possible meaning in a commutative magma, we have no possibility. Suppose $k\_i=2$. So we have a relation yielding, in the associative case, $xxy=xxy$, i.e., the relation $x(xy)=(xx)y$. By the same argument, this is the only relation of length $3$ we can get. But then the equational theory is satisfied by the non-associative commutative model $\{x,y,z,X,Y,Z,u,v,0\}$, with (commutativity being implicit) $xy=Z$, $yz=X$, $xz=Y$, $xX=u$, $yY=v$, all other products being zero.


Note: one can't always remove from an infinite equational theory, a non-redundant one. However one can (a) choose a single representative for all equation (modulo changing variables, reversing equalities. For instance if we have $x\_1x\_2=x\_2x\_1$ and $x\_1x\_3=x\_3x\_1$ we merge into a single one. After this, we have only finitely many equalities of the same total length (say $s=t$ has length the sum of lengths of $s$ and $t$), and then we can remove redundant ones of total length $\le 6$. Then the above argument works, since it only assumes that there is no redundant equality of length $\le 6$.


