I'm going through this book right now, and I don't understand something about exercise 9 from chapter 1:
Determine all elements of t he free algebra $F_V(X_2)$ for the following varieties: (i) The variety of right semigroups $RZ=Mod\{x_1x_2\approx x_2\}$ (iii) The variety of zero semigroups $Z=Mod\{x_1x_2\approx x_3x_4\}$
So in (i) I've been able to come to the conclusion that $F_V(X_2)=\{x_1,x_2\}$ and that $x_1x_2=x_2,\, x_2x_1=x_1$, which is reasonable and cool.
But in (iii), I don't seem to understand where the $x_3$ and $x_4$ come from. From what i understand, because we are talking about $X_2$, there should only be $x_1$ and $x_2$, and no $x_3$ and $x_4$.
Thank you for your help :)
I assume $X_2$ is a set of two elements, that will be the generators of the free algebra.
Part (i) is good, but you should also mention $x_1x_1=x_1$ and $x_2x_2=x_2$.
Your confusion seems to arise from the notation, as you set $X_2=\{x_1,x_2\}$ and the same letters are used to describe the variety where the free algebra on $X_2$ lives.
So, I propose to rename the elements of $X_2$, say $X_2=\{a,b\}$, and for (ii), any interpretation of its variables (via a function $\{x_1,x_2,x_3,x_4\}\to F_V(X_2)$) must satisfy the given equation.
Ultimately, the free algebra will contain the generators, $a$ and $b$. No equation forces them to be equal (to be proved). Moreover, it will contain elements like $a^2$, $ab$, $ba$, $b^2$. These now are forced to be equal to each other in this variety (this is given), but are not forced to be equal to either $a$ or $b$ (to be proved). So, this is a third element, let's call it '$0$'. Now, it is easy to check that no more new elements will be generated by repeatedly applying the operation. So, the free algebra is $\{a,b,0\}$.
Finally, to prove the two remaining statements ('some things are not forced to be equal'), consider this concrete $3$ element zerosemigroup $U:=\{a,b,0\}$ and the obvious homomorphism $F_V(X_2)\to U$. If an equation was forced in $F_V(X_2)$, it would also hold in $U$.