Cristopher Knox

2022-06-29

Let $\mathbf{A}$ be the algebra given by the following multiplication table
$\begin{array}{ccccc}\cdot & 0& 1& 2& 3\\ 0& 0& 0& 0& 0\\ 1& 0& 0& 0& 1\\ 2& 0& 0& 1& 2\\ 3& 0& 1& 2& 3\end{array}$
I need to prove that the variety generated by $\mathbf{A}$ is exactly the variety of commutative semigroups satisfying ${x}^{3}\approx {x}^{4}$.

Jamarcus Shields

Let $V$ be the variety of all commutative semigroups satisfying ${x}^{3}\approx {x}^{4}$. Let $V\left(\mathbf{A}\right)$ be the subvariety generated by $\mathbf{A}$. If $V\left(\mathbf{A}\right)$ is a proper subvariety of $V$, then there is a law that holds in $\mathbf{A}$ that does not hold throughout $V$. Using the identities of $V$, we may reduce any such law to one of the form $s\approx t$ where $s={x}_{1}^{{a}_{1}}\cdots {x}_{k}^{{a}_{k}}$, $t={x}_{1}^{{b}_{1}}\cdots {x}_{k}^{{b}_{k}}$, and ${a}_{i},{b}_{i}\in \left\{0,1,2,3\right\}$ for all $i$. Here a power of the form ${x}_{i}^{0}$, with exponent $0$, should be interpreted as the identity element of $\mathbf{A}$, which is $3$.
Since $s\approx t$ does not hold in $V$, there must be some index where the variables in these words have different exponents, say ${a}_{j}\ne {b}_{j}$. Substitute the identity element $3\in A$ for all variables except the jth, and substitute $2$ for ${x}_{j}$. You obtain from $s\approx t$ that ${2}^{{a}_{j}}={2}^{{b}_{j}}$. But the possible powers of $2$ are all distinct: ${2}^{0}=3,{2}^{1}=2,{2}^{2}=1,{2}^{3}=0$. This makes it impossible to have ${a}_{i},{b}_{i}\in \left\{0,1,2,3\right\}$, ${a}_{j}\ne {b}_{j}$, and ${2}^{{a}_{j}}={2}^{{b}_{j}}$.

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