Let $\mathbf A$ be the algebra given by the following multiplication table \begin{array}{c|ccc} \style{font-family:inherit}{\cdot} & 0 & 1 & 2 & 3 \\\hline 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 $x^3\approx x^4$.
For one direction, it is easy: since it is checkable that $\cdot$ is commutative and associative in $\mathbf A$ and $\forall x\in A,x^3\approx x^4$, so the variety generated by $\mathbf A$ is in the variety of commutative semigroups satisfying $x^3\approx x^4$.
For the other direction, I am thinking about decomposing any such commutative semigroup into product of smaller ones, so as to show they lie in $\mathsf{HSP}(\mathbf A)=\cal V(\mathbf A)$ by Birkhoff's Theorem.
For $1$-element commutative semigroups, there is only one case \begin{array}{c|ccc} \style{font-family:inherit}{\cdot} & 0 \\\hline 0 & 0 \end{array} and it is clearly a subalgebra of $\mathbf A$.
For $2$-element ones, I found that up to isomorphism there should be two types satisfying the identity $x^3\approx x^4$: \begin{array}{c|ccc} \style{font-family:inherit}{\cdot} & 0 & 1\\\hline 0 & 0 & 0 \\ 1 & 0 & 0 \end{array} and \begin{array}{c|ccc} \style{font-family:inherit}{\cdot} & 0 & 1\\\hline 0 & 0 & 0 \\ 1 & 0 & 1 \end{array}
They are also isomorphic to some subalgebras of $\mathbf A$.
However, starting from $3$-element commutative semigroups, it seems hard to characterise types satisfying the identity. On the other hand, the subalgebras in $\mathbf A$ of $3$ elements seem to contain only $\{0,1,2\}$ and $\{0,1,3\}$.
So I wonder whether there is a better way to prove this direction?
Thank you very much for any help!
Let $V$ be the variety of all commutative semigroups satisfying $x^3\approx x^4$. Let $V({\mathbf A})$ be the subvariety generated by ${\mathbf A}$. If $V({\mathbf A})$ 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 $a_i, b_i\in\{0,1,2,3\}$ 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 $a_j\neq b_j$. Substitute the identity element $3\in A$ for all variables except the $j$th, 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 $a_i, b_i\in\{0,1,2,3\}$, $a_j\neq b_j$, and $2^{a_j}=2^{b_j}$.