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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!

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    $\begingroup$ Any variety is generated by its subdirectly irreducible members, so you should try to work out what the subdirectly irreducible semigroupgs of your variety are, and then try to show that these are generated by $\mathbf A$ (disclaimer : I'm not sure this works, but this is how I would approach the question) $\endgroup$ Dec 7, 2019 at 12:36
  • $\begingroup$ @amrsa yes I would like to prove it, and could you please explain why the result is doubtful? $\endgroup$
    – user771160
    Dec 7, 2019 at 18:17
  • $\begingroup$ @amrsa I see. Could you please make it clear how you in general find a finite subdirectly irreducible commutative semigroup with $x^3\approx x^4$? $\endgroup$
    – user771160
    Dec 7, 2019 at 19:51

1 Answer 1

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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 $2^{a_j}=2^{b_j}$.

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  • $\begingroup$ That is exactly what I am looking for. Thank you very much! $\endgroup$
    – user771160
    Dec 8, 2019 at 18:44

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