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Summary

There are semigroups of finite essentially arity which are not strongly abelian. This follows from my original answer and a result I cite in the addition below that I have not read in detail.

Added. Here is a self-contained answer. Take the semigroup $S=\langle x\mid x^3=x^4\rangle$. Then consider the term $t(x_1,x_2)=x_1x_2$. One has $t(x,x^3)=x^3=t(x^2,x^3)$ but $t(x,x)=x^2\neq x^3=t(x^2,x)$. So $S$ is not strongly abelian. But the essential arity of $S$ is $3$ since the value of any word of length $3$ in $S$ is $x^3$.

Original Answer

A finite semigroup $S$ has finite essential arity iff it is locally trivial, that is, $eSe=e$ for all idempotents $e\in S$.

Pf. First suppose that $S$ is not locally trivial. The either $S$ contains a subsemigroup isomorphic to the 2-element semilattice $\lbrace 0,1\rbrace$ with multiplication or it contains a non-trivial group $G$ as a subsemigroup.

In the first case, the variety generated by $S$ contains the free semilattice on any set (i.e. the power set under union with singletons as free generators) and so the term function given by a word on m-letters depends on all those letters. If $S$ contains a non-trivial group $G$, then it contains a cyclic group of prime order $p$ and hence vector spaces over $\mathbb F_p$. But then again the term function depending on any word in m-letters depends on all letters by considering a vector space of dimension m.

So finite essential arity implies locally trivial.

Conversely, if $S$ is locally trivial, then it is known that $S$ satisfies an identity of the form $$x_1\cdots x_myz_1\cdots z_n = x_1\cdots x_mz_1\cdots z_n$$ and so any term operation coming from a word depends only on at most $m+n$ variables (the prefix of length $m$ and the suffix of length $n$).

In particular, no non-trivial monoid has finite essential arity.

Added. If I understood correctly the results of Stepanova, A. A.(RS-FARE-IMC); Trikashnaya, N. V.(RS-FARE-IMC) Abelian and Hamiltonian varieties of groupoids, Algebra Logic 50 (2011), no. 3, 272–278 then a locally trivial semigroup is strongly abelian iff it is an inflation of a rectangular band. Most locally trivial semigroups are not inflations of a rectangular band. If $S$ is an inflation of a rectangular band, then $S^2=S^3$. But if one takes the semigroup $\langle x\mid x^3=x^4\rangle$, then it is locally trivial (hence essentially finite arity) but it is not strongly abelian.

Summary

There are semigroups of finite essentially arity which are not strongly abelian. This follows from my original answer and a result I cite in the addition below that I have not read in detail.

Added. Here is a self-contained answer. Take the semigroup $S=\langle x\mid x^3=x^4\rangle$. Then consider the term $t(x_1,x_2)=x_1x_2$. One has $t(x,x^3)=x^3=t(x^2,x^3)$ but $t(x,x)=x^2\neq x^3=t(x^2,x)$. So $S$ is not strongly abelian. But the essential arity of $S$ is $3$ since the value of any word of length $\geq 3$ in $S$ is $x^3$.

Original Answer

A finite semigroup $S$ has finite essential arity iff it is locally trivial, that is, $eSe=e$ for all idempotents $e\in S$.

Pf. First suppose that $S$ is not locally trivial. The either $S$ contains a subsemigroup isomorphic to the 2-element semilattice $\lbrace 0,1\rbrace$ with multiplication or it contains a non-trivial group $G$ as a subsemigroup.

In the first case, the variety generated by $S$ contains the free semilattice on any set (i.e. the power set under union with singletons as free generators) and so the term function given by a word on m-letters depends on all those letters. If $S$ contains a non-trivial group $G$, then it contains a cyclic group of prime order $p$ and hence vector spaces over $\mathbb F_p$. But then again the term function depending on any word in m-letters depends on all letters by considering a vector space of dimension m.

So finite essential arity implies locally trivial.

Conversely, if $S$ is locally trivial, then it is known that $S$ satisfies an identity of the form $$x_1\cdots x_myz_1\cdots z_n = x_1\cdots x_mz_1\cdots z_n$$ and so any term operation coming from a word depends only on at most $m+n$ variables (the prefix of length $m$ and the suffix of length $n$).

In particular, no non-trivial monoid has finite essential arity.

Added. If I understood correctly the results of Stepanova, A. A.(RS-FARE-IMC); Trikashnaya, N. V.(RS-FARE-IMC) Abelian and Hamiltonian varieties of groupoids, Algebra Logic 50 (2011), no. 3, 272–278 then a locally trivial semigroup is strongly abelian iff it is an inflation of a rectangular band. Most locally trivial semigroups are not inflations of a rectangular band. If $S$ is an inflation of a rectangular band, then $S^2=S^3$. But if one takes the semigroup $\langle x\mid x^3=x^4\rangle$, then it is locally trivial (hence essentially finite arity) but it is not strongly abelian.

Summary

There are semigroups of finite essentially arity which are not strongly abelian. This follows from my original answer and a result I cite in the addition below that I have not read in detail.

Original Answer

A finite semigroup $S$ has finite essential arity iff it is locally trivial, that is, $eSe=e$ for all idempotents $e\in S$.

Pf. First suppose that $S$ is not locally trivial. The either $S$ contains a subsemigroup isomorphic to the 2-element semilattice $\lbrace 0,1\rbrace$ with multiplication or it contains a non-trivial group $G$ as a subsemigroup.

In the first case, the variety generated by $S$ contains the free semilattice on any set (i.e. the power set under union with singletons as free generators) and so the term function given by a word on m-letters depends on all those letters. If $S$ contains a non-trivial group $G$, then it contains a cyclic group of prime order $p$ and hence vector spaces over $\mathbb F_p$. But then again the term function depending on any word in m-letters depends on all letters by considering a vector space of dimension m.

So finite essential arity implies locally trivial.

Conversely, if $S$ is locally trivial, then it is known that $S$ satisfies an identity of the form $$x_1\cdots x_myz_1\cdots z_n = x_1\cdots x_mz_1\cdots z_n$$ and so any term operation coming from a word depends only on at most $m+n$ variables (the prefix of length $m$ and the suffix of length $n$).

In particular, no non-trivial monoid has finite essential arity.

Added. If I understood correctly the results of Stepanova, A. A.(RS-FARE-IMC); Trikashnaya, N. V.(RS-FARE-IMC) Abelian and Hamiltonian varieties of groupoids, Algebra Logic 50 (2011), no. 3, 272–278 then a locally trivial semigroup is strongly abelian iff it is an inflation of a rectangular band. Most locally trivial semigroups are not inflations of a rectangular band. If $S$ is an inflation of a rectangular band, then $S^2=S^3$. But if one takes the semigroup $\langle x\mid x^3=x^4\rangle$, then it is locally trivial (hence essentially finite arity) but it is not strongly abelian.

Summary

There are semigroups of finite essentially arity which are not strongly abelian. This follows from my original answer and a result I cite in the addition below that I have not read in detail.

Added. Here is a self-contained answer. Take the semigroup $S=\langle x\mid x^3=x^4\rangle$. Then consider the term $t(x_1,x_2)=x_1x_2$. One has $t(x,x^3)=x^3=t(x^2,x^3)$ but $t(x,x)=x^2\neq x^3=t(x^2,x)$. So $S$ is not strongly abelian. But the essential arity of $S$ is $3$ since the value of any word of length $3$ in $S$ is $x^3$.

Original Answer

A finite semigroup $S$ has finite essential arity iff it is locally trivial, that is, $eSe=e$ for all idempotents $e\in S$.

Pf. First suppose that $S$ is not locally trivial. The either $S$ contains a subsemigroup isomorphic to the 2-element semilattice $\lbrace 0,1\rbrace$ with multiplication or it contains a non-trivial group $G$ as a subsemigroup.

In the first case, the variety generated by $S$ contains the free semilattice on any set (i.e. the power set under union with singletons as free generators) and so the term function given by a word on m-letters depends on all those letters. If $S$ contains a non-trivial group $G$, then it contains a cyclic group of prime order $p$ and hence vector spaces over $\mathbb F_p$. But then again the term function depending on any word in m-letters depends on all letters by considering a vector space of dimension m.

So finite essential arity implies locally trivial.

Conversely, if $S$ is locally trivial, then it is known that $S$ satisfies an identity of the form $$x_1\cdots x_myz_1\cdots z_n = x_1\cdots x_mz_1\cdots z_n$$ and so any term operation coming from a word depends only on at most $m+n$ variables (the prefix of length $m$ and the suffix of length $n$).

In particular, no non-trivial monoid has finite essential arity.

Added. If I understood correctly the results of Stepanova, A. A.(RS-FARE-IMC); Trikashnaya, N. V.(RS-FARE-IMC) Abelian and Hamiltonian varieties of groupoids, Algebra Logic 50 (2011), no. 3, 272–278 then a locally trivial semigroup is strongly abelian iff it is an inflation of a rectangular band. Most locally trivial semigroups are not inflations of a rectangular band. If $S$ is an inflation of a rectangular band, then $S^2=S^3$. But if one takes the semigroup $\langle x\mid x^3=x^4\rangle$, then it is locally trivial (hence essentially finite arity) but it is not strongly abelian.

A finite semigroup $S$ has finite essential arity iff it is locally trivial, that is, $eSe=e$ for all idempotents $e\in S$.

Pf. First suppose that $S$ is not locally trivial. The either $S$ contains a subsemigroup isomorphic to the 2-element semilattice $\lbrace 0,1\rbrace$ with multiplication or it contains a non-trivial group $G$ as a subsemigroup.

In the first case, the variety generated by $S$ contains the free semilattice on any set (i.e. the power set under union with singletons as free generators) and so the term function given by a word on m-letters depends on all those letters. If $S$ contains a non-trivial group $G$, then it contains a cyclic group of prime order $p$ and hence vector spaces over $\mathbb F_p$. But then again the term function depending on any word in m-letters depends on all letters by considering a vector space of dimension m.

So finite essential arity implies locally trivial.

Conversely, if $S$ is locally trivial, then it is known that $S$ satisfies an identity of the form $x_1\cdots x_myz_1\cdots z_n = x_1\cdots x_mz_1\cdots z_n$ and so any term operation coming from a word depends only on at most $m+n$ variables (the prefix of length $m$ and the suffix of length $n$).

In particular, no non-trivial monoid has finite essential arity.

Summary

There are semigroups of finite essentially arity which are not strongly abelian. This follows from my original answer and a result I cite in the addition below that I have not read in detail.

Original Answer

A finite semigroup $S$ has finite essential arity iff it is locally trivial, that is, $eSe=e$ for all idempotents $e\in S$.

Pf. First suppose that $S$ is not locally trivial. The either $S$ contains a subsemigroup isomorphic to the 2-element semilattice $\lbrace 0,1\rbrace$ with multiplication or it contains a non-trivial group $G$ as a subsemigroup.

In the first case, the variety generated by $S$ contains the free semilattice on any set (i.e. the power set under union with singletons as free generators) and so the term function given by a word on m-letters depends on all those letters. If $S$ contains a non-trivial group $G$, then it contains a cyclic group of prime order $p$ and hence vector spaces over $\mathbb F_p$. But then again the term function depending on any word in m-letters depends on all letters by considering a vector space of dimension m.

So finite essential arity implies locally trivial.

Conversely, if $S$ is locally trivial, then it is known that $S$ satisfies an identity of the form $$x_1\cdots x_myz_1\cdots z_n = x_1\cdots x_mz_1\cdots z_n$$ and so any term operation coming from a word depends only on at most $m+n$ variables (the prefix of length $m$ and the suffix of length $n$).

In particular, no non-trivial monoid has finite essential arity.

Added. If I understood correctly the results of Stepanova, A. A.(RS-FARE-IMC); Trikashnaya, N. V.(RS-FARE-IMC) Abelian and Hamiltonian varieties of groupoids, Algebra Logic 50 (2011), no. 3, 272–278 then a locally trivial semigroup is strongly abelian iff it is an inflation of a rectangular band. Most locally trivial semigroups are not inflations of a rectangular band. If $S$ is an inflation of a rectangular band, then $S^2=S^3$. But if one takes the semigroup $\langle x\mid x^3=x^4\rangle$, then it is locally trivial (hence essentially finite arity) but it is not strongly abelian.