| Russian Mathematical Surveys |
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Brief communications On the intersection of finitely generated varieties of monoids E. W. H. Lee Nova Southeastern University, Fort Lauderdale, FL, USA
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    By the theorem of Oates and Powell [1], the variety $\langle G \rangle$ generated by any finite group $G$ is a Cross variety. It follows that the intersection $\langle G_1 \rangle \cap \langle G_2 \rangle$ of varieties generated by any finite groups $G_1$ and $G_2$ is finitely generated. But this result does not hold for more general algebras. For instance, the intersection $\langle S_1 \rangle_{\mathsf{sem}} \cap \langle S_2 \rangle_{\mathsf{sem}}$ of varieties of semigroups generated by the multiplicative matrix semigroups
$$
\begin{equation*}
S_1=\biggl\{\begin{bmatrix}0&0\\0&0\end{bmatrix}, \begin{bmatrix}0&1\\0&0\end{bmatrix}, \begin{bmatrix}1&0\\0&1\end{bmatrix}\biggr\} \quad\text{and}\quad S_2=\biggl\{\begin{bmatrix}0&0\\0&0\end{bmatrix}, \begin{bmatrix}1&0\\0&0\end{bmatrix}, \begin{bmatrix}0&1\\0&0\end{bmatrix}, \begin{bmatrix}0&0\\0&1\end{bmatrix}\biggr\}
\end{equation*}
\notag
$$
is non-finitely generated ([2], Theorem 1.7). Given how similar monoids are to semigroups, it is quite surprising that the following question has remained open.
Question 1. Do finite monoids $M_1$ and $M_2$ exist so that the intersection $\langle M_1 \rangle_{\mathsf{mon}} \cap \langle M_2 \rangle_{\mathsf{mon}}$ of varieties of monoids is non-finitely generated? The main goal of this article is to exhibit finite monoids $M_1$ and $M_2$ to affirmatively answer Question 1. Although this question is concerned with monoids, it makes no difference if it is addressed within the context of semigroups because the intersections $\langle M_1 \rangle_{\mathsf{mon}} \cap \langle M_2 \rangle_{\mathsf{mon}}$ and $\langle M_1\rangle_{\mathsf{sem}}\cap\langle M_2\rangle_{\mathsf{sem}}$ are simultaneously finitely generated; see the appendix. However, due to the large volume of results on semigroups available in the literature, it is more convenient to work with semigroups than with monoids. Recall that a locally finite variety or algebra is inherently non-finitely based if every locally finite variety containing it is non-finitely based. It turns out that for any finite monoid $M$, the varieties $\langle M \rangle_{\mathsf{mon}}$ and $\langle M \rangle_{\mathsf{sem}}$ are simultaneously inherently non-finitely based ([3], Theorem 4.3). Two important examples of inherently non-finitely based semigroups, due to Sapir [4], are the multiplicative matrix semigroup
$$
\begin{equation*}
B_2^1=\biggl\{\begin{bmatrix}0&0\\0&0\end{bmatrix}, \begin{bmatrix}1&0\\0&0\end{bmatrix},\begin{bmatrix}0&1\\0&0\end{bmatrix}, \begin{bmatrix}0&0\\1&0\end{bmatrix},\begin{bmatrix}0&0\\0&1\end{bmatrix}, \begin{bmatrix}1&0\\0&1\end{bmatrix}\biggr\}
\end{equation*}
\notag
$$
and the Rees quotient $Q$ of the free semigroup $\{x_1,x_2,x_3,\ldots\}^+$ modulo the ideal of all words that are not factors of any of the Zimin words $w_1, w_2, w_3,\ldots$ given by $w_1=x_1$ and $w_n=w_{n-1} x_n w_{n-1}$ for all $n \geqslant 2$. The semigroups $B_2^1$ and $Q$ are aperiodic in the sense that all their subgroups are trivial.
Lemma 2 (Jackson [5], Theorem 1.4). Suppose that $S$ is any finite semigroup that is either aperiodic or of order at most $55$. Then $S$ is inherently non-finitely based if and only if $B_2^1 \in \langle S \rangle_{\mathsf{sem}}$. Lemma 3 (Sapir [4], Theorem 3A). The variety $\langle Q\rangle_{\mathsf{sem}}$ is the only minimal inherently non-finitely based variety of semigroups that is not generated by groups. Theorem 4. Let $\mathcal{K}$ be the class of all inherently non-finitely based finite semigroups $K$ such that $B_2^1 \notin \langle K \rangle_{\mathsf{sem}}$. Then for any $K \in \mathcal{K}$, the intersection $\langle B_2^1 \rangle_{\mathsf{sem}} \cap \langle K \rangle_{\mathsf{sem}}$ is non-finitely generated. Proof. Since every finite group generates a Cross variety of semigroups [1], the class $\mathcal{K}$ does not contain any groups. Therefore, by Lemma 3, the variety $\langle Q \rangle_{\mathsf{sem}}$ is contained in $\langle B_2^1 \rangle_{\mathsf{sem}}$ and $\langle K \rangle_{\mathsf{sem}}$, so that $\langle B_2^1 \rangle_{\mathsf{sem}} \cap \langle K \rangle_{\mathsf{sem}}$ is inherently non-finitely based. Suppose that $\langle B_2^1 \rangle_{\mathsf{sem}} \cap \langle K \rangle_{\mathsf{sem}}= \langle S \rangle_{\mathsf{sem}}$ for some finite semigroup $S$. Then $S$ is inherently non-finitely based. Since $B_2^1$ is aperiodic, $S$ is also aperiodic. It then follows from Lemma 2 that $B_2^1 \in \langle S \rangle_{\mathsf{sem}}$, whence the contradiction $B_2^1 \in \langle K \rangle_{\mathsf{sem}}$ is deduced. $\Box$
Question 1 is thus affirmatively answered by letting $M_1=B_2^1$ and choosing any monoid $M_2$ from $\mathcal{K}$. An example of a monoid in $\mathcal{K}$, due to Volkov and Gol’dberg [6], is the semigroup of $n \times n$ upper triangular matrices over the $q$-element field under usual matrix multiplication, where $n \geqslant 4$ and $q \geqslant 3$. This example has a rather large order of $q^{n(n+1)/2} \geqslant 3^{10}$. Smaller examples in $\mathcal{K}$ have also been exhibited by Jackson [5]. Specifically, he constructed from any finite centreless group $G$ a monoid $G^* \in \mathcal{K}$ of order $9|G|+2$. Since the dihedral group $D_3 \in \mathcal{K}$ of order $6$ is the smallest centreless group, the monoid $D_3^*$ of order $56$ is the smallest example obtained by his construction. In fact, it follows from Lemma 2 that $56$ is the smallest possible order of a semigroup in $\mathcal{K}$. AppendixLemma 5 (Jackson [7], Lemma 1.1). Suppose that $S$ is any semigroup and $M$ is any monoid such that $S \in \langle M \rangle_{\mathsf{sem}}$. Then the smallest monoid $S^1$ containing $S$ belongs to $\langle M \rangle_{\mathsf{mon}}$. Theorem 6. Let $\mathcal{M}=\langle M_1 \rangle_{\mathsf{mon}} \cap \langle M_2 \rangle_{\mathsf{mon}}$ and $\mathcal{S}= \langle M_1 \rangle_{\mathsf{sem}} \cap \langle M_2 \rangle_{\mathsf{sem}}$, where $M_1$ and $M_2$ are any monoids. Consequently, the varieties $\mathcal{M}$ and $\mathcal{S}$ are simultaneously finitely generated. Proof. (i) Suppose that $\mathcal{M}=\langle M \rangle_{\mathsf{mon}}$ for some monoid $M$. If $S \in \mathcal{S}$, then by Lemma 5, one has $S^1 \in \mathcal{M} \subseteq \langle M \rangle_{\mathsf{sem}}$, whence $S \in \langle M \rangle_{\mathsf{sem}}$. Thus $\mathcal{S} \subseteq \langle M \rangle_{\mathsf{sem}}$. Conversely, if $S \in \langle M \rangle_{\mathsf{sem}}$, then by Lemma 5, one has $S^1 \in \langle M \rangle_{\mathsf{mon}}= \mathcal{M} \subseteq \mathcal{S}$, whence $S \in \mathcal{S}$. Therefore, $\langle M \rangle_{\mathsf{sem}} \subseteq \mathcal{S}$.
(ii) Suppose that $\mathcal{S}=\langle S \rangle_{\mathsf{sem}}$ for some semigroup $S$. Since $S \in \mathcal{S}$ by assumption, $\langle S^1 \rangle_{\mathsf{mon}} \subseteq \mathcal{M}$ by Lemma 5. Conversely, since $\mathcal{M} \subseteq \mathcal{S} \subseteq \langle S^1 \rangle_{\mathsf{sem}}$ by assumption, it follows from Lemma 5 that $\mathcal{M} \subseteq \langle S^1 \rangle_{\mathsf{mon}}$. $\Box$ 
Citation in format AMSBIB
\Bibitem{Lee25} |
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