On free subsemigroups of associative algebras

TL;DR

This paper investigates conditions under which associative algebras contain free subsemigroups, confirming Klein's conjecture in several classes of noncommutative domains over countable fields.

Contribution

It identifies specific algebraic conditions ensuring the existence of free subsemigroups, extending Klein's conjecture to new classes of associative algebras.

Findings

Free subsemigroups exist in algebras satisfying polynomial identities and noncommutative modulo prime radical.

Presence of nonartinian primitive subquotients guarantees free subsemigroups.

Uncountable base fields and noncommutativity modulo Jacobson radical also imply free subsemigroups.

Abstract

In 1992, following earlier conjectures of Lichtman and Makar-Limanov, Klein conjectured that a noncommutative domain must contain a free, multiplicative, noncyclic subsemigroup. He verified the conjecture when the center is uncountable. In this note we consider the existence (or not) of free subsemigroups in associative $k$-algebras $R$, where $k$ is a field not algebraic over a finite subfield. We show that $R$ contains a free noncyclic subsemigroup in the following cases: (1) $R$ satisfies a polynomial identity and is noncommutative modulo its prime radical. (2) $R$ has at least one nonartinian primitive subquotient. (3) $k$ is uncountable and $R$ is noncommutative modulo its Jacobson radical. In particular, (1) and (2) verify Klein's conjecture for numerous well known classes of domains, over countable fields, not covered in the prior literature.

Equations

$$d:=a_{1}b{c}_1+a_{2}b{c}_2+\cdots +a_{t}b{c}_t$$

Full text

In 1977, Lichtman conjectured that the group of units of a noncommutative division algebra always contains a noncyclic free subgroup [12]. Since then, an extensive, broad, and ongoing literature has developed on the existence of free subobjects of associative algebras. The reader is referred, e.g., to [9] for an overview of this literature and, e.g., to [7] and references therein for more recent results.

Our focus in this note is on free subsemigroups of associative algebras. (Henceforth, references to free subsemigroups and free subgroups will assume them to be multiplicative and noncyclic.) In 1984, Makar-Limanov conjectured that a noncommutative division algebra must contain a free subsemigroup, and he proved the same for division algebras over uncountable fields [14]. In 1992, Klein proved that a noncommutative domain with an uncountable center must contain a free subsemigroup, and he conjectured the same for all noncommutative domains [10]. Chiba proved in 1995 that a polynomial extension of a division algebra must contain a free subsemigroup (and also that a division algebra over an uncountable field must contain a free subgroup) [5]. In 1996, Reichstein showed that an algebra over an uncountable field contains a free subsemigroup (or subalgebra) if the same holds true after an extension of the scalar field [18] (cf. Smoktunowicz’s constructions [19]).

Our primary aim, then, is to establish the existence of free subsemigroups for some well-known classes of algebras not covered in the prior literature.

We begin with an elementary but useful “specialization” lemma, adapted from Passman [16, §2].

The proof of the following employs a reduction to the $k$-affine case, and I am grateful to Ken Brown for this approach.

Next, since $R$ is semiprime and PI, it follows that $RbR$ cannot be a nil ideal (see, eg., [15, 13.2.6i]). Consequently, there exist $a_{1},c_{1},a_{2},c_{2},\ldots,a_{t},c_{t}\in R$ such that

[TABLE]

is not nilpotent.

Set $R^{\prime}=k\langle x,y,b,a_{1},c_{1},\ldots,a_{t},c_{t}\rangle$, a $k$-affine (not necessarily semiprime) PI algebra. Since $d$ is not nilpotent, the ideal $R^{\prime}bR^{\prime}$ of $R^{\prime}$ is not nil. But it was proved by Amitsur that the Jacobson radical of a $k$-affine PI algebra must be nil [1]. (Nilpotency was later established by Braun [3].) Therefore, $R^{\prime}bR^{\prime}$ cannot be contained in the Jacobson radical of $R^{\prime}$. Hence $xy-yx$ is not contained in the Jacobson radical of $R^{\prime}$, and we can conclude that $R^{\prime}$ is noncommutative modulo its Jacobson radical.

We now know that there exists at least one primitive ideal $P^{\prime}$ of $R^{\prime}$ such that $R^{\prime}/P^{\prime}$ is noncommutative, and so by Kaplansky’s Theorem (see, e.g., [15, 13.3.8]), $R^{\prime}/P^{\prime}$ must be a noncommutative central simple algebra, finite dimensional over a field extension of $k$. Therefore, by (3), $R^{\prime}/P^{\prime}$ contains a free subsemisubgroup and, by (2), $R$ contains a free subsemigroup. ∎

We can also use (5) to show that Makar-Limanov’s conjecture, over $k$, is equivalent to an a priori more general statement.

We conclude with a modest application to algebras over uncountable fields.