Free subgroups in algebras of polynomial growth

Question

What is known about free non-abelian subgroups in finitely generated associative algebras of polynomial growth (e.g., over finite fields, to avoid finite-dimensional free subgroups)? For example, are there examples of ideals I in the group algebra A of the free group F such that A/I has polynomial growth but the natural map F\to A/I is injective?

Answer 1

Let $p$ be a prime number. Consider the algebra $A:=M_2((\mathbb Z/p\mathbb Z)[t])$ and the group $G:=(\mathbb Z/p \mathbb Z) \ast (\mathbb Z/p \mathbb Z) = \langle a,b \mid a^p,b^p \rangle$.

Look at the homomorphism $\varphi \colon G \to A^{\times}$ which is defined by $$\varphi(a):=\left( \begin{matrix} 1& t \\ 0 & 1 \end{matrix} \right), \quad \mbox{and} \quad \varphi(b):=\left( \begin{matrix} 1& 0 \\ t & 1 \end{matrix} \right).$$

It is easy to see that $\varphi$ is injective. Indeed, if for example $w = a^{n_1}b^{m_1} \cdots a^{n_k}b^{m_k}$ with $n_i,m_i \in \{1,\dots,p-1\}$, then the polynomial in the upper left corner of the matrix $\varphi(w)$ has degree $2k$ and leading coefficient equal to $n_1 \cdots n_k m_1 \cdots m_k \neq 0.$ The other cases are similar.

Note that $A$ is of polynomial growth and if $p \geq 3$, then $G$ contains a free subgroup.