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?



Free subgroups in algebras of polynomial growth
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.


