Published online by Cambridge University Press: 20 November 2018
We investigate the problem of determining when   $\text{IA}({{F}_{n}}({{\mathbf{A}}_{m}}\mathbf{A}))$  is finitely generated for all
 $\text{IA}({{F}_{n}}({{\mathbf{A}}_{m}}\mathbf{A}))$  is finitely generated for all   $n$  and
 $n$  and   $m$ , with
 $m$ , with   $n\ge 2$  and
 $n\ge 2$  and   $m\ne 1$ . If
 $m\ne 1$ . If   $m$  is a nonsquare free integer then
 $m$  is a nonsquare free integer then   $\text{IA}({{F}_{n}}({{\mathbf{A}}_{m}}\mathbf{A}))$  is not finitely generated for all
 $\text{IA}({{F}_{n}}({{\mathbf{A}}_{m}}\mathbf{A}))$  is not finitely generated for all   $n$  and if
 $n$  and if   $m$  square free integer then
 $m$  square free integer then   $\text{IA}({{F}_{n}}({{\mathbf{A}}_{m}}\mathbf{A}))$  is finitely generated for all
 $\text{IA}({{F}_{n}}({{\mathbf{A}}_{m}}\mathbf{A}))$  is finitely generated for all   $n$ , with
 $n$ , with   $n\ne 3$ , and
 $n\ne 3$ , and   $\text{IA}({{F}_{3}}({{\mathbf{A}}_{m}}\mathbf{A}))$  is not finitely generated. In case
 $\text{IA}({{F}_{3}}({{\mathbf{A}}_{m}}\mathbf{A}))$  is not finitely generated. In case   $m$  is square free, Bachmuth and Mochizuki claimed in ([7], Problem 4) that
 $m$  is square free, Bachmuth and Mochizuki claimed in ([7], Problem 4) that   $\text{TR}({{\mathbf{A}}_{m}}\mathbf{A})$  is 1 or 4. We correct their assertion by proving that
 $\text{TR}({{\mathbf{A}}_{m}}\mathbf{A})$  is 1 or 4. We correct their assertion by proving that   $\text{TR}({{\mathbf{A}}_{m}}\mathbf{A})=\infty$ .
 $\text{TR}({{\mathbf{A}}_{m}}\mathbf{A})=\infty$ .