p.139 In Remark 8.12, choose a generator $\alpha_i$ for each $A_i$ but drop the claim that $\alpha=(\alpha_i)_i$ generates $A$. In fact, $A$ need not be monogenic as an $F$-algebra, as can be seen already for the $\mathbb{F}_3$-algebra $\mathbb{F}_3\times\mathbb{F}_3\times\mathbb{F}_3$. [Suppose that a generator $u=(u_{1},u_{2},u_{3})$ is given. By the cardinality of the set of generators of the field $\mathbb{F}_3$ as an $\mathbb{F}_3$-algebra, we know that two of the elements in the tuple $u$ are equal. We may suppose $u_{1}=u_{2}$. Then clearly $u$ cannot generate $A$ as an $\mathbb{F}_3$-algebra because for whatever $\mathbb{F}_3$-algebra operation beginning with $u$ you are going to generate always a tuple $v=(v_{1},v_{2},v_{3})$ with $v_{1}=v_{2}$. (Alejandro González Nevado).