p.56 The notation in Exercise 3-2 is that in Example 3.22 (p.50). See also Proposition 5.8.

p.139 In Remark 8.12, choose a generator $\alpha_i$ for each $A_i$ but omit 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).

p.172. The solution to Exercise 1-6 only shows that there does not exist a monic polynomial of degree $\geq2$ that is irreducible modulo all primes (if $f$ is not monic, then it may become linear modulo $p$, and so having a root in $\mathbb{F}{}_{p}$ does not imply that it is reducible).
Here is a corrected solution. Suppose that the polynomial \[ f(X)=a_{m}X^{m}+\cdots+a_{2}X^{2}+a_{1}X+a_{0}\in\mathbb{Z}[X],\quad m\geq2,\quad a_{m}\neq0, \] is irreducible (in particular, nonconstant) modulo $p$ for all primes $p$. Let $S$ be the (finite) set of primes dividing all the coefficients $a_{m},\ldots,a_{2}% $. If $p\in S$, then $f(X)$ modulo $p$ is a polynomial of degree $1$, and so there exists an $n_{p}\in\mathbb{Z}{}$ such that $f(n_{p})$ is not divisible by $p$. By the Chinese Remainder Theorem, there exists an $n\in\mathbb{Z}{}$ such that \[ n\equiv n_{p}\text{ modulo }p,\quad\text{all }p\in S. \] In fact, there is an infinite collection of such $n$, and so we may choose one that is not a root of $(f(X)+1)(f(X)-1)$. Now $f(n)$ is not $\pm1$ and is not divisible by any prime in $S$, and so it is divisible by a prime $p$ not in $S$. Then $f(X)$ mod $p$ has a root in $\mathbb{F}_{p}$ and has degree $\geq2$, and so it is reducible. (Ilya Tyomkin)

p.183. Exercise A-52. Both (a) and (b) are true. Let $|F|=p^m$ and $|E|=p^n$. For (a), we can map $E$ isomorphically onto a subfield of an algebraic closure of $F=\mathbb{F}_{p^m}$ and apply Proposition 4.23. Statement (b) is true because both groups are cyclic and $p^m-1|p^n-1$ if and only if $m|n$.