Let be $F = \mathbb{Z} / 5 \mathbb{Z}$ and $P(X) = X^3 + 2X + 1 \in F[X]$.
(a) Let be $\alpha$ a root of $P(X)$ on a splitting field of $P(X)$ over $F$ and let be $K = F[\alpha]$. Compute the cardinality of $K$.
(b) Prove that $F[\beta] = K$ for every $\beta \in K \ \backslash \ F$.
(c) Prove that $K$ contains a splitting field of $X^{31} - 1$ over $F$.
I would like to know if my solution of item $a$ is correct and some hints on items $b$ and $c$. This is what I thought:
(a) Let be the homomorphism $v_{\alpha}: F[X] \longrightarrow F[\alpha]$ given by $v_{\alpha}(P(X)) := P(\alpha)$. By the First Isomorphism Theorem,
$$F[X]/ \text{Ker} \ v_{\alpha} \cong F[\alpha]$$
I tried to show that $\text{Ker} \ v_{\alpha} = (P(X))$. On the one hand $v_{\alpha}(P(X)) = P(\alpha) = 0$ since $\alpha$ is a root of $P(X)$ by hypothesis, then $(P(X)) \subset \text{Ker} \ v_{\alpha}$. On the other hand, $F = \mathbb{Z} / 5 \mathbb{Z}$ is a field, therefore $F[X]$ is an Euclidean domain, in particular, it's a Principal Ideal Domain, then $\text{Ker} \ v_{\alpha} = (A(X))$ for some $A(X) \in \text{Ker} \ v_{\alpha}$. Applying the Euclide's algorithm for division, exist $Q(X), R(X) \in F[X]$ such that
$$A(X) = Q(X)P(X) + R(X), \ R \equiv 0 \ \text{or} \ \text{degree} (R(X)) < \text{degree} (P(X)) = 3$$
I know that $\text{degree} (R(X)) \neq 0,1$, otherwise, $\alpha \in F$, which is an absurd since $P(X)$ is irreducible in $F[X]$ (it's just apply Eisenstein's criterion for $P(X-1)$ for $p = 3$ to observe this) and $\text{degree} (R(X)) \neq 2$, otherwise, $\alpha^2 \in F$, which imply $\alpha = \alpha^6 = (\alpha^2)^3 \in F$, which is an absurd for the same reason explained previously, then $P(X)$ divides $A(X)$, therefore $\text{Ker} \ v_{\alpha} = (A(X)) \subset (P(X))$ and
$$F[X] / (P(X)) \cong F[\alpha].$$
Looking the quotient, I know that every $R(X) \in F[\alpha]$ has the form $aX^2 + bX + c$, where $a, b, c \in F$, then $|K| = 5^3 = 125$.
(b) Since $\beta \in K \ \backslash F$ and $F \subset F[\alpha]$, it's clear that $F[\beta] \subset F[\alpha]$. Observing exist $B(X) = a_n X^n + \cdots + a_1 X + a_0$ for some $a_i \neq 0$ with $i \in \{ 1, \cdots, n \}$ such that $\beta = B(\alpha) = a_n (\alpha)^n + \cdots + a_1 (\alpha) + a_0$ by the fact that $\beta \in K \ \backslash \ F$, but I don't know how to use this to show that $\alpha \in F[\beta]$.
(c) I don't sure how to start this exercise, but I imagine that I need use item $b$.
$\textbf{EDIT:}$
Following the hints, I tried do the following for items $b$ and $c$:
(b) Since $F[\beta]$ it's a subspace of $K$ over $F$, $[F[\beta] : F] \leq [F[\alpha] : F] = 3$. Let be $m_{\beta}(X)$ the minimal polynomial of $\beta$ over $F$, then $\text{degree} \ (m_{\beta}(X)) \neq 1$, otherwise, $\beta \in F$, which is an absurd since $\beta \in K \backslash F$ by hypothesis.
(c) $K - \{ 0 \}$ it's a group with order |K - { 0 }| = 5^3 - 1 = 124. By Sylow's theorem, we ensure the existence and uniqueness of a $31$-Sylow subgroup $S_{31}$ and $S_{31}$ it's a cylic group, therefore exist only one $x \in K - \{ 0 \}$ such that $o(x) = 31$, all the roots of $X^{31} - 1$ are in $S_{31} = \langle x \rangle$, then $K \supset K - \{ 0 \} \supset \langle x \rangle$, i.e., $K$ it's a field extension that contains all roots of $X^{31} - 1$. Since the splitting field of $X^{31} - 1$ is the smallest field extension that contains all roots of $X^{31} - 1$, $K$ must be contains the splitting field of $X^{31} - 1$.
For part a, once you've shown that $P(X)$ is irreducible you can be done rather quickly; you know that $F[X]$ is a principal ideal domain so $\ker v_a=(A(X))$ for some $A(X)\in F[X]$, and you know that $P(X)\in\ker v_a$ so $P(X)=Q(X)A(X)$ for some $Q(X)\in F[X]$. Since $P(X)$ is irreducible it follows that $A(X)=uP(X)$ or $A(X)=u$ for some unit $u\in F$. If $A(X)=u$ then $$\ker v_a=(A(X))=(u)=F[X],$$ which is clearly false, so $A(X)=uP(X)$ and hence $\ker v_a=(uP(X))=(P(X))$. And then indeed every element of the quotient $F[X]/\ker v_a$ is of the form $aX^2+bX+c$, so $|K|=5^3$.
For part b, consider the minimal polynomial of $\beta$ over $F$. This is an irreducible polynomial with a zero in $K$, hence its degree is at most $3$. Show that it cannot be less than $3$.
Fort part c, note that $K-\{0\}$ is a multiplicative (abelian) group of order $5^3-1$. How many elements of order $31$ does it contain?