Krdytkyu

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 $P(X)inker v_a$ 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 $uin F$. 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?