Prove that if $\alpha \in E$ and $f(\alpha) = 0$, then $E = E^H(\alpha)$.

Question

Let $F \leq C$. Suppose that $f(x) \in F[x]$ is monic, irreducible over $F$ and $\deg f(x) = 6$.

Let $E$ be the splitting field of $f(x)$ over $F$ and let $G = \text{Gal}(E/F)$.

Assume that $[E:F]=12$ and assume that there exists $\sigma\in G$ such that $|\sigma|=3$.

Let $H=⟨σ⟩$ and $K=E^H$.

a) Prove that if $\alpha \in E$ and $f(\alpha) = 0$, then $E = K(\alpha)$.

b) Determine how $f(x)$ factors as a product of irreducible polynomials in $K[x]$. What is the number of irreducible factors of $f(x)$ in $K[x]$ and what are their degrees?

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My attempt was to use the Fundamental Theorem of Galois Theory, so $[E:K] = |H| = |\sigma| = 3$, and $[K:F] = \frac{|G|}{|H|} = 12/3 = 4$.

Then by Tower law, $[E:K] = [E:K(\alpha)][K(\alpha):K]=3$, so I want to try to prove $[K(\alpha):K]=3$, but have no luck so far. Does it have something to do with $f(x)$ (which is also $m_{\alpha, F}(x)$)?

Any hints would be much appreciated, thanks!

Answer 1

Answer the following questions, and combine them:

• Is it possible that $\alpha\in K$? What does the answer to this question tell you about the degree $[K(\alpha):K]$?
• Why is $K(\alpha)$ a subfield of $E$? Apply the tower law to $E/K(\alpha)/K$. What alternatives does this leave to the degree $[K(\alpha):K]$?

Answer 2

For part (a) my attempt:
Assume $\alpha \in E$ and $f(\alpha) = 0$.
Know that $f(x) \in F[x]$ is monic, irreducible over $F$.
Then, $f(x) = m_{\alpha, F}(x)$ which is the minimal polynomial of $\alpha$ over $F$.
So, $[F(\alpha):F] = \text{deg } f(x) = \text{deg } m_{\alpha, F}(x) = 6$.
Notice that $|G| = [E:F] = 12$.
Also, $f(x) \in F[x]$ and $E$ be the splitting field of $f(x)$ over $F$ and $G = Gal(E/F)$.
Then, we know that $G$ is a Galois group of $f(x)$ over $F$.
Since, $\text{deg } f(x) = 6 \geq 2$. Then, $G$ is isomorphic to a subgroup of $S_{4}$.
Which notice that $G$ is isomorphic to $A_{4}$ (i.e. the alternating group with $|A_4| = 12$).
Since, $\sigma \in G$ such that $|\sigma| = 3$ and $H = \langle \sigma \rangle$.
Then, $|H| = 3$. Since, we have $K = E^H$.
Then, by the Fundamental Theorem of Galois Theory, we know that $H = Gal(E/K)$, and $|H| = [E:K] = 3$ and $[K:F] = [G:H] = \frac{|G|}{|H|} = \frac{12}{3} = 4$.
Then, from the Answer given by @Jyrki Lahtonen before and by Tower Law, $[E:K]=3=[E:K(\alpha)][K(\alpha):K]$.
Since, $3$ is prime. So, we can have either $[E:K(\alpha)] = 3$ and $[K(\alpha):K] = 1$ or vise versa.
But notice that $[K(\alpha):K] = 1 = [K:K]$ means that $K(\alpha) = K$.
By Tower Law, $4 = [K:F] = [K:F(\alpha)][F(\alpha):F] = [K:F(\alpha)] \times 6$.
Which is impossible since $6 \nmid 4$.
So, $[E:K(\alpha)] = 3$ and $[K(\alpha):K] = 1$ is false.
Hence, $[E:K(\alpha)] = 1$ and $[K(\alpha):K] = 3$ is correct.
So, By Tower Law, $[E:F] = 12 = [E:K][K:F] = 3 \cdot 4 = [K(\alpha):K][K:F]$.
Therefore, $E=K(\alpha)$.

$\textbf{Is my prove correct ?}$
$\textbf{If not, please help me to edit it to make it look better ! Thanks !}$