Field Extension, Splitting Field and Galois Theory

Question

Let $f(x)\in \mathbb{Q}[x]$ be an irreducible polynomial of degree $n\geq 3$. Let $L$ be the splitting field of $f$, and let $\alpha \in L$ be a root of $f$. Given that $[L:\mathbb{Q}]=n!$, prove that $\mathbb{Q}(\alpha^4)=\mathbb{Q}(\alpha).$

This was a past exam question from the qualifying exam of UC Berkeley

Answer 1

Let $\alpha_{1} = \alpha, \alpha_{2}, \dots, \alpha_{n}$ be the roots of $f$, and let $\Omega = \{ \alpha_{1}, \dots, \alpha_{n} \}$. The Galois group $G \cong S_{n}$ acts transitively on $\Omega$, so $\mathbb{Q}(\alpha)'$ (the subgroup of $G$ corresponding to $\mathbb{Q}(\alpha)$ in the Galois correspondence) is $S_{n-1}$. Now $S_{n}$ is $2$-transitive as $n \ge 2$, and thus primitive, so the $1$-point stabilizer $S_{n-1}$ is maximal in $S_{n}$.

It follows that there are two possibilities for $\mathbb{Q}(\alpha^4) \subseteq \mathbb{Q}(\alpha)$: either it is $\mathbb{Q}(\alpha)$, or it is $\mathbb{Q}$.

If the latter holds, then $\alpha$ is a root of a polynomial $g = x^{4} - c$, for some $c \in \mathbb{Q}$, and the roots of $g$ are $\alpha, -\alpha, i \alpha, -i \alpha$. Clearly $g$ has no roots in $\mathbb{Q}$ (otherwise $\pm \alpha \notin \mathbb{Q}$, or $i \alpha \in \mathbb{Q}$, so that $\alpha^{2} \in \mathbb{Q}$, and in any case $n < 3$), so either it splits in $\mathbb{Q}[x]$ as a product of two irreducibles of degree $2$ (which is excluded by assumption), or it is irreducible in $\mathbb{Q}[x]$, and thus $g = f$.

But then the group $G$ is not $S_{4}$, but rather of order at most $8$, as $L = \mathbb{Q}(\alpha, i)$, so that $\lvert L : \mathbb{Q} \rvert \le \lvert \mathbb{Q}(\alpha)(i) : \mathbb{Q}(\alpha) \rvert \cdot \lvert \mathbb{Q}(\alpha) : \mathbb{Q} \rvert \le 2 \cdot 4 = 8$.

Answer 2

The proof is shorter in the case $n \geq 4$ than for $n=3$.

Let the roots of $f$ be $\alpha=\alpha_1, \alpha_2, \dots, \alpha_n$.

The numbers $\alpha_1^4, \alpha_2^4, \dots, \alpha_n^4$ are all conjugates of $\alpha^4$. If we can prove that they're distinct, then $\alpha^4$ will have $n$ conjugates, from which we deduce that $[\mathbf{Q}(\alpha^4):\mathbf{Q}]=n$ and the result follows.

If there were two $\alpha_j^4$'s that weren't distinct (say $\alpha_1^4 = \alpha_2^4$), then we'd have $\alpha_2 \in \{-\alpha_1, i\alpha_1, -i\alpha_1 \}$. Now since $[L:\mathbf{Q}] = n!$, every permutation of the roots $\alpha_1, \alpha_2, \dots, \alpha_n$ corresponds to an automorphism of $L$. In particular, we can fix $\alpha_1$ and still, over $\mathbf{Q}(\alpha_1)$, $\alpha_2$ has $n-1$ conjugates, hence degree $n-1$ over $\mathbf{Q}(\alpha_1)$. But each of the numbers $-\alpha_1$, $i\alpha_1$, $-i\alpha_1$ has degree $1$ or $2$ over $\alpha_1$.

In view of the above, the only possibility left is $n = 3$ and $\alpha_2 = \pm i\alpha_1$, say $\alpha_2 = i\alpha_1$. In that case, the cyclic automorphism $\sigma$ determined by $\alpha_1 \mapsto \alpha_2 \mapsto \alpha_3 \mapsto \alpha_1$ has order $3$, but also maps $i = \frac{\alpha_2}{\alpha_1} \mapsto \frac{\alpha_3}{\alpha_2} \mapsto \frac{\alpha_1}{\alpha_3} \mapsto \frac{\alpha_2}{\alpha_1}$. Since $\sigma$ has order 3, we have $\sigma(i) \ne -i$, so $\sigma$ must fix $i$. Then we obtain successively $\alpha_2 = i\alpha_1$, $\alpha_3 = -\alpha_1$, $\alpha_1 = -i\alpha_1$, a contradiction.