Degree of Splitting Field Extension is Less Than Factorial of Polynomial Degree

Theorem

Let \(f \in \mathbb{F}[X]\) and \(\mathbb{K}\) be the splitting field of \(f\), with \(n = \deg(f)\). Then \([\mathbb{K} : \mathbb{Q}] \leq n!\).

Proof

Let \(\alpha_1, \dots, \alpha_n\) be the roots of \(f\). Note that if the extension is not separable, then the roots may not be completely distinct, but the list above should at least be exhaustive. With this notation \(\mathbb{K} = \mathbb{F}(\alpha_1, \dots, \alpha_n)\).

First note that because the minimal polynomial divides any polynomial with has the same root, we have that the minimal polynomial of \(\alpha_1\) has degree less than \(n\), and therefore by this result \([\mathbb{F}(\alpha_1) : \mathbb{F}] \leq n\). We now proceed to generalise this by induction. That is, assume that

\[ [\mathbb{F}(\alpha_1, \dots, \alpha_k) : \mathbb{F}] \leq \frac{n!}{(n - k)!}.\]

Now, note that we have a factorisation

\[ f(X) = (X - \alpha_1) \dots (X - \alpha_k) g(X)\]

where \(g(X) \in \mathbb{F}(\alpha_1, \dots, \alpha_k)[X]\). As such, we have that the degree of the minimal polynomial of \(\alpha_{k + 1}\) over \(\mathbb{F}(\alpha_1, \dots, \alpha_k)\) is less than the \(\deg(g) = \deg(f) - k = n - k\). This corresponds with the degree of \(\mathbb{F}(\alpha_1, \dots, \alpha_{k + 1})\) over \(\mathbb{F}(\alpha_1, \dots, \alpha_k)\), and hence

\[ [\mathbb{F}(\alpha_1, \dots, \alpha_{k + 1}) : \mathbb{F}(\alpha_1, \dots, \alpha_k)] \leq n - k.\]

Using this result and the inductive hypothesis we can deduce by the multiplicativity of extension degrees

\[\begin{align*} [\mathbb{F}(\alpha_1, \dots, \alpha_{k + 1}) : \mathbb{F}] &= [\mathbb{F}(\alpha_1, \dots, \alpha_{k + 1}) : \mathbb{F}(\alpha_1, \dots, \alpha_k)] [\mathbb{F}(\alpha_1, \dots, \alpha_k) : \mathbb{F}] \\ &\leq (n - k) \times \frac{n!}{(n - k)!} \\ &= \frac{n!}{(n - k - 1)!} \\ &= \frac{n!}{(n - (k + 1))!}. \end{align*}\]