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

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 $[\mathbb{F}({\alpha }_1):\mathbb{F}]\le n$. We now proceed to generalise this by induction. That is, assume that

Now, note that we have a factorisation

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 $\mathrm{deg}(g)=\mathrm{deg}(f)-k=n-k$. This corresponds with the degree of $\mathbb{F}({\alpha }_1,\dots ,{\alpha }_{k+1})$ over $\mathbb{F}({\alpha }_1,\dots ,{\alpha }_k)$, and hence

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