Degree bounds for splitting fields

Let $K$ be a field and let $f(x)\in K[x]$ be a separable polynomial of degree $n$ (equivalently, $f$ has distinct roots in an algebraic closure ; see separable ⇔ distinct roots ). Let $L$ be the splitting field of $f$ over $K$. Then $L/K$ is finite, normal, and separable, hence Galois (see separable + normal = Galois ).

Theorem (factorial bound). If $f$ is separable of degree $n$, then

One conceptual proof: separability gives $n$ distinct roots in $L$, and the Galois group $\mathrm{Gal}(L/K)$ acts faithfully on this set of roots, yielding an injective homomorphism $\mathrm{Gal}(L/K)\hookrightarrow S_n$. Using degree = group order for finite Galois extensions gives $[L:K]=|\mathrm{Gal}(L/K)|\le |S_n|=n!$.

A useful refinement: if $f=\prod_i f_i$ with $f_i$ separable and $\deg(f_i)=d_i$, then writing $L_i$ for the splitting field of $f_i$, one has

since $L$ is contained in the compositum of the $L_i$.

Examples

1. Sharp bound for a cubic.
   Over $\mathbb{Q}$, $f=x^3-2$ is separable of degree $3$. Its splitting field is $\mathbb{Q}(\sqrt[3]{2},\zeta_3)$, and $[L:\mathbb{Q}]=6=3!$.

2. A quartic with smaller degree than $4!$.
   For $f=x^4-2\in \mathbb{Q}[x]$, the splitting field is $\mathbb{Q}(2^{1/4}, i)$, which has degree $8$, far below $24$.

3. Product of quadratics.
   For $f=(x^2-2)(x^2-3)\in \mathbb{Q}[x]$, the splitting field is $\mathbb{Q}(\sqrt2,\sqrt3)$ and $[L:\mathbb{Q}]=4$, matching the refined bound $(2!)(2!)=4$.