Degree bounds for splitting fields - Imported Knowlpedia Content

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

[L:K] \le n!.

Remarks

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

[L:K] \le \prod_i [L_i:K] \le \prod_i (d_i)!,

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

Examples

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 $[L:\mathbb{Q}]=6=3!$ .

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 $24$ .

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 $(2!)(2!)=4$ .