# $[L:K]$ dividing $n!$ with $L=K(\alpha_1,…\alpha_n)$, and $L$ a splitting field of $f$ with $\alpha_1..\alpha_n$ zeros of $f$

I have just proven the following statement:

Let $$K$$ be a field and $$L$$ the splitting field of a separable polynomial $$f\in K[X]$$ of degree $$n$$. Denote the zeros of $$f$$ in $$L$$ by $$\alpha_1,\alpha_2,...,\alpha_n$$. Then for $$i=1,2,...,n$$: $$[K(\alpha_1,...,\alpha_i):K]\leq n(n-1)...(n-i+1).$$

I now need to show that $$[L:K]$$ divides $$n!$$.

What I figured out and I think I have to use:

• $$|\text{Gal}(L/K)|=[L:K]$$
• $$L/K$$ a normal and separable extension
• $$|\text{Gal}(L'/K)|$$ equals the number of distinct zeros of $$f$$ which lie in $$L$$ if $$L'/K$$ is a simple extension
• And as a hint: The Galoisgroup permutes zeros of polynomials

Denote by $$A=\{\alpha_1,\dots,\alpha_n\}$$ the set of zeroes of $$f(X)$$. Note that for $$\sigma\in Gal(L/K)$$ you have $$\sigma\upharpoonright A\in Sym(A)$$. Indeed, $$0= \sigma(0)= \sigma(f(\alpha_i))= f(\sigma(\alpha_i))$$, so $$\sigma(\alpha_i)\in A$$ for every $$i$$. Since $$\sigma$$ is 1-1 and $$A$$ is finite you get that $$\sigma$$ is a permutation on $$A$$.
Further, if $$\sigma,\tau\in Gal(K/L)$$ are such that $$\sigma\upharpoonright A=\tau\upharpoonright A$$, then $$\sigma=\tau$$. This follows as any element $$\beta$$ of $$L$$ may be written as $$\beta= p(\alpha_1,\dots,\alpha_n)/q(\alpha_1,\dots,\alpha_n)$$ for some polynomials $$p(X_1,\dots,X_n)$$ and $$q(X_1,\dots,X_n)$$ over $$K$$, so $$\sigma(\beta)=\tau(\beta)$$.
Thus, $$Gal(K/L)$$ is naturally identified with a subgroup of $$Sym(A)\cong S_n$$ which has $$n!$$ elements. The conclusion follows by Lagrange.
• @TheCodingWombat You need the second paragraph to conclude that this identification $Gal(L/K)\to Sym(A)$ given by $\sigma\mapsto \sigma\upharpoonright A$ is 1-1. – SMM Apr 3 at 12:01