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.

  • $\begingroup$ Why do we need the second paragraph? In the first paragraph you confirmed that any element of Gal(L/K) is a permutation, so then we can already make the conclusion that it is isomorphic to a subgroup of S_n right? $\endgroup$ – The Coding Wombat Apr 3 at 11:44
  • $\begingroup$ @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. $\endgroup$ – SMM Apr 3 at 12:01

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