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Let $f(x)\in F[x]$ be a polynomial of degree $n$. Let $K$ be a splitting field of $f(x)$ over $F$. Then [K:F] must divides $n!$.

I only know that $[K:F] \le n!$, but how can I show that $[K:F]$ divides $n!$?

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Hint: Can you embed $Gal(K/F)$ into $S_n$? –  Soarer Sep 8 '11 at 6:58
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@soarer: nice hint, but it only works for Galois extensions. –  Georges Elencwajg Sep 8 '11 at 8:30
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1 Answer

up vote 6 down vote accepted

Hint: Try induction on $n$. The base case is clear; in the inductive step, we will want to start with a degree $n+1$ polynomial $f$, and somehow reduce to the case of a degree $\leq n$ polynomial. There are two cases: $f$ is irreducible, and $f$ is reducible.

Suppose $f$ is reducible. Let $p$ be an irreducible factor of $f$, so that $1\leq \deg(p)\leq n$, and let $L$ be the splitting field of $p$ over $F$. Then $K$ is the splitting field of $\frac{f}{p}$ over $L$, and $\deg(\frac{f}{p})=\deg(f)-\deg(p)$. Note that $a!\times b!$ always divides $(a+b)!$ (this is equivalent to the binomial coefficients being integers).

Suppose $f$ is irreducible. Then letting $L=K[x]/(f)\cong K(\alpha)$ for some root $\alpha$ of $f$, we have that $[L:F]=n+1$. Now consider $\frac{f}{x-\alpha}$ (which is of degree $n$) as a polynomial over $L$.

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