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Suppose $f \in K[x]$ is a polynomial with degree $n$, $f = (x-\alpha_1)...(x-\alpha_n)$ over the algebraic colsure. Let $L=K(\alpha_1,...,\alpha_n)$ be the splitting field of $f$. Prove that $[L:K]$ divides $n!$.
I was able to prove it for the case where $f$ is seperable. In this case, $L/K$ is a Galois extension and therefore $[L:K] = |Gal(L/K)|$ and since $Gal(L/K)$ is isomoprhic to a subgroup of $S_n$, the result follows. How do I prove it for the general case?

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Proceed by induction on $n = [L:K]$. If $[L:K] = 1$ then the claim is trivial.

Assume $[L:K] > 1$. We consider separately the cases where $f$ is irreducible and where $f$ is not irreducible.

Suppose first that $f$ is irreducible. Let $\alpha$ be a root of $f$ in a splitting field. Then $f$ is the minimum polynomial of $\alpha$ over $K$, so $[K(\alpha):K] = n$. Then $[L : K(\alpha)] < [L :K]$ and $L$ is the splitting field of $g(x) = f(x)/(x - \alpha)$ over $K(\alpha)$. The degree of $g$ is $n - 1$, so by induction $[L : K(\alpha)] \mid (n-1)!$, and the result follows by the tower law.

Suppose now that $f$ is not irreducible. Then $f = pg$ for $p, g \in K[x]$ with $p$ irreducible. If $L$ is a splitting field for $p$ over $K$, then we are done by the previous paragraph. If not, we may take $K \subset M \subset L$ such that $M$ is a splitting field for $p$ over $K$ (just adjoin the roots of $p$ in L).

Then $M$ is a splitting field for $p$ over $K$, and $L$ is a splitting field for $g$ over $M$. Since our tower of fields has strict inclusions, $[L:K] = [L:M][M:K]$ is a proper factorisation, so by induction $[L:M] \mid (\deg g)!$ and $[M :K] \mid (\deg p)!$. If we define $k = \deg p$, then $n - k = \deg g$, so we have $$ [L:K] = [L:M][M:K] \mid k!(n-k)! $$

And the result follows because $k!(n-k)! \mid n!$ (which is true because binomial coefficients are integers).

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