Let \(K\) be a field, let \(f\in K[x]\) have degree \(n\ge1\), and let \(F\) be its splitting field. Prove that \[ [F:K]\mid n!. \] In particular, \([F:K]\le n!\).
Solution
We prove the divisibility by induction on \(n=\deg f\).
If \(n=1\), the splitting field is \(K\).
Suppose first that \(f\) is irreducible of degree \(n\). Choose a root \(\alpha\in F\). Then \[ [K(\alpha):K]=n. \] Over \(K(\alpha)\) we can write \[ f=(x-\alpha)g, \qquad \deg g=n-1, \] and \(F\) is the splitting field of \(g\) over \(K(\alpha)\). By induction, \[ [F:K(\alpha)]\mid (n-1)!. \] Hence \[ [F:K]=[F:K(\alpha)]\,[K(\alpha):K]\mid (n-1)!\,n=n!. \]
Now suppose \(f\) is reducible. Write \[ f=gh \] with \(g,h\in K[x]\) nonconstant, say \(\deg g=d\) and \(\deg h=n-d\) with \(1\le d<n\). Let \(L\) be the splitting field of \(g\) over \(K\). By induction, \[ [L:K]\mid d!. \] Since \(g\) already splits over \(L\), the splitting field \(F\) of \(f\) is exactly the splitting field of \(h\) over \(L\). Applying induction again, now over the base field \(L\), gives \[ [F:L]\mid (n-d)!. \] Therefore \[ [F:K]=[F:L][L:K]\mid d!(n-d)!. \] Finally, \[ \frac{n!}{d!(n-d)!}=\binom nd\in\mathbb Z, \] so \(d!(n-d)!\mid n!\). Hence \[ \boxed{[F:K]\mid n!}. \] In particular, \([F:K]\le n!\).