Show that if \(f\in K[x]\) has degree \(n\) and \(F\) is a splitting field of \(f\) over \(K\), then \([F:K]\) divides \(n!\).
Solution
We argue by strong induction on \(n=\deg f\).
1. The claim holds for \(n\le1\).
If \(n=0\) or \(1\), the polynomial already splits over \(K\), so \(F=K\) and \([F:K]=1\), which divides \(n!\).
Assume \(n\ge2\) and that the assertion holds for every polynomial of degree strictly smaller than \(n\) over every field.
2. If \(f\) is irreducible over \(K\), then \([F:K]\) divides \(n!\).
Choose a root \(\alpha\in F\). Irreducibility gives \[ [K(\alpha):K]=n. \] Over \(K(\alpha)\) one may write \[ f(x)=(x-\alpha)g(x), \qquad \deg g=n-1. \] The field \(F\) is a splitting field of \(g\) over \(K(\alpha)\): adjoining the roots of \(g\) to \(K(\alpha)\) adjoins all roots of \(f\).
By the induction hypothesis, \[ [F:K(\alpha)]\mid(n-1)!. \] The tower law therefore gives \[ [F:K] =[F:K(\alpha)]\,[K(\alpha):K] =n[F:K(\alpha)] \mid n(n-1)!=n!. \]
3. If \(f\) is reducible over \(K\), then \([F:K]\) divides \(n!\).
Write \[ f(x)=g(x)h(x) \] with \[ r=\deg g>0, \qquad s=\deg h>0, \qquad r+s=n. \] Let \(E\subseteq F\) be the subfield generated over \(K\) by all roots of \(g\). Then \(E\) is a splitting field of \(g\) over \(K\), so by induction \[ [E:K]\mid r!. \]
Over \(E\), the field \(F\) is obtained by adjoining the roots of \(h\), hence is a splitting field of \(h\) over \(E\). Again by induction, \[ [F:E]\mid s!. \] Thus \[ [F:K]=[F:E][E:K]\mid r!s!. \] Finally, \[ \frac{n!}{r!s!}=\binom{n}{r}\in\ZZ, \] so \(r!s!\mid n!\). Hence \([F:K]\mid n!\).
4. Therefore the assertion holds for every degree \(n\).
Every polynomial is either irreducible or reducible. The two cases are covered by 2 and 3, completing the strong induction.