Let \(F\) be a field and \(g(x) \in F[x]\) have degree \(n > 0\).
Define “\(E\) is a splitting field for \(g(x)\) over \(F\).”
Prove, without using the Fundamental Theorem of Galois Theory, that if \(E\) is a splitting field for \(g(x)\) over \(F\) then \([E : F] \le n!\).
Solution
1. A splitting field of \(g\) over \(F\) is an extension \(E/F\) such that\[ g(x)=a\prod_{i=1}^{n}(x-\alpha_i) \quad\text{with }a\in F^\times,\ \alpha_i\in E, \qquad E=F(\alpha_1,\ldots,\alpha_n). \]The roots may repeat.
2. Every degree-\(n\) polynomial over any field has a splitting fieldof degree at most \(n!\) over that field; in particular the given \(E\)satisfies \([E:F]\leq n!\).
We prove the degree bound for every splitting field by induction on \(n\), uniformly over all coefficient fields. Existence follows by successively adjoining roots, using a quotient by an irreducible factor at each step.
For \(n=1\), the only root already lies in the coefficient field, and the generation condition gives \(E=F\). Thus \([E:F]=1=1!\).
Suppose \(n\geq2\) and the bound is known in degree \(n-1\). Choose a root \(\alpha\in E\) and put \(F_1=F(\alpha)\). The minimal polynomial of \(\alpha\) over \(F\) divides \(g\), so \[ [F_1:F]\leq n. \] Division by \(x-\alpha\) in \(F_1[x]\) gives \[ g(x)=(x-\alpha)h(x),\qquad \deg h=n-1. \] All roots of \(h\) lie in \(E\), and adjoining them to \(F_1\) generates \(E\), because together with \(\alpha\) they include all the roots of \(g\). Thus \(E\) is a splitting field of \(h\) over \(F_1\). The induction hypothesis gives \([E:F_1]\leq(n-1)!\). Applying the tower law yields \[ [E:F]=[E:F_1][F_1:F]\leq(n-1)!\,n=n!. \]