Existence of Splitting Field with degree less than n!

If K is a field and f\in K[X] has degree n\geq 1, then there exists a splitting field F of f with [F:K]\leq n!.


We use induction on n=\deg f.

Base case: If n=1, or if f splits over K, then F=K is a splitting field with [F:K]=1\leq 1!.

Induction Hypothesis: Assume the statement is true for degree n-1, where n>1.

If n=\deg f>1 and f does not split over K, let g\in K[X] be an irreducible factor of f with \deg g>1. Let u be a root of g, then \displaystyle [K(u):K]=\deg g>1.

Write f=(x-u)h with h\in K(u)[X] of degree n-1. By induction hypothesis, there exists a splitting field F of h over K(u) with [F:K(u)]\leq(n-1)!.

That is, h=u_0(x-u_1)\dots(x-u_{n-1}) with u_i\in F and F=K(u)(u_1,\dots,u_{n-1})=K(u,u_1,\dots,u_{n-1}). Thus f=u_0(x-u)(x-u_1)\dots(x-u_{n-1}), so f splits over F.

This shows F is a splitting field of f over K of dimension
\begin{aligned}  [F:K]&=[F:K(u)][K(u):K]\\  &\leq (n-1)!(\deg g)\\  &\leq n!  \end{aligned}

Author: mathtuition88


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