## 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 $[F:K]\leq n!$.

Proof:

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} 