Question from Section -5 Subsection -Splitting Fields Of Hungerford Algebra I am trying some questions from Textbook hungerford algebra and was unable to solve this particular problem. I am self studying so can't ask anyone else.
If $f \in K[x]$ has degree n and F is a spitting field of f over K, then [F : K] divides
n!.
I let degree of f =1 then it splits in K and hence assertion is true. but induction can't be used(clearly).
Can you please tell how to approach this question? I am at loss of ideas on why should always this happen.
Thank You!!
 A: We can use strong induction on $\mathrm{deg}(f)$.
Assume that we know that for all $k \leq n$, we have that the splitting field of a degree $k$ polynomial divides $k!$.
Let $\mathrm{deg}(f)=n+1$. If $f$ is not irreducible, we may factor $f=gh$ where $\mathrm{deg}(g)=a,\mathrm{deg}(h)=b$ satisfy $a,b \leq n$, so by the induction hypothesis gives that if $F'$ is splitting field of $g$ over $K$ and $F$ is the splitting field of $h$ over $F'$, then $F$ is also the splitting field of $f$ over $K$. We have by the induction hypothesis $[F:K]=[F:F'][F':K] \mid a!b! \mid (a+b)!=(n+1)!$. Here we used that the binomial coefficient $\binom{a+b}{a}=\frac{(a+b)!}{a!b!}$ is an integer.
So suppose that $f$ is irreducible over $K$. Let $\alpha$ be a root of $K$ in $F$. Then we can write $f=(x-\alpha)g$ in $K[\alpha][x]$ where $g$ has degree $n$. As $f$ is irreducible, we have $[K[\alpha]:K]=n+1$. $F$ is the splitting field of $g$ over $K[\alpha]$, so by the induction hypothesis, we have $[F:K[\alpha]] \mid n!$, thus $[F:K]=[F:K[\alpha]][K[\alpha]:K] \mid n!(n+1)=(n+1)!$
