# the degree of a splitting field of a polynomial

Let $f(x)\in F[x]$ be a polynomial of degree $n$. Let $K$ be a splitting field of $f(x)$ over $F$. Then [K:F] must divides $n!$.

I only know that $[K:F] \le n!$, but how can I show that $[K:F]$ divides $n!$?

• Hint: Can you embed $Gal(K/F)$ into $S_n$? – Soarer Sep 8 '11 at 6:58
• @soarer: nice hint, but it only works for Galois extensions. – Georges Elencwajg Sep 8 '11 at 8:30
• @GeorgesElencwajg sorry, this might be stupid, but when is a splitting field not a Galois extension? – Alex Reinking Feb 19 '15 at 5:58
• @Alex: when it is not separable (which can only happen in characteristic $p$) . – Georges Elencwajg Feb 19 '15 at 7:53

Hint: Try induction on $n$. The base case is clear; in the inductive step, we will want to start with a degree $n+1$ polynomial $f$, and somehow reduce to the case of a degree $\leq n$ polynomial. There are two cases: $f$ is irreducible, and $f$ is reducible.
Suppose $f$ is reducible. Let $p$ be an irreducible factor of $f$, so that $1\leq \deg(p)\leq n$, and let $L$ be the splitting field of $p$ over $F$. Then $K$ is the splitting field of $\frac{f}{p}$ over $L$, and $\deg(\frac{f}{p})=\deg(f)-\deg(p)$. Note that $a!\times b!$ always divides $(a+b)!$ (this is equivalent to the binomial coefficients being integers).
Suppose $f$ is irreducible. Then letting $L=F[x]/(f)\cong F(\alpha)$ for some root $\alpha$ of $f$, we have that $[L:F]=n+1$. Now consider $\frac{f}{x-\alpha}$ (which is of degree $n$) as a polynomial over $L$.
• May I please ask where does $a!\times b!$ comes from? Which field do we consider here? May I please ask for some more explicit explaination here? – PropositionX Apr 27 '17 at 2:53
• You want to show $[K:F]$ divides $n!$, where $n=\deg(f)$. If $f$ has a factor $p$ with $\deg(p)=a$, and we let $b=n-a$ and we let $L$ be the splitting field of $p$ over $F$, then $K$ is the splitting field over $L$ of a polynomial of degree $b<n$ (namely the polynomial $\frac{f}{p}$), and $L$ is the splitting field over $F$ of a polynomial of degree $a<n$ (namely the polynomial $p$), so by induction we have that $[K:L]$ divides $b!$ and $[L:F]$ divides $a!$, and therefore their product $[K:F]=[K:L][L:F]$ divides $a!\times b!$, which itself divides $(a+b)!=n!$. – Zev Chonoles Apr 27 '17 at 3:38
• Sorry I am still confused of the irreducible case. I think in this case we cannot say that $L=F(\alpha)$ is the splitting field of $x-\alpha$ as $F(\alpha)$ has degree greater then 1 so $[F(\alpha):F]$ does not divide $1!$. It seems to be something subtle. May I please ask how can I deal with that? – PropositionX Apr 27 '17 at 7:04
• Is that the fact that: $[K:F(\alpha)][F(\alpha):F]$ where $[K:F(\alpha)]$ divides $n!$ by inductive hypothesis and $[F(\alpha):F]=n+1$, we have $[K:F]$ divides $(n+1)!$. – PropositionX Apr 27 '17 at 7:24
• This is probably very elementary but why does $a!*b!$ divide $(a+b)!$ ? – Dude1662 Feb 26 '19 at 0:26