14
$\begingroup$

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 $n!$?

$\endgroup$
  • 3
    $\begingroup$ Hint: Can you embed $Gal(K/F)$ into $S_n$? $\endgroup$ – Soarer Sep 8 '11 at 6:58
  • 11
    $\begingroup$ @soarer: nice hint, but it only works for Galois extensions. $\endgroup$ – Georges Elencwajg Sep 8 '11 at 8:30
  • 1
    $\begingroup$ @GeorgesElencwajg sorry, this might be stupid, but when is a splitting field not a Galois extension? $\endgroup$ – Alex Reinking Feb 19 '15 at 5:58
  • 5
    $\begingroup$ @Alex: when it is not separable (which can only happen in characteristic $p$) . $\endgroup$ – Georges Elencwajg Feb 19 '15 at 7:53
21
$\begingroup$

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 $L$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – PropositionX Apr 27 '17 at 2:53
  • 3
    $\begingroup$ 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 $(a+b)!=n!$. $\endgroup$ – Zev Chonoles Apr 27 '17 at 3:38
  • $\begingroup$ 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? $\endgroup$ – PropositionX Apr 27 '17 at 7:04
  • 1
    $\begingroup$ 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 $(n+1)!$. $\endgroup$ – PropositionX Apr 27 '17 at 7:24
  • $\begingroup$ This is probably very elementary but why does $a!*b!$ divide $(a+b)!$ ? $\endgroup$ – Dude1662 Feb 26 '19 at 0:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.