divisor

**apple2009** · April 18th 2010, 06:28 PM

If p(x) is a polynomial of degree n in F[x] and if E is the splitting field of p(x) over F, then (E:F) is a divisor of n!

**aliceinwonderland** · April 19th 2010, 12:16 AM

Originally Posted by apple2009

If p(x) is a polynomial of degree n in F[x] and if E is the splitting field of p(x) over F, then (E:F) is a divisor of n!

If the polynomial p(x) is separable or characteristic of F is 0, then it can be shown directly like this.

If E is the splitting field of p(x) in F[x], then E is the Galois extension of F. Thus |Gal(E/F)| = [E:F]. Since Gal(E/F) is the subgroup of S_n (link), [E:F] should be a divisor of n! by Lagrange's theorem.

**NonCommAlg** · April 19th 2010, 06:33 AM

prove it by induction over $n.$ it's obvious for $n=1.$ for $n > 1$ consider two cases:

case 1: $p(x)$ is irreducible. in this case $p(x)=(x-a)q(x),$ for some $a \in E$ and $q(x) \in F(a)[x].$ we have $[F(a):F]=n$ because $p(x) \in F[x]$ is irreducible. also note that $E/F(a)$ is a splitting

field for $q(x)$ and thus, by induction hypothesis, $[E:F(a)]$ divides $(n-1)!.$ therefore $[E:F]=[E:F(a)][F(a):F]=n[E:F(a)] \mid n(n-1)!=n!.$

case 2: $p(x)$ is reducible. then $p(x)=q(x)r(x),$ for some $q(x),r(x) \in F[x]$ with $1 \leq \deg q(x)=m < n$ and $1 \leq \deg r(x)=k < n.$ we also have $m+k=n.$ let $E_1/F$ be the splitting field for

$q(x)$ which is contained in $E.$ then $E/E_1$ would be a splitting field for $r(x) \in E_1[x].$ thus, by induction hypothesis: $[E:F]=[E:E_1][E_1:F] \mid k! m! \mid (m+k)! = n!.$

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