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!

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- Apr 18th 2010, 06:28 PMapple2009divisor
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!

- Apr 19th 2010, 12:16 AMaliceinwonderland
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. - Apr 19th 2010, 06:33 AMNonCommAlg
prove it by induction over $\displaystyle n.$ it's obvious for $\displaystyle n=1.$ for $\displaystyle n > 1$ consider two cases:

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

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

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

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