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Math Help - divisor

  1. #1
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    divisor

    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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  2. #2
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    Quote Originally Posted by apple2009 View Post
    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.
    Last edited by aliceinwonderland; April 19th 2010 at 02:24 AM.
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  3. #3
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    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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