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Finite Field Extensions

  1. May 18, 2009 #1
    1. The problem statement, all variables and given/known data
    Let F be a field, and suppose that alpha is algebraic over F. Prove that if [F(alpha):F] is odd, then F(a^2)=F(a). {For those unfamiliar with notation [] denotes degree of extension and F(alpha) means F adjoined with alpha.)


    3. The attempt at a solution
    Since [F(alpha):F] is odd the basis of F(alpha) consists of an odd number of elements. We also know that b in F(alpha) has the form c_0+c_1alpha+c_2alpha^2+...+c_n-1alpha^(n-1). I really have no idea how to proceed from here. Help would be much appreciated. Thanks.
     
  2. jcsd
  3. May 18, 2009 #2
    First of all, have you learned the Tower Law yet (i.e., if [tex] L \subseteq K \subseteq M [/tex] are fields, then [tex] [M : L] = [M : K] [K : L] [/tex])?

    If you understand this, then the problem is relatively easy. It should be pretty obvious that [tex] F(\alpha^2) \subseteq F(\alpha) [/tex]. Furthermore, [tex] F(\alpha^2) = F(\alpha) [/tex] if and only if [tex] [F(\alpha) : F(\alpha^2) ] = 1 [/tex]. Why is it true that if [tex] [F(\alpha) : F(\alpha^2)] \neq 1 [/tex], then [tex] [F(\alpha) : F(\alpha^2)] = 2 [/tex]? Using the tower law, why would the latter contradict your assumptions about [tex] [F(\alpha):F] [/tex]?
     
  4. May 18, 2009 #3
    Thanks for the help, but sorry we haven't learned about the "tower law." Can you help me with the part where you said that [F(a):F(a^2)]=2 if not 1.
     
  5. May 19, 2009 #4
    Well, for any field [tex] K [/tex] and any [tex] \beta [/tex], the degree of [tex] K(\beta) [/tex] over [tex] K [/tex] is the degree of the minimal polynomial of [tex] \beta [/tex] over [tex] K [/tex]. We know that [tex] \alpha [/tex] satisfies the polynomial [tex] f(x) = x^2 - \alpha^2 [/tex] with coefficients in [tex] F(\alpha^2) [/tex]; hence, the minimal polynomial of [tex] \alpha [/tex] over [tex] F(\alpha^2) [/tex] must divide [tex] f(x) [/tex]. Since the degree of [tex] f(x) [/tex] is 2, this implies that the degree of [tex] \alpha [/tex] over [tex] F(\alpha^2) [/tex] is less than or equal to 2.

    The tower law is the following: For any fields [tex] L \subseteq K \subseteq M [/tex], we have [tex] [M:L] = [M:K] [K:L] [/tex]. (I'm a little surprised that you were assigned this question if you haven't covered this, to be honest; maybe they want you to derive a special case of the tower law yourself?)
     
  6. May 19, 2009 #5

    matt grime

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    There is no need to invoke the tower law - you can prove this entirely constructively. It would be by proving elementary results like this that would lead you to conjecturing and proving the tower law in fact. Invoking the tower law would be exactly like invoking Lagrange's theorem to show that if |G| is odd then o(x)=o(x^2) for x in G.
     
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