• Support PF! Buy your school textbooks, materials and every day products Here!

Finite Field Extensions

  • #1

Homework Statement


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.)


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.
 

Answers and Replies

  • #2
139
12
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]?
 
  • #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.
 
  • #4
139
12
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?)
 
  • #5
matt grime
Science Advisor
Homework Helper
9,395
3
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.
 

Related Threads for: Finite Field Extensions

  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
3
Views
774
Replies
0
Views
825
  • Last Post
Replies
2
Views
950
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
9
Views
2K
Top