How Do Subfields of Finite Fields Relate to Their Elements?

In summary, the first conversation discusses how to show that a finite field of p^n elements has exactly one subfield of p^m elements for each m that divides n. The solution involves using field extensions and the fact that the degree of the subfield must divide the degree of the larger field. The second conversation discusses finding distinct roots of a polynomial using a root of unity. By plugging in different powers of the root of unity, we can show that they are all distinct roots.
  • #1
Mathmajor2010
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Homework Statement


Show that a finite field of [tex] p^n [/tex] elements has exactly one subfield of [tex] p^m [/tex] elements for each m that divides n.

Homework Equations


If [tex] F \subset E \subset K [/tex] are field extensions of [tex] F [/tex], then [tex] [K:F] = [E:F][K:F] [/tex] . Also, a field extension over a finite field of p elements has p^n elements, where n is the number of basis vectors in the extension field.

The Attempt at a Solution


We have a field K of p^n elements, which has degree n because it has n basis vectors. Then, if we have a field extension E of F s.t. [tex] F \subset E \subset K [/tex], then the degree of E has to divide K. That is, the degree of E, call it m, has to divide n. Therefore, E has p^m elements, where m divides it. Does this sound right?Another question I'm stuck on is the following:

Homework Statement


Let [tex] \Phi_p (x) = \frac{x^p -1}{x-1} = x^{p-1} + x^{p-2} + ... x + 1 [/tex]. This polynomial is irreducible over the rationals for every prime p. Let [tex] \alpha [/tex] be a zero of [tex] \Phi_p [/tex] . Show that the set [tex] \{\alpha, \alpha^2, ... \alpha^{p-1} [/tex] are distinct roots of [tex] \Phi_p [/tex] .

The Attempt at a Solution


I'm not really sure how to approach this. It seems to me that since [tex] \alpha [/tex] is a root, we can find an extension field. So, [tex] Q(\alpha) [/tex] is spanned by [tex] 1, \alpha, \alpha^2, ... \alpha^p-2 [/tex]. But, I don't see how to show that each of these is a root. Any suggestions of hints for this one? Thanks!
 
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  • #2
Mathmajor2010 said:

The Attempt at a Solution


We have a field K of p^n elements, which has degree n because it has n basis vectors. Then, if we have a field extension E of F s.t. [tex] F \subset E \subset K [/tex], then the degree of E has to divide K. That is, the degree of E, call it m, has to divide n. Therefore, E has p^m elements, where m divides it. Does this sound right?

You have shown neither that E exists nor that it is unique. All that you've said is that subfields have size a power of p.

Another question I'm stuck on is the following:

Homework Statement


Let [tex] \Phi_p (x) = \frac{x^p -1}{x-1} = x^{p-1} + x^{p-2} + ... x + 1 [/tex]. This polynomial is irreducible over the rationals for every prime p. Let [tex] \alpha [/tex] be a zero of [tex] \Phi_p [/tex] . Show that the set [tex] \{\alpha, \alpha^2, ... \alpha^{p-1} [/tex] are distinct roots of [tex] \Phi_p [/tex] .


Have you tried just plugging in [tex]\alpha^2[/tex] and see what you get? You'll get a sum of a bunch of different values for [tex]\alpha[/tex]. Keep in mind that [tex]\alpha[/tex] is a root of unity, so you know for example that [tex]\alpha^p=1[/tex]. See if you can use this to rewrite those powers of [tex]\alpha[/tex]
 
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