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!