Finite Fields in Algebra

  • #1

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!
 

Answers and Replies

  • #2
Office_Shredder
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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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