# Finite Fields in Algebra

## Homework Statement

Show that a finite field of $$p^n$$ elements has exactly one subfield of $$p^m$$ elements for each m that divides n.

## Homework Equations

If $$F \subset E \subset K$$ are field extensions of $$F$$, then $$[K:F] = [E:F][K:F]$$ . 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. $$F \subset E \subset K$$, 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 $$\Phi_p (x) = \frac{x^p -1}{x-1} = x^{p-1} + x^{p-2} + ... x + 1$$. This polynomial is irreducible over the rationals for every prime p. Let $$\alpha$$ be a zero of $$\Phi_p$$ . Show that the set $$\{\alpha, \alpha^2, .... \alpha^{p-1}$$ are distinct roots of $$\Phi_p$$ .

## The Attempt at a Solution

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

Office_Shredder
Staff Emeritus
Gold Member

## 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. $$F \subset E \subset K$$, 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 $$\Phi_p (x) = \frac{x^p -1}{x-1} = x^{p-1} + x^{p-2} + ... x + 1$$. This polynomial is irreducible over the rationals for every prime p. Let $$\alpha$$ be a zero of $$\Phi_p$$ . Show that the set $$\{\alpha, \alpha^2, .... \alpha^{p-1}$$ are distinct roots of $$\Phi_p$$ .

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