cyclotomic polynomials

**Sampras** · January 9th 2010, 07:11 AM

We know that $\Phi_{p}(x) = \frac{x^p-1}{x-1} = x^{p-1}+x^{p-2}+ \dots + x+1$ is irreducible over $\mathbb{Q}$ for every prime $p$ . Suppose $\gamma$ is a zero of $\Phi_{p}(x)$ and consider the field $\mathbb{Q}(\gamma)$ .

(a) Show that $\gamma, \gamma^2, \dots, \gamma^{p-1}$ are distinct zeros of $\Phi_{p}(x)$ and show that they are all the zeros of $\Phi_{p}(x)$ .

(b) Show that $G(Q(\gamma)/ \mathbb{Q})$ is abelian of order $p-1$ .

(c) Show that the fixed field of $G(\mathbb{Q}(\gamma)/ \mathbb{Q})$ is $\mathbb{Q}$ .

For (a), these are the nth roots of unity? For (b), consider two automorphisms of $\mathbb{Q}(\gamma)$ and show that they commute? For (c), isn't this by definition?

**HallsofIvy** · January 9th 2010, 07:25 AM

Originally Posted by Sampras

We know that $\Phi_{p}(x) = \frac{x^p-1}{x-1} = x^{p-1}+x^{p-2}+ \dots + x+1$ is irreducible over $\mathbb{Q}$ for every prime $p$ . Suppose $\gamma$ is a zero of $\Phi_{p}(x)$ and consider the field $\mathbb{Q}(\gamma)$ .

(a) Show that $\gamma, \gamma^2, \dots, \gamma^{p-1}$ are distinct zeros of $\Phi_{p}(x)$ and show that they are all the zeros of $\Phi_{p}(x)$ .

First, the zero of $\Phi{p}(x)$ are just the zeros of $x^p- 1$ , excluding 1. If if $\gamma$ is such a 0, then $\gamma^p= 1$ so that $(\gamma^n)^p= (\gamma^p)^n= 1$ . The only 'hard' part is showin that $\gamma^n$ is not 1 for n= 1, 2, ..., p-1. It is important that p is prime. here.
Suppose $\gamma^i= \gamma^j$ for i< j< p. Then $\gamma^{j- i}= 1$ . Show that that is impossible.

[tex](b) Show that $G(Q(\gamma)/ \mathbb{Q})$ is abelian of order $p-1$ .

(c) Show that the fixed field of $G(\mathbb{Q}(\gamma)/ \mathbb{Q})$ is $\mathbb{Q}$ .

For (a), these are the nth roots of unity? For (b), consider two automorphisms of $\mathbb{Q}(\gamma)$ and show that they commute? For (c), isn't this by definition?[/QUOTE]

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