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A complex number $z=\cos(\alpha)+i\sin(\alpha)=e^{i\alpha}$ is called an $n$th root of unity if $z^n-1=0$, and is a primitive $n$th root of unity if in addition $z^k-1\neq0$ for $1\leq k. The $n$th cyclotomic polynomial, $\Phi_n(x)$, is the monic polynomial of which the zeros are the distinct primitive $n$th roots of unity. Take $n>1$.

(a) Show that $\zeta=e^{2\pi i/n}$ is a primitive $n$th root of unity.

(b) Show that $\Phi_n(x)=\prod_{\substack{1\leq j\leq n\\ (j,n)=1}}(x-\zeta^j),$ and that $x^n-1=\prod_{d\mid n}\Phi_d(x)$.

(c) By applying the inversion formula to $\log \Phi_n(x)$, or otherwise, show that $\Phi_n(x)=\prod_{d\mid n}(x^{n/d}-1)^{\mu(d)}.$ In particular, conclude that $\Phi_n\in\mathbb{Z}[x]$.

(d) Suppose that $p$ is prime, $p\nmid n$, and $\Phi_n(a)\equiv 0\pmod p$. Show that
$\qquad$ (i) $\hskip0.11in p\nmid a$;
$\qquad$ (ii) $\hskip0.07in t\mid n$, where $t=\operatorname{ord}_p a$;
$\qquad$ (iii) $\hskip0.03in$if $a^{n/d}\equiv 1\pmod p$, then $d\mid\frac{n}{t}$;
$\qquad$ (iv) $\hskip0.03in$if $p^s\| (a^t-1)$, then $p^s\|(a^{mt}-1)$ when $p\nmid m$;
$\qquad$ (v) $\hskip0.07in$if $n/t>1$, then $p\nmid \Phi_n(a)$ [use (c)];
$\qquad$ (vi) $\hskip0.03in$the congruence $\Phi_n(x)\equiv0\pmod p$ is solvable if and only if $p\equiv 1\pmod n$.

(e) Show that there are infinitely many primes $p\equiv 1\pmod n$. [Hint: Consider $\Phi_n(np_1\cdots p_ry)$, where the $p_i$ are $\equiv1\bmod n$.]

These are a bunch of exercises on cyclotomic polynomial from Fundamentals of Number theory, p. 131, William J. Leveque. I have thought really hard, but still cannot prove most of them. Could anyone talk something about them or give me some reference books on that? I am really curious about their proofs. Thanks in advance.

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    @GerryMyerson Yes!2011-11-21

2 Answers 2

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My algebra notes (http://www.math.umn.edu/~garrett/m/algebra/) prove many of the iconic basic facts about cyclotomic polynomials and cyclotomic fields (rather than leaving them for exercises).

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We've established in the comments that the difficulties start with (d)(i). So: $\Phi_n(a)=\prod_{d\mid n}(a^{n/d}-1)^{\mu(d)}$ from (c). Since $p$ is, by hypothesis, a prime, $\Phi_n(a)\equiv0\pmod p$ implies at least one of the factors $a^{n/d}-1$ is zero mod $p$. But if $p$ divides $a$, then each of these factors is $-1$ mod $p$, contradiction, hence, $p$ doesn't divide $a$.

EDIT: For (d)(ii), as noted above, at least one of the factors $a^{n/d}-1$ is zero mod $p$, so, if $t$ is the order of $a$ mod $p$, then $t$ divides $n/d$ for that $d$; a fortiori, $t$ divides $n$.