10.4 Cyclotomic Polynomials and Wedderburn's Theorem
1.
a. 
c. 
e. 
Now 
.
.
Finally 
3. Since n is odd, d|2n if and only if either d|n or d = 2b, b|n. Thus, by induction
Now observe that Φ2(x) = − Φ1(− x), and (by induction) Φ2d(x) = Φd(− x) if d < n. Hence
Thus Φn(− x) = Φ2n(x), as required.
5. If 
, these fields are 
and 
respectively. Now 
and 
, so 
. But gcd (m, n) = 1, so write 1 = rm + sn; r, 
. Then 
, so 
. Thus 
, these fields are and respectively. Now and , so . But gcd (m, n) = 1, so write 1 = rm + sn; r, . Then , so . Thus
6. a. If S ⊆ R is a finite subring of the division ring S, then char S ≠ 0, say char S = p. Let 
and let 
. Then if 0 ≠ s 
S, 1, s, . . ., sn are not linearly independent over 
, so r0 + r1s + 
+ rnsn = 0, 
. We may assume r0 ≠ 0 (can cancel s) so
and let . Then if 0 ≠ s S, 1, s, . . ., sn are not linearly independent over , so r0 + r1s + + rnsn = 0, . We may assume r0 ≠ 0 (can cancel s) so
7. Write σ(n) = ∑ d|nμ(d). If n = 1, σ(n) = μ(1) = 1. If 
then μ(d) = 0 for any d|n with 
. If we write m = p1p2 pr, then σ(n) = σ(m). If d|m, then μ(d) = 1 if and only if d is the product of an even number (possibly 0) of the pi, and μ(d) = − 1 otherwise. Since half the divisors d are in each category, σ(m) = 0.
then μ(d) = 0 for any d|n with . If we write m = p1p2 pr, then σ(n) = σ(m). If d|m, then μ(d) = 1 if and only if d is the product of an even number (possibly 0) of the pi, and μ(d) = − 1 otherwise. Since half the divisors d are in each category, σ(m) = 0.
8. a. The sums are equal by replacing d by n/d throughout. Use the hint:
But ∑d|nμ(d) = 1 if and only if n = c by the preceding exercise. We have xn − 1 = ∏ d|nΦd(x). Fix x and take a formal logarithm:
Let σ(n) = log (xn − 1) and β(n) = log (Φn(x)). By M öbius we get
The result follows.
If formal logarithms are distasteful, repeat Exercise 8(a) with Σ replaced by Π and all coefficients replaced by exponents.
9. Write n/m = k so n = km. Thus d|m if and only if kd|n. Now Exercise 8(b) gives
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