Brian Bi

## Section 16.10. Roots of Unity

Exercise 16.10.1 The degree of $$\zeta_3$$ over $$\Q$$ is 2, therefore its degree over $$\Q(\zeta_7)$$ is either 2 or 1. Suppose it is 1. Then $$\zeta_3 \in \Q(\zeta_7)$$, so $$\Q(\zeta_7)$$ contains the subfield $$\Q(\sqrt{-3})$$. But by Theorem 16.10.12, the unique quadratic extension of $$\Q$$ contained in $$\Q(\zeta_7)$$ is $$\Q(\sqrt{-7})$$, so a contradiction has been reached. Therefore $$[\Q(\zeta_3, \zeta_7) : \Q(\zeta_7)] = 2$$, and $[\Q(\zeta_3, \zeta_7) : \Q(\zeta_3)] = \frac{[\Q(\zeta_3, \zeta_7) : \Q(\zeta_7)][\Q(\zeta_7) : \Q]} {[\Q(\zeta_3) : \Q]} = 6$

Exercise 16.10.2 In Example 16.10.3, the field $$L_2$$ was defined as the fixed field of $$\sigma^4$$, which maps $$\zeta$$ to $$\zeta^{3^4} = \zeta^{-4}$$. Let \begin{align*} \alpha_1 &= \zeta + \zeta^{-4} + \zeta^{-1} + \zeta^4 \\ \alpha_2 &= \zeta^3 + \zeta^5 + \zeta^{-3} + \zeta^{-5} \\ \alpha_3 &= \zeta^{-8} + \zeta^{-2} + \zeta^8 + \zeta^2 \\ \alpha_4 &= \zeta^{-7} + \zeta^2 + \zeta^{-8} + \zeta^{-2} \end{align*} Clearly $$\alpha_1$$ is fixed by $$\sigma^4$$, so $$\Q(\alpha_1) \subseteq L_2$$. In addition, the set $$\{\alpha_1, \alpha_2, \alpha_3, \alpha_4\}$$ is the orbit of $$\alpha_1$$ under $$\sigma$$, so by Theorem 16.5.2, the minimal polynomial for $$\alpha_1$$ over $$\Q$$ has degree 4, so $$[\Q(\alpha_1) : \Q] = 4$$. But $$L_2$$ is also a quartic extension, so $$\alpha_1$$ in fact generates $$L_2$$ over $$\Q$$.

Exercise 16.10.3 By Proposition 16.10.2, $$\Gal(\Q(\zeta)/\Q)$$ is isomorphic to $$C_6$$ and is generated by $$\sigma = \zeta \mapsto \zeta^3$$ since 3 is a primitive root modulo 7. By the fundamental theorem, there are two intermediate fields, namely the fixed fields of the $$C_2$$ subgroup generated by $$\sigma^3 = \zeta \mapsto \zeta^{-1}$$ and the $$C_3$$ subgroup generated by $$\sigma^2 = \zeta \mapsto \zeta^2$$. By the fundamental theorem, the former field has degree 3 over $$\Q$$, so we will call it $$L_3$$. We observe that the elements $$\zeta^2 + \zeta^{-2}$$ and $$\zeta^3 + \zeta^{-3}$$ belong to $$L_3$$, and that since $$\{1, \zeta, \zeta^2, \zeta^3, \zeta^{-3}, \zeta^{-2}\}$$ form a $$\Q$$-basis of $$\Q(\zeta)$$, the three numbers $$1$$, $$\zeta^2 + \zeta^{-2}$$, and $$\zeta^3 + \zeta^{-3}$$ are linearly independent over $$\Q$$ and must therefore form a basis of $$L_3$$. Similarly, $$\zeta + \zeta^2 + \zeta^{-3}$$ belongs to $$L_2$$, and is linearly independent from 1, therefore it generates $$L_2$$ over $$\Q$$.

It follows that an element of $$\Q(\zeta)$$ belongs to $$L_3$$ if and only if, when written in the basis $$\{1, \zeta, \zeta^2, \zeta^3, \zeta^{-3}, \zeta^{-2}\}$$, it takes the form $$p + q(\zeta^2 + \zeta^{-2}) + r(\zeta^3 + \zeta^{-3})$$, and belongs to $$L_2$$ if and only if it takes the form $$p + q(\zeta + \zeta^2 + \zeta^{-3})$$ in the same basis, where $$p$$ and $$q$$ are rational. These results allow us to easily determine the degrees of the following elements of $$\Q(\zeta)$$:

1. This is not in either form, so the field it generates must be the full field $$\Q(\zeta)$$; the degree is 6.
2. This can be rewritten as $$\zeta^3 + \zeta^{-3}$$, which satisfies the criterion to be a member of $$L_3$$. Therefore its degree is 3.
3. This can be rewritten as $$\zeta^3 + \zeta^{-2} + (-1 - \zeta - \zeta^2 - \zeta^3 - \zeta^{-3} - \zeta^{-2}) = -1 - \zeta - \zeta^2 - \zeta^{-3}$$, which satisfies the criterion to be a member of $$L_2$$. Therefore its degree is 2.

Exercise 16.10.4 This exercise can be solved by the same methods as Exercise 16.10.3.

Exercise 16.10.5 This exercise can be solved by the same methods used to characterize the subfields of $$\Q(\zeta_7)$$ in Exercise 16.10.3.

Exercise 16.10.6

1. We will follow the proof in the text closely. We will assume that for every prime $$p$$ there is a primitive root $$g$$ modulo $$p$$. There are various proofs of this theorem, but we won't get into them here.

Let $$p$$ be a prime other than 2, let $$g$$ be a primitive root modulo $$p$$, and let $$\sigma = \zeta \mapsto \zeta^g$$; $$\sigma$$ generates the Galois group $$G$$ of $$\Q(\zeta)/\Q$$. Let $$H$$ denote the subgroup $$\langle \sigma^2 \rangle$$. The $$G$$-orbit of $$\zeta$$ splits into two $$H$$-orbits $$O_1, O_2$$, with $$\zeta \in O_1$$. Let $$\alpha_1, \alpha_2$$ respectively denote the sums over the orbits $$O_1, O_2$$. The set $$\{\alpha_1, \alpha_2\}$$ is a $$G$$-orbit, so by Theorem 16.5.2, the elements $$\alpha_1, \alpha_2$$ have degree 2 over the fixed field of $$G$$, which is $$\Q$$, and their minimal polynomial is $$(x - \alpha_1) (x - \alpha_2)$$. The linear coefficient is $$-(\alpha_1 + \alpha_2) = -\sum_{i=1}^{p-1} \zeta^i = 1$$. We wish to compute the constant coefficient, $$\alpha_1 \alpha_2 = \prod_{x \in O_1} \prod_{y \in O_2} xy$$.

In general, $$O_1 = \{\zeta, \sigma^2(\zeta), \sigma^4(\zeta), \ldots, \sigma^{p-3}(\zeta)\}$$ and $$O_2 = \{\sigma(\zeta), \sigma^3(\zeta), \ldots, \sigma^{p-2}(\zeta)\}$$. Let $$\tau = \sigma^{(p-1)/2}$$. Then $$\tau^2 = \sigma^{p-1}$$ is the identity automorphism, but $$\tau$$ is not the identity. Therefore $$\tau$$ has order 2, and must map $$\zeta$$ to $$\zeta^{-1}$$.

If $$p = 4k + 1$$, then $$\tau = \sigma^{2k}$$, and $$\zeta^{-1} = \tau(\zeta)$$ occurs in $$O_1$$. Therefore for each $$\zeta^j$$ that occurs in $$O_1$$, $$\zeta^{-j}$$ also occurs. The product $$\alpha_1 \alpha_2$$ contains $$(2k) \cdot (2k) = 4k^2$$ terms when fully expanded, and none of them are equal to 1 since $$\zeta^j$$ and $$\zeta^{-j}$$ always occur in the same orbit. Since $$\alpha_1\alpha_2$$ is rational, each of the $$4k$$ powers of $$\zeta$$ different from 1 must occur the same number of times in the product. Therefore, each one occurs $$k$$ times, and $$\alpha_1 \alpha_2 = k \sum_{i=1}^{p-1} \zeta^i = -k$$. The minimal polynomial is $$x^2 + x - k$$, and the discriminant is $$4k + 1 = p$$, therefore the quadratic intermediate field is $$\Q(\sqrt{p})$$.

If $$p = 4k + 3$$, then $$\tau = \sigma^{2k + 1}$$, and $$\zeta^{-1} = \tau(\zeta)$$ occurs in $$O_2$$. Therefore if $$\zeta^j$$ occurs in one orbit, then $$\zeta^{-j}$$ occurs in the other, and of the $$(2k + 1)^2$$ terms in the product $$\alpha_1\alpha_2$$, $$2k + 1$$ of them are equal to 1. In the remaining $$2k(2k + 1)$$ terms, each of the $$4k + 2$$ powers of $$\zeta$$ different from 1 must occur the same number of times, that is, $$k$$ times. So $$\alpha_1 \alpha_2$$ is the sum of the contribution $$2k + 1$$ from the 1's in the product, and $$-k$$ from the other terms, for a total of $$k + 1$$. The minimal polynomial is $$x^2 + x + (k + 1)$$, and the discriminant is $$-4k - 3 = -p$$, therefore the quadratic intermediate field is $$\Q(\sqrt{-p})$$.

Exercise 16.10.7

1. The polynomial $$x^n - 1$$ has $$n$$ roots, given by $$\zeta_n^j$$ for $$j \in \{0, 1, \ldots, n-1\}$$, and therefore splits completely in the field $$\Q(\zeta_n)$$. This field extension of $$\Q$$ is also clearly generated by the roots of $$x^n - 1$$. Thus $$K$$ is a splitting field over $$\Q$$, and equivalently a Galois extension.

2. An element of $$\Gal(K/\Q)$$ is completely determined by the image of $$\zeta_n$$. Since $$\zeta_n$$ is a primitive $$n$$th root of unity, its image under a $$\Q$$-automorphism must also be a primitive $$n$$th root of unity, that is, it must map $$\zeta_n$$ to $$\zeta_n^j$$ where $$j \in U$$. The composition of such automorphisms is multiplicative in $$j$$, so $$\Gal(K/\Q)$$ is isomorphic to a subgroup of $$U$$.

3. For $$n = 6$$, $$\Gal(K/\Q)$$ contains the identity and the complex conjugation operation, which maps $$\zeta_6$$ to $$\zeta_6^{-1}$$, and $$U$$ also has order 2.

For $$n = 8$$, $$K = \Q(i, \sqrt{2})$$ and there are four automorphisms according to $$i \mapsto \pm i$$ and $$\sqrt{2} \mapsto \pm \sqrt{2}$$. The order of $$U$$ is also 4.

For $$n = 12$$, $$K = \Q(i, \sqrt{3})$$ and there are four automorphisms according to $$i \mapsto \pm i$$ and $$\sqrt{3} \mapsto \pm \sqrt{3}$$. The order of $$U$$ is also 4.

Exercise 16.10.8 The splitting field of $$x^8 - 1$$ is the field $$\Q(\zeta_8)$$ discussed in 16.10.7, so its Galois group is the group of units in $$\Z/8\Z$$, which has order 4. Since $$1 \equiv 1^2 \equiv 3^2 \equiv 5^2 \equiv 7^2 \pmod 8$$, this group must be isomorphic to the Klein four-group (and not the cyclic group of order 4).

Similarly, the Galois group of $$x^{12} - 1$$ has order 4, and since $$1 \equiv 1^2 \equiv 5^2 \equiv 7^2 \equiv 11^2 \pmod{12}$$, the Galois group is again isomorphic to $$V_4$$.

By Exercise 12.4.18, the cyclotomic polynomial $$\Psi_9(x)$$ is irreducible, which implies that the Galois group of the polynomial $$x^9 - 1$$ has order at least equal to that of $$U$$, and is hence again isomorphic to $$U$$. In this case, there is a primitive root 2, so the Galois group is cyclic of order 6.

Exercise 16.10.9

1. By the product rule, we explicitly have $$f'(\alpha_i) = \prod_{j \ne i} (\alpha_i - \alpha_j)$$ with all other terms vanishing since they contain the factor $$x - \alpha_i$$ evaluated at $$x = \alpha_i$$. Consequently, $$\prod_{i=1}^n f'(\alpha_i)$$ contains a total of $$n(n-1)$$ factors, where $$\alpha_i - \alpha_j$$ occurs together with $$\alpha_j - \alpha_i$$ exactly once for each pair $$1 \le i < j \le n$$. Their product is $$-(\alpha_i - \alpha_j)^2$$. The product of the $$n(n-1)/2$$ such squares in the determinant, and there are $$n(n-1)/2$$ minus signs, so the determinant differs in sign from $$\prod_{i=1}^n f'(\alpha_i)$$ precisely when $$n(n-1)/2$$ is odd, or more precisely, when $$n$$ is congruent to 2 or 3 (modulo 4).

2. The case $$p = 2$$ is simple and can be dealt with separately. For the polynomial $$x^p - 1$$ with $$p$$ an odd prime, the roots are the $$p$$th roots of unity and the derivative is $$px^{p-1}$$, which, when evaluated at $$x$$ a $$p$$th root of unity, becomes $$px^{-1}$$. The product of all such values is $$p^p$$ since each root except 1 cancels with its reciprocal. If $$p$$ is congruent to 1 modulo 4, then by part (a), the discriminant is $$p^p$$. The cyclotomic field contains the square root of the discriminant, so it contains the quadratic subfield $$\Q(\sqrt{p^p}) = \Q(\sqrt{p})$$. If $$p$$ is congruent to 3 modulo 4, then by part (a), the discriminant is $$-p^p$$, so similarly, the cyclotomic field contains $$\Q(\sqrt{-p^p}) = \Q(\sqrt{-p})$$.

Exercise 16.10.10 Let $$\alpha_1, \ldots, \alpha_p$$ be the roots of a complex monic polynomial $$P$$ of degree $$p$$ and suppose that for $$i \in \{1, \ldots, p - 1\}$$, the quantities $$\gamma_i = \sum_{j=1}^p \zeta^{(i-1)(j-1)} \alpha_j$$ vanish. (Note that we have relabelled the $$\gamma$$'s from the question text, for notational convenience.) We will show that $$P$$ is reducible.

Suppose $$\alpha_1 + \ldots + \alpha_p = \delta$$. Then we have a system of $$p$$ linear equations in the $$p$$ unknowns $$\alpha_1, \ldots, \alpha_p$$, which can be written $$\gamma_i = \sum_{j=1}^p M_{ij} \alpha_j$$ where $$M_{ij} = \zeta^{(i-1)(j-1)}$$, $$\gamma_2 = \ldots = \gamma_p = 0$$, and $$\gamma_1 = \delta$$. It can be shown that $$M^{-1} = \frac{1}{p}M^*$$, so $$\alpha_i = \frac{1}{p} \sum_{j=1}^p \zeta^{-(i-1)(j-1)} \gamma_j$$ Since only $$\gamma_1$$ is nonzero, this reduces to $$\alpha_i = \delta/p$$ for each $$i$$. This implies that $$P(x) = (x - \delta/p)^p$$, and $$P$$ is not irreducible.