7. Cyclotomic Extensions（分圆扩张）
An \(n\)th root of unity is an element \(\omega\) of a field with \(\omega^n = 1\). For instance, the complex number \(e^{2\pi i/n}\) is an \(n\)th root of unity. We have seen roots of unity arise in various examples. In this section, we investigate the field extension \(F(\omega)/F\), where \(\omega\) is an \(n\)th root of unity. Besides being interesting extensions in their own right, these extensions will play a role in applications of Galois theory to ruler and compass constructions and to the question of solvability of polynomial equations.
Definition 7.1. If \(\omega \in F\) with \(\omega^n = 1\), then \(\omega\) is an \(n\)th root of unity. If the order of \(\omega\) is \(n\) in the multiplicative group \(F^*\), then \(\omega\) is a primitive \(n\)th root of unity. If \(\omega\) is any root of unity, then the field extension \(F(\omega)/F\) is called a cyclotomic extension.
We point out two facts about roots of unity. First, if \(\omega \in F\) is a primitive \(n\)th root of unity, then we see that \(\operatorname{char}(F)\) does not divide \(n\) for, if \(n = pm\) with \(\operatorname{char}(F) = p\), then \(0 = \omega^n - 1 = (\omega^m - 1)^p\). Therefore, \(\omega^m = 1\), and so the order of \(\omega\) is not \(n\). Second, if \(\omega\) is an \(n\)th root of unity, then the order of \(\omega\) in the group \(F^*\) divides \(n\), so the order of \(\omega\) is equal to some divisor \(m\) of \(n\). The element \(\omega\) is then a primitive \(m\)th root of unity.
The \(n\)th roots of unity in a field \(K\) are exactly the set of roots of \(x^n - 1\). Suppose that \(x^n - 1\) splits over \(K\), and let \(G\) be the set of roots of unity in \(K\). Then \(G\) is a finite subgroup of \(K^*\), so \(G\) is cyclic by Lemma 6.1. Any generator of \(G\) is then a primitive \(n\)th root of unity.
To describe cyclotomic extensions, we need to use the Euler phi function. If \(n\) is a positive integer, let \(\phi(n)\) be the number of integers between \(1\) and \(n\) that are relatively prime to \(n\). The problems below give the main properties of the Euler phi function. We also need to know about the group of units of the ring \(\mathbb{Z}/n\mathbb{Z}\). Recall that if \(R\) is a commutative ring with \(1\), then the set
is a group under multiplication; it is called the group of units of \(R\). If \(R = \mathbb{Z}/n\mathbb{Z}\), then an easy exercise shows that
Therefore, \(|(\mathbb{Z}/n\mathbb{Z})^*| = \phi(n)\).
We now describe cyclotomic extensions of an arbitrary base field.
Proposition 7.2. Suppose that \(\operatorname{char}(F)\) does not divide \(n\), and let \(K\) be a splitting field of \(x^n - 1\) over \(F\). Then \(K/F\) is Galois, \(K = F(\omega)\) is generated by any primitive \(n\)th root of unity \(\omega\), and \(\operatorname{Gal}(K/F)\) is isomorphic to a subgroup of \((\mathbb{Z}/n\mathbb{Z})^*\). Thus, \(\operatorname{Gal}(K/F)\) is Abelian and \([K : F]\) divides \(\phi(n)\).
Proof. Since \(\operatorname{char}(F)\) does not divide \(n\), the derivative test shows that \(x^n - 1\) is a separable polynomial over \(F\). Therefore, \(K\) is both normal and separable over \(F\); hence, \(K\) is Galois over \(F\). Let \(\omega \in K\) be a primitive \(n\)th root of unity. Then all \(n\)th roots of unity are powers of \(\omega\), so \(x^n - 1\) splits over \(F(\omega)\). This proves that \(K = F(\omega)\). Any automorphism of \(K\) that fixes \(F\) is determined by what it does to \(\omega\). However, any automorphism restricts to a group automorphism of the set of roots of unity, so it maps the set of primitive \(n\)th roots of unity to itself. Any primitive \(n\)th root of unity in \(K\) is of the form \(\omega^t\) for some \(t\) relatively prime to \(n\). Therefore, the map \(\theta : \operatorname{Gal}(K/F) \to (\mathbb{Z}/n\mathbb{Z})^*\) given by \(\sigma \mapsto t + n\mathbb{Z}\), where \(\sigma(\omega) = \omega^t\), is well defined. If \(\sigma, \tau \in \operatorname{Gal}(K/F)\) with \(\sigma(\omega) = \omega^s\) and \(\tau(\omega) = \omega^t\), then \((\sigma\tau)(\omega) = \sigma(\omega^t) = \omega^{st}\), so \(\theta\) is a group homomorphism. The kernel of \(\theta\) is the set of all \(\sigma\) with \(\sigma(\omega) = \omega\); that is, \(\ker(\theta) = \{\mathrm{id}\}\). Thus, \(\theta\) is injective, so \(\operatorname{Gal}(K/F)\) is isomorphic to a subgroup of the Abelian group \((\mathbb{Z}/n\mathbb{Z})^*\), a group of order \(\phi(n)\). This finishes the proof. □
Example 7.3. The structure of \(F\) determines the degree \([F(\omega) : F]\) or, equivalently, the size of \(\operatorname{Gal}(F(\omega)/F)\). For instance, let \(\omega = e^{2\pi i/8}\) be a primitive eighth root of unity in \(\mathbb{C}\). Then \(\omega^2 = i\) is a primitive fourth root of unity. The degree of \(\mathbb{Q}(\omega)\) over \(\mathbb{Q}\) is \(4\), which we will show below. If \(F = \mathbb{Q}(i)\), then the degree of \(F(\omega)\) over \(F\) is \(2\), since \(\omega\) satisfies the polynomial \(x^2 - i\) over \(F\) and \(\omega \notin F\). If \(F = \mathbb{R}\), then \(\mathbb{R}(\omega) = \mathbb{C}\), so \([\mathbb{R}(\omega) : \mathbb{R}] = 2\). In fact, if \(n \ge 3\) and if \(\tau\) is any primitive \(n\)th root of unity in \(\mathbb{C}\), then \(\mathbb{R}(\tau) = \mathbb{C}\), so \([\mathbb{R}(\tau) : \mathbb{R}] = 2\).
Example 7.4. Let \(F = \mathbb{F}_2\). If \(\omega\) is a primitive third root of unity over \(F\), then \(\omega\) is a root of \(x^3 - 1 = (x - 1)(x^2 + x + 1)\). Since \(\omega \ne 1\) and \(x^2 + x + 1\) is irreducible over \(F\), we have \([F(\omega) : F] = 2\) and \(\min(F, \omega) = x^2 + x + 1\). If \(\rho\) is a primitive seventh root of unity, then by factoring \(x^7 - 1\), by trial and error or by computer, we get
The minimal polynomial of \(\omega\) is then one of these cubics, so \([F(\omega) : F] = 3\). Of the six primitive seventh roots of unity, three have \(x^3 + x + 1\) as their minimal polynomial, and the three others have \(x^3 + x^2 + 1\) as theirs. This behavior is different from cyclotomic extensions of \(\mathbb{Q}\), as we shall see below, since all the primitive \(n\)th roots of unity over \(\mathbb{Q}\) have the same minimal polynomial.
We now investigate cyclotomic extensions of \(\mathbb{Q}\). Let \(\omega_1, \ldots, \omega_r\) be the primitive \(n\)th roots of unity in \(\mathbb{C}\). Then
so there are \(\phi(n)\) primitive \(n\)th roots of unity in \(\mathbb{C}\). In Theorem 7.7, we will determine the minimal polynomial of a primitive \(n\)th root of unity over \(\mathbb{Q}\), and so we will determine the degree of a cyclotomic extension of \(\mathbb{Q}\).
Definition 7.5. The \(n\)th cyclotomic polynomial is \(\Phi_n(x) = \prod_{i=1}^{r} (x - \omega_i)\), the monic polynomial in \(\mathbb{C}[x]\) whose roots are exactly the primitive \(n\)th roots of unity in \(\mathbb{C}\).
For example,
Moreover, if \(p\) is prime, then all \(p\)th roots of unity are primitive except for the root \(1\). Therefore,
From this definition of \(\Phi_n(x)\), it is not clear that \(\Phi_n(x) \in \mathbb{Q}[x]\), nor that \(\Phi_n(x)\) is irreducible over \(\mathbb{Q}\). However, we verify the first of these facts in the next lemma and then the second in Theorem 7.7, which shows that \(\Phi_n(x)\) is the minimal polynomial of a primitive \(n\)th root of unity over \(\mathbb{Q}\).
Lemma 7.6. Let \(n\) be any positive integer. Then \(x^n - 1 = \prod_{d|n} \Phi_d(x)\). Moreover, \(\Phi_n(x) \in \mathbb{Z}[x]\).
Proof. We know that \(x^n - 1 = \prod (x - \omega)\), where \(\omega\) ranges over the set of all \(n\)th roots of unity. If \(d\) is the order of \(\omega\) in \(\mathbb{C}^*\), then \(d\) divides \(n\), and \(\omega\) is a primitive \(d\)th root of unity. Gathering all the \(d\)th root of unity terms together in this factorization proves the first statement. For the second, we use induction on \(n\); the case \(n = 1\) is clear since \(\Phi_1(x) = x - 1\). Suppose that \(\Phi_d(x) \in \mathbb{Z}[x]\) for all \(d < n\). Then from the first part, we have
Since \(x^n - 1\) and \(\prod_{d|n} \Phi_d(x)\) are monic polynomials in \(\mathbb{Z}[x]\), the division algorithm, Theorem 3.2 of Appendix A, shows that \(\Phi_n(x) \in \mathbb{Z}[x]\). □
We can use this lemma to calculate the cyclotomic polynomials \(\Phi_n(x)\) by recursion. For example, to calculate \(\Phi_8(x)\), we have
so
The next theorem is the main fact about cyclotomic polynomials and allows us to determine the degree of a cyclotomic extension over \(\mathbb{Q}\).
Theorem 7.7. Let \(n\) be any positive integer. Then \(\Phi_n(x)\) is irreducible over \(\mathbb{Q}\).
Proof. To prove that \(\Phi_n(x)\) is irreducible over \(\mathbb{Q}\), suppose not. Since \(\Phi_n(x) \in \mathbb{Z}[x]\) and is monic, \(\Phi_n(x)\) is reducible over \(\mathbb{Z}\) by Gauss' lemma. Say \(\Phi_n = f(x)h(x)\) with \(f(x), h(x) \in \mathbb{Z}[x]\) both monic and \(f\) irreducible over \(\mathbb{Z}\). Let \(\omega\) be a root of \(f\). We claim that \(\omega^p\) is a root of \(f\) for all primes \(p\) that do not divide \(n\). If this is false for a prime \(p\), then since \(\omega^p\) is a primitive \(n\)th root of unity, \(\omega^p\) is a root of \(h\). Since \(f(x)\) is monic, the division algorithm shows that \(f(x)\) divides \(h(x^p)\) in \(\mathbb{Z}[x]\). The map \(\mathbb{Z}[x] \to \mathbb{F}_p[x]\) given by reducing coefficients mod \(p\) is a ring homomorphism. For \(g \in \mathbb{Z}[x]\), let \(\bar{g}\) be the image of \(g(x)\) in \(\mathbb{F}_p[x]\). Reducing mod \(p\) yields \(\overline{\Phi_n(x)} = \bar{f} \cdot \bar{h}\). Since \(\overline{\Phi_n(x)}\) divides \(x^n - 1\), the derivative test shows that \(\overline{\Phi_n(x)}\) has no repeated roots in any extension field of \(\mathbb{F}_p\), since \(p\) does not divide \(n\). Now, since \(a^p = a\) for all \(a \in \mathbb{F}_p\), we see that \(\overline{h(x^p)} = \overline{h}(x)^p\). Therefore, \(\bar{f}\) divides \(\bar{h}^p\), so any irreducible factor \(\bar{q} \in \mathbb{F}_p[x]\) of \(\bar{f}\) also divides \(\bar{h}\). Thus, \(\bar{q}^2\) divides \(\bar{f}\bar{h} = \overline{\Phi_n(x)}\), which contradicts the fact that \(\overline{\Phi_n}\) has no repeated roots. This proves that if \(\omega\) is a root of \(f\), then \(\omega^p\) is also a root of \(f\), where \(p\) is a prime not dividing \(n\). But this means that all primitive \(n\)th roots of unity are roots of \(f\), for if \(\alpha\) is a primitive \(n\)th root of unity, then \(\alpha = \omega^t\) with \(t\) relatively prime to \(n\). Then \(\alpha = \omega^{p_1 \cdots p_r}\), with each \(p_i\) a prime relatively prime to \(n\). We see that \(\omega^{p_1}\) is a root of \(f\), so then \((\omega^{p_1})^{p_2} = \omega^{p_1 p_2}\) is also a root of \(f\). Continuing this shows \(\alpha\) is a root of \(f\). Therefore, every primitive \(n\)th root of unity is a root of \(f\), so \(\Phi_n(x) = f\). This proves that \(\Phi_n(x)\) is irreducible over \(\mathbb{Z}\), and so \(\Phi_n(x)\) is also irreducible over \(\mathbb{Q}\). □
If \(\omega\) is a primitive \(n\)th root of unity in \(\mathbb{C}\), then the theorem above shows that \(\Phi_n(x)\) is the minimal polynomial of \(\omega\) over \(\mathbb{Q}\). The following corollary describes cyclotomic extensions of \(\mathbb{Q}\).
Corollary 7.8. If \(K\) is a splitting field of \(x^n - 1\) over \(\mathbb{Q}\), then \([K : \mathbb{Q}] = \phi(n)\) and \(\operatorname{Gal}(K/\mathbb{Q}) \cong (\mathbb{Z}/n\mathbb{Z})^*\). Moreover, if \(\omega\) is a primitive \(n\)th root of unity in \(K\), then \(\operatorname{Gal}(K/\mathbb{Q}) = \{\sigma_i : \gcd(i, n) = 1\}\), where \(\sigma_i\) is determined by \(\sigma_i(\omega) = \omega^i\).
Proof. The first part of the corollary follows immediately from Proposition 7.2 and Theorem 7.7. The description of \(\operatorname{Gal}(K/\mathbb{Q})\) is a consequence of the proof of Proposition 7.2. □
If \(\omega\) is a primitive \(n\)th root of unity in \(\mathbb{C}\), then we will refer to the cyclotomic extension \(\mathbb{Q}(\omega)\) as \(\mathbb{Q}_n\).
Example 7.9. Let \(K = \mathbb{Q}_7\), and let \(\omega\) be a primitive seventh root of unity in \(\mathbb{C}\). By Corollary 7.8, \(\operatorname{Gal}(K/\mathbb{Q}) \cong (\mathbb{Z}/7\mathbb{Z})^*\), which is a cyclic group of order \(6\). The Galois group of \(K/\mathbb{Q}\) is \(\{\sigma_1, \sigma_2, \sigma_3, \sigma_4, \sigma_5, \sigma_6\}\), where \(\sigma_j(\omega) = \omega^j\). Thus, \(\sigma_1 = \mathrm{id}\), and it is easy to check that \(\sigma_3\) generates this group. Moreover, \(\sigma_i \circ \sigma_j = \sigma_{ij}\), where the subscripts are multiplied modulo \(7\). The subgroups of \(\operatorname{Gal}(K/\mathbb{Q})\) are then
whose orders are \(1\), \(2\), \(3\), and \(6\), respectively. Let us find the corresponding intermediate fields. If \(L = \mathcal{F}(\sigma_3^3) = \mathcal{F}(\sigma_6)\), then \([K : L] = |\langle \sigma_6 \rangle| = 2\) by the fundamental theorem. To find \(L\), we note that \(\omega\) must satisfy a quadratic over \(L\) and that this quadratic is
Expanding, this polynomial is
Therefore, \(\omega + \omega^6 \in L\). If we let \(\omega = \exp(2\pi i/7) = \cos(2\pi/7) + i\sin(2\pi/7)\), then \(\omega + \omega^6 = 2\cos(2\pi/7)\). Therefore, \(\omega\) satisfies a quadratic over \(\mathbb{Q}(\cos(2\pi/7))\); hence, \(L\) has degree at most \(2\) over this field. This forces \(L = \mathbb{Q}(\cos(2\pi/7))\). With similar calculations, we can find \(M = \mathcal{F}(\sigma_3^2) = \mathcal{F}(\sigma_2)\). The order of \(\sigma_2\) is \(3\), so \([M : \mathbb{Q}] = 2\). Hence, it suffices to find one element of \(M\) that is not in \(\mathbb{Q}\) in order to generate \(M\). Let
This element is in \(M\) because it is fixed by \(\sigma\). But, we show that \(\alpha\) is not in \(\mathbb{Q}\) since it is not fixed by \(\sigma_6\). To see this, we have
If \(\sigma_6(\alpha) = \alpha\), this equation would give a degree \(6\) polynomial for which \(\omega\) is a root, and this polynomial is not divisible by
a contradiction. This forces \(\alpha \notin \mathbb{Q}\), so \(M = \mathbb{Q}(\alpha)\). Therefore, the intermediate fields of \(K/\mathbb{Q}\) are
Example 7.10. Let \(K = \mathbb{Q}_8\), and let \(\omega = \exp(2\pi i/8) = (1 + i)/\sqrt{2}\). The Galois group of \(K/\mathbb{Q}\) is \(\{\sigma_1, \sigma_3, \sigma_5, \sigma_7\}\), and note that each of the three nonidentity automorphisms of \(K\) has order \(2\). The subgroups of this Galois group are then
Each of the three proper intermediate fields has degree \(2\) over \(\mathbb{Q}\). One is easy to find, since \(\omega^2 = i\) is a primitive fourth root of unity. The group associated to \(\mathbb{Q}(i)\) is \(\langle \sigma_5 \rangle\), since \(\sigma_5(\omega^2) = \omega^{10} = \omega^2\). We could find the two other fields in the same manner as in the previous example: Show that the fixed field of \(\sigma_3\) is generated over \(\mathbb{Q}\) by \(\omega + \sigma_3(\omega)\). However, we can get this more easily due to the special form of \(\omega\). Since \(\omega = (1 + i)/\sqrt{2}\) and \(\omega^{-1} = (1 - i)/\sqrt{2}\), we see that \(\sqrt{2} = \omega + \omega^{-1} \in K\). The element \(\omega + \omega^{-1} = \omega + \omega^7\) is fixed by \(\sigma_7\); hence, the fixed field of \(\sigma_7\) is \(\mathbb{Q}(\sqrt{2})\). We know \(i \in K\) and \(\sqrt{2} \in K\), so \(\sqrt{-2} \in K\). This element must generate the fixed field of \(\sigma_3\). The intermediate fields are then
The description of the intermediate fields also shows that \(K = \mathbb{Q}(\sqrt{2}, i)\).