Cyclotomic Extensions

§Solvability by Radicals, ⁋Definition 2, we introduced roots of unity and primitive roots of unity. By definition, an \(n\)-th root of unity \(\zeta\) is a solution to \(\x^n-1=0\), but if it is not a primitive root, then for a divisor of \(n\), say \(d\), it is also a solution to \(\x^d-1=0\). For example, over \(\mathbb{R}\), when adjoining solutions to the equation \(\x^4-1=0\),

\[\x^4-1=(\x-1)(\x+1)(\x^2+1)\]

and among these, since the solutions of \(\x-1\) and \(\x+1\) are not primitive roots of unity, the only equation that actually yields primitive roots is \(\x^2+1\). In this post, we will examine polynomials whose roots are precisely such primitive \(n\)-th roots of unity, and as a result, show that these polynomials are irreducible polynomials with integer coefficients, and in particular, that when \(\mathbb{K}=\mathbb{Q}\), the Galois group of \(\mathbb{Q}(\zeta)/\mathbb{Q}\) is precisely \((\mathbb{Z}/n\mathbb{Z})^\times\). As in the previous post, all fields appearing in this post are assumed to have characteristic \(0\) unless specified otherwise.

Cyclotomic Polynomials

A rigorous definition of the polynomial discussed above is as follows.

Definition 1 For a field \(\mathbb{K}\) and a natural number \(n\geq1\), the monic polynomial whose roots in \(\overline{\mathbb{K}}\) are all primitive \(n\)-th roots of unity,

\[\Phi_n(\x)=\prod_{\zeta}(\x-\zeta)\in\overline{\mathbb{K}}[\x]\]

is called the \(n\)-th cyclotomic polynomial. Here, the product is taken over all generators of \(\mu_n\), denoted \(\zeta\).

Fix a generator of \(\mu_n\), denoted by \(\zeta\). Then, since we have already verified that the generators of \(\mu_n\) are precisely those with \(\gcd(m,n)=1\), namely \(\zeta^m\), denoting the number of \(1\leq m\leq n\) coprime to \(n\) by \(\varphi(n)\), we know that

\[\Phi_n(\x)=\prod_{\substack{1\leq m\leq n\\ \gcd(m,n)=1}}(\x-\zeta^m),\qquad \deg\Phi_n=\varphi(n)\]

holds. For instance, since \(\mu_1=\{1\}\) and \(\mu_2=\{\pm1\}\), we have \(\Phi_1=\x-1\) and \(\Phi_2=\x+1\); and from the fact that the generators of \(\mu_4\) are \(\pm i\), we have \(\Phi_4=\x^2+1\). For a prime \(p\), every element except \(1\) in \(\mu_p\) is a generator, so

\[\Phi_p(\x)=\frac{\x^p-1}{\x-1}=\x^{p-1}+\x^{p-2}+\cdots+\x+1\]

holds.

By definition, \(\Phi_n\) is an element of \(\overline{\mathbb{K}}[\x]\), but the following proposition shows that this polynomial actually has integer coefficients.

Proposition 2 For a natural number \(n\geq1\), the following hold:

1. \(\x^n-1=\prod_{d\mid n}\Phi_d(\x)\).
2. \(\Phi_n(\x)\in\mathbb{Z}[\x]\). In particular, \(\Phi_n\) does not depend on the choice of \(\mathbb{K}\).

Proof

First, we show the first result. Considering an element in \(\mu_n\) of order \(d\), \(\omega\), the subgroup generated by \(\omega\), namely \(\langle\omega\rangle\), is a cyclic group of order \(d\) and satisfies \(\omega^d=1\). That is, \(\omega\) is a primitive \(d\)-th root of unity. Conversely, given any divisor of \(n\), say \(d\), a primitive \(d\)-th root of unity is an element of \(\mu_d\subseteq\mu_n\) and has order \(d\); thus \(\mu_n\) is partitioned, over the divisors of \(n\), denoted \(d\), into the sets of primitive \(d\)-th roots of unity. Then the first result is simply obtained by grouping

\[\x^n-1=\prod_{\omega\in\mu_n}(\x-\omega)\]

according to this partition.

Now we show the second result. First, if \(n=1\), this is trivial since \(\Phi_1=\x-1\in\mathbb{Z}[\x]\). Assume \(n>1\), and suppose that the cyclotomic polynomials have integer coefficients for all natural numbers less than \(n\). Now, setting

\[g(\x)=\prod_{\substack{d\mid n\\ d<n}}\Phi_d(\x)\]

then \(g\) is monic and belongs to \(\mathbb{Z}[\x]\) by the induction hypothesis, and from the first result,

\[\x^n-1=\Phi_n(\x) g(\x)\]

holds. Meanwhile, since \(g\) is monic, by [Ring Theory] §Polynomial Rings, ⁋Proposition 5 and the discussion immediately following it, the condition

\[\x^n-1=qg+r,\qquad \deg r<\deg g\]

is uniquely satisfied by \(q,r\in\mathbb{Z}[\x]\), so this \(q\) must be \(\Phi_n\), and thus \(\Phi_n\) also becomes a polynomial with integer coefficients.

It should be noted that the coefficients of \(\Phi_n(\x)\) need not all be \(0\) or \(\pm 1\); the smallest counterexample is \(\Phi_{105}\). (Wikipedia, Cyclotomic Polynomial)

Irreducibility over the Rational Field

By Proposition 2, we have \(\Phi_n(\x)\in\mathbb{Z}[\x]\), so \(\Phi_n\) can be viewed, over the smallest field containing \(\mathbb{Z}\), namely \(\mathbb{Q}\), as an element of the polynomial ring \(\mathbb{Q}[\x]\); since \(\mathbb{Q}\) is the prime field of characteristic \(0\) and is contained in every field \(\mathbb{K}\) of characteristic zero, there is no issue with doing so. In this case specifically, we first show that \(\Phi_n\) is the minimal polynomial of \(\zeta\). For this, it suffices to show that \(\Phi_n\) is irreducible in \(\mathbb{Q}[\x]\). (§Algebraic Extensions, ⁋Theorem 15)

Lemma 3 If a monic polynomial \(f\in\mathbb{Z}[\x]\) is written, for monic polynomials in \(\mathbb{Q}[\x]\), \(g,h\), as \(f=gh\), then \(g,h\in\mathbb{Z}[\x]\).

Proof

Since the two monic polynomials \(g,h\) are elements of \(\mathbb{Q}[\x]\), we can ensure that the scalar multiples \(ag\) and \(bh\) belong to \(\mathbb{Z}[\x]\) by choosing minimal positive integers \(a,b\). Then by the minimality of \(a\) and \(b\), the greatest common divisors of the coefficients of \(ag\) and \(bh\) must both be \(1\).

Now \(abf=(ag)(bh)\), and our claim is that \(ab=1\). Suppose for contradiction that \(ab>1\), and let a prime divisor of \(ab\), say \(p\), be given. Then since in \(\mathbb{F}_p[\x]\) we have \(ab=0\), this equality becomes

\[0=\overline{ag}\cdot \overline{bh}\]

and since \(\mathbb{F}_p[\x]\) is an integral domain ([Ring Theory] §Polynomial Rings, ⁋Proposition 4), we must have \(\overline{ag}=0\) or \(\overline{bh}=0\). That is, the greatest common divisor of the coefficients of \(ag\) or \(bh\) must be divisible by \(p\), which is a contradiction, yielding the desired result.

Then we can show the following.

Theorem 4 (Gauss) For any natural number \(n\geq1\), \(\Phi_n\) is irreducible in \(\mathbb{Q}[\x]\).

Proof

Fix a primitive \(n\)-th root of unity \(\zeta\in\overline{\mathbb{Q}}\). Then the minimal polynomial of \(\zeta\), \(f\in \mathbb{Q}[\x]\), must divide both \(\Phi_n\) and \(\x^n-1\) (§Algebraic Extensions, ⁋Theorem 15), and from this, if we write \(\x^n-1=fg\), then since both \(f,g\) are monic, we have \(f,g\in\mathbb{Z}[\x]\) by Lemma 3.

The core of the proof is that given any root of \(f\), say \(\omega\), and any prime not dividing \(n\), say \(p\), \(\omega^p\) is also a root of \(f\). Suppose for contradiction that \(f(\omega^p)\neq0\). First, by definition \(\omega^n=1\), so \(\omega^p\) is a root of \(\x^n-1\), and therefore \(g(\omega^p)=0\). That is, \(\omega\) is a root of \(g(\x^p)\in\mathbb{Z}[\x]\). Meanwhile, since \(f\) is a minimal polynomial, it is monic and irreducible, and because \(f(\omega)=0\), by §Algebraic Extensions, ⁋Theorem 15, \(f\) is the minimal polynomial of \(\omega\), and thus \(f\) divides \(g(\x^p)\). Here, since \(f\) is monic, just as in the proof of Proposition 2, the quotient of \(g(\x^p)\) divided by \(f\) also belongs to \(\mathbb{Z}[\x]\).

Now let us reduce this relation to \(\mathbb{F}_p[\x]\). In \(\mathbb{F}_p\), every element \(c\) satisfies \(c^p=c\), and since the Frobenius endomorphism is a ring homomorphism (§Fields, ⁋Theorem 10), writing the image of \(g\) as \(\bar{g}(\x)=\sum c_i\x^i\in \mathbb{F}_p[\x]\), we have

\[\bar{g}(\x)^p=\sum c_i^p\x^{ip}=\sum c_i(\x^p)^i=\bar{g}(\x^p)\]

holds. Thus \(\bar{f}\) divides \(\bar{g}^p\). Meanwhile, since \(\deg\bar{f}=\deg f\geq1\), if we choose an irreducible factor of \(\bar{f}\), say \(h\), then \(h\) divides \(\bar{g}^p\). Since \(\mathbb{F}_p[\x]\) is a UFD ([Ring Theory] §Polynomial Rings, ⁋Theorem 16), by [Ring Theory] §Integral Domains, ⁋Proposition 17, \(h\) is prime and thus must divide \(\bar{g}\). Hence \(h^2\) divides \(\bar{f}\bar{g}=\x^n-1\).

However, since \(p\nmid n\), in \(\mathbb{F}_p\) we have \(n\neq0\), and therefore \(\x^n-1\) and its derivative \(n\x^{n-1}\) have no common roots. That is, \(\x^n-1\) has no multiple roots in \(\overline{\mathbb{F}_p}\), so it is a contradiction that \(h^2\) divides \(\x^n-1\).

Now we take an \(m\) satisfying \(\gcd(m,n)=1\) and \(1\leq m\leq n\), and let \(m=p_1\cdots p_r\) be its prime factorization. Then each \(p_i\) is coprime to \(n\) and hence does not divide \(n\), so by applying the above claim successively to

\[\zeta,\zeta^{p_1},\zeta^{p_1p_2},\ldots\]

we see that \(\zeta^m\) is a root of \(f\). That is, \(f\) has all \(\varphi(n)\) primitive \(n\)-th roots of unity as roots, and since \(f\) divides \(\Phi_n\) and both are monic, we have \(f=\Phi_n\). Therefore \(\Phi_n\) is a minimal polynomial, and hence irreducible.

Then in particular, the following holds.

Corollary 5 For a primitive \(n\)-th root of unity \(\zeta\), \(\Phi_n\) is the minimal polynomial of \(\zeta\) over \(\mathbb{Q}\), and \([\mathbb{Q}(\zeta):\mathbb{Q}]=\varphi(n)\).

Proof

Since \(\Phi_n\) is monic, irreducible by Theorem 4 (Gauss), and \(\Phi_n(\zeta)=0\), by §Algebraic Extensions, ⁋Theorem 15, \(\Phi_n\) is the minimal polynomial of \(\zeta\), and the dimension of \(\mathbb{Q}[\zeta]=\mathbb{Q}(\zeta)\) over \(\mathbb{Q}\) is \(\deg\Phi_n=\varphi(n)\).

Galois Group of Cyclotomic Extensions

Now we examine the extension obtained by adjoining a primitive root of unity.

Definition 6 For a field \(\mathbb{K}\) and a primitive \(n\)-th root of unity \(\zeta\), the extension \(\mathbb{K}(\zeta)/\mathbb{K}\) is called the \(n\)-th cyclotomic extension. In particular, when \(\mathbb{K}=\mathbb{Q}\), we call \(\mathbb{Q}(\zeta)\) the \(n\)-th cyclotomic field, and fixing a generator denoted by \(\zeta_n\), we write this as \(\mathbb{Q}(\zeta_n)\).

For two generators \(\zeta,\zeta'\), since \(\zeta'\in\langle\zeta\rangle\) and \(\zeta\in\langle\zeta'\rangle\), we have \(\mathbb{K}(\zeta)=\mathbb{K}(\zeta')\). That is, a cyclotomic extension does not depend on the choice of generator, and \(\mathbb{K}(\zeta)=\mathbb{K}(\mu_n)\) holds.

For an arbitrary field \(\mathbb{K}\), it is not easy to say anything specific about \(\mathbb{K}(\zeta)/\mathbb{K}\). For example, if we consider \(\Phi_4\) over \(\mathbb{C}\), since the solutions \(\x=\pm i\) of \(\x^4-1=0\) are already contained in \(\mathbb{C}\), we have \([\mathbb{K}(\zeta):\mathbb{K}]=1\). However, the following theorem shows that in the minimal case where \(\mathbb{K}=\mathbb{Q}\), they behave just as we would expect.

Theorem 7 \(\mathbb{Q}(\zeta_n)/\mathbb{Q}\) is a finite degree Galois extension, and the injective homomorphism

\[\Gal(\mathbb{Q}(\zeta_n)/\mathbb{Q})\rightarrow(\mathbb{Z}/n\mathbb{Z})^\times\]

of §Solvability by Radicals, ⁋Proposition 3 is an isomorphism.

Proof

That \(\mathbb{Q}(\zeta_n)/\mathbb{Q}\) is a finite degree Galois extension and the given homomorphism is injective is §Solvability by Radicals, ⁋Proposition 3. Meanwhile, since \(\mathbb{Q}(\zeta_n)/\mathbb{Q}\) is a Galois extension, by §Fundamental Theorem of Galois Theory, ⁋Proposition 2 and Corollary 5, we have

\[\lvert\Gal(\mathbb{Q}(\zeta_n)/\mathbb{Q})\rvert = [\mathbb{Q}(\zeta_n):\mathbb{Q}] = \varphi(n) = \lvert(\mathbb{Z}/n\mathbb{Z})^\times\rvert\]

Since an injective map between two finite sets of the same size is surjective, the given injective homomorphism is an isomorphism.

In particular, \(\Gal(\mathbb{Q}(\zeta_n)/\mathbb{Q})\) is an abelian group of order \(\varphi(n)\). Then, by §Fundamental Theorem of Galois Theory, ⁋Theorem 1, the subextensions of \(\mathbb{Q}(\zeta_n)\) correspond bijectively to the subgroups of \((\mathbb{Z}/n\mathbb{Z})^\times\), and since every subgroup of an abelian group is normal, by §Fundamental Theorem of Galois Theory, ⁋Corollary 6, all these subextensions are Galois extensions of \(\mathbb{Q}\).

For example, when \(n=8\), we have \((\mathbb{Z}/8\mathbb{Z})^\times\cong\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}\), so there exist exactly three non-trivial proper subextensions. Indeed, for \(\zeta=\zeta_8=(1+i)/\sqrt{2}\), since

\[\zeta^2=i,\qquad \zeta+\zeta^{-1}=\sqrt{2},\qquad \zeta-\zeta^{-1}=i\sqrt{2}=\sqrt{-2}\]

\(\mathbb{Q}(\zeta_8)\) contains three distinct quadratic fields \(\mathbb{Q}(i), \mathbb{Q}(\sqrt{2}), \mathbb{Q}(\sqrt{-2})\), and by the Galois correspondence, these are all the non-trivial subextensions of \(\mathbb{Q}(\zeta_8)\). That is, we can see that simply by adjoining an \(8\)-th root of \(1\) to \(\mathbb{Q}\), irrational numbers such as \(\sqrt{2}\) are also naturally obtained.

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References

[Bou] N. Bourbaki. Algebra II: Chapters 4–7. Springer, 2003.
[Lan] S. Lang. Algebra. Graduate texts in mathematics. Springer, 2002.