6 Cyclotomic extensions

6.1 Roots of unity

Since we are very interested in solutions by radicals, a natural type of field extension to consider is a simple radical extension $K(\sqrt[n]{a})/K$, obtained by adjoining the $n$th root of some element $a\in K$. As we have already seen, the special case $n=2$ and $a\in K^{\times}\setminus{K^\times}^2$ (i.e. $a$ not a square in $K$) produces a Galois extension of degree $2$. For $n\geq 3$, however, one doesn’t always obtain a Galois extension this way. For instance $\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q}$ is not normal, but we rectified this unpleasant situation by including a cube root of unity $\omega=e^{2\pi i/3}$ in our fields. By setting $K=\mathbb{Q}(\omega)$, we obtain a Galois extension $K(\sqrt[3]{2})/K$ of degree $3$ with cyclic Galois group isomorphic to $\mathbb{Z}/3$.

In this chapter, we’ll look extensions of $\mathbb{Q}$ arising by adjoining roots of unity (cyclotomic extensions), and in the following chapter, we’ll classify Galois extensions with cyclic Galois group (cyclic extensions), provided the base field contains suitable roots of unity.

To begin with, we need some terminology and basic results concerning roots of unity.

Definition 6.1 Given a field $K$, an element $\zeta\in K$ is called a primitive $n$-th root of unity if $\zeta^n=1$ but $\zeta^m\neq 1$ for all $0<m<n$. In other words, $\zeta$ has order $n$ in the multiplicative group $K^\times$.

Example 6.1 $(a)$ Every field has a (unique) primitive $1$st root of unity, namely $\zeta=1$.

$(b)$ If $\ch(K)\neq 2$, then $\zeta=-1$ is a primitive $2$nd root of unity. On the other hand, if $\ch(K)=2$, then $\zeta=-1=+1$ is only a primitive $1$st root of unity.

$(c)$ If $\ch(K)=p$ where $p$ is a prime integer, then $K$ doesn’t contain any primitive $p$-th roots of unity. Indeed, $\zeta^p=1$ implies $\zeta^p-1=(\zeta-1)^p$ so $\zeta=1$.

$(d)$ If $K=\mathbb{C}$, then $\zeta_n=e^{2\pi i/n}$ is a primitive $n$-th root of unity.

Proposition 6.1 Let $\zeta\in K$ be a primitive $n$-th root of unity and $m\in\mathbb{Z}$. Then $\zeta^m$ is a primitive $(n/d)$-th root of unity where $d=\gcd(m,n)$. In particular,

$\;\;(i)$ $\zeta^m=1$ if and only if $m$ is an integer multiple of $n$,

$\;(ii)$ $\zeta^m$ is a primitive $n$-th root of unity if and only if $\gcd(m,n)=1$.

Proof. Given $d=\gcd(m,n)$, we have $m/d,n/d\in\mathbb{Z}$ and \[(\zeta^m)^{n/d}=(\zeta^n)^{m/d}=1.\] Hence $\zeta^m$ is an $(n/d)$-th root of unity. Suppose it’s not a primitive $(n/d)$-th root of unity, so that $(\zeta^m)^k=1$ for some $1\leq k<n/d$. By the Division Algorithm, there exist $q,r\in\mathbb{Z}$ with $mk=qn+r$ where $0\leq r<n$. Then \[\zeta^r=(\zeta^n)^q\zeta^r=(\zeta^m)^k=1\quad\implies\quad r=0\] since $\zeta$ is a primitive $n$-th root of unity. Hence \[mk=qn\quad\implies\quad (m/d)k=(n/d)q \quad\implies (n/d)\mid (m/d)k.\] But $n/d$ and $m/d$ are coprime so $(n/d)\mid k$. This contradicts $1\leq k<n/d$.

The other two statements follow from the cases $d=n$ and $d=1$.

An interesting invariant of a field $K$ is the set of all roots of unity in $K$. This is clearly a subgroup of the multiplicative group $K^\times$ since it contains $1$ and is closed under multiplication.

Definition 6.2 Given a field $K$, we write $\mu(K)$ for the subgroup of all roots of unity in $K^\times$.

If we suppose that $|\mu(K)|$ is finite, then Theorem 5.2 implies that $\mu(K)$ is actually a cyclic group. In other words, if there is a primitive $n$th root of unity $\zeta\in K$ with $n$ maximal, then $\mu(K)$ is generated by $\zeta$ and every root of unity is a suitable power of $\zeta$, i.e. $\mu(K)=\{1,\zeta_n,...,\zeta_n^{n-1}\}$.

Example 6.2 $(a)$ For a finite field $\mathbb{F}_q$, every non-zero element is a root of unity and in the last chapter we saw that $\mathbb{F}_q^{\,\times}$ is always cyclic. In particular, we have \[\mu(\mathbb{F}_q)=\mathbb{F}_q^{\,\times}\cong\mathbb{Z}/(q-1).\]

$(b)$ $\mu(\mathbb{Q})=\{\pm 1\}\cong\mathbb{Z}/2$ with generator $\zeta_2=-1$.

$(c)$ $\mu(\mathbb{Q}(i))=\{\pm 1,\pm i\}\cong\mathbb{Z}/4$ with generator $\zeta_4=i$.

$(d)$ The field $\mathbb{Q}(\sqrt{-3})$ contains the familiar cube root of unity $\omega=e^{2\pi i/3}\in\mathbb{Q}(\sqrt{-3})$. It also contains the $6$th root of unity $\zeta_6=(1+\sqrt{-3})/2$ and $\zeta_6^2=\omega$, $\zeta_6^3=-1$. We’ll see in the next section that $\zeta_n$ for $n\geq 6$ has too high a degree to be in this quadratic field. Thus $\zeta_6$ generates all roots of unity in this field and \[\mu(\mathbb{Q}(\sqrt{-3}))=\{\pm 1,\pm\omega,\pm\omega^2\}\cong\mathbb{Z}/6.\]

$(e)$ $\mu(\mathbb{C})$ is not cyclic. However, it is the union of infinitely many cyclic subgroups of $\mathbb{C}^\times$, for instance, those generated by $\zeta_n=e^{2\pi i/n}$ for each $n$.

6.2 Cyclotomic polynomials

Let $p$ be a prime and consider the primitive $p$-th root of unity $\zeta_p=e^{2\pi i /p}\in\mathbb{C}$. This is evidently a root of \[f(x)=\frac{x^p-1}{x-1}=x^{p-1}+x^{p-2}+\cdots+x+1\] and back in Section 2.3.3, we showed that $f(x)$ is irreducible over $\mathbb{Q}$ (by applying Eisenstein’s criterion to $f(x+1)$ at the prime $p$). In particular, $f(x)$ is the minimal polynomial of $\zeta_p$ over $\mathbb{Q}$.

We would like to similarly find the minimal polynomial of a primitive $n$-th root of unity $\zeta_n=e^{2\pi i/n}\in\mathbb{C}$ for general $n$. Bearing in mind that all primitive $n$-th roots of unity should have the same algebraic properties, we should expect that the roots of this minimal polynomial are all the primitive $n$-th roots of unity. From the previous section, the set of all primitive $n$-th roots of unity in $\mathbb{C}$ is \[P_n=\{\zeta_n^k\;\mid\; \text{$1\leq k<n\;$ and $\;\gcd(k,n)=1$}\}\subset\mathbb{C}.\] We define the $n$-th cyclotomic polynomial by \[\Phi_n(x)=\prod_{\zeta\in P_n} (x-\zeta) =\prod_{\substack{1\leq k<n \\ \gcd(k,n)=1}} (x-\zeta_n^k).\] First let’s consider the number of primitive $n$-th roots of unity, i.e. the degree of $\Phi_n(x)$. From the definition, this is the number of integers between $1$ and $n$ which are coprime to $n$ \[\varphi(n)=|\{k\in\mathbb{Z} \;\mid\; 1\leq k<n\;\;\;\text{and}\;\;\;\gcd(k,n)=1\}|=|(\mathbb{Z}/n)^\times|. \] Those who did Elementary Number Theory II last year will recognise this is Euler’s $\varphi$-function. We state (without proof) some of the properties of this function which can be used to easily calculate it.

Lemma 6.1 $(i)$ For a prime number $p$, we have $\varphi(p)=p-1$.

$(ii)$ For a prime power $p^r$, we have $\varphi(p^r)=p^r-p^{r-1}$.

$(iii)$ If $m$ and $n$ are coprime, then $\varphi(mn)=\varphi(m)\varphi(n)$.

In particular, if we factorise $n=p_1^{r_1}\cdots p_s^{r_s}$ where $p_1,...,p_s$ are distinct primes, then \[\varphi(n)=\prod_{j=1}^s (p_j^{r_j}-p_j^{r_j-1}) =n\prod_{\substack{p\,\mid\, n \\ p\;\text{prime}}} \left(1-\frac{1}{p}\right).\] For instance, $120=2^3\cdot 3\cdot 5$ so \[\varphi(120)=\varphi(2^3)\varphi(3)(\varphi(5)=(2^3-2^2)(3-1)(5-1)=32.\]

Theorem 6.1 For each $n\geq 1$, $\Phi_n(x)$ is a monic polynomial with integer coefficients and degree $\varphi(n)$. Furthermore, \[x^n-1=\prod_{m\mid n} \;\Phi_m(x) \] where the product is over all divisors $m$ of $n$.

Proof. Clearly $\Phi_n(x)$ is monic and by the above discussion, $\deg\Phi_n(x)=\varphi(n)$.

The roots of $x^n-1$ are all the $n$-th roots of unity (not just primitive ones) and \[x^n-1=\prod_{0\leq k<n}\;(x-\zeta_n^k).\] Given $0\leq k<n$, let $d=\gcd(k,n)$ so that $k=dl$ and $n=dm$ for $l, m\in\mathbb{Z}$ with $\gcd(l,m)=1$.

Since $\zeta_n^k=\zeta_{dm}^{dl}=\zeta_m^l$, all factors of the form $x-\zeta_n^k$ with $\gcd(k,n)=d$ will contribute to give $\Phi_m(x)$. The product of all factors will then be $x^n-1=\prod_{m\mid n}\Phi_m(x)$.

To see that $\Phi_n(x)$ has integer coefficients, we use induction on $n$. It’s obvious for $n=1$, and for $n>1$ the induction hypothesis implies \[x^n-1=\Phi_n(x)\prod_{\substack{m\mid n \\ m<n}} \;\Phi_m(x)=\Phi_n(x)g(x)\] where $g(x)\in\mathbb{Z}[x]$. Now by the division algorithm (for monic polynomials in $\mathbb{Z}[x]$), it follows that $\Phi_n(x)\in\mathbb{Z}[x]$.

The factorisation in the above Theorem gives an elementary way to compute $\Phi_n(x)$ explicitly for any $n$. First consider the special case where $n=p$ is prime. Then the Theorem says $x^p-1=\Phi_1(x)\Phi_p(x)$ and since $\Phi_1(x)=x-1$, we recover the $p$-th cyclotomic polynomial from earlier: \[\Phi_p(x)=\frac{x^p-1}{x-1}=x^{p-1}+x^{p-2}+\cdots+x+1.\] Taking $n=p^2$, we have $x^{p^2}-1=\Phi_1(x)\Phi_p(x)\Phi_{p^2}(x)=(x^p-1)\Phi_{p^2}(x)$ and so \[\Phi_{p^2}(x)=\frac{x^{p^2}-1}{x^p-1}=x^{p^2-p}+x^{p^2-2p}+\cdots+x^p+1=\Phi_p(x^p).\] In a similar way, by induction one obtains the $p^r$-th cyclotomic polynomial \[\Phi_{p^r}(x)=\frac{x^{p^r}-1}{x^{p^{r-1}}-1}=\Phi_p(x^{p^{r-1}}).\] When $n$ is not a prime power, we can work similarly, just using all the divisors of $n$. For instance, we have \[x^6-1=\Phi_1(x)\Phi_2(x)\Phi_3(x)\Phi_6(x)=(x-1)(x+1)(x^2+x+1)\Phi_6(x)\] which leads to $\Phi_6(x)=x^2-x+1$. You’ll maybe notice that $\Phi_6(x)=\Phi_3(-x)$ - this is not an accident. When $m$ is an odd integer, we can generally show that \[\Phi_{2m}(x)=\Phi_m(-x).\] Indeed, suppose that $\zeta$ is a primitive $m$-th root of unity. The roots of the polynomial $\Phi_{2m}(x)$ are precisely the primitive $2m$-th roots of unity and so we just need to show that $-\zeta$ is a primitive $2m$-th root of unity. Let $n$ be the (multiplicative) order of $-\zeta$ in $\mathbb{C}^\times$. Now \[\zeta^{2n}=(-\zeta)^{2n}=1\quad\implies\quad m\mid 2n\] and since $m$ is odd, this means $n=km$ for some integer $k$. Furthermore, \[(-\zeta)^{2m}=\zeta^{2m}=(\zeta^m)^2=1\quad\implies\quad n=km\mid 2m\] and so $k$ equals $1$ or $2$, i.e. $n=m$ or $n=2m$. However, $n=m$ is not possible since $(-\zeta)^m=-\zeta^m=-1$.

Putting together everything we’ve calculated so far, here are the first $10$ cyclotomic polynomials: \[\begin{align*} \Phi_1(x)&=x-1 & \Phi_2(x)&=x+1 \\ \Phi_3(x)&=x^2+x+1 & \Phi_4(x)&=x^2+1 \\ \Phi_5(x)&=x^4+x^3+x^2+x+1 & \Phi_6(x)&=x^2-x+1 \\ \Phi_7(x)&=x^6+x^5+x^4+x^3+x^2+x+1 & \Phi_8(x)&=x^4+1 \\ \Phi_9(x)&=x^6+x^3+1 & \Phi_{10}(x)&=x^4-x^3+x^2-x+1 \\ \end{align*}\]

Remark. In all the examples above, the coefficients are always $0$ or $\pm 1$. This is not generally true but you have to work quite hard to find a counterexample! The first time it fails to happen is $\Phi_{105}(x)$ which has a term $-2x^{41}$ and it’s possible to show that the coefficients of $\Phi_n(x)$ can be arbitrarily large.

As mentioned earlier, for $p$ prime, the $p$-th cyclotomic polynomial is irreducible over $\mathbb{Q}$ via Eisenstein’s Criterion. In fact, the same trick shows that \[\Phi_{p^r}(x)=\Phi_p(x^{p^{r-1}})=\frac{x^{p^r}-1}{x^{p^{r-1}}-1}=\sum_{j=0}^{p-1}\;x^{j\,p^{r-1}}.\] is also irreducible over $\mathbb{Q}$. Check that the non-leading coefficients of $\Phi_{p^r}(x+1)$ are all multiples of $p$ and the constant coefficient is $\Phi_{p^r}(1)=p$. More generally, we have to work harder but still have the following very important fact:

Theorem 6.2 For $n\geq 1$, the $n$-th cyclotomic polynomial $\Phi_n(x)$ is irreducible over $\mathbb{Q}$. In particular, $\Phi_n(x)$ is the minimal polynomial of $\zeta_n$ over $\mathbb{Q}$.

Proof. Let $f(x)\in\mathbb{Q}[x]$ be an irreducible factor of $\Phi_n(x)$ and write \[\Phi_n(x)=f(x)g(x).\] By Gauss’s Lemma, we can assume that $f(x), g(x)$ are monic polynomials in $\mathbb{Z}[x]$, since $\Phi_n(x)$ is.

We wish to show that $f(\zeta_n^k)=0$ for all $1\leq k<n$ and $\gcd(k,n)=1$ as then $\Phi_n(x)=f(x)$ is irreducible.

Each such $k$ is a product of primes not dividing $n$ so it will suffice to show that for any given primitive $n$-th root of unity $\zeta$, we have \[f(\zeta)=0\quad\implies\quad f(\zeta^p)=0 \qquad\text{for all primes $p\nmid n$.}\] Suppose on the contrary that $f(\zeta)=0$ and $f(\zeta^p)\neq 0$. Now $\zeta^p$ is also a primitive $n$-th root of unity since $\gcd(p,n)=1$, so $\Phi_n(\zeta^p)=0$ and hence $g(\zeta^p)=0$. In particular, $\zeta$ is a root of $f(x)$ and of $g(x^p)$, so these two polynomials have a non-trivial common factor in $\mathbb{Q}[x]$ and, by Gauss’s Lemma, we can take it in $\mathbb{Z}[x]$. We will show this factor gives a contradiction by reducing modulo $p$.

Write $\overline{f}(x)$, $\overline{g}(x)$ for the corresponding reductions in $\mathbb{F}_p[x]$, and notice that by the Child’s Binomial Theorem and Fermat’s Little Theorem we have \[\overline{g}(x^p)=\overline{g}(x)^p.\] In particular, since $\overline{f}(x)$ and $\overline{g}(x^p)$ have a non-trivial common factor in $\mathbb{F}_p[x]$, so must $\overline{f}(x)$ and $\overline{g}(x)$ and therefore $\overline{f}(x)\overline{g}(x)$ has a multiple root. But this implies that $x^n-1$, regarded as a polynomial in $\mathbb{F}_p[x]$, has a multiple root.

This is the required contradiction: as $p\nmid n$, the polynomial $x^n-1$ and $D(x^n-1)=nx^{n-1}$ (considered as polynomials in $\mathbb{F}_p[x]$) have no common roots and so $x^n-1\in\mathbb{F}_p[x]$ cannot have a multiple root.

6.3 Cyclotomic extensions and their Galois groups

A cyclotomic extension of a field $K$ is obtained by adjoining a primitive $n$-th root of unity to $K$. We will mainly be interested in extensions where we adjoin roots of unity to the rational numbers $K=\mathbb{Q}$.

Definition 6.3 Let $n\geq 1$ and $\zeta_n=e^{2\pi i/n}\in\mathbb{C}$. Then $\mathbb{Q}(\zeta_n)$ is the $n$-th cyclotomic extension of $\mathbb{Q}$.

Note that if we chose a different primitive $n$-th root of unity, we would obtain the same extension since they all have the same minimal polynomial $\Phi_n(x)\in\mathbb{Q}[x]$. We can describe the first few easily:

$\underline{n=1}:\;$ $\zeta_1=1$ so $\mathbb{Q}(\zeta_1)=\mathbb{Q}$.
$\underline{n=2}:\;$ $\zeta_2=-1$ so $\mathbb{Q}(\zeta_2)=\mathbb{Q}$.
$\underline{n=3}:\;$ $\zeta_3=(-1+\sqrt{-3})/2$ is a root of $\Phi_3(x)=x^2+x+1$ and $\mathbb{Q}(\zeta_3)=\mathbb{Q}(\sqrt{-3})$.
$\underline{n=4}:\;$ $\zeta_4=i$ is a root of $\Phi_4(x)=x^2+1$ and $\mathbb{Q}(\zeta_4)=\mathbb{Q}(i)$.
$\underline{n=5}:\;$ $\zeta_5$ is a root of $\Phi_5(x)=x^4+x^3+x^2+x+1$ and $[\mathbb{Q}(\zeta_5):\mathbb{Q}]=4$.
$\underline{n=6}:\;$ $\zeta_6=(1+\sqrt{-3})/2$ is a root of $\Phi_6(x)=x^2-x+1=\Phi_3(-x)$ and $\mathbb{Q}(\zeta_6)=\mathbb{Q}(\sqrt{-3})=\mathbb{Q}(\zeta_3)$.
$\underline{n=7}:\;$ $\zeta_7$ is a root of $\Phi_7(x)=x^6+x^5+x^4+x^3+x^2+x+1$ and $[\mathbb{Q}(\zeta_7):\mathbb{Q}]=6$.
$\underline{n=8}:\;$ $\zeta_8=(1+i)/\sqrt{2}$ is a root of $\Phi_8(x)=x^4+1$ and $\mathbb{Q}(\zeta_8)=\mathbb{Q}(i,\sqrt{2})$.
$\underline{n=9}:\;$ $\zeta_9$ is a root of $\Phi_9(x)=x^6+x^3+1$ and $[\mathbb{Q}(\zeta_9):\mathbb{Q}]=6$. We can write $\mathbb{Q}(\zeta_9)=\mathbb{Q}(\sqrt[3]{\zeta_3})$.
$\underline{n=10}:\;$ $\zeta_{10}$ is a root of $\Phi_{10}(x)=x^4-x^3+x^2-x+1=\Phi_5(-x)$ and $\mathbb{Q}(\zeta_{10})=\mathbb{Q}(\zeta_5)$.

Since $\Phi_n(x)$ is irreducible over $\mathbb{Q}$, we generally have \[ [\mathbb{Q}(\zeta_n):\mathbb{Q}]=\varphi(n)\] as one can verify in the above examples. One might ask - what if we adjoin multiple primitive roots of unity of different orders to $\mathbb{Q}$? And what about intersections of different cyclotomic fields? In each case, we just get another cyclotomic extension:

Lemma 6.2 If $\gcd(m,n)=1$, then $\mathbb{Q}(\zeta_n,\zeta_m)=\mathbb{Q}(\zeta_{mn})$ and $\mathbb{Q}(\zeta_m)\cap\mathbb{Q}(\zeta_n)=\mathbb{Q}$.

More generally, if $\gcd(m,n)=d$, then $\mathbb{Q}(\zeta_n,\zeta_m)=\mathbb{Q}(\zeta_{mn/d})$ and $\mathbb{Q}(\zeta_m)\cap\mathbb{Q}(\zeta_n)=\mathbb{Q}(\zeta_d)$.

Proof. Suppose $\gcd(m,n)=1$. Since $\zeta_m=\zeta_{mn}^n$ and $\zeta_n=\zeta_{mn}^n$, we have $\mathbb{Q}(\zeta_m,\zeta_n)\subset\mathbb{Q}(\zeta_{mn})$. For the reverse inclusion, the Euclidean Algorithm gives \[rm+sn=\gcd(m,n)=1\qquad\text{for some $r,s\in\mathbb{Z}$}.\] Then \[\zeta_{mn}=\zeta_{mn}^{rm+sn}=(\zeta_{mn}^m)^r(\zeta_{mn}^n)^s=\zeta_n^r\zeta_m^s\] and so $\mathbb{Q}(\zeta_{mn})\subset\mathbb{Q}(\zeta_m,\zeta_n)$.

For the intersection, consider the degrees: \[[\mathbb{Q}(\zeta_{mn}):\mathbb{Q}]=\varphi(mn)=\varphi(m)\varphi(n)=[\mathbb{Q}(\zeta_{m}):\mathbb{Q}] [\mathbb{Q}(\zeta_{n}):\mathbb{Q}]\]

We leave the more general statements as an exercise.

It’s also possible to determine which roots of unity lie in $\mathbb{Q}(\zeta_n)$. One has \[\mu\left(\mathbb{Q}(\zeta_n)\right)=\begin{cases} \;\;\;\langle\,\zeta_n\,\rangle\cong\mathbb{Z}/n\quad\;\,\text{if $n$ is even,}\\ \langle\,-\zeta_{n}\,\rangle\cong\mathbb{Z}/2n\quad\text{if $n$ is odd.}\end{cases}\]

We won’t prove this, but point out that if $\mathbb{Q}(\zeta_n)$ contains a primitive $N$-th root of unity, then $\mathbb{Q}(\zeta_N)\subset\mathbb{Q}(\zeta_n)$ and so by comparing degrees, one has $\varphi(N)\leq\varphi(n)$. Now one can show that $\varphi(N)\to\infty$ as $N\to\infty$ (albeit erratically) so there is a maximal $N$ with $\mathbb{Q}(\zeta_N)\subset\mathbb{Q}(\zeta_n)$. In particular, $\mu\left(\mathbb{Q}(\zeta_n)\right)$ is a finite and hence cyclic subgroup of $\mathbb{Q}(\zeta_n)^\times$. Examining the Euler $\varphi$-function more carefully then gives the result.

It’s easy to see that $\mathbb{Q}(\zeta_n)/\mathbb{Q}$ is always a Galois extension. Indeed, $\mathbb{Q}(\zeta_n)$ is the splitting field of the irreducible polynomial $\Phi_n(x)$ over $\mathbb{Q}$, hence normal, and the characteristic is zero so the extension is separable. Furthermore, we know the Galois group has order $\deg\Phi_n(x)=\varphi(n)=|(\mathbb{Z}/n)^\times|$.

Theorem 6.3 For $\sigma\in\Gal(\mathbb{Q}(\zeta_n)/\mathbb{Q})$, there is a unique integer $1\leq k_\sigma<n$ with $\gcd(k_\sigma,n)=1$ such that $\sigma(\zeta_n)=\zeta_n^{k_\sigma}$. Furthermore, $\sigma\mapsto k_\sigma\bmod n$ gives an isomorphism \[\Gal(\mathbb{Q}(\zeta_n)/\mathbb{Q}) \;\tilde{\longrightarrow}\; (\mathbb{Z}/n)^\times.\]

Proof. An automorphism $\sigma\in \Gal(\mathbb{Q}(\zeta_n)/\mathbb{Q})$ is uniquely determined by $\sigma(\zeta_n)$, and since this also a root of $\Phi_n(x)$, we must have \[\sigma(\zeta_n)=\zeta_n^{k_\sigma}\quad\text{where $1\leq k_\sigma<n\;$ and $\;\gcd(k_\sigma,n)=1$}.\] Also notice this $k_\sigma$ doesn’t depend on the initial choice of primitive $n$-th root of unity $\zeta_n$. Indeed, if $\zeta$ is another primitive $n$-th root of unity, then $\zeta=\zeta_n^m$ where $\gcd(m,n)=1$ and \[\sigma(\zeta)=\sigma(\zeta_n^m)=\sigma(\zeta_n)^m=\zeta_n^{mk_\sigma}=\zeta^{k_\sigma}.\] We obtain a map \[\begin{align*} \pi\;:\; \Gal(\mathbb{Q}(\zeta_n)/\mathbb{Q}) &\longrightarrow(\mathbb{Z}/n)^\times \\ \sigma &\longmapsto k_\sigma\bmod n \end{align*}\] and we just need to check it is an isomorphism of groups. Suppose $\pi(\sigma_1)=k_1\mod n$ and $\pi(\sigma_2)=k_2\mod n$ where $k_1,k_2\in\mathbb{Z}$ are coprime to $n$. Then \[\sigma_1\sigma_2(\zeta_n)=\sigma_1(\zeta_n^{k_2})=\sigma_1(\zeta_n)^{k_2}=\zeta_n^{k_1k_2}\] so $\pi(\sigma_1\sigma_2)=k_1k_2\bmod n=\pi(\sigma_1)\pi(\sigma_2)$ and $\pi$ is a group homomorphism.

Given $\sigma\in\ker(\pi)$, we have $\sigma(\zeta_n)=\zeta_n$ and $\sigma=\id$ so $\pi$ is an injection. Finally, $\sigma$ is an injection between finite groups of the same order \[|\Gal(\mathbb{Q}(\zeta_n)/\mathbb{Q})|=\deg\Phi_n(x)=\varphi(n)=|(\mathbb{Z}/n)^\times|\] and so $\pi$ is an isomorphism.

We remark that since the Galois group $\Gal(\mathbb{Q}(\zeta_n)/\mathbb{Q})$ is abelian, any subgroup is normal and by the Fundamental Theorem, any subfield of $\mathbb{Q}(\zeta_n)$ is automatically normal and Galois over $\mathbb{Q}$.

Before moving on to further Galois theoretic aspects, let’s pause to consider the structure of the multiplicative group $(\mathbb{Z}/n)^\times$ of invertible elements modulo $n$. If $n=m_1m_2$ where $\gcd(m_1,m_2)=1$, then the Chinese Remainder Theorem for rings implies there is an isomorphism of rings \[\mathbb{Z}/n\;\cong\;\mathbb{Z}/m_1\,\times\,\mathbb{Z}/m_2.\] By taking the subgroups of invertible elements, we obtain an isomorphism of groups \[(\mathbb{Z}/n)^\times\;\cong\;(\mathbb{Z}/m_1)^\times\,\times\,(\mathbb{Z}/m_2)^\times.\] Note that considering the orders of these groups recovers the identity $\varphi(n)=\varphi(m_1)\varphi(m_2)$.

Applying the above decomposition repeatedly, if $n=p_1^{r_1}\cdots p_s^{r_s}$ where $p_1,...,p_s$ are distinct primes, then \[(\mathbb{Z}/n)^\times=(\mathbb{Z}/p_1^{r_1})^\times\times\cdots\times(\mathbb{Z}/p_s^{r_s})^\times.\] For the factors on the right, we give the following facts without proof:

If $p\neq 2$, then $(\mathbb{Z}/p^r)^\times$ is a cyclic group of order $\varphi(p^r)=p^r-p^{r-1}$.
If $p=2$, the groups $(\mathbb{Z}/2)^\times=\{\id\}$ and $(\mathbb{Z}/4)^\times\cong\mathbb{Z}/2$ are both cyclic. However, for $r\geq 3$, the group $(\mathbb{Z}/2^r)^\times$ is not cyclic - try it for $(\mathbb{Z}/8)^\times$. In fact, \[(\mathbb{Z}/2^r)^\times\;\cong\;\mathbb{Z}/2\,\times\,\mathbb{Z}/2^{r-2}.\]

Let’s now see what this tells us about some Galois groups of cyclotomic extensions.

Example 6.3 $(a)\;$ We already know that $\Gal(\mathbb{Q}(\zeta_3)/\mathbb{Q})\cong(\mathbb{Z}/3)^\times\cong\mathbb{Z}/2$, since $\mathbb{Q}(\zeta_3)=\mathbb{Q}(\sqrt{-3})$.

Similarly, for any prime $p$, we have \[\Gal(\mathbb{Q}(\zeta_p)/\mathbb{Q})\cong(\mathbb{Z}/p)^\times\cong\mathbb{Z}/(p-1).\]

$(b)\;$ When $n$ is not prime, it’s possible for $\Gal(\mathbb{Q}(\zeta_n)/\mathbb{Q})$ to be cyclic. For instance, \[\Gal(\mathbb{Q}(\zeta_4)/\mathbb{Q})\cong(\mathbb{Z}/4)^\times\cong\mathbb{Z}/2\] and \[\Gal(\mathbb{Q}(\zeta_9)/\mathbb{Q})\cong(\mathbb{Z}/9)^\times\cong\mathbb{Z}/6.\] To construct this last isomorphism explicitly, check that $(\mathbb{Z}/9)^\times$ is generated by $2\bmod 9$. Indeed, looking modulo $9$, \[2^0\equiv 1,\;\;\;2^1\equiv 2,\;\;\;2^2\equiv 4,\;\;\;2^3\equiv 8,\;\;\;2^4\equiv 7,\;\;2^5\equiv 5\] gives all six integers $1\leq k<9$ with $\gcd(k,9)=1$. Thus $\Gal(\mathbb{Q}(\zeta_9)/\mathbb{Q})$ is generated by the automorphism with $\sigma(\zeta_9)=\zeta_9^2$.

$(c)\;$ When $n$ is not prime, it’s also possible for $\Gal(\mathbb{Q}(\zeta_n)/\mathbb{Q})$ to not be cyclic. For instance, \[\Gal(\mathbb{Q}(\zeta_8)/\mathbb{Q})\cong(\mathbb{Z}/8)^\times\cong\mathbb{Z}/2\times\mathbb{Z}/2.\] We actually already knew this - recall that $\zeta_8=(1+i)/\sqrt{2}$ and $\mathbb{Q}(\zeta_8)=\mathbb{Q}(i,\sqrt{2})$. Methods from last term showed that $\Gal(\mathbb{Q}(i,\sqrt{2})/\mathbb{Q})\cong\mathbb{Z}/2\times\mathbb{Z}/2$.

Remark. A Galois extension with abelian Galois group is called an abelian extension. From the above, the cyclotomic extensions $\mathbb{Q}(\zeta_n)/\mathbb{Q}$ are always abelian. There is a remarkable and deep theorem by Kronecker and Weber saying that every abelian extension of $\mathbb{Q}$ is contained in some cyclotomic extension $\mathbb{Q}(\zeta_n)$.

Further properties when $n$ is an odd prime

When $n=p>2$ is an odd prime, we have the cyclic Galois group \[G=\Gal(\mathbb{Q}(\zeta_p)/\mathbb{Q})\cong(\mathbb{Z}/p)^\times\cong\mathbb{Z}/(p-1).\] A generator of $G$ is given by $\zeta_p\mapsto\zeta_p^k$ where $1\leq k<p$ and $k\bmod p$ has maximal order $p-1$ in $(\mathbb{Z}/p)^\times$.

Now $\mathbb{Q}(\zeta_p)/\mathbb{Q}$ is simple of degree $\deg\Phi_p(x)=p-1$, so the usual choice for a $\mathbb{Q}$-basis would be \[\{1,\zeta,...,\zeta^{p-2}\}.\] However, this isn’t very natural from the point of view of the Galois group - this basis isn’t invariant under the action of $G$. Instead, it’s often better to use the set \[\{\zeta,...,\zeta^{p-1}\}.\] This is a basis since \[\Phi_p(\zeta)=1+\zeta+\cdots+\zeta^{p-2}+\zeta^{p-1}=0\] and now any $\sigma\in G$ maps basis elements to basis elements. We can write any $\alpha\in\mathbb{Q}(\zeta_p)$ using this basis \[\alpha=\sum_{j=1}^{p-1}\, a_j\zeta_p^j\qquad\text{where $a_j\in\mathbb{Q}$}.\] (Notice that $\alpha\in\mathbb{Q}$ precisely when $a_1=a_2=\cdots=a_{p-1}$.) Given $\alpha$ written in this way, we can determine the minimal polynomial $f(x)\in\mathbb{Q}[x]$ of $\alpha$ over $\mathbb{Q}$ as follows. For any $\tau\in G$, the conjugate $\tau(\alpha)\in\mathbb{Q}(\zeta_p)$ is also a root of $f(x)$. These conjugates will all have the form \[\sum_{j=1}^{p-1}\, a_j\zeta_p^{jr}\] for some $1\leq r\leq p-1$. Now these $p-1$ elements may not be distinct, but if we find the distinct conjugates $\alpha=\alpha^{(1)}, \alpha^{(2)},...,\alpha^{(d)}$, then the minimal polynomial of $\alpha$ over $\mathbb{Q}$ equals \[f(x)=(x-\alpha^{(1)})(x-\alpha^{(2)})\cdots(x-\alpha^{(d)}).\]

Example 6.4 The extension $\mathbb{Q}(\zeta_5)/\mathbb{Q}$ has cyclic Galois group \[G=\Gal(\mathbb{Q}(\zeta_5)/\mathbb{Q})\cong(\mathbb{Z}/5)^\times\cong\mathbb{Z}/4\] and a generator $\sigma\in G$ is given by $\sigma(\zeta_5)=\zeta_5^2$. (To see this, check that $2\bmod 5$ has multiplicative order $4$ in $(\mathbb{Z}/5)^\times$ and so $\sigma^2\neq\id$, $\sigma^4=\id$.)

Now consider the element $\eta=\zeta_5+\zeta_5^4\in\mathbb{Q}(\zeta_5)$. We calculate \[\sigma(\eta)=\zeta_5^2+\zeta_5^8=\zeta_5^2+\zeta_5^3\quad\text{and}\quad\sigma^2(\eta)=\zeta_5^4+\zeta_5^6=\eta.\] Note $\sigma(\eta)\neq\eta$ since $\{\zeta_5,\zeta_5^2,\zeta_5^3,\zeta_5^4\}$ is a basis of $\mathbb{Q}(\zeta_5)/\mathbb{Q}$ and its elements are linearly independent over $\mathbb{Q}$. In particular, $\mathbb{Q}(\eta)$ is the fixed field of the order $2$ subgroup $\langle\,\sigma^2\,\rangle\subset G$ and $[\mathbb{Q}(\eta):\mathbb{Q}]=2$. Furthermore, the minimal polynomial of $\eta$ over $\mathbb{Q}$ is now \[\begin{align*} (x-\eta)(x-\sigma(\eta))&=(x-\zeta_5-\zeta_5^4)(x-\zeta_5^2-\zeta_5^3) \\ &=x^2-(\zeta_5+\zeta_5^4+\zeta_5^2+\zeta_5^3)x+(\zeta_5^3+\zeta_5^4+\zeta_5^6+\zeta_5^7) \\ &=x^2-(\zeta_5+\zeta_5^2+\zeta_5^3+\zeta_5^4)x+(\zeta_5+\zeta_5^2+\zeta_5^3+\zeta_5^4) \\ &=x^2+x-1 \end{align*}\] where we’ve used $\zeta_5+\zeta_5^2+\zeta_5^3+\zeta_5^4=-1$.

An amusing corollary of this appears if we notice \[\begin{align*} \eta&=\zeta_5+\zeta_5^{-1}=e^{2\pi i/5}+e^{-2\pi i/5}=2\cos(2\pi/5)\quad\text{and} \\ \sigma(\eta)&=\zeta_5^2+\zeta_5^{-2}=e^{4\pi i/5}+e^{-4\pi i/5}=2\cos(4\pi/5) \end{align*}\]

Since $x^2+x-1$ has roots $(-1\pm\sqrt{5})/2$, we’ve deduced that \[\cos\left(\frac{2\pi}{5}\right)=\frac{-1+\sqrt{5}}{4}\qquad\text{and}\qquad \cos\left(\frac{4\pi}{5}\right)=\frac{-1-\sqrt{5}}{4}.\] Working similarly in the field $\mathbb{Q}(\zeta_{17})$, Gauss showed that \[\begin{multline*} \qquad\quad\cos\left(\frac{2\pi}{17}\right)=-\frac{1}{16}+\frac{1}{16}\sqrt{17}+\frac{1}{16}\sqrt{34-2\sqrt{17}} \\ +\frac{1}{8}\sqrt{17+3\sqrt{17}-\sqrt{34-2\sqrt{17}}-2\sqrt{34+2\sqrt{17}}}.\qquad\qquad \end{multline*}\] See the Problem Sheets for further details! Since this algebraic number is expressed as a sequence of repeated quadratic radical operations, one is then able to construct a regular $17$-sided polygon using only ruler and compass.

Back to general odd prime $p>2$, we have \[G=\Gal(\mathbb{Q}(\zeta_p)/\mathbb{Q})\cong(\mathbb{Z}/p)^\times\cong\mathbb{Z}/(p-1)\] generated by $\sigma(\zeta_p)=\zeta_p^k$ where $1\leq k<p$ and $k\bmod p$ has maximal order $p-1$ in $(\mathbb{Z}/p)^\times$.

Since this is cyclic, there is a unique cyclic subgroup for each divisor $d\mid p-1$. The Fundamental Theorem thus tells us that $\mathbb{Q}(\zeta_p)$ contains exactly one subfield $M$ with $[\mathbb{Q}(\zeta_p):M]=d$. Writing $p-1=dm$, we see that $[M:\mathbb{Q}]=m$ where $m$ is also a divisor of $p-1$. As $p$ is odd, taking $m=2$ shows that $\mathbb{Q}(\zeta_p)$ contains a unique quadratic extension of $\mathbb{Q}$. This must be of the form $\mathbb{Q}(\sqrt{D_p})$ for some integer $D_p$ and one can actually find an explicit formula for it. Define \[\Theta=\zeta_p-\sigma(\zeta_p)+\sigma^2(\zeta_p)-\sigma^3(\zeta_p)+\cdots+\sigma^{p-3}(\zeta_p)-\sigma^{p-2}(\zeta_p).\] This special element in $\mathbb{Q}(\zeta_p)$ behaves like $\sqrt{D_p}$ for some $D_p\in\mathbb{Q}$. Indeed, using $\sigma^{p-1}=\id$, we find $\sigma(\Theta)=-\Theta$ and $\sigma^{2}(\Theta)=\Theta$. Furthermore, $\Theta$ is non-zero since the $\sigma^j(\zeta_p)$ are the elements of the basis $\{\zeta_p,...,\zeta_p^{p-1}\}$. In other words, $\Theta$ is invariant with respect to the action of the index $2$ subgroup $\langle\,\sigma^2\,\rangle$ of $G$, and $\Theta$ changes sign under the induced action of $G/\langle\,\sigma^2\,\rangle\cong\mathbb{Z}/2$.

Fact: We have $D_p=\Theta^2=(-1)^{(p-1)/2}p$.

For small $p$, this fact can be obtained by direct computation, and for general $p$, it can be proved using techniques from Elementary Number Theory (as in certain proofs of Quadratic Reciprocity.)

There are various symmetry tricks we can use to simplify explicit calculations. For instance, when multiplying out $\Theta^2$, one obtains a linear combination of the form $a_0+a(\zeta_p+\zeta_p^2+\cdots+\zeta_p^{p-1})=a_0-a\in\mathbb{Q}$. Thus one only needs to follow the coefficients of $\zeta_p^0=1$ and $\zeta_p$.

Example 6.5 When $p=5$, using the generator $\sigma(\zeta_5)=\zeta_5^2$, we have \[\Theta=\zeta_5-\zeta_5^2+\zeta_5^4-\zeta_5^3\] and multiplying out leads to $\Theta^2=4-(\zeta_5+\zeta_5^2+\zeta_5^3+\zeta_5^4)=5$. We recover the fact that $\mathbb{Q}(\zeta_5)$ contains the unique quadratic extension $\mathbb{Q}(\sqrt{5})/\mathbb{Q}$.

Similarly, when $p=7$ a generator of the Galois group is $\sigma(\zeta_7)=\zeta_7^3$ and we have \[\Theta=\zeta_7-\zeta_7^3+\zeta_7^2-\zeta_7^6+\zeta_7^4-\zeta_7^5.\] Multiplying out leads to $\Theta^2=-6+(\zeta_7+\zeta_7^2+\zeta_7^3+\zeta_7^4+\zeta_7^5+\zeta_7^6)=-7$ and we find that $\mathbb{Q}(\zeta_7)$ contains the unique quadratic extension $\mathbb{Q}(\sqrt{-7})/\mathbb{Q}$.