Any complex number \(z = x + iy\) may be written in the “polar” form

where \(r = |z| = \sqrt{x^2 + y^2}\) is the distance of \(z\) to the origin, and \(\varphi \) is the angle measured from the positive \(x\)-axis. The \(n\)-th roots of unity are therefore of the form

since for all \(k\)

We obtain these roots geometrically by inscribing a regular \(n\)-gon into the unit circle. Note that \(\lambda _k = \zeta ^k\) for all \(k\), where \(\zeta = e^{\frac{2\pi i}{n}}\). Thus the \(n\)-th roots of unity form a cyclic group \(\{ \zeta , \zeta ^2, \dots , \zeta ^{n-1}, \zeta ^n = 1\} \) of order \(n\).

Our first ingredient comes from a blend of linear algebra and basic group theory. For an arbitrary element \(s \in R\), let \(C_s\) be the set \(\{ x \in R : xs = sx\} \) of elements which commute with \(s\); \(C_s\) is called the centralizer of \(s\). Clearly, \(C_s\) contains \(0\) and \(1\) and is a sub-division ring of \(R\). The center \(Z\) is the set of elements which commute with all elements of \(R\), thus \(Z = \bigcap _{s \in R} C_s\). In particular, all elements of \(Z\) commute, \(0\) and \(1\) are in \(Z\), and so \(Z\) is a finite field. Let us set \(|Z| = q\).

We can regard \(R\) and \(C_s\) as vector spaces over the field \(Z\) and deduce that \(|R| = q^n\), where \(n\) is the dimension of the vector space \(R\) over \(Z\), and similarly \(|C_s| = q^{n_s}\) for suitable integers \(n_s \ge 1\).

Now let us assume that \(R\) is not a field. This means that for some \(s \in R\) the centralizer \(C_s\) is not all of \(R\), or, what is the same, \(n_s {\lt} n\).

On the set \(R^* := R \setminus \{ 0\} \) we consider the relation

It is easy to check that \(\sim \) is an equivalence relation. Let

be the equivalence class containing \(s\). We note that \(|A_s| = 1\) precisely when \(s\) is in the center \(Z\). So by our assumption, there are classes \(A_s\) with \(|A_s| \ge 2\). Consider now for \(s \in R^*\) the map \(f_s : x \mapsto x^{-1}sx\) from \(R^*\) onto \(A_s\). For \(x, y \in R^*\) we find

for \(C_s^* := C_s \setminus \{ 0\} \), where \(C_s^* x = \{ zx : z \in C_s^*\} \) has size \(|C_s^*|\). Hence any element \(x^{-1}sx\) is the image of precisely \(|C_s^*| = q^{n_s} - 1\) elements in \(R^*\) under the map \(f_s\), and we deduce \(|R^*| = |A_s| |C_s^*|\). In particular, we note that

We know that the equivalence classes partition \(R^*\). We now group the central elements \(Z^*\) together and denote by \(A_1, \dots , A_t\) the equivalence classes containing more than one element. By our assumption we know \(t \ge 1\). Since \(|R^*| = |Z^*| + \sum _{k=1}^t |A_k|\), we have proved the so-called class formula

where we have \(1 {\lt} \frac{q^n - 1}{q^{n_k} - 1} \in \mathbb {N}\) for all \(k\).

With ?? we have left abstract algebra and are back to the natural numbers. Next we claim that \(q^{n_k} - 1 \mid q^n - 1\) implies \(n_k \mid n\). Indeed, write \(n = a n_k + r\) with \(0 \le r {\lt} n_k\), then \(q^{n_k} - 1 \mid q^{a n_k + r} - 1\) implies

and thus \(q^{n_k} - 1 \mid q^{(a-1)n_k + r} - 1\), since \(q^{n_k}\) and \(q^{n_k} - 1\) are relatively prime. Continuing in this way we find \(q^{n_k} - 1 \mid q^r - 1\) with \(0 \le r {\lt} n_k\), which is only possible for \(r = 0\), that is, \(n_k \mid n\). In summary, we note

Now comes the second ingredient: the complex numbers \(\mathbb {C}\). Consider the polynomial \(x^n - 1\). Its roots in \(\mathbb {C}\) are called the \(n\)-th roots of unity. Since \(\lambda ^n = 1\), all these roots \(\lambda \) have \(|\lambda | = 1\) and lie therefore on the unit circle of the complex plane. In fact, they are precisely the numbers \(\lambda _k = e^{\frac{2k\pi i}{n}} = \cos (2k\pi /n) + i \sin (2k\pi /n)\), \(0 \le k \le n - 1\). Some of the roots \(\lambda \) satisfy \(\lambda ^d = 1\) for \(d {\lt} n\); for example, the root \(\lambda = -1\) satisfies \(\lambda ^2 = 1\). For a root \(\lambda \), let \(d\) be the smallest positive exponent with \(\lambda ^d = 1\), that is, \(d\) is the order of \(\lambda \) in the group of the roots of unity. Then \(d \mid n\), by Lagrange’s theorem (“the order of every element of a group divides the order of the group”). Note that there are roots of order \(n\), such as \(\lambda _1 = e^{\frac{2\pi i}{n}}\).

Now we group all roots of order \(d\) together and set

Note that the definition of \(\phi _d(x)\) is independent of \(n\). Since every root has some order \(d\), we conclude that

Here is the crucial observation: The coefficients of the polynomials \(\phi _n(x)\) are integers (that is, \(\phi _n(x) \in \mathbb {Z}[x]\) for all \(n\)), where in addition the constant coefficient is either \(1\) or \(-1\). Let us carefully verify this claim. For \(n = 1\) we have \(1\) as the only root, and so \(\phi _1(x) = x - 1\). Now we proceed by induction, where we assume \(\phi _d(x) \in \mathbb {Z}[x]\) for all \(d {\lt} n\), and that the constant coefficient of \(\phi _d(x)\) is \(1\) or \(-1\). By ??,

where \(p(x) = \prod _{d \mid n, d {\lt} n} \phi _d(x) = \sum _{j=0}^{n-\ell } p_j x^j\), \(\phi _n(x) = \sum _{k=0}^{\ell } a_k x^k\), with \(p_0 = 1\) or \(p_0 = -1\). Since \(-1 = p_0 a_0\), we see \(a_0 \in \{ 1, -1\} \). Suppose we already know that \(a_0, a_1, \dots , a_{k-1} \in \mathbb {Z}\). Computing the coefficient of \(x^k\) on both sides of ?? we find

By assumption, all \(a_0, \dots , a_{k-1}\) (and all \(p_j\)) are in \(\mathbb {Z}\). Thus \(p_0 a_k\) and hence \(a_k\) must also be integers, since \(p_0\) is \(1\) or \(-1\).

We are ready for the coup de grâce. Let \(n_k \mid n\) be one of the numbers appearing in ??. Then

We conclude that in \(\mathbb {Z}\) we have the divisibility relations

Since ?? holds for all \(k\), we deduce from the class formula ??

but this cannot be. Why? We know \(\phi _n(x) = \prod (x - \lambda )\) where \(\lambda \) runs through all roots of \(x^n - 1\) of order \(n\). Let \(\lambda = a + ib\) be one of those roots. By \(n {\gt} 1\) (because of \(R \ne Z\)) we have \(\lambda \ne 1\), which implies that the real part \(a\) is smaller than \(1\). Now \(|\lambda |^2 = a^2 + b^2 = 1\), and hence

and so \(|q - \lambda | {\gt} q - 1\) holds for all roots of order \(n\). This implies

which means that \(\phi _n(q)\) cannot be a divisor of \(q - 1\), contradiction and end of proof.