The polygon an equation forces

The equation $z^n = 1$ has $n$ solutions in the complex numbers, and the solutions are not scattered. They are the corners of a regular $n$-sided polygon inscribed in the unit circle, with one corner at $1$.

That is the picture, and the rest of this essay is what can be read off it. The answer is: a group, an identity, a factorisation with whole-number coefficients, and a connection between the divisors of $n$ and the arithmetic of the polygon that nobody would guess from the drawing.

Why they are evenly spaced #

The reason comes from what multiplication does to the plane.

A complex number can be written by its length and its angle, and multiplying two of them multiplies the lengths and adds the angles. So raising $z$ to the $n$th power raises its length to the $n$th power and multiplies its angle by $n$. For $z^n = 1$ the length must be one, since no other positive number has an $n$th power equal to one; and the angle multiplied by $n$ must be a whole number of turns.

So the angle is a whole number of $n$th parts of a turn, and there are exactly $n$ distinct such angles. The solutions are

$$\zeta_k = \cos\frac{2\pi k}{n} + i \sin\frac{2\pi k}{n}, \qquad k = 0, 1, \dots, n-1,$$

the corners of a regular polygon. And they multiply the way angles add, which makes them a group under multiplication: a cyclic one, generated by any solution whose angle is a full $n$th of a turn.

The cyclic structure is the reason the whole subject is arithmetic rather than geometry. Multiplying two roots adds their indices modulo $n$; taking powers multiplies the index; and the group of $n$th roots of unity is the integers modulo $n$ under addition, wearing a circle.

They add to nothing #

The most quoted fact about the roots is the one the picture makes obvious.

The $n$ roots of unity add to zero for $n > 1$. Three proofs, and each is worth having because each generalises differently.

By symmetry. The set of roots is unchanged by multiplication by $\zeta_1$, which rotates it. So the sum $S$ satisfies $\zeta_1 S = S$, and since $\zeta_1 \ne 1$ this forces $S = 0$.

By the polynomial. The roots are the roots of $z^n - 1$, whose coefficient of $z^{n-1}$ is zero, and the sum of the roots of a monic polynomial is minus that coefficient. This is Vieta’s relation, and it says the same thing with no geometry.

By the picture. The corners of a regular polygon average to its centre, which is the origin. Laying the arrows end to end closes them into a loop, which is what the figure draws.

The identity is not decoration. It is why the discrete Fourier transform works: summing $\zeta^{jk}$ over $j$ gives zero unless $k$ is a multiple of $n$, and that orthogonality is what lets a signal be split into frequencies and put back together. The Fourier series is the same identity in the continuous limit, with an integral in place of the sum.

The factorisation with whole numbers in it #

Now the part that is not visible in the polygon at all.

Each root has an order: the smallest number of times it must be multiplied by itself to reach $1$. The root at $k$ steps round has order $n/\gcd(k, n)$, so the orders are exactly the divisors of $n$, and each divisor $d$ has $\varphi(d)$ roots of that order — where $\varphi$ counts how many numbers below $d$ share no factor with it.

Group the roots by order and take, for each divisor $d$, the polynomial whose roots are exactly the roots of order $d$:

$$\Phi_d(x) = \prod_{\substack{1 \le k \le d \\ \gcd(k,d)=1}} \bigl(x - \zeta_d^{\,k}\bigr).$$

That is the $d$th cyclotomic polynomial, and since every root of $x^n - 1$ has exactly one order,

$$x^n - 1 = \prod_{d \mid n} \Phi_d(x).$$

The surprise is that each $\Phi_d$ has whole-number coefficients, though its definition is a product over complex numbers. The reason is an induction: $\Phi_1 = x - 1$ has whole coefficients, and dividing $x^n - 1$ by the product of the earlier ones is a division of one monic integer polynomial by another, which leaves an integer polynomial. The figures do exactly that division, in whole numbers, and refuse to proceed if a step leaves a remainder — which is a check on the claim rather than an implementation of it.

Two further facts about these polynomials are worth knowing and neither is obvious. Each $\Phi_d$ is irreducible over the rationals — it does not factor further with rational coefficients — which is a theorem of Gauss and is what makes the roots of order $d$ genuinely inseparable from one another by rational means. And the coefficients are not always $0$ and $\pm 1$: the first counterexample is $\Phi_{105}$, whose $x^7$ coefficient is $-2$, and the smallest index is $105$ precisely because it is the smallest product of three distinct odd primes.

The Möbius function, arriving from a polygon #

Add up only the primitive roots — those of order exactly $n$ — and something odd happens.

The sum is $1$ when $n$ is $1$, and for larger $n$ it is $-1$ when $n$ is a product of an odd number of distinct primes, $+1$ when it is a product of an even number, and $0$ when $n$ has a repeated prime factor. That is the Möbius function $\mu(n)$, an object from the theory of divisors with no evident business in a picture about angles.

The figure computes both sides for every $n$ up to twelve and requires them to agree. The reason they do is a piece of bookkeeping worth following. Summing over all roots gives zero for $n > 1$; the roots split by order; so the sums over primitive roots of each divisor add to zero. That relation, held for every $n$, determines the primitive sums uniquely from the trivial case, and the solution is exactly $\mu$ — because $\mu$ is defined by having that relation, which is Möbius inversion.

So the appearance of $\mu$ is not a coincidence discovered but an identity forced: the sums satisfy the recursion that defines the function, so they are the function. That style of argument — two things satisfy the same recursion and agree at the start, so they agree everywhere — is the workhorse of the subject, and here it converts a statement about angles into a statement about prime factorisations.

Gauss’s periods, and how seventeen splits #

The constructibility argument in the next section is a degree count, and a degree count proves possibility without showing how. What Gauss actually did with the seventeen-gon is worth a paragraph, because it is the roots of unity being used rather than described.

The sixteen primitive seventeenth roots can be arranged in a single cycle: pick a number whose powers run through all sixteen non-zero residues modulo seventeen — three is one — and list the roots in the order $\zeta^{3^0}, \zeta^{3^1}, \zeta^{3^2}, \dots$. Now split that list alternately into two halves of eight and add each half. The two sums, called periods, are irrational, but their sum and their product are whole numbers, so each is a root of a quadratic with integer coefficients and each is expressible with one square root.

Split each half alternately again, into four groups of four; each of the four sums is a root of a quadratic over the field the first square root generated. Split again into eight groups of two, and once more. After four splittings the groups have one member each, and $\zeta$ itself has been expressed by four nested square roots — which is a construction, since a square root is a compass step.

The mechanism is the cyclic structure doing the work: the group of the sixteen primitive roots is cyclic of order sixteen, and $16 = 2^4$ means it has a chain of subgroups each of index two in the last. Each step down the chain is a quadratic. For a heptagon the corresponding group has order six, whose only chain has a step of index three, and a cubic is not a compass step — which is the same obstruction as a cube that cannot be doubled, arriving in a different problem.

What the polygon can be drawn with #

The last consequence brings the subject back to geometry and is the reason it was studied.

Constructing a regular $n$-gon means constructing the point $\zeta_n$, which is possible exactly when that number lies in a field reachable by repeated square roots — a tower whose degree over the rationals is a power of two. The degree of $\zeta_n$ is $\varphi(n)$, by the irreducibility of the cyclotomic polynomial, so the polygon is constructible exactly when $\varphi(n)$ is a power of two. Written in terms of prime factors, that says $n$ is a power of two times a product of distinct Fermat primes.

Gauss proved the seventeen-sided polygon constructible at nineteen and treated it as the achievement of his life. The proof is not a construction; it is the observation that $\varphi(17) = 16$ and that the sixteen primitive roots can be split into halves, quarters and eighths by sums that satisfy quadratic equations over one another. The impossibility half — that a heptagon cannot be built — is the same computation read the other way, and it needs the degree argument rather than a failed search.

Where else the same numbers turn up #

Three appearances outside algebra, each of which is the orthogonality identity in a different costume.

Splitting a sequence into frequencies. The discrete Fourier transform of a list of $n$ numbers is the list of sums $\sum_j x_j \zeta^{-jk}$, and inverting it works because $\sum_j \zeta^{j(k-k')}$ is $n$ when $k = k'$ and zero otherwise — which is the sum identity applied to the $n$th roots of a rotated unity. Every fast transform algorithm is an exploitation of the fact that the $n$th roots contain the $(n/2)$th roots, so a transform of even length is two transforms of half the length plus a twiddle.

Counting with symmetry. To count arrangements up to rotation, average over the rotations, which is Burnside’s method; the character sums that appear are sums of roots of unity, and the vanishing of those sums for non-trivial rotations is what makes the average come out whole. The same device counts necklaces, and the counting is the whole content of the argument that a prime divides a certain difference.

Deciding divisibility. A whole number is divisible by $n$ exactly when a certain sum of roots of unity vanishes, and taking that seriously turns divisibility conditions into statements about vanishing sums — the technique behind the circle method, where the number of ways of writing an integer in some form is extracted as an integral of a generating function against a root of unity.

In all three the roots are doing one job: they are the characters of a cyclic group, functions that turn addition of indices into multiplication of values, and that sum to zero unless they are trivial. Everything above is that sentence applied.

Where it fails, and what it costs #

The picture is only about the unit circle. Roots of unity are the solutions of one very special equation, and nothing here says anything about the roots of a general polynomial, which need not be evenly spaced or on a circle or expressible at all.

Order is not size. The root at one step round and the root at $n-1$ steps are adjacent on the circle and have the same order; the root at $n/2$ steps is diametrically opposite $1$ and has order two. Reading arithmetic properties off the geometry of the polygon works for the group structure and fails for almost everything else.

Irreducibility is a real theorem. The cyclotomic polynomial is irreducible over the rationals, and that fact does the heavy lifting in the constructibility argument. It is not visible in any figure and its proof is a genuine piece of algebra — the standard one goes through reduction modulo a prime.

What the pictures cannot show #

The polygon is drawn for one $n$ at a time, and the interesting statements are about how the structure varies with $n$: which divisors appear, how the degrees split, when the coefficients stop being $\pm 1$. None of that is in a picture of a polygon, and the figures that carry it are tables.

The claim that the cyclotomic coefficients are whole numbers is checked by performing the divisions exactly, which is evidence at the sizes computed rather than a proof; the proof is the induction above. And the coefficient $-2$ in $\Phi_{105}$ is beyond anything drawable here, since the polynomial has degree forty-eight.

The ladder from here #

Below: multiplication as turning, which is the fact the whole structure rests on, and what the coefficients already know, which gives the sum identity without geometry. Sideways: which polygons can be drawn, where the totient decides, and the group drawn as a map, which is what the cyclic structure looks like when the group rather than the circle is the object. Above: cyclotomic fields, Gauss sums, and the reciprocity laws that grew out of studying exactly these numbers.

What is worth carrying away #

The roots of unity are the smallest object on which arithmetic, geometry and algebra say the same thing, and the value of that is not the individual facts but the translation.

A statement about the divisors of $n$ becomes a statement about which corners of a polygon return to the start soonest. A statement about factorising a polynomial over the rationals becomes a statement about grouping those corners. A statement about ruler-and-compass constructions becomes a statement about whether a count of coprime residues is a power of two. Nothing in the list is deducible from the others by staring; each translation has to be built. But once built, a question can be asked in whichever of the three languages makes it easiest, and that is the whole reason the object sits at the centre of so much of the subject.