Sums of roots of unity
This note, which I’ll probably expand upon later, is about roots of unity. I was reading Shafarevich’s excellent Discourses on Algebra, and wondered what the identity at the end of section 2.4 meant for the polynomial xn − 1, the defining polynomial for the nth roots of unity (where n > 1).
The reason I wondered was that the roots of unity are very symmetric: they lie evenly along the unit circle in the complex plane. As an example, the ninth roots of unity (i.e., the roots of the polynomial when n is 9) are represented in figure 1.
Really, I started thinking about the identity (as discussed below), but then I wondered what other ways there are to show that the sum of all of the roots of unity is 0. This note has a number of ways of showing this fact.
1. The intuitive way
The nth roots of unity lie evenly on the unit circle, so their center of mass better be at the origin. So, the sum of the complex numbers as vectors is zero.
2. The direct way
The most direct way to find the sum of the nth roots of unity is as follows. Let x = ω0 + ⋯ + ωn − 1 be the sum of all n of the roots of unity. Since roots of unity have unit length, since ωin = 1 for all i, and since 1/ωi is also an nth root of unity, ωix must be exactly x (ωi just permutes the components of the sum). That is, (1 − ωi)x = 0, which means either x = 0 or ωi = 1, and because we can choose an i such that ωi ≠ 1, we conclude that x = 0.
3. By representation theory
The direct way reminded me of one-dimensional representations of the cyclic group Cn. Take an irreducible complex one-dimensional representation ρ of Cn = ⟨α⟩, and first note that the value of ρα completely characterizes ρ since ραk = (ρα)k.
One can show that all of the irreducible representations of Cn are one-dimensional from the fact that Cn is abelian, and furthermore, using the fact that the sum of the squares of the dimensions of the irreducible representations is equal to |Cn|, we can conclude that each root of unity corresponds to a different irreducible representation of Cn.
Next, we compute x = 1 + α + α2 + ⋯ + αn − 1. Since ρx is an endomorphism of complex representations from irreducible ρ, it must be some constant in C. We can see, as before, that 1 − α is in the kernel of ρx, so either ρ is the trivial representation or x = 0, and since we can choose ρ not to be trivial, this completes the proof.
4. By polynomial theory
This kind of method should be well known to anyone who did high-school competition math. We take the polynomials f(x) = xn − 1 and g(x) = (x − ω0)(x − ω1)⋯(x − ωn − 1) and note that they are the same because they are both degree n, have the same n roots, and are monic. We use the fact they are monic by seeing that dividing each polynomial through by x − ωi results in a monic polynomial, and so dividing through by all such monomials will result in 1 for each of them, proving f = g. Now, the xn − 1 term has 0 as a coefficient according to f and −ω0 − ω1 − ⋯ − ωn − 1 according to g, which completes the proof.
5. By Shafarevich’s identity
This method is the reason I thought about any of this to begin with. It’s not very direct, but I think it is kind of interesting. First, we need the identity from the end of section 2.4 of Discourses on Algebra, which I will derive here. To do this, we will develop the theory of the interpolating polynomial. The idea is that, given a set of n + 1 values vi at distinct values xi, what is the nth degree polynomial f with f(xi) = vi for all i?
Let F(x) = (x − x1)(x − x2)⋯(x − xn + 1), and let Fi(x) be the polynomial obtained by dividing F through by x − xi. We define
so that fi(xi) = 1 and fi(xj) = 0 when i ≠ j. This provides a convenient basis for polynomials: observe that
is the interpolating polynomial as discussed.
If k is a natural number not exceeding n, if we have vi = xik, then since f(x) must be xk, we have
|xn + 1k|
|Fn + 1(xn + 1)|
and since each Fi is a monic degree-n polynomial, this means that if k < n,
|xn + 1k|
|Fn + 1(xn + 1)|
and if k = n, then
|xn + 1n|
|Fn + 1(xn + 1)|
These are the identities at the end of section 2.4.
We won’t be using these identities directly, but rather in spirit. We let xi = ωi and vi = 1, where ωi is an nth root of unity (with ω1 = 1). Notice that each Fi is an (n − 1)-degree polynomial, so since both f(x) and g(x) = 1 are at-most-(n − 1)-degree polynomials which agree at n points, f = g. By taking the leading xn − 1 coefficient, this means
Notice that Fi(ωi) is the product of all ωi − ωj for j ≠ i, and ωi − ωj = ωi(1 − ωi−1ωj). Then, we can see Fi(ωi) = ωin − 1F1(ω1) since ωi−1 is just permuting the roots of unity. Because (ωin − 1)−1 = ωi, we have
and so ω1 + ⋯ + ωn = 0.