Gauss and Regular Polygons: Gaussian Periods

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Introduction

In order to solve the equation $ z^{n} - 1 = 0$ Gauss introduced some sums of the $ n^{th}$ roots of unity which he called periods, and using these periods he was able to reduce the solution of $ z^{n} - 1 = 0$ to a sequence of solutions of equations of lower degrees. The technique offered by Gauss is extremely beautiful and completely novel and it uses the symmetry between the various $ n^{th}$ roots of unity to achieve the final solution.

Reduction to the Case when $ n$ is Prime

Before we proceed to show what Gauss did we can first restrict ourselves to the case when $ n$ is prime. This will not be a problem as the following result shows.
If $ \xi_{1}, \xi_{2}, \ldots, \xi_{r}$ are the $ r^{th}$ roots of unity and $ \eta_{1}, \eta_{2}, \ldots, \eta_{s}$ are the $ s^{th}$ roots of unity then the $ (rs)^{th}$ roots of unity are given by the expression $ \xi_{i}\sqrt[r]{\eta_{j}}$ for $ i = 1, 2, \ldots, r,\,\, j = 1, 2, \ldots, s$.

To establish this we only need to observe that $$z^{rs} - 1 = (z^{r})^{s} - 1 = \prod_{j = 1}^{s}(z^{r} - \eta_{j})$$ Henceforth we shall therefore focus only on the case when $ n$ is prime and we will denote it by the letter $ p$. In what follows $ p$ will be a prime unless otherwise explicitly stated.

Gaussian Periods

The roots of the equation $ z^{p} - 1 = 0$ are $ 1, \zeta, \zeta^{2}, \ldots, \zeta^{p - 1}$ where $$\zeta = \cos\left(\frac{2\pi}{p}\right) + i\sin\left(\frac{2\pi}{p}\right)$$ Moreover all the roots except $ 1$ are primitive $ p^{th}$ roots of unity (since $ \phi(p) = p - 1$). Gauss arranged these primitive roots into a particular order through which he could exploit the symmetry between them to the full extent. To do so we need a result about prime numbers (also proved by Gauss) which we state (we don't prove it here to avoid any digression, it can be found online)as follows:

For any given prime $ p$, there exists a positive integer $ g$ such that $ g^{p - 1} \equiv 1 \mod p$, but $ g^{i} \not\equiv 1 \mod p$ for any positive integer $ i < (p - 1)$.
Such a number $ g$ is called a primitive root of $ p$. There are exactly $ \phi(p - 1)$ primitive roots of $ p$.

It follows from the above result that the numbers $ 1, g, g^{2}, \ldots, g^{p - 2}$ when taken modulo $ p$ are all different from one another and non-zero. Therefore they are equal to $ 1, 2, \ldots, (p -1)$ in some order.

Let us now choose any primitive root $ g$ of $ p$. Then since $ \zeta^{p} = 1$, we can arrange the primitive $ p^{th}$ roots of unity in the following order:
$$\zeta^{g^{0}}, \zeta^{g^{1}}, \zeta^{g^{2}}, \ldots \zeta^{g^{p - 2}}$$ To simplify notation we will write $ \zeta_{i} = \zeta^{g^{i}}$ so that the primitive $ p^{th}$ roots of unity are $ \zeta_{0}, \zeta_{1}, \zeta_{2}, \ldots, \zeta_{p - 2}$. This ordering of the roots is permuted cyclically each time we replace $ \zeta$ by $ \zeta^{g}$. Thus we have $$\zeta \rightarrow \zeta^{g} \Rightarrow \zeta_{0} \rightarrow \zeta_{1} \rightarrow \zeta_{2} \rightarrow \cdots \rightarrow \zeta_{p - 2} \rightarrow \zeta_{0}$$ This is the symmetry which we will exploit along the lines of Gauss.

Let $ e, f$ be two positive integers such that $ ef = (p - 1)$, then we define $ e$ complex numbers each called a period of $ f$ terms as follows: \begin{align} \eta_{0} &= \zeta_{0} + \zeta_{e} + \zeta_{2e} + \cdots + \zeta_{e(f - 1)}\notag\\ \eta_{1} &= \zeta_{1} + \zeta_{e + 1} + \zeta_{2e + 1} + \cdots + \zeta_{e(f - 1) + 1}\notag\\ \eta_{2} &= \zeta_{2} + \zeta_{e + 2} + \zeta_{2e + 2} + \cdots + \zeta_{e(f - 1) + 2}\notag\\ \cdots &= \cdots\notag\\ \eta_{e - 1} &= \zeta_{e - 1} + \zeta_{2e - 1} + \zeta_{3e - 1} + \cdots + \zeta_{p - 2}\notag \end{align} As can be easily seen that a period of $ 1$ term is a root $ \zeta_{0}$ whereas a period of $ (p - 1)$ terms is the sum of all roots i.e. $ \zeta_{0} + \zeta_{1} + \cdots + \zeta_{p - 2} = -1$ which is rational. Thus the structure of a period with lesser number of terms is more complex than that of a period of larger number of terms. We will now study the properties of these periods which will ultimately lead us to the radical expressions for the $ p^{th}$ primitive root $ \zeta$.

Automorphisms of Field $ \mathbb{Q}(\zeta)$

To set up things more formally, we define $ \mathbb{Q}(\zeta)$ to be the set of all rational expressions in $ \zeta$ with rational coefficients. This set forms a field and moreover every member of $ \mathbb{Q}(\zeta)$ can be expressed uniquely in the following way: $$ a_{0} + a_{1}\zeta + a_{2}\zeta^{2} + \cdots + a_{p - 2}\zeta^{p - 2}$$ where $ a_{i}$ are rational. This is again a very standard result from field theory and not difficult to prove. Moreover it is not specific to $ \zeta$, but rather holds good for any algebraic number $ \zeta$ whose minimal polynomial is of degree $ (p - 1)$ and $ p$ need not be necessarily prime for this result.

Coming back to the $ p^{th}$ primitive root $ \zeta$ and prime $ p$, we observe that since $ \Phi_{p}(\zeta) = 0$ we have \begin{align} &1 + \zeta + \zeta^{2} + \cdots + \zeta^{p - 1} = 0\notag\\ &\Rightarrow a_{0} = - a_{0}(\zeta + \zeta^{2} + \cdots + \zeta^{p - 1})\notag\\ &\Rightarrow a_{0} + a_{1}\zeta + a_{2}\zeta^{2} + \cdots + a_{p - 2}\zeta^{p - 2}\notag\\ &\,\,\,\,\,\,\,\,= (a_{1} - a_{0})\zeta + (a_{2} - a_{0})\zeta^{2} + \cdots + (a_{p - 2} - a_{0})\zeta^{p - 2} + (-a_{0})\zeta^{p - 1}\notag \end{align} Thus we are able to express every $ u \in \mathbb{Q}(\zeta)$ as a linear combination of powers of $ \zeta$ with rational coefficients as: $$ a_{1}\zeta + a_{2}\zeta^{2} + \cdots + a_{p - 1}\zeta^{p - 1}$$ and this expression is unique. Changing the indices if necessary we can see that the above can be rearranged to get $$ a_{o}\zeta_{0} + a_{1}\zeta_{1} + a_{2}\zeta_{2} + \cdots + a_{p - 2}\zeta_{p - 2}$$ We can now setup a mapping, say $ \sigma$,  from $ \mathbb{Q}(\zeta)$ onto itself defined by $ \sigma(1) = 1$ and \begin{align} &\sigma(a_{o}\zeta_{0} + a_{1}\zeta_{1} + a_{2}\zeta_{2} + \cdots + a_{p - 3}\zeta_{p - 3} + a_{p - 2}\zeta_{p - 2})\notag\\ &\,\,\,\,= a_{o}\zeta_{1} + a_{1}\zeta_{2} + a_{2}\zeta_{3} + \cdots + a_{p - 3}\zeta_{p - 2} + a_{p - 2}\zeta_{0}\notag \end{align} Thus the mapping $ \sigma$ permutes the $ \zeta$'s cyclically. The above mapping is one-one and onto and we will show that it is an automorphism of field $ \mathbb{Q}(\zeta)$ by showing that $ \sigma(uv) = \sigma(u)\sigma(v)$ (it is obvious that $ \sigma(u + v) = \sigma(u) + \sigma(v)$).

From the definition of $ \sigma$ it is clear that $ \sigma(\zeta_{i}) = \zeta_{i}^{g}$ so that $ \sigma(\zeta_{i}\zeta_{j}) = \sigma(\zeta_{i})\sigma(\zeta_{j})$. Since $ \sigma$ is linear i.e. $ \sigma(au + bv) = a\sigma(u) + b\sigma(v)$ for rationals $ a, b$, and if we observe that members $ u, v \in \mathbb{Q}(\zeta)$ are linear combinations of $ \zeta$'s, it follows that $ \sigma(uv) = \sigma(u)\sigma(v)$.

It is thus established that $ \sigma$ is an automorphism of $ \mathbb{Q}(\zeta)$ and moreover it leaves the rationals and only the rationals fixed i.e. $ \sigma(u) = u \Rightarrow u \in \mathbb{Q}$. Now we can compose $ \sigma$ with itself to get other automorphisms. Then for any positive integer $ e$, the mapping $ \sigma^{e}$ (i.e. $ \sigma$ applied $ e$ times) is an automorphism of $ \mathbb{Q}(\zeta)$. It turns out that these automorphisms are linked with the properties of the Gaussian periods and ultimately enable us to express them in the form of radicals thereby finding radical expressions for the primitive $ p^{th}$ roots of unity. This we will finalize in the next post to keep the length of current post manageable.

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3 comments :: Gauss and Regular Polygons: Gaussian Periods

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  1. Hi again, one thing is still bothering me, why is the sum of the primitive roots = -1?

  2. @Kazbich-the-bandit,
    Note that if we take all $n^{\text{th}}$ roots of unity then these are roots of $x^{n} - 1 = 0$ and hence the sum of these roots is $0$. One of these roots is $x = 1$ and hence the sum of other roots (except $x = 1$) is $-1$. Now in this post we have replaced $n$ with $p$ where $p$ is prime. When $p$ is prime then all roots except $x = 1$ are primitive roots. Hence sum of primitive roots is $-1$. If $p$ was not prime then sum of primitive roots will not be $-1$.

  3. OK, I get it. Thanks