### Properties of Gaussian Periods

In this post we are going to establish the following properties of the Gaussian periods which will ultimately lead to a solution of the equation $ z^{p} - 1 = 0$. Again as in previous post, $ p$ is to be considered a prime unless otherwise stated. In the following we have $ e, f$ as two positive integers with $ ef = (p - 1)$.**Any period of $ f$ terms can be expressed as a polynomial in any other period of $ f$ terms with rational coefficients.****If $ g$ divides $ (p - 1)$ and $ f$ divides $ g$, then any period of $ f$ terms is a root of a polynomial equation of degree $ g / f$ whose coefficients are rational expressions of a period of $ g$ terms.**

The second property is the key to solving the equation $ z^{p} - 1 = (z - 1)\Phi_{p}(z) = 0$. If we just look at the periods of $ 1$ term, then clearly they are the roots $ \zeta_{0}, \zeta_{1}, \zeta_{2}, \ldots, \zeta_{p - 2}$ of the equation $$\Phi_{p}(z) = z^{p - 1} + z^{p - 2} + \cdots + z + 1 = 0$$ These are the roots which we wish to obtain. So we can have a chain of divisors $ f_{0}, f_{1}, f_{2}, \ldots, f_{k}$ of $ (p - 1)$ such that $$1 = f_{0} < f_{1} < f_{2} < \cdots < f_{k - 1} < f_{k} = (p - 1)$$ where $ f_{i} \mid f_{i + 1}$ and, say $ d_{i} = f_{i + 1} / f_{i}$. Then by the property (2) we can observe that a period of $ f_{i}$ terms can be found out as root of equation of degree $ d_{i}$ whose coefficients are rational expressions of a period of $ f_{i + 1}$ terms.

As an example let $ p = 13$. Then we can have the chain of divisors of $ p - 1 = 12$ as $ 1, 2, 6, 12$. Now a period of $ 12$ terms is basically the sum of all roots of $ \Phi_{13}(z) = 0$ and hence $ -1$. Therefore a period of $ 6$ terms, say $ \eta_{1}$, is a root of a quadratic equation (because $ 12 / 6 = 2$) with rational coefficients. And then a period of $ 2$ terms, say $ \eta_{2}$, is a root of a cubic equation (because $ 6 / 2 = 3$) whose coefficients are rational expressions in $ \eta_{1}$. Finally a period of one term, say $ \zeta_{0}$, is a root of a quadratic equation whose coefficients are rational expressions in $ \eta_{2}$. Since all the equations involved are solvable using standard formulas consisting of radicals, it is easily seen that the desired root $ \zeta_{0}$, which is a primitive $ 13^{th}$ root of unity, is expressible in radicals.

What is happening here is that if $ \eta_{f_{i}}$ is a period of $ f_{i}$ terms then the fields formed by combining them with rationals form a nested sequence where one includes the next as follows: $$\mathbb{Q}(\eta_{f_{0}}) \supset \mathbb{Q}(\eta_{f_{1}}) \supset \mathbb{Q}(\eta_{f_{2}}) \supset\cdots \supset \mathbb{Q}(\eta_{f_{k - 1}}) \supset \mathbb{Q}(\eta_{f_{k}}) =\mathbb{Q}$$ More technically we say that

**field $ \mathbb{Q}(\eta_{f_{i}})$ is a finite extension of field $ \mathbb{Q}(\eta_{f_{i + 1}})$ of degree $ d_{i} = f_{i + 1} / f_{i}$.**(This is much stronger than what we are going to prove in the post. It requires that the polynomials being talked about in property (2) be irreducible over the field of their coefficients. This is true but rather difficult to prove.) This is illustrated for the case of $ p = 13$ below: $$1 \mid 2 \mid 6 \mid 12 = (p - 1)$$ $$\mathbb{Q}(\zeta_{0}) = \mathbb{Q}(\eta_1) \supset \mathbb{Q}(\eta_{2}) \supset \mathbb{Q}(\eta_{6}) \supset \mathbb{Q}(\eta_{12}) = \mathbb{Q}$$ The really interesting case occurs when $ (p - 1)$ is a power of $ 2$, say $ (p - 1) = 2^{m}$. Then one possible chain of divisors is $ 1, 2, 2^{2}, 2^{3}, \ldots, 2^{m}$. And here we have the polynomial degrees $ d_{i} = 2$ so that we can obtain the primitive $ p^{th}$ root of unity by solving a series of quadratic equations. It thus follows that the expression for the $ p^{th}$ primitive root of unity will consist of a radical expression involving square roots only. And therefore a regular polygon of $ p$ sides is constructible by Euclidean tools. In this case $ p = 2^{m} + 1$ and since $ p$ is prime, $ m$ must be a power of $ 2$, so that $ p$ must be a Fermat prime. Thus we have been able to establish the following:

**If $ p$ is a Fermat prime then a regular polygon of $ p$ sides can be constructed using Euclidean tools only.**

This is the key result which Gauss established and thus settled the problem of constructing regular polygons with ruler and compass. We will prove the complete result about polygon construction later and now focus on the proofs of the properties of Gaussian periods. To do so it is better to talk in the language of field automorphisms and fixed fields and then analyze the field $ \mathbb{Q}(\zeta)$ where $ \zeta$ is a primitive $ p^{th}$ root of unity (i.e. a period of $ 1$ term).

### Field Automorphisms and Fixed Fields

From the previous post we take note that the function $ \sigma$ defined by $ \sigma(1) = 1$ from $ \mathbb{Q}(\zeta)$ onto itself defined by \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} is an automorphism. Also we noticed that it leaves the rationals and only the rationals fixed so that the*fixed field*for the automorphism $ \sigma$ is the $ \mathbb{Q}$.

Let $ e, f$ be positive integers with $ ef = (p - 1)$ and we will now analyze the fixed field $ K_{f}$ of the automorphism $ \sigma^{e}$. We establish the following:

*Every element $ a \in K_{f}$ can be expressed uniquely as a linear combination of $ e$ periods of $ f$ terms with rational coefficients.*

*Proof:*First we note that since $ a \in \mathbb{Q}(\zeta)$ we can express it uniquely in the form $$a = a_{0}\zeta_{0} + a_{1}\zeta_{1} + \cdots + a_{p - 2}\zeta_{p - 2}$$ which can be written in more expanded form as \begin{align} a &= a_{0}\zeta_{0} + a_{1}\zeta_{1} + \cdots + a_{e - 1}\zeta_{e - 1}\notag\\ &\,\,\,\,+ a_{e}\zeta_{e} + a_{e + 1}\zeta_{e + 1} + \cdots + a_{2e - 1}\zeta_{2e - 1}\notag\\ &\,\,\,\,+ \,\, \cdots\notag\\ &\,\,\,\,+ a_{e(f - 1)}\zeta_{e(f - 1)} + a_{e(f - 1) + 1}\zeta_{e(f - 1) + 1} + \cdots + a_{p - 2}\zeta_{p - 2}\notag \end{align} Since $ a \in K_{f}$, therefore $ \sigma^{e}(a) = a$, and then we have \begin{align} a &= \sigma^{e}(a) = a_{0}\zeta_{e} + a_{1}\zeta_{e + 1} + \cdots + a_{e - 1}\zeta_{2e - 1}\notag\\ &\,\,\,\,+ a_{e}\zeta_{2e} + a_{e + 1}\zeta_{2e + 1} + \cdots + a_{2e - 1}\zeta_{3e - 1}\notag\\ &\,\,\,\,+\,\, \cdots\notag\\ &\,\,\,\,+ a_{e(f - 1)}\zeta_{0} + a_{e(f - 1) + 1}\zeta_{1} + \cdots + a_{p - 2}\zeta_{e - 1}\notag \end{align} Comparing the coefficients we get \begin{align} a_{0} &= a_{e} = a_{2e} = \cdots = a_{e(f - 1)}\notag\\ a_{1} &= a_{e + 1} = a_{2e + 1} = \cdots = a_{e(f - 1) + 1}\notag\\ \cdots &= \cdots\notag\\ a_{e - 1} &= a_{2e - 1} = a_{3e - 1} = \cdots = a_{p - 2}\notag \end{align} Therefore \begin{align} a &= a_{0}(\zeta_{0} + \zeta_{e} + \cdots + \zeta_{e(f - 1)})\notag\\ &\,\,\,\,+ a_{1}(\zeta_{1} + \zeta_{e + 1} + \cdots + \zeta_{e(f - 1) + 1})\notag\\ &\,\,\,\,+\,\,\cdots\notag\\ &\,\,\,\,+ a_{e - 1}(\zeta_{e - 1} + \zeta_{2e - 1} + \cdots + \zeta_{p - 2})\notag \end{align} Thus $ a$ has been expressed as a linear combination of the periods of $ f$ terms and this expression is unique (since $ a \in \mathbb{Q}(\zeta)$ has such a unique expression).

From the above result it follows that the periods form a basis of $ K_{f}$ over $ \mathbb{Q}$ and so the dimension of $ K_{f}$ is $ e$. And therefore any linearly independent set (over rationals) of members of $ K_{f}$ of size $ e$ is a basis. We can now show that if $ \eta$ is a period of $ f$ terms then the numbers $ 1, \eta, \eta^{2}, \ldots, \eta^{e - 1}$ form a linearly independent set over $ \mathbb{Q}$. First we note that $ \eta \in K_{f}$ so that all these numbers are in $ K_{f}$. If we have $$a_{0} + a_{1}\eta + a_{2}\eta^{2} + \cdots + a_{e - 1}\eta^{e - 1} = 0\tag{1}$$ then $ \eta$ is a root of polynomial equation $ P(z) = a_{0} + a_{1}z + \cdots + a_{e - 1}z^{e - 1} = 0$. By applying the automorphisms $ \sigma, \sigma^{2}, \ldots, \sigma^{e - 1}$ on $ (1)$ we see that the $ a_{i}$ are left unchanged and therefore $ \eta, \sigma(\eta), \sigma^{2}(\eta), \ldots, \sigma^{e - 1}(\eta)$ are the roots of the polynomial equation $ P(z) = 0$.

Thus $ P(z) = 0$ has $ e$ distinct roots $ \eta, \sigma(\eta), \sigma^{2}(\eta), \ldots, \sigma^{e - 1}(\eta)$ (note that these are the $ e$ distinct periods of $ f$ terms). This is impossible as $ P(z)$ is of degree at most $ (e - 1)$, unless all the coefficients $ a_{0}, a_{1}, \ldots, a_{e - 1}$ are zero. It follows that the numbers $ 1, \eta, \eta^{2}, \ldots, \eta^{e - 1}$ form a linearly independent set over $ \mathbb{Q}$ and they form a basis of $ K_{f}$ over $ \mathbb{Q}$. Thus any member $ a \in K_{f}$ can be expressed as a unique linear combination of $ 1, \eta, \eta^{2}, \ldots, \eta^{e - 1}$ over the rationals i.e. $ K_{f} = \mathbb{Q}(\eta)$. So the fixed field of $ \sigma^{e}$ is the field obtained by combining a period of $ f$ terms to the rationals.

It now follows that if $ \eta^{\prime}$ is another period of $ f$ terms then $ \eta^{\prime} \in K_{f}$ and therefore $$ \eta^{\prime} = a_{0} + a_{1}\eta + a_{2}\eta^{2} + \cdots + a_{e - 1}\eta^{e - 1}$$ And thus we see that

**any period of $ f$ terms can be expressed as a polynomial (with rational coeffieicients) in any other period of $ f$ terms.**The first property of the periods is thus established.

We now need to establish the second property. Let us take two positive integers $ f, g$ which are divisors of $ (p - 1)$ and also assume that $ f$ divides $ g$. We also introduce numbers positive integers $ e, h$ such that $$ ef = gh = (p - 1)$$ Also let $ k = g / f = e / h$. Let $ K_{f}$ and $ K_{g}$ be the fixed fields of automorphisms $ \sigma^{e}$ and $ \sigma^{h}$. Then we note that $$ \sigma^{e} = \sigma^{hk} = (\sigma^{h})^{k}$$ and therefore any element fixed by $ \sigma^{h}$ is also left fixed by $ \sigma^{e}$ and thus we have $ K_{g} \subset K_{f}$. We then prove the following:

*Every element in $ K_{f}$ is a root of a polynomial equation of degree $ g / f$ with coefficients in $ K_{g}$.*

*Proof:*Let $ a \in K_{f}$ so that $ \sigma^{e}(a) = a$ and consider the polynomial equation $$ P(z) = (z - a)(z - \sigma^{h}(a))(z - \sigma^{2h}(a)) \cdots (z - \sigma^{h(k - 1)}(a)) = 0$$ This is polynomial equation of degree $ k = g / f$ and if we apply the map $ \sigma^{h}$ over it, the factors are permuted cyclically (since $ \sigma^{h}(\sigma^{h(k - 1)}(a)) = \sigma^{hk}(a) = \sigma^{e}(a) = a$) and therefore the polynomial remains unchanged. This means that the coefficients of the polynomial are left unchanged by the automorphism $ \sigma^{h}$ and therefore the coefficients are in $ K_{g}$.

It is thus established that

**any period of $ f$ terms is a root of a polynomial equation of degree $ g / f$ where the coefficients themselves are polynomials in a period of $ g$ terms.**And the second property of the periods is established. Generalizing the above we can see that if we have a chain of divisors $ f_{0}, f_{1}, f_{2}, \ldots, f_{m}$ of $ (p - 1)$ such that $$ 1 = f_{0} < f_{1} < f_{2} < \cdots < f_{m - 1} < f_{m} = (p - 1)$$ where $ f_{i} \mid f_{i + 1}$ then the corresponding fields $ K_{f_{i}}$ have the following relationship among themselves: $$ \mathbb{Q} = K_{f_{m}} \subset K_{f_{m - 1}} \subset \cdots \subset K_{f_{2}} \subset K_{f_{1}} \subset K_{f_{0}} = \mathbb{Q}(\zeta)$$ and moreover each of the fields $ K_{f_{i}}$ is a finite extension of $ \mathbb{Q}$ of degree $ e_{i} = (p - 1) / f_{i}$. Since $ K_{f_{i}} = \mathbb{Q}(\eta_{f_{i}})$ where $ \eta_{f_{i}}$ is a period of $ f_{i}$, it follows that the above sequence of fields is the same as the one we talked earlier which are obtained by combining the periods with the rationals. A $ p^{th}$ primitive root of unity can now be obtained by solving a series of polynomial equations of degrees $ d_{i} = f_{i + 1} / f_{i}$. Gauss proved that these algebraic equations were solvable using radicals even if the degree $ d_{i} > 4$. We shall not prove this here as it is not needed for the equations involved in polygon constructions.

To reiterate the achievement of this series of posts till now, we just mention the following:

**If $ p$ is a Fermat prime then a regular polygon of $ p$ sides is constructible using Euclidean tools.**

We will prove the general result and its converse in the next post.

*Note:*The treatment of theory of Gaussian periods is taken from "Galois Theory of Algebraic Equations" by Jean-Pierre Tignol. The interested reader should grab a copy of this exceedingly wonderful book and look into further details. Somehow this theory of Gaussian periods is not considered significant after the development of Galois Theory and modern textbooks almost entirely avoid their discussion. But I believe that the theory of periods (or for that matter anything contained in Gauss'

*Disquisitiones Arithmeticae*) is exactly the thing that excites a budding mathematician.

**Print/PDF Version**

Truly no matter if someone doesn’t know then its up to other users that they will help, so here it happens.

Eduardo

April 9, 2013 at 9:31 AMI am seeking to grasp the theorem on an intuitive level, so please bear with me. Firstly, the second property of Gaussian periods, with particular application to p = 13, states that a period of two terms is a root of cubic equation (that is obvious) with coefficients that are rational expressions in a period of 6 terms. Would I be correct if I paraphrase the last statement as saying that the each coefficient of the cubic equation can be written as expression of 6 terms, ie as an element of Q(nu6)?

Secondly, this particular property must be a direct consequence of the fact that, as in the example, 2 divides 6, which in turn divides 12. However, I fail to see how it follows. I mean, why is it true?

Jon Grech

July 10, 2014 at 3:02 AM@Kazbich-the-bandit,

First a small typo in your comment. Symbol $\eta$ in $\mathbb{Q}(\eta_{6})$ is called "eta" and not "nu" and "nu" is denoted by $\nu$. Next your interpretation that "a period of two terms is a root of a cubic with coefficients that are elements of $\mathbb{Q}(\eta_{6})$" is correct. The proof of this is given in the post. See the result "Every element in $K_{f}$ is a root of a polynomial ..." and its proof thereafter. Let me know if you have any specific trouble in understanding this proof. You have to focus on the part which deals with automorphisms $\sigma$ and its powers.

Paramanand

July 10, 2014 at 9:34 AMIn the proof of the first property it is claimed that eta, a period of f terms is a member of Kf, the fixed field of sigma repeated e times. So that sigma^e(eta) = eta, right? As I understand it, the operation should send the terms zeta(0) to zeta(e), zeta(1) to zeta(e+1),...,zeta(f-1) to zeta(e+(f-1)). I am to understand that the operation leaves eta fixed because Kf is of dimension e - so that (zeta(0),...,zeta(f-1)) denotes the same vector as (zeta(e),...,zeta(e+f-1))?

Jon Grech

July 11, 2014 at 11:43 AM@Kazbich-the-bandit,

By definition $eta$, a period of $f$ terms always looks like $$\eta = \zeta_{r} + \zeta_{e + r} + \zeta_{2e + r} + \cdots + \zeta_{e(f - 1) + r}$$ and clearly if we apply $\sigma^{e}$ to $\eta$ each term above is replaced by next term and last term is replaced by first term thereby the sum $\eta$ remains fixed. Hence by definition of $\eta$ it belongs to the field $K_{f}$.

Paramanand

July 12, 2014 at 3:47 PMDear Paramanand

I was reading your contributions with delight. I totally agree that the theory of Gaussian periods has been forgotten at the expense of more modern Theories. At university students are confronted with abstract terms about wonderful objects that they cannot see with their beautiful 'colors' or cannot feel with their amazing 'shape'.

I use the heptadecagon as a lighthouse for all my Algebra-Studies, because it's viewpoint is so enlightening.

I have still not fully understood what Gauss intuitively saw when he separated the 2 subgroups of order 8.

All I know is this:

The first sub-group {3,10,5,11,14,7,2,16} are all the primitive roots of 17

The second sub-group {1,9,13,15,16,8,4,2} are all the quadratic residues of 17

2^p-1 with q = (p-1)/2 can be factorized into 2^q+1 × 2^q-1

Is this a natural way to divide these 2 periods?

Many thanks in advance for further clarification

Gerald Schmidt

October 30, 2014 at 10:09 PM@Gerald Schmidt

There is a typo in the subgroups you have mentioned. The partitioning is with $\{3, 10, 5, 11, 14, 7, 12, 6\}$ and $\{1, 9, 13, 15, 16, 8, 4, 2\}$. In terms of Modern Algebra, the second set containing $1$ is a subgroup and the other one is a coset. Together these cosets form the quotient group of order $2$. Thus the subgroup $\{1, 9, 13, 15, 16, 8, 4, 2\}$ is normal subgroups of the full group $\{1, \ldots, 16\}$. The same technique (based on a primitive root of $p$) would work for any Fermat prime $p$ apart from $17$.

Paramanand

October 31, 2014 at 11:18 AM