Constructibility of Regular n-Gons

Ebony Ann Harvey

2007; revised with xy-pic in 2010

Abstract

Regular n-gons are constructible with a straightedge and compasswhen the field extension of constructible elements over the field of Ra-tional numbers is a power of 2. Using field theory we can understandhow the degree of the minimal cyclotomic polynomial, whose roots arethe primitive nth roots of unity, relates to the field extension. Then,Galois theory will give us the link between the subgroups of automor-phisms and the subfields of the extension fields, enabling us explicitlyfind the quadratic extension field.

1 Introduction

Given a straightedge1 and a compass what regular n-gons can one construct?The answer may surprise you, but only some n-gons2 can be constructed. Forexample, we can construct an 8-gon, but we cannot construct a 7-gon. It ispossible to construct certain n-gons where n is a prime number, i.e., n =3, 5, 17,3 but not when n = 19. These and other questions on constructibilityintrigued mathematicians for centuries until German mathematician CarlFriedrich Gauss4 formulated a proof to the constructibility of a regular 17-gon based of the properties of certain polynomial equations we will discuss.To understand why some n-gons are constructible and other n-gons are not,

1A ruler without unit markings.2Polygons with n-number vertices of equal angles and n-number of sides of equal length.3Greek mathematician Euclid of Alexandria had given constructions for the regular

triangle, square, regular pentagon, etc.41777-1885; contributed to many mathematical and scientific fields.

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it is necessary to translate the geometric constructions into algebraic terms.Then we will apply field theory and Galois theory to understand what ishappening algebraically. In the final step we will translate the algebra backto geometry and make constructions for ourselves.

2 Geometric and Algebraic Picture of n-Gons

Let us inscribe the n-gon in the unit circle, defined as x2 + y2 = 1, on thecomplex plane so that both have center at the origin (0,0) as shown below.

C

(0, 0) Pn = (1, 0)

P1 = (x1, y1)

P2 = (x2, y2)

Pn−1 = (xn−1, yn−1)

. . .

The vertices of the n-gon lie on the circumference of the unit circle and are ofequal distance apart with a vertex at point (1,0). The remaining n−1 verticesare represented in Cartesian5 coordinates by (x1, y1), (x2, y2), . . . , (xn−1, yn−1),labeled counterclockwise from (xn, yn)=(1,0). If we can construct the pointsPk = (xk, yk) where k = 1, 2, . . . , n − 1, then we can construct the corre-sponding n-gon. It will be more beneficial to represent the vertices via polarcoordinates because then we can think of dividing the unit circle into n slices

5French mathematician Rene Descartes’s (1596-1650) invention of Cartesian coordi-nates revolutionized mathematics by providing a link Euclidean geometry and algebra.

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of equal size6. In general, given a non-zero complex number z = a+ bi ∈ C,let r =

√a2 + b2 be the distance of z from the origin and let θ be the angle

formed by z and the x-axis. Then by Euler’s formula7

z = reiθ = r(cos θ + i sin θ), r > 0, 0 ≤ θ < 2π.

Over C there are n distinct solutions8 to the equation xn = 1, the points wechoose as the vertices Pk. Given that r = 1 we have the elements

Pk = e2πki/n = cos2kπ

n+ i sin

2kπ

n, for k = 1, 2 . . . , n.

We will use the greek letter “zeta” ζkn instead of Pk when referring to thekth vertex of a particular n-gon in polar coordinates. This is only to be inagreement with the notation that is already well recognized. For all k,

(ζkn)n =(

cos2kπ

n+ i sin

2kπ

n

)n= cos 2kπ + i sin 2kπ = 1 + i · 0 = 1.

This equality is given by De Moivre’s formula9 and leads us to the followingdefinition.

Definition The collection {ζn, ζ2n, . . . , ζn−1n , 1} is called the nth roots of unityin C because they are the roots, x ∈ C, of the polynomial equation xn = 1.

This is not just some arbitrary set of values. The nth roots of unity havea particular structure and valuable properties that we will repeatedly use.Here we state the two fundamental properties.

6The notion of measuring the length of a chord for an angle had been used since antiq-uity but the present day conception was formalized by Belgian mathematician Gregoirede Saint-Vincent (1584-1667) and Italian mathematician Bonaventura Cavalieri indepen-dently.

7Swiss mathematician and physicist Leonhard Euler (1707-1783) made important dis-coveries in diverse fields of science. This particular formula shows the relationship betweenthe complex exponential function and trigonometric functions.

8This is a nontrivial fact and is consequence of the Fundamental Theorem of Algebra.Using the language that will be introduced later, we are essentially saying that everypolynomial over the reals splits completely over the complex numbers.

9French mathematician Abraham de Moivre’s (1667-1754) claim to fame. He provedthis for all n ∈ N and Euler proved it for all n ∈ R using Euler’s formula.

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1. A group is a pair–a set and an operation–such that the set is closedunder the operation and satisfies three properties called group axioms.These axioms state that there must be (i) a unique identity element,(ii) every element must have a unique inverse, and (iii) the elements areassociative. This may seem a little abstract, but here is the punch line:the nth roots of unity form a group under the operation multiplication.We denote this group by the greek letter “mu” and write µn. Theidentity element is 1 and the inverse of ζkn is ζ−kn . The reader shouldcheck the following statements and associativity:

If (ζkn)n = 1 then (ζ−kn )n = 1. Further, if (ζjn)n = 1 then (ζknζjn)n = 1

2. The group µn is cyclic. This means that at least one element cangenerate every other element in the group by taking successive powers.This element is ζkn for k relatively prime to n, i.e. gcd(k, n) = 1.Generators are called primitive nth roots of unity. For all d dividingn, dth roots of unity are nth roots of unity. This notion agrees withour geometric intuition because the vertices of the triangle are alsovertices of the hexagon and both are vertices of the dodecagon, etc.Algebraically, the proof is simple: ζnd = (ζdd )n/d = 1. Therefore, wehave the subgroup µd 6 µn for all d|n.

We will talk more about this group later; however, now that you have ageometric and algebraic sense of what we would like to discuss, we will moveon to how the straightedge and compass are used.

2.1 Rules of Construction

With the straightedge and compass we are allowed four types of operations:

1. Connect two given points by a straight line

2. Find the point of intersection of two straight lines

3. Draw a circle of given radius and center

4. Find the point(s) of intersection of a straight line and a circle or thepoints of intersection of two circles.

We can use these four operations to make increasingly large fields of numbers.

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2.2 Constructing A Tower of Fields

2.2.1 The Natural Numbers, N and the Integers, Z

Given a line ` make an arbitrary point on it and call the point a = 0. Makea circle with the center at a and label the intersection on the right as b = 1.Draw a circle with center at b with radius the length of the segment ab. Labelthe rightmost intersection of this circle and ` as c = 2. Continuing in thismanner we can construct the set of natural numbers. Similarly, we can buildpoints to the left of a = 0 to designate negatives and construct the set ofintegers as shown.

•b = 1•a = 0 •c = 2 · · · `•d = −1 •

It follows immediately that we can build the Cartesian plane and considercoordinate points (x, y) with integer coefficients.

2.2.2 Operations and the Rational Numbers, Q

Since the lengths a, b ∈ Z are constructible, we can perform the binaryoperations of addition and subtraction, a±b, multiplication, ab, and divisiona/b, the latter two via similar triangles. Par consequence, we construct thefield of rational numbers Q = {p/q | p, q ∈ Z, q 6= 0}.

a/b

a

b

1c 1

√c

5

If c ∈ Q is given, construct√c as follows: draw a circle with diameter 1+c

and erect the perpendicular to the diameter as shown above. The length ofthe perpendicular is

√c and this length is not necessarily in Q. The collection

of elements that can be obtained from a finite sequence of these operations iscalled the field of constructible elements and this field is a proper subfield ofC. We will denote this field by K and describe tit in more detail later. Twofamous straightedge and compass problems that were proven to be impossibleto do because certain values could not be constructed. The first problemknown as “squaring the circle” calls for the construction of a square with thesame area as a given circle. The second problem called “doubling the cube”asks for a cube with twice the volume of a given cube. There constructionsare impossible because they require the values

√π and 3

√2 respectively which

we will see cannot be constructed with a ruler and compass. Below we willclassify the field of constructible elements more precisely.

2.2.3 The Tower

Above we mentioned that the field of constructible elements K is strictlylarger than Q but a proper subset of C.

Definition The field K is said to be an extension field of Q. This is inter-changeably denoted by Q ⊂ K, K/Q (read “K over Q”) or by the diagrambelow

K

Q

The notation emphasizes that Q is a subfield of K and may be called thebase field. To reiterate, a point (x, y) becomes constructible if the lengths xand y lie in the field of constructible elements, i.e. x, y ∈ K. The question is,“Which points lie in K and hence are constructible?” To answer we considerthe algebra of the straightedge and compass constructions.

2.3 Field Extensions

By operation 1 under the rules of construction, the straight line connectingtwo points will be defined algebraically by the linear equation (y − y1) =

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m(x− x1) where m = (y2 − y1)/(x2 − x1) in some field, F . By operation 2,the point of intersection of two lines in F is a solution in the same field F

ax+ by − c = a′x+ b′y − c′

(a− a′)x+ (b− b′)y = (c+ c′).

By operation 3, circles with center (h, k) and radius r are defined by thequadratic equation (x − h)2 + (y − k)2 = r2 with h, k, r ∈ F . And byoperation 4, the intersection between a line and circle, we solve the linearequation for y and substitute into the quadratic equation

ax+ by = c⇒ y = mx+ C

(x− h)2 + ((mx+ C)− k)2 = r2.

The result is an equation that is quadratic in terms of x and linear in termsof y. The intersection of two circles yields an equation that is linear

2x(h′ − h) + 2y(k′ − k) = r2 − h2 − k2 − r′2 + h′2 + k′2.

Thus, the four operations with straightedge and compass produce elementsin a field that is at most what is called a quadratic extension of Q. In otherwords, the field of constructible elements K is the smallest field extension of Qthat is closed under square root and complex conjugation. Given any a ∈ Qwe can take successive square roots and that element will be constructible.The number 3

√2 is a solution to the cubic equation x3 − 2 and cannot be

computed with square roots and hence not constructible. The number π’stranscendence over Q means that it cannot be computed from some rationalnumber by addition, subtraction, multiplication, division, or square roots–

the only tools that we have. But a number like√

23 +√−6 is constructible.

Field extensions have a property called the degree of the field extension.We will make a formal definition for the degree in the next section but weforeshadow now what is to come for motivation. It is a fact of field theorythat extension degrees are multiplicative in towers and if the extension degreefactors the field can be factored into subfields. For example if F ⊆ L ⊆ Kare field extensions then the degree of K/F is equal to the degree of K/Ltimes the degree of L/F . The take away for us is that quadratic extensionshave degree equal to 2. For straightedge and compass constructions, sincewe can make at most a quadratic extension, the degree of a field extensioncontaining some constructible elements over the field of rational numbers Qmust be a power of two, 2m.

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2.4 Degree and Dimension

We now know that a is constructible if and only if there is a tower of fieldsQ ⊂ K1 ⊂ K2 ⊂ · · · ⊂ Km such that a ∈ Km and the degree of each fieldextension is 2. Let us formally define degree and relate the degree of thefield extension to the nth roots of unity and the polynomial equation xn = 1.Denote the field of rational numbers plus the nth roots of unity by Q(ζn),which is read as “Q adjoined ζn”. We can consider Q(ζn) as a Q-vector spaceby simply considering the action of any a ∈ Q on Q(ζn) as multiplication inQ(ζn) by a. This leads us to our next definition.

Definition The degree of a field extension K/F , denoted [K : F ], is thedimension of K as a vector space over F , [K : F ] = dimF K

Linear Algebra tells us that the dimension of a finite vector space is thecardinality of any basis. Therefore we seek to find a basis for Q(ζn) as avector space over Q to find the degree of the extension. Before stating thetheorem we make a definition.

Definition A nonzero, nonconstant polynomial p(x) ∈ Q[x] is irreducible ifand only if given any factorization, say p(x) = m(x)n(x), either m(x) or n(x)is a nonzero constant polynomial.

Theorem 1 Let p(x) ∈ Q[x] be an irreducible polynomial of degree d overfield Q such that p(x) has a root in Q(ζn) be as above. Let x ≡ x mod p(x).Then the elements 1, x, x2, . . . , xd−1 are a basis for Q(ζn) as a vector spaceover Q and the degree of the extension is d, i.e. [Q(ζn) : Q] = d.

Proof We must show the elements 1, x, x2, . . . , xd−1 form a spanning setand are linearly independent. Let a(x) ∈ Q[x] be any nonzero polynomial.Dividing a(x) by p(x) we obtain:

a(x) = q(x)p(x) + r(x) q(x), r(x) ∈ Q[x] with deg r(x) < d.

Note a(x) ≡ r(x) mod p(x), thus a(x) is congruent to a polynomial withdegree less than d. This means that a(x) can be written as a linear combi-nation of the elements 1, x, x2, . . . , xd−1. Hence, 1, x, x2, . . . , xd−1 span Q(ζn)as a vector space over Q. We must show that these elements are linearly in-dependent in Q(ζn). Suppose the elements 1, x, x2, . . . , xd−1 are not linearlyindependent. Then there is a linear combination

b0 + b1x+ b2x2 + · · ·+ bd−1x

d−1 = 0

8

in Q(ζn) with b0, b1, . . . , bd−1 ∈ Q, not all 0. This is equivalent to

b0 + b1x+ b2x2 + · · ·+ bd−1x

d−1 ≡ 0 mod p(x)

meaningp(x) divides b0 + b1x+ b2x

2 + · · ·+ bd−1xd−1

in Q[x]. But this is a contradiction, because p(x), being degree d, cannotdivide a polynomial that has degree less than d. Thus 1, x, x2, . . . , xd−1 area basis for Q(ζn)/Q and [Q(ζn) : Q] = d.

�

Theorem 1 says that if we find a polynomial that is irreducible over therationals with roots in some extension field, then we can determine the degreeof the extension field–it is simply the degree of the irreducible polynomialsatisfying the above of minimal degree. We call this the minimal polynomialbecause among all nonzero polynomials with the required roots we take theone of smallest degree. The polynomial xn − 1 is not irreducible over Q(ζn)because x−1 can always be factored out. But there will be an irreducible partand the degree of that polynomial will be the degree of the field extension.

2.5 Galois Theory

Translating straightedge and compass constructions to the language of fieldtheory gave complete answers to the possibility of impossibility of certainconstructions as we have seen. We now introduce Galois theory to laterunderstand the relationship between the field Q(ζn)/Q and any intermediatefields, Kj, that may exist. Evariste Galois (1811-1832) was a young Frenchmathematician whose work was not seen as significant or profound untilmany years after his death. His most famous contribution was proving thethere was no explicit formula for finding the solutions of a general quinticpolynomial as there is for quadratic, cubic and quartic polynomials. Wewill introduce just enough field theory to state the Fundamental Theoremof Galois Theory which shows a relationship between fields and groups. Forfurther reading and a complete treatment see [2].

2.5.1 The Basics

Throughout these sections let K be a field.

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Definition A map, σ, that preserves addition and multiplication of K withitself is called an automorphism of K. This means that given x, y ∈ K,

σ(x+ y) = σ(x) + σ(y)

σ(xy) = σ(x)σ(y)

The collection of automorphisms of K is denoted Aut(K). An automorphismσ ∈Aut(K) is said to fix an element x ∈ K if σ(x) = x. If F is a subfield ofK then σ is said to fix the field F if σ(x) = x for all x ∈ F .

Any field has at least one automorphism called the trivial automorphism,denoted by 1. For any automorphism σ(1) = 1 and σ(0) = 0. The setAut(K) is a group under the operation composition of functions.

Definition Let K/F be an extension field. Define Aut(K/F ) to be the groupof automorphisms of K which fix F . The group Aut(K/F ) is a subgroup ofAut(K).

Given a subgroup H ⊂ Aut(K) the fixed field of H is the set of all elementsx ∈ K such that all elements of H fix x.

Definition Suppose that the number of automorphisms in Aut(K/F ) isequal to the degree of the field extension K/F , i.e. |Aut(K/F )|=[K : F ].Then K/F is a called a Galois extension or equivalently K is said to beGalois over F . If this is the case the group of automorphisms Aut(K/F ) iscalled the Galois group of K/F , and denoted Gal(K/F ).

Definition A field extension K/F is called a called a splitting field for thepolynomial p(x) ∈ F [x] if it is the smallest field containing the roots, x ∈ Cof p(x). This means that p(x) factors–or splits–as a product of linear factorsin the larger field K[x].

Corollary 1 Let p(x) ∈ F [x] be a polynomial with no repeated roots. If K/Fis the splitting field of p(x), meaning p(x) can be factored completely into itsdistinct linear components, then K/F is Galois.

Now we present the Fundamental Theorem of Galois Theory.

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2.5.2 The Fundamental Theorem

Theorem 2 Let K/F be a Galois extension with Galois group G = Gal(K/F ).There is a 1-1 correspondence between the subgroups of G and the interme-diary fields of K/F . A subgroup H ⊂ G maps to the fixed field of H and theinverse maps the intermediary field L ⊂ K to the subgroup Aut(K/L).

We will not prove this result as we have not introduced enough materialto do so; we only introduced enough to understand the statement of thetheorem and use it later. Invoking this theorem will allow us to you bothgroup and field knowledge and observations to understand the fields we needfor construction.

Above we denoted the field containing the nth roots of unity as Q(ζn). Bydefinition Q(ζn) is a splitting field for f(x) = xn−1. We constructed it to beso! The polynomial f(x) = xn − 1 has n distinct roots, i.e. the nth roots ofunity, thus by the corollary, the field extension Q(ζn)/Q is a Galois extensionwith some Galois group G = Gal (Q(ζn)/Q). Each automorphism σ ∈ G isdetermined by its actions on ζn. Since a primitive nth root of unity must mapto another primitive root10 define σa(ζn) = ζan, for a ∈ Z/nZ, gcd(a, n) = 1.The Galois group G consists of all these σa. This gives the following groupisomorphism:

(Z/nZ)× → G

a 7→ σa.

The Euler totient function φ(n) computes the number of elements a less thann, satisfying gcd(a, n) = 1. Therefore the number of elements in G is φ(n)which allows us to conclude that [Q(ζn) : Q] = φ(n) as well. Earlier wesaid that n was not the degree of this field extension because the polynomialxn − 1 was not irreducible over Q. The polynomial that is irreducible is thepolynomial with degree φ(n), called the cyclotomic polynomial.

10The Galois group permutes the roots of f(x). The group action is both transitive andfaithful.

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2.6 The Cyclotomic Polynomial

Definition The nth cyclotomic polynomial Φn(x) ∈ C[x] is the polynomialwhose roots are the primitive nth roots of unity:

Φn(x) =∏

ζn primitive

(x− ζn) =∏

1≤a<n(a,n)=1

(x− ζan).

Note that we are just saying the same thing in two ways with the equality toreally emphasize the fact that primitive roots of unity occur precisely whenthe power is relatively prime to n.

Theorem 3 The nth cyclotomic polynomial Φn(x) has degree φ(n) whereφ(n) = |(Z/nZ)×|. That is, the number of elements a ∈ Z/nZ such thatgcd(a, n) = 1.

Proof The roots of the polynomial xn − 1 are the nth roots of unity, so wehave the factorization

xn − 1 =∏ζ∈µn

(x− ζ).

Grouping together the linear factors of xn− 1 where ζ is a primitive dth rootof unity for all d dividing n we obtain

xn − 1 =∏d|n

∏ζ primitive∈µd

(x− ζ)

where the inside product is Φd(x) by definition. This implies

xn − 1 =∏d|n

Φd(x)

and we see by comparison that the degree n =∑

d|n φ(d). Thus the degree

of the cyclotomic polynomial Φn(x) is the number of units in n.

�

Now we have a polynomial of the required degree with the required roots.We move now to show that the polynomial has rational coefficients and isirreducible as required.

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Theorem 4 The cyclotomic polynomial Φn(x) is an irreducible monic poly-nomial in Z[x] of degree φ(n).

Proof Suppose that Φn(x) is not irreducible. Then we have a nontrivialfactorization

Φn(x) = f(x)g(x) with f(x), g(x) monic in Z[x].

Take f(x) to be an irreducible factor of Φn(x) because every polynomial canbe written as the product of irreducible polynomial factors. Let ζn be aprimitive nth root of unity which is a root of f(x), making f(x) the minimalpolynomial for ζn over Q. Let p be any prime that does not divide n. Fromabove we know that ζpn is again a primitive nth root of unity and a root ofeither f(x) or g(x).

Suppose g(ζpn) = 0. Then ζn is a root of g(xp) and since f(x) is theminimal polynomial for ζn, f(x) must divide g(xp) in Z[x]:

g(xp) = f(x)h(x), h(x) ∈ Z[x].

If we reduce this equation mod p, we obtain

g(xp) = f(x)h(x) in Fp[x]

where Fp denotes a finite field of p elements, like Z/pZ for example. By atheorem of field theory11,

g(xp) = (g(x))p

and using this substitution we get

(g(x))p = f(x)h(x) in Fp[x].

Thus, f(x) and g(x) have a factor in common in Fp[x]. From Φn(x) =f(x)g(x) we see by reducing mod p that Φn(x) = f(x)g(x), and so fromabove it follows that Φn(x) ∈ Fp[x] has a multiple root–the factor f(x) andg(x) had in common. But then xn−1 would have a multiple root in Fp sinceΦn(x) divides it. This is a contradiction since xn − 1 has n distinct roots.

Hence ζpn must be a root of f(x) (we assumed it was a root of g(x)).Thus, ζan is a root of f(x) for every integer a relatively prime to n. Writea = p1p2 · · · pk as a product of (not necessarily distinct) primes not dividingn so that ζp1 is a root of f(x), (ζp1)px is a root of f(x), etc. But this meansthat every primitive nth root of unity is a root of f(x), so f(x) has degreeφ(n). Therefore, f(x) = Φn(x), and Φn(x) is irreducible.

11Uses the Frobenius automorphism of F, a 7→ ap.

13

�

An awesome consequence of this section is that we have a recursive for-mula for computing Φn(x) for any n:

Φ1(x) = x− 1

Φ2(x) =x2 − 1

Φ1(x)= x+ 1

Φ3(x) =x3 − 1

Φ1(x)= x2 + x+ 1

Φ4(x) =x4 − 1

Φ1(x)Φ2(x)= x2 + 1

...

Note that for the special case n = p, p prime we have the polynomial

Φp(x) =xp − 1

Φ1(x)= xp−1 + xp−2 + · · ·+ x+ 1.

Given any n-gon we will simply consider the nth roots of unity and theirsplitting field Q(ζn)/Q. If the degree of the extension, i.e. φ(n), is a powerof 2, then that n-gon is constructible with a straightedge and compass. Whenφ(n) is computed, one sees that n can be written as n = 2kp1 · · · pl where thepi are distinct odd primes such that pi − 1 is a power of 2. Primes with thischaracteristic are called Fermat primes and have the form pi = 22r + 1 forsome nonnegative integer r. The first four Fermat primes are

21 + 1 = 3, 22 + 1 = 5, 24 + 1 = 17, 28 + 1 = 257.

Take note, 232+1 is not prime and it is not known if there are infinitely manyFermat primes [2]. This information allows us to completely characterizeconstructible n-gons.

Theorem 5 (Constructible n−Gons) The regular n-gon can be constructedby straightedge and compass if and only if n = 2kp1 · · · pl is the product of apower of 2 and distinct Fermat primes.

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3 Constructing n-gons

3.1 The triangle and square

The equilateral triangle is the simplest polygon defined when n = 3. Thevertices are the values ζk3 = exp(2πki/3) for k = 1, 2, 3 because these are theroots of the equation x3 = 1. Consider the field extension Q(ζ3)/Q whichcontains the 3rd roots of unity. Theorem 1 and Theorem 4 tell us the degree ofthe extension is the degree of the minimal polynomial Φ3(x). By the formula

Φ3(x) =x3 − 1

Φ1(x)=x3 − 1

x− 1= x2 + x+ 1

is a degree two polynomial. This polynomial is irreducible by definition so[Q(ζ3) : Q] = 2 is a power of 2; thus, we know the triangle is constructible.

Using the quadratic formula we find that the complex roots of x2 + x+ 1are

ζ3 =−1 +

√3i

2and ζ23 =

−1−√

3i

2.

This tells us that the real part of both P1, P2 lie on the line x = −1/2.Construct the unit circle along with the x-axis and y-axis. Note this willmost often be our given first step.

15

Now construct a unit circle using (−1, 0) as the center. Draw a perpendicularbisector at the midpoint of the segment (-1,0) to (0, 0), where the unit circlesmeet as shown.

Now simply connect the segments to build the triangle

P3 = (1, 0)

P1 = (−12,√32i)

P2 = (−12,−√32i)

Constructing the square is a trivial matter. All one must do is draw the

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four segments connecting the x- and y-axis in a diamond as seen below.

P4 = (1, 0)

P1 = (0, i)

P2 = (−1, 0)

P3 = (0,−i)

However, we will still show that the calculations of the field extension andminimal polynomial agree with the work above. The four vertices are theroots to the equation x4 − 1, thus taking one primitive fourth root of unityζ4 we again consider the degree of the extension Q(ζ4)/Q. The minimalpolynomial is the degree 2 polynomial

Φ4(x) =x4 − 1

Φ1(x)Φ2(x)= x2 + 1

which is irreducible over Q. The roots are ±i which we can easily construct.

3.2 The hexagon, octagon and more

Given that we are able to bisect angles with a straightedge and compass oncea particular n-gon is constructed we may immediately construct the 2n-gon,4n-gon, 8n-gon, etc. by bisecting the previous n-gon’s angles as shown.

17

From above work given n = 3 we can further construct the regular hexagon,dodecagon, 24-gon, 48-gon, 96-gon etc. For n = 4 we get the octagon, 16-gon,32-gon, 64-gon, 128-gon and the list goes on.

3.3 The pentagon

Here is where our computations get more interesting and we must be morerefined with our techniques. The vertices of the pentagon or 5-gon satisfyx5 = 1. The minimal polynomial for the extension Q(ζ5)/Q is

Φ5(x) = x4 + x3 + x2 + x+ 1.

This tells us that [Q(ζ5) : Q] = 4. Four is a power of 2, so the 5-gon is aconstructible n-gon. Recall from above that the degree of any field extensionis multiplicative in towers. The degree of this field extension can be factored,4 = 2 × 2, into two quadratic extensions thus there is an intermediate fieldL satisfying

Q(ζ5)

|2L

|2Q

We would like to determine what the subfield L is.

3.3.1 The group µ5

Each ζkn is a primitive 5th root of unity (why?) thus any one root can generatethe others. Recall that we defined automorphisms σa of the Galois groupG = Gal (Q(ζ5)/Q) by σa(ζn) = ζan. The five roots are

ζ5 = e2πi5 , ζ25 = e

4πi5 , ζ35 = e

6πi5 , ζ45 = e

8πi5 , ζ55 = 1.

We can “rewrite” the polynomial equation Φ5(x) = x4 + x3 + x3 + x+ 1 as

ζ45 + ζ35 + ζ25 + ζ5 + 1 = 0. (1)

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Now, for a = (Z/5Z)×, σa(ζk5 ) = ζakn . When a = 1 each point maps to itself,

so σ1 is the trivial automorphism and we will denote it by 1. Checking howσ2, σ3 and σ4 map elements:

σ2(1) = 1 σ3(1) = 1 σ4(1) = 1

σ2(ζ5) = ζ25 σ3(ζ5) = ζ35 σ4(ζ5) = ζ45σ2(ζ

25 ) = ζ45 σ3(ζ

25 ) = ζ5 σ4(ζ

25 ) = ζ35

σ2(ζ35 ) = ζ5 σ3(ζ

35 ) = ζ45 σ4(ζ

35 ) = ζ25

σ2(ζ45 ) = ζ35 σ3(ζ

45 ) = ζ25 σ4(ζ

45 ) = ζ5

we see the automorphisms permute the roots as expected. According to thefundamental theorem the subfield L corresponds to a subgroup H ⊂ G whichhas two elements because [L : Q] = 2. One of these elements is the trivialautomorphism, but the other is to be determined. Since G ∼= (Z/5Z)× wewill look at this cyclic group to gather information about G. The cyclic group(Z/5Z)× is generated by 2,

21 = 2, 22 = 4, 23 = 3, 24 = 1.

There are three subgroups: {1},{4, 1}, and {2, 4, 3, 1}. The first becauseone is trivial and the second because powers of four will either look likeone or look like four modulo 5 and the last because it is the whole group.Translating this information back to G we find that the nontrivial subgroupis H = {1, σ4}, and H corresponds to L.

The elements of H must fix the elements of the field L. Let x ∈ L writex as the linear combination of the basis elements x = ζ5 + ζ45 . Then x is fixedby σ4,

σ4(x) = σ4(ζ5) + σ4(ζ45 )

= ζ45 + ζ5

= x

showing x is indeed an element of the subfield, L. Looking at equation1 above, it is possible to show a relation between the minimal polynomialΦ5(x) and the minimal polynomial for L/Q.

1 = 1

x = ζ5 + ζ45x2 = (ζ5 + ζ45 )2 = ζ25 + 2 + ζ35

x2 + x− 1 = ζ25 + 2 + ζ35 + ζ5 + ζ45 − 1 = 0

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The minimal polynomial for L/Q is x2 +x−1. One can check that this poly-nomial is irreducible by the rational root test. Using the quadratic equationthe roots of this polynomial are

x =1±√

5

2.

These values reside in the field L(√D) where D = b2− 4c. Thus the subfield

L is the field Q(√

5) and the tower is,

Q(ζ5)

|2Q(√

5)

|2Q

3.3.2 Explicit Construction

The complex number C = cos θ + i sin θ has multiplicative inverse, C−1 =cos θ−i sin θ, since the product of the two is CC−1 = cos2 θ+sin2 θ = 1. Notethat C +C−1 = 2 cos θ is a real number. Instead of determining the complexroot of unity ζ5 = cos 2π/5 + i sin 2π/5, we can determine the real-valuedcombinations of roots

ζ5 + ζ−15 = 2 cos 2π/5

which will be much easier. Let η0 = ζ5 + ζ45 and η1 = ζ25 + ζ35 . Minorcalculation shows that η0 and η1 add up to -1 and have the product 1, andthus are the roots of the minimal polynomial

x2 + (η0 + η1)x+ (η0η1) = x2 + x− 1 = 0.

Using the quadratic formula we obtain,

η0 =−1 +

√5

2, η1 =

−1−√

5

2.

Using algebra and geometry one can now construct a regular 5-gon. Con-struct the unit circle and plane as above then draw the midpoint of `. Draw

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the line segment from the midpoint to 1 and then bisect the angle. Wherethe angle bisector crosses the x-axis is the value η0.

`

η0

`

Construct the perpendicular bisector to the x-axis at η0 and where this lineand the unit circle meet is ζ5 and ζ45 .

ζ5

ζ45

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Since η0 = 2 cos 2π/5 we have cos θ = (±√

5 − 1)/4 which gives sin θ =

(±√

10± 2√

5)/4 giving the cartesian coordinates of the pentagon.

P5 = (1, 0)

P1 =(√

5−14, 10+2

√5

4

)

P2 =(−√5−14, 10−2

√5

4

)

P3 =(−√5−14,−10−2

√5

4

)

P4 =(√

5−14, 10−2

√5

4

)

As mentioned before by bisecting angles repeatedly, the construction of thedecagon, 20-gon, 40-gon, etc. soon follow.

3.4 Impossible Constructions

In the introduction we mentioned that the heptagon (n=7) and the nonagon(n=9) are impossible constructions and now we are able to explain why.First consider the case for n = 7. The cyclotomic polynomial Φ7(x) =x6 + x5 + x4 + x3 + x2 + x + 1 is a degree six polynomial which means that

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the field extension Q(ζ7)/Q has two possibilities for towers

Q(ζ7)

3

Q(ζ7)

2

L

2

or L

3

Q Q

because the degrees are multiplicative. Either way, one of the extensions isnot quadratic and therefore the elements in that field cannot be constructedby straightedge and compass.

Similarly, when n = 9,

Φ9 =x9 − 1

Φ1Φ3

=x9 − 1

(x− 1)(x2 + x+ 1)=x9 − 1

x3 − 1= x6 + x3 + 1

a degree six polynomial which again means that the values needed do not liein the constructible field. Some other n-gons that are not constructible witha compass and straightedge are n = 11, 13, 14, , 18, 19, 21, 22, 23, 25 . . . . Aswe mentioned the discovery of the constructibility of the 17-gon was a breakthrough for the field and we introduce the algebra for that construction now.

3.5 Gauss’s 17-gon

Let ζ = ζ17. The field extension Q(ζ)/Q is degree 16. The tower of subfieldsLi = LHi and the corresponding subgroups Hi ⊂ Gal (Q(ζ)/Q) is

Q(ζ)

2

←→ {1}

L3

2

←→ H3

L2

2

←→ H2

L1

2

←→ H1

Q ←→ Gal (Q(ζ)/Q) .

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Let us first consider L1 and H1. The index of the subgroups [G : H1] = 2 andthe automorphism σ2 generates this subgroup H1. Starting with ζ successiveapplications of σ2 yields,

ζ, ζ2, ζ4, ζ8, ζ16, ζ15, ζ13, ζ9

and the remaining eight roots of unity are obtained in the sequence

ζ3, ζ6, ζ12, ζ7, ζ14, ζ11, ζ5, ζ10.

Define the values, called periods, η0 and η1 as

η0 = ζ + ζ2 + ζ4 + ζ8 + ζ9 + ζ13 + ζ15 + ζ16

η1 = ζ3 + ζ5 + ζ6 + ζ7 + ζ10 + ζ11 + ζ12 + ζ14.

First, note that η0, η1 are real numbers because a root and its inverse areadded together, canceling the imaginary part. This is the same technique weused when n = 5 above. Second, we have η0 + η1 = −1. To see this considerΦ17(x). Next, the product η0η1 = −4. Putting this together we see that η0and η1 are the roots of x2 − (η0 + η1)x + η0η1 = 0 which is to say η0 and η1satisfy x2 + x− 4 = 0.

Starting with ζ make a sequence of fourth powers then do the same forζ3. Essentially, this means taking the alternating values as they appear inthe sequences above and make the following periods

η′0 = ζ + ζ4 + ζ13 + ζ16 η′2 = ζ3 + ζ5 + ζ12 + ζ14

η′1 = ζ2 + ζ8 + ζ9 + ζ15 η′3 = ζ6 + ζ7 + ζ10 + ζ11.

Here again each η′i is a real number and one can check that η′0η′1 = η′2η

′3 = −1.

Therefore η′0, η′1 are roots to the equation x2−η0x−1 = 0 and η′2, η

′3 are roots

to the equation x2 − η1x− 1 = 0.Take the last two periods as

η′′0 = ζ + ζ16 and η′′1 = ζ4 + ζ13.

These are real numbers that satisfy the equation x2−η′0x+η′2 = 0. The reader

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should now verify that the fundamental theorem gives the correspondence

Q(ζ)

2

←→ {1}

Q(ζ + ζ16)

2

←→ 〈σ16〉

Q(ζ + ζ4 + ζ13 + ζ16)

2

←→ 〈σ4〉

Q (ζ + ζ2 + ζ4 + ζ8 + ζ9 + ζ13 + ζ15 + ζ16)

2

←→ 〈σ2〉

Q ←→ Gal (Q(ζ)/Q)

with each subgroup Hi index 2 over Hi+1 and defined by the generator.The successive subgroups being index 2 reflects the fact that the periodsfor Hi satisfy a quadratic equation whose coefficients involve periods forHi+1. Given the minimal polynomials which are all quadratic we can computeexplicit values for all the values of η with care to choose the correct squareroot given that η′′0 = 2 cos 2π/17 and η′′1 = 2 cos 8π/17. We omit writing allof the values but show a couple below:

η0 =

√17− 1

2

η′0 =η02

+

√1 +

η204

=

√17− 1

4+

1

4

√34− 2

√17.

For instructions on how to construct the 17-gon see [1] or [2]. Now one isa few constructions away from the 34-gon, 51-gon, 54-gon, 85-gon, even the255-gon and 257-gon!

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References

[1] L.E. Dickson. Monographs on Topics of Mathematics Relevant to theElementary Field, chapter VIII Constructions with Ruler and Compass;Regular Polygons, pages 351–386. Longmans, Green and Co., 1911.

[2] David S. Dummit and Richard M. Foote. Abstract Algebra. John Wiley& Sons Inc., Hoboken, NJ, third edition, 2004.

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