1.10. Constructions with straight-edge and compass.

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The Greeks understood that

integers and the rational numbers. They were surprised to find that the length of the

diagonal of a square of side 1, namely

√

2, is not rational. They thus realized that they needed

to extend their number system. They then hoped that the “constructible” numbers would

suffice. Suppose we are given a length, which we call 1, a straight-edge, and a compass (device

for drawing circles). A number (better a length) is constructible if it can be constructed by

forming successive intersections of

• lines drawn through two points already constructed, and

• circles with centre a point already constructed and radius a constructed length.

This led them to three famous problems that they were unable to solve: is it possible

to duplicate the cube, trisect an angle, or square the circle by straight-edge and compass

constructions? We’ll see that the answer to all three is negative.

Let F be a subfield of R. The F-plane is F × F ⊂ R × R. We make the following

definitions:

A line in the F-plane is a line through two points in the F-plane. Such a line is given by

an equation:

ax + by + c = 0, a, b, c ∈ F.

10 J.S. MILNE

A circle in the F-plane is a circle with centre an F-point and radius an element of F. Such

a circle is given by an equation:

(x − a)2 + (y − b)2 = c2, a, b, c ∈ F.

Lemma 1.24. Let L _= L_ be F-lines, and let C _= C_ be F-circles.

(a) L ∩ L_ = ∅ or consists of a single F-point.

(b) L ∩ C = ∅ or consists of one or two points in the F[

√

e]-plane, some e ∈ F.

(c) C ∩ C_ = ∅ or consists of one or two points in the F[

√

e]-plane, some e ∈ F.

Proof. The points in the intersection are found by solving the simultaneous equations,

and hence by solving (at worst) a quadratic equation with coefficients in F.

Lemma 1.25. (a) If c and d are constructible, then so also are c Ѓ} d, cd, and c

d (d _= 0).

(b) If c > 0 is constructible, then so also is

√

c.

Proof. First show that it is possible to construct a line perpendicular to a given line

through a given point, and then a line parallel to a given line through a given point. Hence

it is possible to construct a triangle similar to a given one on a side with given length. By

an astute choice of the triangles, one constructs cd and c−1. For (b), draw a circle of radius

c+1

2 about ( c+1

2 , 0), and draw a vertical line through the point A = (1, 0) to meet the circle

at P. The length AP is

√

c. (For more details, see for example, Rotman, Galois Theory,

Appendix 3.)

Theorem 1.26. (a) The set of constructible numbers is a field.

(b) A number α is constructible if and only if it is contained in field of the form

Q[

√

a1, . . . ,

√

ar], ai ∈ Q[

√

a1, . . . ,

√

ai−1].

Proof. (a) Immediate from (a) of Lemma 1.25.

(b) From (a) we know that the set of constructible numbers is a field containing Q, and

it follows from (a) and Lemma 1.25 that every number in Q[

√

a1, . . . ,

√

ar] is constructible.

Conversely, it follows from Lemma 1.24 that every constructible number is in a field of the

form Q[

√

a1, . . . ,

√

ar].

Now we can apply the (not quite elementary) result Proposition 1.10 to obtain:

Corollary 1.27. If α is constructible, then α is algebraic over Q, and [Q[α] : Q] is a

power of 2.

Proof. We know that [Q[α] : Q] divides [Q[

√

a1, . . . ,

√

ar] : Q] = 2r.

Corollary 1.28. It is impossible to duplicate the cube by straight-edge and compass

constructions.

Proof. The problem is to construct a cube with volume 2. This requires constructing

a root of the polynomial X3 − 2 = 0. But this polynomial is irreducible (by Eisenstein’s

criterion for example), and so [Q[ 3

√

2] : Q] = 3.

Corollary 1.29. In general, it is impossible to trisect an angle by straight-edge and

compass constructions.

FIELDS AND GALOIS THEORY 11

Proof. Knowing an angle is equivalent to knowing the cosine of the angle. Therefore, to

trisect 3α, we have to construct a solution to

cos 3α = 4cos3 α − 3cos α.

For example, take 3α = 60;to construct α, we have to solve 8x3 − 6x − 1 = 0, which is

irreducible.

Corollary 1.30. It is impossible to square the circle by straight-edge and compass constructions.

Proof. A square with the same area as a circle of radius r has side

√

πr. Since π is

transcendental, so also is

√

π.

We now consider another famous old problem, that of constructing a regular polygon.

Note that Xm − 1 is not irreducible;in fact

Xm − 1 = (X − 1)(Xm−1 + Xm−2 + ・ ・ ・ + 1).

Lemma 1.31. If p is prime then Xp−1 + ・ ・ ・ + 1 is irreducible; hence Q[e2πi/p] has degree

p − 1 over Q.

Proof. Consider

f(X + 1) =

(X + 1)p − 1

X

= Xp−1 + ・ ・ ・ + a2X2 + a1X + p,

with ai =

_ p

i+1

_

. Sincep|ai, i = 1, ..., p−2, f(X+1) is irreducible by Eisenstein’s criterion.

In order to construct a regular p-gon, p an odd prime, we need to construct cos 2π

p . But

Q[e

2πi

p ] ⊃ Q[cos 2π

p ] ⊃ Q. The degree of Q[e

2πi

p ] over Q[cos 2π

p ] is 2—the equation

α2 − 2cos

2π

p

・ α + 1 = 0, α= e

2πi

p ,

shows that it is ≤ 2, and it is not 1 because Q[e

2πi

p ] is not contained in R. Hence [Q[cos 2π

p ] :

Q] = p−1

2 .

Thus if the regular p-gon is constructible, then (p − 1)/2 = 2k some k (later, we shall see

a converse), which imples p = 2k+1 + 1. But 2r + 1 can only be a prime if r is a power of 2,

because otherwise r has an odd factor t, and for t odd,

Y t + 1 = (Y + 1)(Y t−1 − Y t−2 + ・ ・ ・ + 1).

Thus if the regular p-gon is constructible, then p = 22k + 1 for some k. Fermat conjectured

that all numbers of the form 22k + 1 are prime, and claimed to show that this is true for

k ≤ 5—for this reason primes of this form are called Fermat primes. For 0 ≤ k ≤ 4, the

numbers p = 3, 5, 17, 257, 65537, are prime but Euler showed that 232 + 1 = 641 ・ 6700417,

and we don’t know of any more Fermat primes.

Gauss showed that

cos

2π

17

= − 1

16

+

1

16

√

17 +

1

16

 

34 − 2

√

17 +

1

8

_

17 + 3

√

17 −

 

34 − 2

√

17 − 2

 

34 + 2

√

17

when he was 18 years old. This success encouraged him to become a mathematician.

12 J.S. MILNE

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