where [Q(Xk+1) : Q(Xk )] = 1 or 2 for 2 k m − 1. Furthermore, each extension field Q(Xk ) is a subfield of R because Q and Xk are sets of real numbers. By dropping each field Q(Xi ) that is a trivial extension of Q(Xi−1), it follows that

Proof. If α is constructible, then α Kn Kn−1 · · · K0 = Q, where Ki is an extension field of degree 2 over Ki−1. By Theorem 11.6,

[Kn : Q(α)][Q(α) : Q] = [Kn : Q]

= [Kn : Kn−1][Kn−1 : Kn−2] · · · [K1 : Q] = 2n.

Corollary 13.7. If [Q(α) : Q] = 2r for some r 0, then α is not constructible.

Corollary 13.6 does not give a sufficient condition for α to be constructible, as shown in Example 13.17 below.

Example 13.8. Can a root of the polynomial x5 + 4x + 2 be constructed using straightedge and compass?

Solution. Let α be a root of x5 + 4x + 2. By Eisenstein’s criterion, x5 + 4x +

Example 13.9. Can a root of the polynomial x4 − 3x2 + 1 be constructed using straightedge and compass?

constructed.

DUPLICATING A CUBE

Let l be the length of the sides of a given cube so√that its volume is l3. A cube with double the volume will have sides of length 3 2 l.

√

Proposition 13.10. 3 2 is not constructible.

√

Since we cannot construct a length of 3 2 l starting with a length l, the ancient problem of duplicating the cube is insoluble.

TRISECTING AN ANGLE

Certain angles can be trisected using straightedge and compass. For example, π, π/2, 3π/4 can be trisected because π/3, π/6, and π/4 can be constructed. However, we show that not all angles are trisectable by proving that π/3 cannot be trisected.

If we are given the angle φ, we can drop a perpendicular from a point a unit distance from the angle to construct the lengths cos φ and sin φ, as shown in Figure 13.4. Conversely, if either cos φ or sin φ is constructible, it is possible to construct the angle φ. Hence, if we are given an angle φ, we can construct all numbers in the extension field Q(cos φ). Of course, if cos φ Q, then Q(cos φ) = Q.

We can now consider those numbers that are constructible from Q(cos φ). This notion of constructibility is similar to our previous notion, and similar results hold, except that the starting field is Q(cos φ) instead of Q.

Theorem 13.11. The angle φ can be trisected if and only if the polynomial 4x3 − 3x − cos φ is reducible over Q(cos φ).

Proof. Let θ = φ/3. The angle θ can be constructed from φ if and only if cos θ can be constructed from cos φ. It follows from De Moivre’s theorem and the binomial theorem that

cos φ = cos 3θ = 4 cos3 θ − 3 cos θ.

Hence cos θ is a root of f (x) = 4x3 − 3x − cos φ.

If f (x) is reducible over Q(cos φ), then cos θ is a root of a polynomial of degree 1 or 2 over Q(cos φ); thus [Q(cos φ, cos θ ) : Q(cos φ)] = 1 or 2. Hence, by Propositions 11.16 and 13.3, cos θ is constructible from Q(cos φ).

1

sin f

f

cos f

Figure 13.4. Constructing sin φ and cos φ from the angle φ.

B

2q Q

Y

Figure 13.6. Trisection of the angle φ using a marked ruler.

Triangle YOQ is isosceles, so angle OQY = 2θ . Also,

φ = angle AOB = angle OXQ + angle OQX = θ + 2θ = 3θ.

Hence θ = φ/3.

SQUARING THE CIRCLE

Given any circle of radius r, its area is π r2, so that a square with the same

√ √

area has sides of length π r. We can square the circle if and only if π is constructible.

√ √

Proposition 13.15. [Q( π ) : Q] is infinite, and hence π is not constructible.

Proof. The proof of this depends on the fact that π is transcendental over Q; that is, π does not satisfy any polynomial equation with rational coefficients. This was mentioned in Chapter 11, and a proof is given in Stewart [35].

scendental, π, π , π , . . . are linearly independent over Q, and [Q(π ) : Q] is infinite. Therefore,

Hence the circle cannot be squared by straightedge and compass.

CONSTRUCTING REGULAR POLYGONS

Another problem that has been of great interest to mathematicians from the time of the ancient Greeks is that of constructing a regular n-gon, that is, a regular polygon with n sides. This is equivalent to constructing the angle 2π/n or the