Let C be constructible. Then there exists a set of complex numbers {1,...,n} such that 12 Q, i2 Q(1,...,i-1)(2 < i < n). and such that Q(1,...,n). Thus, if is constructible, deg(Q()/Q) = 2k for some k. If E is an extension of the field F, then E may be regarded as a vector space over F. The product ·v of a scalar F and a vector v E is just the product of and v, considered as elements of E. The dimension of E over F is called the degree of E over F, denoted deg(E/F). All rational numbers are constructible. The set of rational numbers. If E is a field, then a subfield of E is a subset of E which is also a field with respect to the operations of E. The complex numbers.

### Constructions with Straightedge and Compass

We may not appear to have accomplished anything of significance yet concerning the theory of fields, but we have enough machinery to apply to several significant problems. In this section we will consider the problem of constructing geometrical figures using only a compass and a straightedge. For our purposes, a straightedge is a device which can be used only to draw a line between two points and a plane. It has no measurement capabilities. Such constructions were first considered by the Greek geometers and are probably familiar to the reader from a high school course in geometry. The Greeks posed three very interesting questions concerning constructibility which they were unable to solve:

Problem 1: (Trisecting an Angle) Given an arbitrary angle , construct the angle /3, using only straightedge and compass.

Problem 2: (Duplicating a Cube) Given an arbitrary cube C of volume V, construct a cube having volume 2V, using only straightedge and compass.

Problem 3: (Squaring a Circle) Given an arbitrary circle C of area A, construct a square having area A, using only straightedge and compass.

Our plan in this section is to first discuss the general problem of constructibility. From this discussion will emerge a powerful theory which will provide immediate solutions to Problems 1 and 2 and will point the way to a solution of Problem 3. It will turn out that all three problems have no solution. We will turn our attention to the construction of regular polygons.

Let us first lay down the ground rules for our investigation. Let us restrict ourselves to plane constructions. Further, we will identify the Euclidean plane with the complex numbers, by identifying the point with coordinates (a,b) with the complex number a + bi. Finally, let us assume that in addition to a compass and straightedge, we are given the line segment connecting (0,0) and (1,0), somehow marked off on our plane. (This line segment of length 1 is just used as a comparison device. We could just as well begin our constructions by drawing a line segment of arbitrary length and using it for comparison.) Our problem is to decide which geometric figures can be constructed. Henceforth, when we refer to "construction" we will always mean "construction with straightedge and compass, given the above unit segment as initial data."

Let us now reduce our geometric problem to an algebraic problem. It is clear that each geometric figure which we can construct can be defined by a finite number of points, line segments, and circular arcs. Let us describe a given geometrical figure by means of a family of complex numbers as follows: The finite number of points can be viewed as complex numbers as described above. A line segment is described by the complex numbers corresponding to its endpoints. A circular arc is described by four numbers: two corresponding to the endpoints of the arc, on corresponding to the center of the circle, and one equal to the length of the radius of the circle. It is clear that the numbers described are necessary and sufficient to construct the geometrical figure. Thus, let us henceforth think of a geometrical figure in terms of a collection of complex numbers describing points, lines and arcs in the figure. It is clear that in order to construct a given figure, it is both necessary and sufficient to be able to construct the line segments connecting (0,0) to each of the complex numbers describing the figure. This leads us to the following definition.

Definition 1: A complex number is said to be constructible if the line segment connecting (0,0) to is constructible.

From our above discussion we have

Theorem 2: A geometrical figure is constructible if and only if each of the complex numbers describing it is constructible.

Thus, our original question concerning constructibility of geometric figures is reduced to one concerning constructibility of complex numbers. Let C denote the set of all constructible complex numbers. We will give a more-or-less complete description of C.

Before proceeding further, let us recall some of the possible elementary constructions of Euclidean geometry. For the actual constructions refer back to high school geometry.

Basic Constructions

1. Bisect a given angle.

2. Construct a line perpendicular to a given line segment L at a given point P on L.

3. Construct a line through a given point P which is parallel to a given line L.

4. Given line segments of lengths l and l', construct a line segment of length ll'.

5. Given a line segment of length l, construct a line segment of length 1/l.

6. Given a line segment of length l, construct a line segment of length .

7. Construct an angle equal to the sum of two given angles.

Examples of some of these constructions are given in theappendices.

Theorem 3: C is a subfield of C.

Proof: Our assertion amounts to the following: If and are constructible, then so are + , -, · , 1/ (0). Let us consider each case separately. Recall that addition of complex numbers can be accomplished geometrically via the parallelogram law. Therefore, let us construct and as in Figure 1. Use basic construction 4 above to construct line parallel to On measure off a line segment of length , with one endpoint of the line segment at . Then the other end of the measured segment is + . Thus, + is constructible.

Figure 1: Construction of + .

In order to construct -, construct as in Figure 2. Continue the line segment through the origin and on the continued portion measure off a segment of length , having O as one endpoint. Then the other endpoint of the measured segment is -.

Figure 2: Construction of -

Let us construct · . Without loss of generality, we may assume that 0, 0. Then and can both be written in polar form

= r1(cos1 + i sin1).
= r2(cos2 + i sin2).

By de Moivre's theorem,

· = r1r2[cos(1 + 2) + i sin(1 + 2)]

Thus, the line segment O · has length r1r2 and makes an angle 1 + 2 with the positive X-axis. This is the clue to the construction of · . Refer to Figure 3, where and have been constructed. Construct the line segment making and angle 1 + 2 with the positive X-axis. Measure off a segment of length r1r2 on . This is possible by basic construction 4. Then = · .

Figure 3: Construction of ·

Finally, assume that 0. Then, since sin22 + cos22 = 1 and cos(-2) = cos2, sin(-2) = -sin2, we have

1/ = r2-1(cos2 + i sin2)-1
= r2-1[(cos2 - i sin2)/(cos22 + sin22)]
= r2-1[cos(-2) + i sin(-2)].

Thus, the line segment O-1 has length r2-1 and makes an angle - with the positive X-axis. The construction of this segment uses basic construction 5.

Corollary 4: All rational numbers are constructible.

Proof: We have bee given 1 as part as our initial data. Thus, 1 C. But since C is a field, this implies that Q C.

Proposition 5: Let C be constructible and let denote one of the square roots of . Then is constructible.

Proof: Without loss of generality, assume that 0. By de Moivre's theorem,

= ± [cos(1/2) + i sin(1/2)].

Let us only consider the positive sign. The reasoning for the negative sign is similar. Then the line segment O is a line segment of length which makes an angle of 1/2 with the positive X-axis. Therefore, can be constructed using basic constructions 1 and 6.

Theorem 6: Let 1,...,n be complex numbers such that

12 Q,
i2 Q(1,...,i-1)   (2 < i < n).

Proof: By Corollary 4, every element of Q is constructible and by Proposition 5 and the assumption 12 Q, we see that 1 is constructible. Therefore, since the constructible numbers form a field, every element of Q() is constructible. Thus, the theorem is true for n = 1. Let us proceed by induction on n. Let n > 1 and assume the theorem for n - 1. Then every element of Q(1,...,n-1) is constructible. By hypothesis, n2 Q(1,...,n-1) so that n is a square root of a constructible number and hence is constructible by Proposition 5. But since the constructible numbers form a field, every element of Q(1,...,n-1)(n) = Q(1,...,n) is constructible.

The amazing fact is that the converse is also true. We have

Theorem 7: Let C be constructible. Then there exists a set of complex numbers {1,...,n} such that

12 Q,
i2 Q(1,...,i-1)   (2 < i < n).

and such that Q(1,...,n). Thus, if is constructible, deg(Q()/Q) = 2k for some k.

Before proceeding with the proof of Theorem 7, let us do some preliminary work. Assume that the points 1,2,...,m-1 have been constructed. What points can we construct using 1,2,...,m-1? There are two elementary constructions which we can perform: (a) We can draw a line L connecting i to j (ij). (b) We can draw a circle C with center at i which passes through j (ij). Thus, if m is constructible from 1,2,...,m-1, then m is either the intersection of two lines L1 and L2, the intersection of a line L and circle C, or the intersection of two circles C1 and C2. Let

(1)
j = j + ij   (j,j R, 1 < j < m),
(2)
Fj = Q(i, 1, 1,..., j, j)    (1 < j < m).

We will consider separately three cases:

Case 1: m is the intersection of lines L1 and L2.

The lines L1 and L2 have equations

L1: a1x + b1y + c1 = 0,
(3)
L2: a2x + b2y + c2 = 0,

where a1,b1,c1,a2,b2,c2 Fm-1. The system of equations (3) has the solution x = m, y = m, by assumption. But the solution of the system (3) can be calculated rationally in terms of a1,b1,c1,a2,b2,c2 and therefore lies in Fm-1. Thus we see that

(4)
m Fm-1,    m Fm-1
Fm = Fm-1

since Fm = Fm-1(m,m) if m > 2.

Case 2: m is the intersection of the line L and the circle C.

The line L and the circle C have the equations

L: ax + by + c = 0,
(5)
C: dx2 + ey2 + fx + gy + h = 0,

where a,b,c,d,e,f,g,h Fm-1. On of the solutions of the system of equations (5) is x = m, y = m, by assumption. On the other hand, the system (5) can be solved by substituting the linear relation into the quadratic equation and then solving for, say, x, from which the corresponding value of y can be computed. Therefore, the solutions of system (5) can be computed in terms of the square root of an element , where is computable rationally in terms of a,b,c,d,e,f,g,h. In particular, Fm-1 and m Fm-1(), m Fm-1(). Further,

(6)
Fm Fm-1(),

since Fm = fm-1(m,m) if m > 2.

Case 3: m is the intersection of the circles C1 and C2.

The equations of C1 and C2 are given by

C1: x2 + y2 + a1x + b1y + c1 = 0,
C2: x2 + y2 + a2x + b2y + c2 = 0,

where a1,b1,c1,a2,b2,c2 Fm-1. The points of intersection of C1 and C2 are the same as the points of intersection of C1 with the line whose equation is

(a1 - a2)x + (b1 - b2)y + (c1 - c2) = 0.

Therefore, by case 2, there exists in Fm-1 such that

(7)
Fm Fm-1().

By comparing the results of cases 1-3, we have

Lemma 8: Suppose that the points 1,...,m-1 (m > 2) have been constructed and that m is constructible from 1,...,m-1. Further, suppose that 1 = j + i j, where j and j are real, and let

Fj = Q(i,1,1,...,j,j)    (1 < j < m.).

Then there exists Fm-1 such that Fm Fm-1().

Proof: By (6) and (7) we are done in cases 2 and 3. In case 1 we may set = 1 by (4).

Let us now prove Theorem 7. Suppose that can be constructed by successively constructing 1,2,...,n = , where 1 is constructed from 0 = 1. For m = 1,...,n let Fm be as in the Lemma, and set F0 = Q(i). Let us apply Lemma 8 to each of the sets of numbers 0,1,...,m (1 < m < n) to get that there exists m Fm-1 such that

(8)
Fm Fm-1().    (1 < m < n).

Define

0 = i,  j =    (1 < j < n).

We assert that

(9)
Fm Q(0,1,...,m)   (1 < m < n).

This follows trivially by induction from (8). It is clearly true for m = 1 by (8). Assume that m > 1 and that (9) is true for m - 1. Then, by the induction hypothesis and (8),

Fm Fm-1(m)
Q(0,...,m-1)(m)
= Q(0,...,m).

This completes the induction and hence (9) is proved. It is clear that 02 = -1 Q. Also, by (9), for 1 < m < n, we have

m2 = m Fm-1 Q(0,...,m-1).

Finally, by (9) for m = n, and the fact that n = , we have

= n + i n Fn Q(0,...,n).

This completes the proof of the first assertion of Theorem 7. In order to show that deg(Q()/Q) = 2k for some k, it suffices to show that

deg(Q(0,...,n)/Q) = 2r

for some r. But, if we set i = Q(0,...,i) (0 < i < n), we have deg( i+1/ i) = 1 or 2 since i+12 i and i+1 = i(i+1). Moreover,

deg(Q(0,...,n)/Q)
= deg( n/ n-1) · deg( n-1/ n-2)···deg( 1/ 0)
= 2r

for some r.

Let us now illustrate Theorem 7 by applying it to the problem of trisecting an angle. Not only will we show that there is no general procedure for trisecting an angle with compass and straightedge, but there are particular angles which cannot be trisected, For example, let us prove:

Theorem 9: A 60o angle cannot be trisected using only a compass and a straightedge.

Proof: It is well known that a 60o angle can be constructed using only a compass and a straightedge. Therefore, a 60o angle can be trisected if and only if it is possible to construct a 20o angle using only a compass and a straightedge. But let us see what it means for an angle to be constructible. If is constructible, then we may place one side of on the X-axis and construct the point where the other side of intersects the circle of radius 1 with center at the origin (see Figure 4). But by elementary trigonometry, this point is cos + i sin. Therefore, if angle is constructible, the number cos + i sin is constructible. Conversely, if cos + i sin is constructible, then angle is constructible (see Figure 4). Therefore, a 20o angle is constructible if and only if

cos(20o) + i sin(20o) =

is constructible.

Figure 4: Construction of the Angle .

Let us assume that is constructible. By Theorem 7, deg(Q()/Q) = 2r for some r. However, is a primitive eighteenth root of 1, so by Theorem 1 of examples,

deg(Q()/Q) = (18)
= 6.

Thus, a contradiction is reached and is not constructible. Therefore, an angle of 60o cannot be trisected using only straightedge and compass.

Let us next turn to the problem of duplicating the cube. For simplicity's sake let us consider a cube C of side 1. We wish to construct a cube of volume 2. This is equivalent to constructing a line segment of length . Let us show that the number is not constructible. Note that

deg(Q()/Q) = 3.

Therefore, by Theorem 7, we see that is not constructible. Thus, we have

Theorem 10: It is impossible to duplicate the cube of side 1 using only a straightedge and compass.

Let us now turn to the problem of squaring the circle. For the sake of simplicity, let us consider the circle of radius 1. It has area equal to . Thus, we are asked to construct a square of area . This is equivalent to constructing a line segment of length . Thus, the question of squaring the circle comes down to: Is constructible. We cannot give a complete proof of the fact that the answer is no. However, let us at least give an indication of the idea involved. If is constructible, then deg(Q()/Q) is a power of 2. In particular, is algebraic over Q by Theorem 5 of the section on algebraic and transcendental numbers. Actually, is transcendental over Q. However, the proof of this last statement is very deep and will not be included here.

Let us now take up the problem of the construction of a regular polygon of n sides. Constructing such a polygon is equivalent to constructing such a polygon is equivalent to constructing an angle of 360/n. But as we saw in the proof of Theorem 9, the angle can be constructed if and only if cos + i sin is constructible. Therefore, a regular polygon of n sides can be constructed if and only if

cos(360/n) + i sin(360/n) = n

is constructible. But n is a primitive nth root of 1 and deg(Q(n)/Q) = (n). Therefore, by Theorem 7, if n is constructible, = 2r for some r. Thus, if a regular polygon of n sides is constructible, then (n) = 2r. Actually the converse is also true. Let us investigate the values of n for which (n) is a power of two. By consulting Table 1, we see that this is not always true. For example (7) = 6, so that it is impossible to construct a regular 7-gon.

Table 1: Values of (n)
 n (n) n (n) 2 1 12 4 3 2 13 12 4 2 14 6 5 4 15 8 6 2 16 8 7 6 17 16 8 4 18 6 9 6 19 18 10 4 20 8 11 10 21 12

Let

n = p1r1· p2r2··· ptrt  (ri > 0)

be the decomposition of n into a product of powers of distinct primes p1,..., pt. Then

(10)
(n) = (p1r1) ··· (ptrt),

so that (n) is a power of 2 if and only if (pjrj) is a power of 2 (1 < j < t). Note that

(pjrj) = pjrj-1(pj - 1).

Therefore, if pj = 2, (pjrj is automatically a power of two. Thus, we are reduced to the following question: Let p be an odd prime, r a positive integer. When is pr-1(p - 1) a power of 2? It is clearly necessary and sufficient that r = 1 and p - 1 = 2k for some k. However, if p - 1 = 2k, then k must be a power of 2. For if k is divisible by an odd prime q, we must have k = qu for some positive integer u. But then

p = 2k + 1
= 2qu + 1
= (2u + 1)(2(q-1)u - 2(q-2)u + .. + 1),

which contradicts the fact that p is prime. Thus, we have shown that if pr-1(p - 1) is a power of 2, then r = 1 and p = 22v + 1. And conversely, if p = 22v + 1, then (p) is a power of 2. A prime p of the form

22v + 1

is called a Fermat prime. The first few Fermat primes after 3 are

221 + 1 = 5,
222 + 1 = 17,
223 + 1 = 257,
224 + 1 = 65,537.

It is not true that every integer of the form 22v + 1 is prime. For as Euler showed, 225 + 1 is divisible by 641. The net result of the above discussion is

Theorem 11: In order for a regular polygon of n sides to be constructible using only straightedge and compass, it is necessary and sufficient for n to be of the form

n = 2rp1p2...pw,

where p1p2...pw are distinct Fermat primes.