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Let beta element of C be constructible. Then there exists a set of complex numbers {alpha1,...,alphan} such that alpha12 element of Q, alphai2 element of Q(alpha1,...,alphai-1)(2 < i < n). and such that beta element of Q(alpha1,...,alphan). Thus, if beta is constructible, deg(Q(beta)/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 alpha·v of a scalar alpha element of F and a vector v element of E is just the product of alpha 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 theta, construct the angle theta/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 alpha is said to be constructible if the line segment connecting (0,0) to alpha 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 square root of L.

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 alpha and beta are constructible, then so are alpha + beta, -beta, alpha · beta, 1/beta (betanot equal0). Let us consider each case separately. Recall that addition of complex numbers can be accomplished geometrically via the parallelogram law. Therefore, let us construct segment from origin to alpha and segment from origin to beta as in Figure 1. Use basic construction 4 above to construct line segment from beta to gamma parallel to segment from origin to alpha On segment from beta to gamma measure off a line segment of length segment from origin to alpha, with one endpoint of the line segment at beta. Then the other end of the measured segment is alpha + beta. Thus, alpha + beta is constructible.

figure 1
Figure 1: Construction of alpha + beta.

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

figure 2
Figure 2: Construction of -beta

Let us construct alpha · beta. Without loss of generality, we may assume that alphanot equal0, betanot equal0. Then alpha and beta can both be written in polar form

alpha = r1(costheta1 + i sintheta1).
beta = r2(costheta2 + i sintheta2).

By de Moivre's theorem,

alpha · beta = r1r2[cos(theta1 + theta2) + i sin(theta1 + theta2)]

Thus, the line segment Oalpha · beta has length r1r2 and makes an angle theta1 + theta2 with the positive X-axis. This is the clue to the construction of alpha · beta. Refer to Figure 3, where alpha and beta have been constructed. Construct the line segment segment from origin to gamma making and angle theta1 + theta2 with the positive X-axis. Measure off a segment segment from origin to delta of length r1r2 on segment from origin to gamma. This is possible by basic construction 4. Then delta = alpha · beta.

figure 3
Figure 3: Construction of alpha · beta

Finally, assume that betanot equal0. Then, since sin2theta2 + cos2theta2 = 1 and cos(-theta2) = costheta2, sin(-theta2) = -sintheta2, we have

1/beta = r2-1(costheta2 + i sintheta2)-1
= r2-1[(costheta2 - i sintheta2)/(cos2theta2 + sin2theta2)]
= r2-1[cos(-theta2) + i sin(-theta2)].

Thus, the line segment Obeta-1 has length r2-1 and makes an angle -theta 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 element of C. But since C is a field, this implies that Q subset of C.

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

Proof: Without loss of generality, assume that alphanot equal0. By de Moivre's theorem,

square root of alpha = ± square root of r1[cos(theta1/2) + i sin(theta1/2)].

Let us only consider the positive sign. The reasoning for the negative sign is similar. Then the line segment Osquare root of alpha is a line segment of length square root of r1 which makes an angle of theta1/2 with the positive X-axis. Therefore, square root of alpha can be constructed using basic constructions 1 and 6.

Theorem 6: Let alpha1,...,alphan be complex numbers such that

alpha12 element of Q,
alphai2 element of Q(alpha1,...,alphai-1)   (2 < i < n).

Proof: By Corollary 4, every element of Q is constructible and by Proposition 5 and the assumption alpha12 element of Q, we see that alpha1 is constructible. Therefore, since the constructible numbers form a field, every element of Q(square root of alpha) 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(alpha1,...,alphan-1) is constructible. By hypothesis, alphan2 element of Q(alpha1,...,alphan-1) so that alphan 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(alpha1,...,alphan-1)(alphan) = Q(alpha1,...,alphan) is constructible.

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

Theorem 7: Let beta element of C be constructible. Then there exists a set of complex numbers {alpha1,...,alphan} such that

alpha12 element of Q,
alphai2 element of Q(alpha1,...,alphai-1)   (2 < i < n).

and such that beta element of Q(alpha1,...,alphan). Thus, if beta is constructible, deg(Q(beta)/Q) = 2k for some k.

Before proceeding with the proof of Theorem 7, let us do some preliminary work. Assume that the points beta1,beta2,...,betam-1 have been constructed. What points can we construct using beta1,beta2,...,betam-1? There are two elementary constructions which we can perform: (a) We can draw a line L connecting betai to betaj (inot equalj). (b) We can draw a circle C with center at betai which passes through betaj (inot equalj). Thus, if betam is constructible from beta1,beta2,...,betam-1, then betam 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)
betaj = gammaj + ideltaj   (gammaj,deltaj element of R, 1 < j < m),
(2)
Fj = Q(i, gamma1, delta1,..., gammaj, deltaj)    (1 < j < m).

We will consider separately three cases:

Case 1: betam 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 element of Fm-1. The system of equations (3) has the solution x = alpham, y = deltam, 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)
alpham element of Fm-1,    deltam element of Fm-1
implies Fm = Fm-1

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

Case 2: betam 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 element of Fm-1. On of the solutions of the system of equations (5) is x = alpham, y = deltam, 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 eta, where eta is computable rationally in terms of a,b,c,d,e,f,g,h. In particular, eta element of Fm-1 and alpham element of Fm-1(square root of eta), deltam element of Fm-1(square root of eta). Further,

(6)
Fm subset of Fm-1(square root of eta),

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

Case 3: betam 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 element of 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 eta in Fm-1 such that

(7)
Fm subset of Fm-1(square root of eta).

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

Lemma 8: Suppose that the points beta1,...,betam-1 (m > 2) have been constructed and that betam is constructible from beta1,...,betam-1. Further, suppose that beta1 = alphaj + i deltaj, where alphaj and deltaj are real, and let

Fj = Q(i,alpha1,delta1,...,alphaj,deltaj)    (1 < j < m.).

Then there exists eta element of Fm-1 such that Fm subset of Fm-1(square root of eta).

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

Let us now prove Theorem 7. Suppose that beta can be constructed by successively constructing beta1,beta2,...,betan = beta, where beta1 is constructed from beta0 = 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 beta0,beta1,...,betam (1 < m < n) to get that there exists etam element of Fm-1 such that

(8)
Fm subset of Fm-1(square root of eta m).    (1 < m < n).

Define

alpha0 = i,  alphaj = square root of eta j    (1 < j < n).

We assert that

(9)
Fm subset of Q(alpha0,alpha1,...,alpham)   (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 subset of Fm-1(alpham)
subset ofQ(alpha0,...,alpham-1)(alpham)
= Q(alpha0,...,alpham).

This completes the induction and hence (9) is proved. It is clear that alpha02 = -1 element of Q. Also, by (9), for 1 < m < n, we have

alpham2 = etam element of Fm-1 subset of Q(alpha0,...,alpham-1).

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

beta = alphan + i deltan element of Fn subset of Q(alpha0,...,alphan).

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

deg(Q(alpha0,...,alphan)/Q) = 2r

for some r. But, if we set tilde Fi = Q(alpha0,...,alphai) (0 < i < n), we have deg( tilde Fi+1/ tilde Fi) = 1 or 2 since alphai+12 element of tilde Fi and tilde Fi+1 = tilde Fi(alphai+1). Moreover,

deg(Q(alpha0,...,alphan)/Q)
= deg( tilde Fn/ tilde Fn-1) · deg( tilde Fn-1/ tilde Fn-2)···deg( tilde F1/ tilde F0)
= 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 theta to be constructible. If theta is constructible, then we may place one side of theta on the X-axis and construct the point where the other side of theta intersects the circle of radius 1 with center at the origin (see Figure 4). But by elementary trigonometry, this point is costheta + i sintheta. Therefore, if angle theta is constructible, the number costheta + i sintheta is constructible. Conversely, if costheta + i sintheta is constructible, then angle theta is constructible (see Figure 4). Therefore, a 20o angle is constructible if and only if

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

is constructible.

Contstuction of theta
Figure 4: Construction of the Angle theta.

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

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

Thus, a contradiction is reached and zeta 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 cube root of 2. Let us show that the number cube root of 2 is not constructible. Note that

deg(Q(cube root of 2)/Q) = 3.

Therefore, by Theorem 7, we see that cube root of 2 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 pi. Thus, we are asked to construct a square of area pi. This is equivalent to constructing a line segment of length square root of pi. Thus, the question of squaring the circle comes down to: Is square root of pi 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 square root of pi is constructible, then deg(Q(square root of pi)/Q) is a power of 2. In particular, square root of pi is algebraic over Q by Theorem 5 of the section on algebraic and transcendental numbers. Actually, square root of pi 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 theta can be constructed if and only if costheta + i sintheta is constructible. Therefore, a regular polygon of n sides can be constructed if and only if

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

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


Table 1: Values of phi(n)
n phi(n) n phi(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)
phi(n) = phi(p1r1) ··· phi(ptrt),

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

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

Therefore, if pj = 2, phi(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 phi(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.