Appendix A Constructibility

This book is dedicated to the synthetic (or axiomatic) approach to geometry, di- rectly inspired and motivated by the work of Greek geometers of antiquity. The Greek geometers achieved great work in geometry; but as we have seen, a few prob- lems resisted all their attempts (and all later attempts) at solution: circle squaring, trisecting the angle, duplicating the cube, constructing all regular polygons, to cite only the most popular ones. We conclude this book by proving why these various problems are insolvable via ruler and compass constructions. However, these proofs concerning ruler and compass constructions completely escape the scope of syn- thetic geometry which motivated them: the methods involved are highly algebraic and use theories such as that of fields and polynomials. This is why we these results are presented as appendices. Moreover, we shall freely use (with precise references) various algebraic results developed in [5], Trilogy II. We first review the theory of the minimal polynomial of an algebraic element in a field extension. Furthermore, since it will be essential for the applications, we prove the Eisenstein criterion for the irreducibility of (such) a (minimal) polynomial over the field of rational numbers. We then show how to associate a field with a given geometric configuration and we prove a formal criterion telling us—in terms of the associated fields—when one can pass from a given geometrical configuration to a more involved one, using only ruler and compass constructions.

A.1 The Minimal Polynomial

This section requires some familiarity with the theory of polynomials in one variable over a field K. The reader not familiar with this material will find some essential aspects of the theory in Appendix A of [5], Trilogy II.

Proposition A.1.1 If K ⊆ L is a field extension, then is a K-vector space.

Proof The scalar multiplication of k ∈ K and l ∈ L is their product in L. 

F. Borceux, An Axiomatic Approach to Geometry, DOI 10.1007/978-3-319-01730-3, 355 © Springer International Publishing Switzerland 2014 356 A Constructibility

Under the conditions of Proposition A.1.1, it thus makes sense to speak of the (vectorial) dimension of L over K, which we shall write as dim[L : K].

Proposition A.1.2 Given finite dimensional field extensions K ⊆ L ⊆ M, one has

dim[M : K]=dim[M : L]×dim[L : K].

Proof Let l1,...,lr be a basis of L as a K-vector space and m1,...,ms a basis of M as an L-vector space. It suffices to observe that the rs elements limj constitute a ∈ basis of M over K. Indeed an element μ M can be uniquely written as j λj mj , ∈ = with λj L, while each λj can itself be uniquely written as λj i ki,j li with ki,j ∈ K. The conclusion follows at once. 

Proposition A.1.3 Given a finite dimensional field extension K ⊆ L, every element l ∈ L is a root of a polynomial with coefficients in K.

Proof Suppose that dim[L : K]=n. Then the n + 1 elements of L

1,l,l2, ..., ln cannot be linearly independent over K, proving the existence of scalars ki ∈ K such that 2 n k01 + k1l + k2l +···+knl = 0. This proves that l is a root of the polynomial

2 n p(X) = k0 + k1X + k2X +···+knX ∈ K[X]. 

The situation described in Proposition A.1.3 has been given a name:

Definition A.1.4 An element l ∈ L in a field extension K ⊆ L is algebraic over K when it is a root of a polynomial with coefficients in K. When all elements of L are algebraic over K, the extension K ⊆ L itself is called algebraic.

The main result of this section is then:

Theorem A.1.5 Consider a field extension K ⊆ L and an element l ∈ L which is algebraic over K. There exists a unique polynomial p(X) ∈ K[X] such that 1. p(l) = 0; 2. the leading coefficient of p(X) is 1; 3. p(X) is irreducible in K[X]. Moreover, under these conditions: 1. p(X) divides in K[X] all other polynomials admitting l as a root; 2. the dimension of the subfield K(l) ⊆ L generated by K and l is equal to the degree of the polynomial p(X). A.1 The Minimal Polynomial 357

The polynomial p(X) is called the minimal polynomial of l over K.

Proof Among all the polynomials in K[X] admitting l as a root, choose a non-zero one whose degree is minimal. Possibly dividing by the leading coefficient (which does not change the roots), we can assume that this leading coefficient is 1. This is our polynomial p(X). If q(X) ∈ K[X] is such that q(l) = 0, the Euclidean division of q(X) by p(X) (see Theorem A.2.1 in [5], Trilogy II) yields

q(X) = p(X)d(X) + r(X), degree r(X)< degree p(X).

Since q(l) = 0 = p(l), this implies r(l) = 0. By minimality of the degree of p(X), we conclude that r(X)= 0. Thus p(X) divides q(X). This also proves that p(X) is irreducible in K[X]. Indeed otherwise l would be a root of a proper factor q(X) of p(X), which is a contradiction since by what we have just proved, p(X) would then have to divide q(X), which is of strictly lower degree. The uniqueness of p(X) also follows immediately. Indeed given another irre- ducible polynomial q(X) ∈ K[X] such that q(l) = 0, p(X) divides q(X) which is irreducible, thus q(X) = kp(X) with k ∈ K. Since both have the same leading coefficient 1, necessarily k = 1. It remains to prove the result concerning dimensions. If p(X) has degree n,the sequence of elements − 1,l,l2, ..., ln 1 is linearly independent over K. Indeed otherwise (as in the proof of Proposi- tion A.1.3) there would be a polynomial of degree n − 1 with coefficients in K[X] admitting l as a root; this would contradict the choice of p(X) with minimal degree. It remains to observe that − 1,l,l2, ..., ln 1 generate K(l) as a vector space over K. Writing

n n−1 p(X) = X + kn−1X +···+k1X + k0,ki ∈ K we get at once n n−1 l =−kn−1l −···−k1l − k0. This already proves that the subset n−1 M = α0 + α1l +···+αn−1l |αi ∈ K ⊆ L is not only stable under addition and subtraction, but also under multiplication, since each power ln after multiplication can be replaced by a combination of strictly lower powers; repeating this operation as many times as needed, one ends up with an expression where all powers of l are at most n − 1, thus with an element of M. 358 A Constructibility

Moreover, the inverse in L of every element m ∈ M is still in M, proving that M is in fact a subfield of L; and then trivially, M = K(l). When m ∈ K, this is trivial. Otherwise, m = q(l) for a non-constant polynomial

n−1 q(X) = β0 + β1(X) +···+βn−1X ,βi ∈ K.

But q(X) and p(X) do not have a common non-constant factor (they are “relatively prime”), because p(X) is irreducible and has a larger degree than q(X).BytheBe- zout theorem (see Theorem A.3.2 in [5], Trilogy II), there exist polynomials r(X), s(X) in K[X] such that

r(X)p(X) + s(X)q(X) = 1 with moreover

degree r(X)< degree q(X), degree s(X) < degree p(X) = n.

Taking the values at l we find s(l)q(l) = 1, proving that s(l) is the inverse of m = q(l). Indeed, s(l) ∈ M since degree s(X) < n. 

Let us conclude with the classical terminology for “non-algebraic” elements:

Definition A.1.6 An element l ∈ L in a field extension K ⊆ L which is not alge- braic over K is said to be transcendental over K.

As an immediate consequence of Proposition A.1.3 we have

Proposition A.1.7 If a field extension K ⊆ L contains a transcendental element over K, its dimension is infinite.

A.2 The Eisenstein Criterion

Most impossibility proofs that we shall consider in Appendices B and C will be in terms of algebraic elements over of the field of rational numbers. Therefore, in order to easily identify the corresponding minimal polynomials, it is important to have an efficient criterion telling us when a polynomial in Q[X] is irreducible (see Theorem A.1.5). Of course, multiplying all the coefficients of a polynomial in Q[X] by a common multiple of their denominators, we obtain a polynomial with coefficients in Z, which still has the same roots. But more amazingly:

Proposition A.2.1 Let p(X) ∈ Z[X] be a polynomial which is irreducible in Z[X]. Then p(X) is also irreducible in Q[X]. A.2 The Eisenstein Criterion 359

Proof It is equivalent to prove that when p[X]∈Z[X] is reducible in Q[X],itis reducible in Z[X]. Suppose therefore that p(X) = r(X)s(X) with r(X) and s(X) two non-constant polynomials in Q[X]. Multiplying by a common multiple n of the denominators of the coefficients of r(X) and s(X), we obtain np(X) = r(X)s(X), where r(X) and s(X) now have their coefficients in Z. Let us write

k l r(X)= rkX +···+r0, s(X) = slX +···+s0,ri,sj ∈ Z.

Since p(X) has coefficients in Z, n divides all the coefficients of np(X). There- fore every prime factor d of n also divides all the coefficients of np(X) = r(X)s(X). Let us prove that d necessarily divides all the coefficients of r(X) or all the coeffi- cients of s(X). We prove this by a reductio ad absurdum. Assuming this is not the case, write respectively i and j for the smallest indices such that d does not divide i+j ri and sj . The coefficient of X in np(X) = r(X)s(X) is

risj + ri−1sj+1 + ri−2sj+2 +··· .

By choice of i and j, the prime number d does not divide the first term of this sum (see Proposition 3.7.5), but divides all the other terms. Thus d does not divide the sum itself. Since this sum is a coefficient of np(X), we have a contradiction. So d divides all the coefficients of one of the two polynomials r(X) or s(X).It is thus possible to simplify d on both sides maintain equality np(X) = r(X)s(X) and still maintain equality between polynomials with coefficients in Z. Repeating the operation for all the prime factors of n, we end up with p(X) presented as the product of two polynomials with coefficients in Z[X]. 

The following criterion (a sufficient condition for being irreducible) will prove to be very useful:

Theorem A.2.2 (The Eisenstein criterion) A polynomial

n n−1 p(X) = anX + an−1X +···+a1X + a0,ai ∈ Z is irreducible in Q[X] as soon as there exists a prime number d such that:

1. d does not divide an; 2. d divides an−1,...,a0; 2 3. d does not divide a0.

Proof Once more we use a reductio ad absurdum. Suppose that p(X) is reducible in Q[X]; then by Proposition A.2.1, p(X) is reducible in Z[X]. Write p(X) = r(X)s(X) with

k l r(X)= rkX +···+r0,s(X)= slX +···+s0,ri,sj ∈ Z,k,l>0.

By assumption, d divides a0 = r0s0, thus divides r0 or s0 (see Proposition 3.7.5). Let us say that d divides r0; the other case is analogous. On the other hand d does not 360 A Constructibility divide an = rksl , thus certainly d does not divide rk. Let us write i for the smallest index such that d does not divide ri ; we therefore have 0

ai = ris0 + ri−1s1 + ri−2s2 +··· .

2 By choice of i, d does not divide ri . And since d divides r0 but d does not divide a0 = r0s0, d does not divide s0. Thus d being prime, it does not divide the first term ris0 of the sum (see Proposition 3.7.5). Since by choice of i, d divides all the other terms of the sum, d does not divide the sum. Since k,l > 0, we also have k,l < n and therefore by assumption, d divides ai . This is the expected contradiction. 

Counterexample A.2.3 The Eisenstein criterion is not a necessary condition for irreducibility in Q[X].

Proof Consider the polynomial

p(X) = X2 + X + 2.

The only prime number dividing a0 = 2 is 2 and it does not divide a1 = 1. Thus the Eisenstein criterion does not apply. Nevertheless, the polynomial p(X) is irreducible over Q[X] because it is irre- ducible over Z[X] (see Proposition 3.11.11). Indeed if the polynomial p(X) was reducible in Z[X], it would be the product of two polynomials of degree 1

X2 + X + 2 = (aX + b)(cX + d) = acX2 + (ad + bc)X + bd, a,b,c,d ∈ Z.

From ac = 1 we get at once a =±1 = c and from bd = 2 we obtain b =±2, d =±1 (or the converse). So in any case we end up with ad + bc =±3, which is a contradiction since a1 = 1. 

A.3 Ruler and Compass Constructibility

This section intends only to make precise some common notions of constructibility in the Euclidean plane R2 (see Example 8.5.3).

Definition A.3.1 A geometric configuration is a finite set of points, straight lines and circles in the Euclidean plane R2. A given point of the configuration, or an intersection point of two distinct lines of the configuration—straight or circular—is called a node of the configuration.

Let us thus make very clear that given a line or a circle in a configuration, the points of this line or this circle are generally not nodes. For example, the config- uration comprising two parallel straight lines has no node at all, while the config- uration comprising two intersecting lines has the corresponding intersection point A.3 Ruler and Compass Constructibility 361 as unique node. If you want additional points on these lines to be nodes of the configuration, you have to include these points at once as given elements of the configuration. Let us also recall that Proposition 5.10.5 indicates that we could as well reduce our attention to configurations comprising only of points and circles; but we shall not make this quite unnatural choice.

Lemma A.3.2 A geometric configuration has finitely many nodes.

Proof Two lines intersect in at most one point, two circles in at most two points, a line and a circle in at most two points. This implies the result, because a configu- ration contains only finitely many points, straight lines and circles. 

The problem of ruler and compass constructions is thus to determine, starting from a given geometric configuration, if it is possible to construct with ruler and compass some specified more involved configuration.

Definition A.3.3 An elementary extension of a given geometric configuration is one which is obtained by adding one line to this configuration, namely: the straight line passing through two of the nodes, or the circle with center some node and pass- ing through some other node. A finite extension of a given geometric configuration is one which can be obtained via a finite sequence of successive elementary exten- sions.

And thus as certainly expected:

Definition A.3.4 A point of the Euclidean plane R2 is constructible from a given geometric configuration when it can be presented as a node in some finite extension of that configuration.

Notice that it follows trivially from Definitions A.3.3 and A.3.4 that: • no point at all is constructible from a geometric configuration without any nodes; • if a configuration has a unique node P , only that point P is constructible from the configuration. So the problem of constructible points with respect to a configuration becomes sen- sible only when there are at least two nodes. Of course, there is no loss of generality in choosing a system of coordinates based on these two nodes. It is therefore conve- nient to use the following terminology:

Definition A.3.5 A geometric configuration in standard position is a geometric configuration, together with the choice of an orthonormal system of Cartesian co- ordinates, so that the two points with coordinates (0, 0) and (1, 0) are nodes of the configuration. 362 A Constructibility

Fig. A.1

We shall often refer to the following:

Definition A.3.6 By the basic geometric configuration we mean the configura- tion, in some orthonormal system of Cartesian coordinates, comprising only the two points (0, 0) and (1, 0).

Notice at once that

Lemma A.3.7 Given a geometric configuration in standard position, the point with coordinates (0, 1) is constructible and the two coordinate axes thus belong to some finite extension of the configuration.

Proof The results of Sect. 3.1—which we freely use without further reference— justify at once the following argument; we refer to Fig. A.1. Write A = (0, 0) and B = (1, 0); the line through A and B is the first coordinate axis. The circle with center A and radius AB cuts this first axis at a second point C. The circle with center C and radius CB cuts the circle with center B and radius BC at two points D and E. The line through D and E is perpendicular to the first axis at A and is therefore the second axis. The circle with center A and radius AB cuts this second axis at two points F and G which have the coordinates (0, ±1). 

It will also be useful to use the following terminology:

Definition A.3.8 A real number r is constructible from a given geometric config- uration in standard position when the point with coordinates (r, 0) is constructible from that configuration.

As expected:

Proposition A.3.9 A point P with coordinates (a, b) is constructible from a given geometric configuration in standard position if and only if both real numbers a and b are constructible from that configuration. A.4 Constructibility Versus Field Theory 363

Fig. A.2

Proof We freely use the results of Sect. 3.1, without explicit mention; we refer to Fig. A.2.GivenP = (a, b), one gets the two points (a, 0) and (0,b) when drawing from P perpendiculars to the axis of coordinates. Drawing a circle with center the origin and passing through (0,b) then yields the intersection (b, 0) with the first axis. Conversely, given the points (a, 0) and (b, 0), a circle with center the origin and passing through (b, 0) yields (0,b) as intersection with the second axis. Next drawing through (a, 0) and (0,b)the perpendiculars to the axis yields as intersection the point P = (a, b). 

A.4 Constructibility Versus Field Theory

The key point in handling problems of constructibility is to associate a field to each geometric configuration.

Definition A.4.1 The field of coordinates of a geometric configuration in standard position is the subfield K of R generated by the Cartesian coordinates of all the nodes of the configuration.

Thus if Ai = (ai,bi), i ∈ I , are the various nodes of the configuration, K is the subfield of the reals generated by all the elements ai and bi . So trivially:

Proposition A.4.2 The field of coordinates of the basic configuration is the field Q of rational numbers.

Observe at once that:

Lemma A.4.3 If K ⊆ R is the field of coordinates of a geometric configuration in standard position, all the numbers k ∈ K are constructible from that configuration (see Definition A.3.8).

Proof By Definition A.3.5,thesetS of real numbers appearing as one of the coor- dinates of a node of the configuration contains already the numbers 0 and 1, whose presence in a subfield is compulsory. The field K is then trivially the set of all real numbers which can be obtained via a finite sequence of arithmetical opera- tions (addition, subtraction, multiplication, division) from the numbers in S. It thus suffices to observe that all four arithmetical operations on real numbers r can be 364 A Constructibility performed via ruler and compass constructions on the corresponding points of coor- dinates (r, 0). This is trivial for addition and subtraction. The cases of multiplication and division follow at once from Proposition 3.6.8: Given three quantities α, β, α, one can construct from these, with ruler and compass, the  α = α quantity β such that β β . =  = ×   =  = β  Choosing α 1 we find β β α and choosing α 1, we get β α .

On the other hand:

Lemma A.4.4 Consider a geometric configuration in standard position and its field K of coordinates. Every line joining two nodes, and every circle centered at some node and passing through another node, admit equations whose coefficients are in K.

Proof The line passing through the points (a, b) and (c, d) admits the equation

(d − b)(X − a) = (c − a)(Y − b) while the circle centered at (a, b) and passing through (c, d) has equation

(X − a)2 + (Y − b)2 = (c − a)2 + (d − b)2.

Since by assumption a, b, c, d are in K, so are all the coefficients of these equa- tions. 

We now arrive at the key result for proving the impossibility of some ruler and compass constructions:

Theorem A.4.5 Let K be the field of coordinates of a geometric configuration in standard position. A real number r is constructible from that configuration if and only if there is a finite sequence of field extensions

K = K0 ⊂ K1 ⊂ K2 ···⊂Kn such that r ∈ Kn and each field Ki+1 has dimension 2 over Ki .

Proof Assume first the existence of a finite sequence of field extensions as in the statement. By Proposition A.1.3, every element r ∈ Ki+1 is algebraic over Ki . Since the field Ki(r) is contained in Ki+1, the dimension of Ki(r) over Ki is at most 2. By Theorem A.1.5, the corresponding minimal polynomial p(X) of r over Ki then has degree at most 2. Trivially, r ∈ Ki if and only if the minimal polynomial has degree 1, that is, if p(X) = X − r. So when r ∈ Ki , the minimal polynomial p(X) has degree 2 and

2 p(X) = aX + bX + c, a,b,c ∈ Ki,a= 0. A.4 Constructibility Versus Field Theory 365

Since r is a root of p(X) √ −b ± b2 − 4ac r = . 2a Since the four arithmetical operations can be performed via ruler and com- pass constructions (see the proof of Lemma A.4.3), as well as square roots (see Lemma 3.2.4), the number r is constructible with ruler and compass from the three numbers a, b, c in Ki . Thus given a number r ∈ Kn, it is constructible with ruler and compass from finitely many numbers in Kn−1; each of these numbers is itself constructible from finitely many numbers in Kn−2 and so on. So eventually, r is constructible with ruler and compass from finitely many numbers in K,thatis,viaLemmaA.4.3, from the nodes of the given configuration. Conversely, consider a finite extension of the given configuration and its field L of coordinates. We must prove the existence of a sequence of field extensions as in the statement, ending with Kn = L. Of course it suffices to prove the same result for every elementary extension (see Definition A.3.3). Furthermore, since in an elementary extension one adds finitely many nodes, it suffices to prove the result for the addition of one single node. We shall consider separately the cases of adding a node which is the intersection of two straight lines, of a straight line and a circle, or of two circles. So let us consider a geometric configuration in standard position with field of co- ordinates M. Two intersecting lines joining nodes of the configuration admit equa- tions of degree 1 whose coefficients are elements of M (see Lemma A.4.4). The solution of the corresponding system of two equations aX + bY = c rX + sY = t is simply det cb det ac = ts = rt X ab ,Y ab . det rs det rs These two quantities lie still in M. Next consider the case of a line joining two nodes and intersecting a circle cen- tered at a node and passing through another node. We have to solve a system of equations aX + bY = c (X − r)2 + (Y − s)2 = t2 = = c ∈ with all coefficients in M (see Lemma A.4.4 again). If a 0, we obtain Y b M. Introducing this value into the second equation, it remains an equation of degree 2 in X with coefficients in M. Solving that equation yields the value of X.Ifa = 0, then = 1 − X a (c bY). Introducing this value into the second equation yields an equation of 366 A Constructibility degree 2 in Y , with all coefficients still in M. Solving this equation yields the value = 1 − of Y , from which by further arithmetical operations we get the value X a (c bY). Thus in order to add to M the coordinates of the intersection points it suffices, if needed, to add the solutions of one equation of degree 2 with coefficients in M. The solutions of an equation

αZ2 + βZ + γ = 0 of degree 2 with coefficients in M are given by the well-known formula −β ± β2 − 4αγ . 2α If l = β2 − 4αγ is in M, the solutions of the equation lie in M as well. Otherwise l is a root of the polynomial Z2 −l2. This polynomial has coefficients in M and is ir- reducible in M[X], because l ∈ M. By Theorem A.1.5, it is the minimal polynomial of l over M. Therefore, still by Theorem A.1.5, M(l) is an extension of degree 2 of M and this extension contains the roots of the given second degree equation. This remains the case for two intersecting circles, both centred at some node and passing through a node. We have this time to solve a system of equations (X − a)2 + (Y − b)2 = c2 (X − r)2 + (Y − s)2 = t2 with coefficients in M (see Lemma 3.11.11). Subtracting these two equations yields

−2aX + 2rX + a2 − r2 − 2bY + 2sY + b2 − s2 = c2 − t2.

This is an equation of degree 1 in X and Y with coefficients in M.Itnowremains to solve the system comprising this equation of degree 1 and one of the two orig- inal equations of degree 2; but this is precisely the second case treated above: the intersection of a line and a circle. 

Corollary A.4.6 Consider a geometric configuration in standard position and its field K of coordinates. If a real number r is constructible from that configuration, then it is algebraic over K with a minimal polynomial whose degree is a power of 2.

Proof By Theorem A.4.5 and adopting the notation used there, r belongs to some n extension Kn of K. By Proposition A.1.2,dim[Kn : K]=2 . By Proposition A.1.3, r is algebraic over K. By Theorem A.1.5, the degree of the minimal polynomial of r over K is equal to dim[K(r) : K]. But since K ⊆ K(r) ⊆ Kn, Proposition A.1.2 implies n 2 = dim[Kn : K]=dim Kn : K(r) × dim K(r) : K . So dim[K(r) : K] is a power of 2 as a divisor of a power of 2. 

Let us highlight the fact that Corollary A.4.6 is only a necessary condition. A.4 Constructibility Versus Field Theory 367

Corollary A.4.7 Consider a geometric configuration in standard position and its field K of coordinates. A number r which is transcendental over K is not con- structible from the given configuration.

Proof This follows by Corollary A.2.2, since a transcendental number over K is non-algebraic.  Appendix B The Classical Problems

This appendix is devoted to proving the impossibility—via ruler and compass constructions—of the three so-called “classical problems” of Greek geometry: du- plicating the cube, trisecting the angle and squaring the circle.

B.1 Duplicating the Cube

See Sect. 2.5 for an historical approach to this problem.

Proposition B.1.1 Duplicating the cube is impossible via ruler and compass con- structions.

Proof Taking as unit length the√ edge of the given cube, we must construct a cube√ of volume 2, that is, of edge 3 2. The question is therefore to determine if 3 2 is constructible from the basic configuration (see Definition A.3.6). The field of coordinates of the basic configuration is Q (see Proposition A.4.2). Choosing d = 2 in the Eisenstein criterion (Theorem A.2.2), the polynomial X3 − 2 is irreducible√ over Q. Therefore, by Theorem A.2.2, X3 − 2 is the mini- 3 mal polynomial of 2 over√ Q. Since the degree 3 of that polynomial is not a power of 2, by Corollary A.4.6, 3 2 is not constructible. 

B.2 Trisecting the Angle

See Sect. 2.3 for an historical approach to this problem.

Proposition B.2.1 Trisecting the angle is impossible via ruler and compass con- structions.

F. Borceux, An Axiomatic Approach to Geometry, DOI 10.1007/978-3-319-01730-3, 369 © Springer International Publishing Switzerland 2014 370 B The Classical Problems

Fig. B.1

Proof Start with the basic configuration (see Definition A.3.6) whose field of coor- dinates is the field Q of rational numbers (see Proposition A.4.2). Draw an equilat- eral triangle ABC on the segment joining A = (0, 0) and B = (0, 1) (see Proposi- tion 3.1.5 and Fig. B.1). The measure of the angle at A is 60 degrees. If this angle can be trisected, the trisector will meet the circle of centre A and radius 1 (that is, the circle passing through B) at the point P with coordinates P = (cos 20◦, ± sin 20◦). To prove the impossibility of this trisection, it thus suffices to prove that cos 20◦ is not constructible over the rationals (see Proposition A.3.9). We start from the well-known trigonometric formula cos 3θ = 4 cos3 θ − 3 cos θ. Choosing θ = 20◦, we conclude that cos 20◦ is a root of the polynomial 1 q(X) = 4X3 − 3X − , 2 ◦ = 1 ◦ just because cos 60 2 . Equivalently, cos 20 is a root of the polynomial 2q(X) = 8X3 − 6X − 1. Perform now a change of variables Y + 1 X = ,Y= 2X + 1. 2 The polynomial 2q(X) transforms as r(Y) = Y 3 + 3Y 2 − 3. Putting d = 3 in the Eisenstein criterion (Theorem A.2.2), we get that r(Y) is irreducible in Q[X]. Therefore 2q(X) is already irreducible in Q[X]. By Theo- rem A.1.5 1 3 1 p(X) = 2q(X) = X3 − X − 8 4 8 is then the minimal polynomial of cos 20◦ over Q. By Corollary A.4.6, since the degree 3 of that polynomial is not a power of 2, cos 20◦ is not constructible.  B.3 Squaring the Circle 371

B.3 Squaring the Circle

See Sect. 2.4 for an historical approach to this problem.

Lemma B.3.1 To prove the impossibility of squaring the circle with ruler and com- pass, it suffices to prove that the number π is transcendental over the field Q of rational numbers.

Proof Squaring√ a circle of radius 1, thus of area π, is the construction√ of a square with side π. The problem thus reduces to proving that π is not constructible from the basic geometric configuration (see Definition A.3.6). But since multipli- cations (see Proposition 3.6.8) and square roots√ (see Proposition 3.2.4) can be per- formed by ruler and compass constructions, π is constructible from the basic con- figuration if and only if π is constructible from that configuration. Now the field of coordinates of the basic configuration is the field Q of rational numbers (see Proposition A.4.2). Thus if π is transcendental over Q, by Corollary A.4.7, it is not constructible with ruler and compass from the basic configuration. 

The first proof of the transcendence of π over the rationals is due to Lindemann, in 1882. Many other proofs are known: some of them are rather elegant, using so- phisticated theories which are definitely beyond the scope of this book; others rely on more elementary but highly technical results. We propose such a technical ele- mentary proof, based only on results explicitly proved in this “trilogy”; in fact our proof is fairly close to that of Lindemann. A key ingredient of the proof will be the theory of symmetric polynomials de- veloped in Appendix E of [5], Trilogy II. A polynomial p(X1,...,Xn) with coeffi- cients in Z is symmetric when it remains unchanged under every permutation of the variables. The “elementary” symmetric polynomials with two variables are

X1 + X2,X1X2, with three variables

X1 + X2 + X3,X1X2 + X1X3 + X2X3,X1X2X3, and so on. The so-called “structural theorem” for symmetric polynomials over the integers attests that

Given a symmetric polynomial p(X1,...,Xn,) with coefficients in Z, there exists a poly- nomial q(X1,...,Xn) with coefficients in Z, with degree less than or equal to the degree of p(X1,...,Xn), such that p(X1,...,Xn) = q σ1(X1,...,Xn),...,σn(X1,...,Xn)

where the σi (X1,...,Xn) are the elementary symmetric polynomials with n variables. (See Theorem E.2.1 and Remark E.2.2 in [5], Trilogy II.)

Theorem B.3.2 (Lindemann) The number π is transcendental over Q. 372 B The Classical Problems

Proof We develop the proof by a reductio ad absurdum: that is, we suppose that π is algebraic over Q and we infer a contradiction. √ We show first that the algebraicity of π forces that of iπ, with i = −1 ∈ C.It suffices to consider the field extensions

Q ⊆ Q(π) ⊆ Q(π)(i).

By assumption, π is algebraic over Q and certainly, i is algebraic over Q(π), since it is root of the polynomial X2 + 1 (see Definition A.1.4). By Theorem A.1.5, both extensions are finite dimensional, thus so too is the composite of these, by Propo- sition A.1.2. Therefore the element iπ ∈ Q(π)(i) is algebraic over Q by Proposi- tion A.1.3. Multiplying the corresponding minimal polynomial by a common mul- tiple of the denominators of its coefficients, we obtain an irreducible polynomial q(X) ∈ Z[X]

n n−1 1 q(X) = anX + an−1X +···+a1X + a0,ai ∈ Z such that q(iπ) = 0 (see Theorem A.1.5 again). Viewed as a polynomial q(X) ∈ C[X], we know that q(X) is a product of factors of degree 1 (see Theorem F.3.3 and Corollary A.6.6 in [5], Trilogy II). Let us write

q(X) = an(X − r1)(X − r2) ···(X − rn), ri ∈ C with r1 = iπ and an ∈ Z. We next use the complex exponential function

+ ea bi = ea(cos b + i sin b), a,b ∈ R.

Let us recall that exey = ex+y for x,y ∈ C (see Proposition F.2.12 in [5], Trilogy II). We have in particular

eiπ = cos π + i sin π =−1 that is er1 + 1 = 0.

This implies of course er1 + 1 · er2 + 1 ··· ern + 1 = 0.

Expanding the product, this can be re-written as erj = 0 J ⊆{1,...,n} j∈J with the usual convention that the empty product equals 1. But the property ex+y = exey of the exponential allows us to further rewrite the equality as e( j∈J rj ) = 0. J ⊆{1,...,n} B.3 Squaring the Circle 373

For every integer m ∈{1,...,n}, we shall now prove the existence of a polyno- mial fm(X) ∈ Z[X] whose roots in C are precisely all the quantities rj ,J⊆{1,...,n}, card(J ) = m, j∈J where card(J ) indicates the number of elements of J . Trivially, when m = 1, it suffices to choose f1(X) = q(X) ∈ Z[X]. We shall write further 1 q(X) = (X − r1)(X − r2) ···(X − rn) = q(X). an The coefficients of q(X) are thus rational numbers; but they are also the elemen- tary symmetric expressions in r1,...rn. This proves that all these expressions are rational numbers. 1 2 − The polynomial f2(X) must admit the 2 (n n) roots    = + rjj rj rj ,j

We consider first the polynomial in C[X] = −  ∈ C[ ] f 2(X) (X rjj ) X j

  rjjj  = rj + rj  + rj  ,j

g(X) = f1(X) · f2(X) ···fn(X) with coefficients in Z, whose roots in C are precisely the quantities rj , ∅=J ⊆{1,...,n}. j∈J 374 B The Classical Problems

Of course it can be the case that some of the roots are equal to 0, that is, that g(X) contains some factors X. Canceling all these factors X, we end up with a polynomial h(X), still with coefficients in Z, and whose roots are the non-zero quantities rj = 0, ∅=J ⊆{1,...,n}. j∈J

Notice that r1 = iπ remains one of the roots of h(X). Thus h(X) has degree k ≥ 1; we shall write it as

k k−1 h(X) = bkX + bk−1X +···+b1X + b0,bi ∈ Z.

Notice that b0 = 0 because 0 is not a root of h(X). Of course bk = 0 since we have called k the degree of h(X). Let us now go back to the equality er1 + 1 · er2 + 1 ··· ern + 1 = e( j∈J rj ) = 0. J ⊆{1,...,n} The quantity j∈J rj is equal to zero for all the zero-roots of g(X), but also when J is the empty set. Since e0 = 1, the equality can therefore be re-written as

l er1 + 1 · er2 + 1 ··· ern + 1 = esj + t = 0(∗) j=1 where 0 = t ∈ N and the sj = 0arethek roots of h(X). Let us now consider an arbitrary prime number p and the corresponding polyno- mial pk−1 p−1 p (bk) X (h(X)) αp(X) = . (p − 1)! Observe that

degree αp(X) = (p − 1) + pk thus its derivative of order p + pk is zero. Adding all the successive derivatives of αp, we define further = +  +  +···+ (p−1+pk) βp(X) αp(X) αp(X) αp(X) α . Then compute the derivative −X  = −X  − =− −X e βp(X) e βp(X) βp(X) e αp(X). This implies X −Y −X −0 −X − e αp(Y ) dY = e βp(X) − e βp(0) = e βp(X) − βp(0). 0 B.3 Squaring the Circle 375

A change of variable Y = XZ in the integral yields 1 −XZ −X − e αp(XZ)X dZ = e βp(X) − βp(0). 0

Multiplying both sides by eX, we obtain 1 X−XZ X −X e αp(XZ) dZ = βp(X) − e βp(0). 0

Choosing successively X = s1,...,sk—the roots of h(X)—and adding the results we obtain k 1 k k sj (1−Z) sj − sj e αp(sj Z)dZ = βp(sj ) − e βp(0) j=1 0 j=1 j=1 k = βp(sj ) + tβp(0) j=1 where 0 = t ∈ N is the natural number appearing in the equality (∗) above. To complete the proof, we shall show that when the prime number p tends to infinity, the left hand side of the equality k 1 k sj (1−Z) − sj e αp(sj Z)dZ = βp(sj ) + tβp(0) (∗∗) j=1 0 j=1 tends to zero while the right hand side is a non-zero integer. This will be the expected contradiction. Let us first handle the case of the right hand side. The index j being fixed, the polynomial h(X) ∈ C[X] contains a factor X − sj , thus the polynomial α(X) ∈ (v) C[X] contains at least p factors X − sj . Therefore all its derivatives αp (X) up to the order p − 1 still contain at least one factor X − sj . This proves that

k (v) α (sj ) = 0, 0 ≤ v

(v) Consider now the derivative αp (sj ) of order v for some v ≥ p. Since h(sj ) = 0, when explicitly computing the derivative, only those terms where the factor (h(X))p has been differentiated at least p times will be non-zero. But these successive deriva- tions will have produced a multiplicative scalar p!, which will thus allow simplifi- (v) cation with the denominator (p − 1)!. As a consequence, αp (sj ) is a polynomial expression in sj , with all coefficients in Z. Moreover, all these coefficients are di- 376 B The Classical Problems

pk−1 visible by p and, of course, also by (bk) . This implies that the expression

k 1 δ (s ,...,s ) = α(v)(s ) p,v 1 k pk−1 p j p(bk) j=1 has coefficients in Z and, trivially, is symmetric with respect to s1,...,sk. Since the degree of α is (p − 1) + pk and v ≥ p, the degree of the corresponding polynomial δp,v(X1,...,Xn) is at most pk − 1. Let us now (once more) apply to this polynomial δp,v the structural theorem for symmetric polynomials recalled in the comments preceding the present Theorem. The elementary symmetric expressions in s1,...,sk are the coefficients of the poly- nomial 1 h(X) = (X − s1) ···(X − sk). bk

Thus δp,v(s1,...,sk) can be written as a polynomial expression of degree at most b pk − 1, with coefficients in Z, in terms of the j . bk b1 bk δp,v(s1,...,sn) = εp,v ,..., . bk bk

Since εp,v has degree at most pk − 1, pk−1 b1 bk bk εp,v ,..., bk bk is an integer and therefore

k (v) = pk−1 = b1 bk = αp (sj ) p(bk) δp,v(s1,...,sk) pεp,v ,..., pϕv bk bk j=1 for some integer ϕv ∈ Z. Let us now study the situation at X = 0. It follows at once from the definition of α(X) that ⎧ ⎨⎪0 when vp− 1. As a consequence

k k p−1+pk p−1+pk + = (v) + (v) βp(sj ) tβp(0) αp (sj ) t αp (0) j=0 j=0 v=1 v=1 B.3 Squaring the Circle 377

p−1+pk pk−1 p = pϕv + t(bk) (b0) v=1 pk−1 p = ϕp + t(bk) (b0) ,ϕ∈ Z.

Since k, bk and b0 are non-zero, when the prime number p is such that p>max t,|bk|, |b0| ,

pk−1 p then p cannot possibly divide t(bk) (b0) and therefore does not divide ϕp + pk−1 p t(bk) (b0) either. In particular, this last quantity cannot be zero. We have thus proved that for a sufficiently large value of p

pk−1 p 0 = ϕp + t(bk) (b0) ∈ Z.

We now have to take care of the left hand side of the equality (∗∗). The quantity ψj = sup h(sj Z) ∈ R 0≤Z≤1 is well-defined as the supremum of the continuous function |h(sj Z)| on the compact interval [0, 1]. We then have | |pk−1| |p−1 p bk sj (ψj ) αp(sj Z) ≤ . (p − 1)! On the other hand the quantity 1 k s ( −Z) η = max e j 1 dZ ∈ R j=1 0 is also well-defined, as the supremum of a finite family of definite integrals of con- tinuous functions on a compact interval. But then k 1 k 1 |b |pk−1|s |p−1(ψ )p sj (1−Z) sj (1−Z) k j j sj e αp(sj Z)dZ ≤ sj e dZ (p − 1)! j=1 0 j=1 0 k |b |pk−1|s |p−1(ψ )p 1 k j j sj (1−Z) ≤ sj e dZ (p − 1)! j=1 0

k pk−1 p−1 p |bk| |sj | (ψj ) η ≤ sj . (p − 1)! j=1

From the well-known fact that an lim = 0,n∈ N n→∞ n! 378 B The Classical Problems for every real number a, the last upper bound on the right hand side does indeed tend to zero as p tends to infinity. 

Putting together Lemma B.3.1 and Theorem B.3.2 we conclude that

Corollary B.3.3 Squaring the circle is impossible via ruler and compass construc- tions. Appendix C Regular Polygons

This section is devoted to proving the so-called GaussÐWantzel theorem, which tells us which regular polygons can be constructed with ruler and compass. In 1801, Johann Carl Friedrich Gauss (1777Ð1855) gave a sufficient condition on n for the constructibility of the regular n-gon with ruler and compass. Gauss then conjectured that his condition was also necessary; this was proved in 1837 by Pierre Laurent Wantzel (1814Ð1848). It should nevertheless be made clear that the constructibility condition makes an essential use of the so-called Fermat primes, the prime numbers n of the form 22 + 1, and today, only five such numbers are known! We propose a proof based on field theory, with some flavor of Galois theory.

C.1 What the Greek Geometers Knew

This section lists some elementary results, already known to the Greek geometers.

Proposition C.1.1 If the regular n-gon is constructible with ruler and compass and m ≥ 3 divides n, then the regular m-gon is constructible as well.

= n Proof Construct the regular n-gon with successive vertices S0 to Sn−1. Put k m and join the vertices S0, Sk, S2k, and so on. 

Proposition C.1.2 Let n and m be two relatively prime numbers. If the regular n- gon and m-gon are constructible with ruler and compass, then the regular nm-gon is constructible as well.

Proof By the Bezout theorem, there exist integers r and s such that rn + sm = 1 (see Corollary 3.7.4). Therefore

1 1 1 2π 2π 2π = r + s and thus = r + s . nm m n nm m n

F. Borceux, An Axiomatic Approach to Geometry, DOI 10.1007/978-3-319-01730-3, 379 © Springer International Publishing Switzerland 2014 380 C Regular Polygons

This last formula gives us the angle at the center of the circle intercepting the side of the regular nm-gon, in terms of the corresponding angles for the n-gon and the m-gon. 

Proposition C.1.3 If the regular n-gon is constructible with ruler and compass, so is the regular (2kn)-gon, for every natural number k.

Proof Doubling the number of sides corresponds to bisecting the angle at the center of the circle, intercepting the side of the regular polygon. This can be done with ruler and compass (see Proposition 3.1.11). 

The Greek geometers were able to construct the equilateral triangle (n = 3; see Proposition 4.6.3) and the regular pentagon (m = 5, see Sect. 2.2). By Proposi- tion C.1.2 they could thus construct the regular 15-gon and by Proposition C.1.3, the regular 2k × 3 -gon, 2k × 5 -gon, 2k × 15 -gon,k∈ N.

The Greeks did not discover any of the other constructible regular polygons. The next progress occurred only in 1796 when Gauss—who was then 19 years old—proved the constructibility of the regular heptadecagon (the 17-gon). Gauss proved that 2π 1 √ √ cos = −1 + 17 + 34 − 2 17 17 16 √ √ √ + 2 17 + 3 17 − 34 − 2 17 − 2 34 + 2 17 .

Since arithmetical operations and square roots are constructible with ruler and com- 2π pass (as recalled in the proof of Theorem A.4.5), the angle 17 and thus the hep- tadecagon are constructible with ruler and compass. Five years later, Gauss gener- alized this result as the sufficient condition in our Theorem C.6.4.

C.2 The Problem in Algebraic Terms

Let us think in terms of Cartesian coordinates in R2. We want to inscribe a regular n-gon in the circle of radius 1 centered at the origin. If we choose (1, 0) as first vertex, the next vertex will have the coordinates 2π 2π cos , sin . n n C.2 The Problem in Algebraic Terms 381

If now we view R2 as the usual representation of the field of complex numbers, this second vertex is the representation of the complex number

2π 2π ζ = cos + i sin . n n n The constructibility of the regular n-gon then becomes, in the spirit of our criterion Theorem A.4.5:

Proposition C.2.1 The regular n-gon (n ≥ 3) is constructible with ruler and com- pass if and only if, given the complex number

2π 2π ζ = cos + i sin ∈ C, n n n there exists a finite chain of field extensions

Q = K0 ⊂ K1 ⊂ K2 ⊂···⊂Km such that ζn ∈ Km and each extension Kn ⊂ Kn+1 is 2-dimensional.

2π Proof The regular n-gon is constructible precisely when the angle n is con- structible, that is when the point 2π 2π cos , sin n n is constructible. By Theorem A.4.5, we thus have field extensions 2π 2π 2π Q ⊂ K ⊂···⊂Q cos ⊂ K ⊂···⊂Q cos sin 1 n j n n where each individual√ extension has dimension 2. Consider now i = −1 ∈ C. Since i is not a real number, 2π 2π i ∈ K[ ] = Q cos sin ⊆ R. n n n

2 But i is root of the polynomial X + 1 with coefficients in K[n]. This polynomial 2 X + 1 is irreducible over K[n] since its two roots i and −i are not in K[n]. Thus by 2 Theorem A.1.5, X +1 is the minimal polynomial of i over K[n]. By Theorem A.1.5 we thus have a further extension 2π 2π K[ ] ⊂ K[ ](i) = Q cos sin (i) n n n n of dimension, and of course, ζn belongs to this last extension. 382 C Regular Polygons

Conversely, suppose now that ζn ∈ Km, with Km as in the statement. It is imme- diate that 1 2π 2π = cos − i sin ζn n n and since K is a subfield of C, 1 ∈ K . Therefore m ζn m

1 2π ζn + = 2 cos ∈ Km ζn n

2π ∈ 2π and finally, cos n Km. By Theorem A.4.5, cos n is constructible, thus so too is 2π  the angle n . Therefore the regular n-gon is constructible.

Corollary C.2.2 Let n ≥ 3 be a natural number such that the regular n-gon is constructible with ruler and compass. Then the degree of the minimal polynomial of ζn over Q is a power of 2.

Proof With the notation of Proposition C.2.1,wehave

Q ⊆ Q(ζn) ⊆ Km.

By Proposition A.1.2 dim Km : Q(ζn) × dim Q(ζn) : Q = dim[Km : Q].

Since this last dimension is a power of 2, the same holds for dim[Q(ζn) : Q].By Theorem A.1.5, this is the degree of the minimal polynomial of ζn over Q. 

C.3 Fermat Primes

Since every natural number factors as a product of prime numbers, let us first inves- tigate the case of a regular p-gon, with p a prime number. Notice that since there is no ‘2-gon’, we will not lose anything if we assume that p is odd. In view of Corollary C.2.2, we should first calculate the minimal polynomial of

2π 2π ζ = cos + i sin . p p p

Proposition C.3.1 Let p be a prime number. The minimal polynomial of ζp ∈ C over Q is

p−1 p−2 αp(X) = X + X +···+X + 1. C.3 Fermat Primes 383

Proof Observe immediately that Xp − 1 αp(X) = . X − 1 Since (see Corollary F.2.7 in [5], Trilogy II) 2πp 2πp ζ p = cos + i sin = cos 2π + i sin 2π = 1 p p p it already follows that αp(ζp) = 0. By Theorem A.1.5, it remains to prove that αp(X) is irreducible over Q. We shall prove this via a reduction to the Eisenstein criterion Theorem A.2.2. The polynomial αp(X) is irreducible over Q if and only if the polynomial

β(Y) = αp(1 + Y) is irreducible over Q.But (1 + Y)p − 1 β(Y) = = Y p−1 + γ(Y) Y where γ(Y)has degree p − 2. The coefficients of γ(Y)are the binomial coefficients − − ··· − + p = p(p 1)(p 2) (p k 1) k k(k − 1)(k − 2) ···1 for 0

Corollary C.3.2 Let p be an odd prime number. If the regular p-gon is con- structible, then p = 2n + 1 for some n ∈ N.

Proof Putting together Corollary C.2.2 and Proposition C.3.1, we get that p − 1 must be a power of 2 (see Proposition C.3.4 and Definition C.3.3). 

Definition C.3.3 A Fermat prime is an odd prime number of the form p = 2n + 1, for some natural number n ∈ N.

As their name suggests, the prime numbers involved in Definition C.3.3 were first studied by Pierre de Fermat (1605Ð1665). An important point to observe is:

m Proposition C.3.4 Every Fermat prime has the form 2(2 ) + 1. 384 C Regular Polygons

Proof Let p = 2n + 1 be a Fermat prime. We use a reductio ad absurdum. Suppose that n admits an odd divisor a ≥ 3; let us say, n = ab, so that b

But since a is odd − − − ta + 1 = (t + 1) ta 1 − ta 2 + ta 3 −···−t + 1 .

This proves, t = 2b, that p is divisible by 2b + 1, where 1 < 2b + 1 < 2n + 1 = p. This is a contradiction, because p is prime. 

Having observed that

0 2(2 ) + 1 = 3

1 2(2 ) + 1 = 5

2 2(2 ) + 1 = 17

3 2(2 ) + 1 = 257

4 2(2 ) + 1 = 65537

n are all prime, Fermat conjectured that all numbers of the form 2(2 ) + 1 are prime. This conjecture was disproved in 1732 by Leonhard Euler (1707Ð1783) who found that 5 2(2 ) + 1 = 4 294 967 297 = 641 × 6 700 417. The only Fermat primes which are known at the time of writing are the five numbers n already known to Fermat. It has been checked that all the numbers 2(2 ) + 1 with 5 ≤ n ≤ 32 (and also many others) are composite. But nothing is known about the possible existence of large Fermat primes.

C.4 Elements of Modular Arithmetic

The following example of a field is probably well-known to the reader.

Proposition C.4.1 Let n ≥ 2 be a natural number. 1. The set

Zn ={0, 1, 2,...,n− 1} becomes a commutative ring with unit when provided with the operations

• a +n b is the remainder upon division of a + b by n; C.4 Elements of Modular Arithmetic 385

• a ×n b is the remainder upon division of a × b by n. (see Proposition 3.7.2). 2. Equivalently, Zn can be defined as the quotient of Z by the equivalence relation a ≈ b iff ∃z ∈ Z a − b = zn.

In that case, writing [a] for the equivalence class of a

[a]+n [b]=[a + b], [a]×n [b]=[a × b]. 3. The quotient map

Z −→ Zn,a→ [a] is a ring homomorphism. 4. When n is a prime number, (Zn, +n;×n) is a field.

Zn is called the ring (or field, when n is prime) of integers modulo n.

Proof Assertions 1, 2, 3 in the statement are straightforward to check. Let us prove the existence of inverses when n is prime. Given 1 ≤ a

∃u, v ∈ Z ua + vp = 1.

Then by assertion 3

[u]×n [a]=[1] and [u] is the inverse of [a] modulo p. 

Corollary C.4.2 Let n = 0 be a natural number. There exists a ring homomorphism

Z[X]−→Zn[X],q(X)→ q(X) obtained when replacing each coefficient b of a polynomial q(x) ∈ Z[X] with the corresponding coefficient [b] modulo n (see Proposition C.4.1.2).

Proof The addition or multiplication of two polynomials is performed via additions and/or multiplications of their coefficients. By assertion 3 in Proposition C.4.1, these operations are preserved by the quotient modulo n. 

From now on, when using operations modulo n, we shall simply write + and × instead of +n and ×n.

Proposition C.4.3 When p is a prime number Z∗ ={ ∈ Z | = } p x p x 0 is an abelian group with respect to multiplication modulo p. 386 C Regular Polygons

Proof This follows by assertion 4 in Proposition C.4.1. 

Theorem C.4.4 (Fermat’s little theorem) Let p be a prime number. For every ∈ Z∗ p−1 = number a p, a 1 modulo p. ∈ Z∗ Proof Each a p is invertible modulo p (see Proposition C.4.3). Multiplication 1 by a and a modulo p then induce inverse bijections between the sets Zp ={1, 2,...,p− 1}, a, 2a, ..., (p− 1)a . − Z∗ The second set is thus comprised of p 1 distinct elements of p, which itself has − Z∗ p 1 elements: thus this second set is equal to p. Therefore the two sets are equal and multiplying together the elements in each one of these sets, we get the following equality in Zp: − (p − 1)!=ap 1 · (p − 1)!.

Since Zp is a field (see Proposition C.4.1 again), we can simplify successively by p−1 1, 2,...,p− 1 to get eventually a = 1inZp. 

In view of Theorem C.4.4, it now makes sense to define:

∈ Z∗ Definition C.4.5 Let p be a prime number. The order of an element a p is the smallest exponent k = 0 such that ak = 1 modulo p.

∈ Z∗ Proposition C.4.6 Let p be a prime number. The order k of an element a p is a divisor of p − 1.

Proof The subset i a= a ∈ Zp|1 ≤ i ≤ k Z∗ × = k is a subgroup of ( p, ): indeed it contains 1 a , it is trivially stable under mul- tiplication, and the inverse of ai is ak−i . It follows at once that b b ≈ c iff ∈a c Z∗   is an equivalence relation on p. The subgroup a has precisely k elements. ∈ Z∗ Given b p, the two mappings y a→[b],x→ bx;[b]→a,y→ b are inverse bijections between a and the equivalence class [b] of b. Thus all equiv- alence classes have the same number k of elements. Since these classes constitute a Z∗ − − = partition of p, which has p 1 element, we get p 1 lk, where l is the number of equivalence classes.  C.5 A Flavour of Galois Theory 387

Theorem C.4.7 Let p = 2n + 1 be a Fermat prime. Then the multiplicative group Z∗ − p has an element of order p 1.

Z∗ i Proof By Proposition C.4.6, the order of an element of p has the form 2 for i 0 ≤ i ≤ n. An element a of order 2i with i ≤ n − 1 is such that a(2 ) = 1 modulo p, n−1 thus a fortiori such that a(2 ) = 1. All these elements of order at most 2n−1 are n−1 thus roots of the polynomial X(2 ) − 1. But a polynomial of degree 2n−1 over a field has at most 2n−1 roots (see Corollary A.6.7 in [5], Trilogy II). Thus there are n−1 n−1 Z∗ at most 2 elements of order at most 2 . The remaining elements of p—thus at least 2n−1 other elements—then necessarily have the order 2n. 

C.5 A Flavour of Galois Theory

The reader familiar with the Galois theory of fields will immediately notice that this section proves (part of) the Galois theorem in a very particular case, where the proof becomes very easy. We keep writing

2π 2π ζ = cos + i sin ∈ C. p p p

Proposition C.5.1 Let p be a prime number. 1. The elements 2 p−1 ζp,ζp, ..., ζp constitute a basis of Q(ζ ) as a Q-vector space. 2. The elements of this basis are precisely the p − 1 roots of αp(X) in C. 3. These elements are also all the p-th roots ζ = 1 of 1 in C. ∈ Z∗ 4. For every k p, the operation

i → ki ζp ζp

is a permutation of the elements of this basis.

Proof Since αp(X) is the minimal polynomial of ζp, we know by the proof of The- p−2 orem A.1.5 that 1,ζp,...,ζp constitute a basis of Q(ζp) as a vector field over Q. But since ζp is a root of αp

p−1 1 =−ζp −···−ζp.

Therefore p−1 p−2 −ζp −···−ζp,ζp,..., ζp 388 C Regular Polygons is a basis of Q(ζp). This immediately implies that the elements

p−2 p−1 ζp, ..., ζp ,ζp generate Q(ζp), which is of dimension p − 1. Thus necessarily, these p − 1 gener- ators constitute a basis of Q(ζp) as a Q-vector space. ∈ Z∗ By Proposition C.4.3,givenk p, multiplication by k modulo p is a permuta- Z∗ 1 tion of p, whose inverse is multiplication by k modulo p. In view of that permutation, the two sums p−1 +···+ + k p−1 +···+ k + ζp ζp 1 and ζp ζp 1 are comprised of the same terms, possibly in a different order. Thus, since the first k = − sum is zero, so is the second. This proves that αp(ζp) 0. Since we have p 1 such k elements ζp , they are all the roots of αp(X) (see Corollary F.2.9 in [5], Trilogy II). It is well-known (see Corollary A.6.7 in [5], Trilogy II) that the p-th roots of 1 are 2kπ 2kπ ζ k = cos + i sin 0 ≤ k ≤ p − 1 p p p from which the third assertion follows immediately. 

Proposition C.5.2 Let p be a prime number and k a divisor of p − 1. The mapping

fk : Q(ζp) → Q(ζp), p−1 +···+ → k(p−1) +···+ k ∈ Q ap−1ζp a1ζp ap−1ζp a1ζp,ai is a field homomorphism.

Proof The mapping fk trivially preserves 0, 1, addition and subtraction. Now given two polynomials

p−1 p−1 s(X) = ap−1X +···+a0,t(X)= bp−1X +···+b0 in Q(X), divide their product by αp(X) (see Theorem A.2.1 in [5], Trilogy II):

s(X)t(X) = αp(X)q(X) + r(X), degree r(X)

k By Proposition C.5.1, both ζp and ζp are roots of αp(X). Thus fk s(ζp)t(ζp) = fk r(ζp) = k r ζp = k k s ζp t ζp = fk s(ζp) fk t(ζp) . C.5 A Flavour of Galois Theory 389

The preservation of inverses follows at once from their uniqueness and the preser- vation of the multiplication (a ring homomorphism between fields is a field homo- morphism). 

Proposition C.5.3 Let p be a prime number. For every field homomorphism

f : Q(ζ ) −→ Q(ζp) the elements Fix(f ) = x ∈ Q(ζp)|f(x)= x constitute an intermediate field Q ⊆ Fix(f ) ⊆ Q(ζp).

Proof Fix(f ) is stable under all four arithmetical operations simply because f pre- serves them. 

Now we present the “Galois-type” result for studying the constructibility of a p-gon.

= n + ∈ Z∗ Theorem C.5.4 Let p 2 1 be a Fermat prime. Let k p be an element of order p − 1(see Theorem C.4.7). For every 1 ≤ m ≤ n and with the notation of Proposition C.5.2: : Q = m dim Fix(fk(2m) ) 2 .

Proof By assertions 1 and 4 in Proposition C.5.1, the elements k 1 k p−1 ζp , ..., ζp constitute a basis of Q(ζp) as a Q-vector space. An element b ∈ Q(ζ ) can thus be written in a unique way as = k p−1 +···+ k ∈ Q b ap−1 ζp a1ζp,ai .

Again by assertion 4 in Proposition C.5.1, fk(2m) induces a permutation of the set k p−1 k ζp , ..., ζp .

The element b is thus fixed by fk(2m) precisely when, for every index i, the coeffi- cient ai is equal to the coefficient aik(2m) . − = n Z∗ (2m) Since k is an element of order p 1 2 in p, k is an element of order (n−m)  (2m)⊆Z∗ 2 . As observed in the proof of Proposition C.4.6, the subgroup k p m generated by k(2 ) has 2(n−m) elements, as do each of the corresponding 2m equiva-  (2m)⊆Z∗ ∈ Z∗ ∈ Q lence classes i k p, for all i p. Thus b (ζp) is fixed by fk(2m) when, (2m) for each index i, all the coefficients ai for i in the equivalence class ik  are m equal. Since there are 2 such equivalence classes, Fix(fk(2m) ) is isomorphic to 390 C Regular Polygons the families of 2m rational numbers (one for each equivalence class) and therefore [ : Q]= m  dim Fix(fk(2m) ) 2 .

The reader familiar with Galois theory should keep in mind that an element b ∈ Q (ζp) fixed by fk(2m) is fixed by the group of all powers of this homomorphism, which has 2n−m elements. On the other hand we are in characteristic zero and every field homomorphism fixes the rational numbers.

C.6 The GaussÐWantzel Theorem

We now have all the necessary tools for characterising those regular polygons which are constructible with ruler and compass.

Proposition C.6.1 If p = 2n + 1 is a Fermat prime, then the regular p-gon is constructible with ruler and compass.

− Z∗ Proof We refer to Theorem C.5.4 and choose an element k of order p 1in p. Trivially,  m ≤ m =⇒ Fix(f (2m) ) ⊆ Fix(f  ). k k2(2m ) We thus obtain a chain of subfields

Q = Fix(f 0 ) ⊆ Fix(f 1 ) ⊆···⊆Fix(f ( n) ) = Q(ζ ). k(2 ) k(2 ) k 2 p The first equality holds because 0 Fix(f 0 ) : Q = = , dim k(2 ) 2 1

Fix(f 0 ) = Q thus k(2 ) . The last equality holds because : Q = n = Q : Q dim Fix(fk(2n) ) 2 dim (ζp) = Q implies Fix(fk(2n) ) (ζp). Some readers may be puzzled by the fact that, in the proof of Theorem C.5.4,we (20) = have instead proved that the elements of Fix(fk ) Fix(fk) are those of the form k p−1 +···+ k ∈ Q a ζp ζp ,a . But do not forget that k p−1 +···+ k =− ζp ζp 1 k because ζp is a root of αp(X) (see Proposition C.5.1.2). For the last equality, one (2n) = could equivalently have observed that fk(2n) is the identity mapping, since k 1 modulo p. C.6 The GaussÐWantzel Theorem 391

Now by Proposition A.1.2 and Theorem C.5.4 again Fix(f i+1 ) : Q = Fix(f i+1 ) : Fix(f i ) × Fix(f i ) : Q dim k(2 ) dim k(2 ) k(2 ) dim k(2 ) thus + [Fix(f i+1 ) : Q] i 1 dim k(2 ) 2 Fix i+1 : Fix i = = = dim (f (2 ) ) (f (2 ) ) i 2. k k [Fix i : Q] dim (fk(2 ) ) 2 The proof is completed by Proposition C.2.1. 

The last step towards the solution of the problem is to investigate the con- structibility of a p2-gon.

Proposition C.6.2 Let p be a prime number. The minimal polynomial of ζp2 over Q is − − γ(X)= Xp(p 1) + Xp(p 2) +···+X + 1.

Proof We write as usual 2π 2π ζ 2 = cos + i sin . p p2 p2 Observe at once that

2 (Xp)p − 1 X(p ) − 1 γ(X)= = . Xp − 1 Xp − 1 By Corollary F.2.7 in [5], Trilogy II,

2 2 2 2πp 2πp ζ p = cos + i sin = cos 2π + i sin 2π = 1. p2 p2 p2 On the other hand 2πp 2πp 2π 2π ζ p = cos + i sin = cos + i sin = 1. p2 p2 p2 p p = It follows at once that γ(ζp2 ) 0. By Theorem A.1.5, it remains to prove that γ(X) is irreducible over Q. We shall prove this via a reduction to the Eisenstein criterion Theorem A.2.2. The polynomial γ(X)is irreducible over Q if and only if the polynomial

δ(Y) = γ(1 + Y) is irreducible over Q. As in the proof of Proposition C.3.1, considering the form of the binomial coefficients, we get

2 2 2 (1 + Y)(p ) − 1 (1 + pε(Y)+ Y (p )) − 1 pε(Y)+ Y (p ) δ(Y) = = = (1 + Y)p − 1 (1 + pϕ(Y)+ Y p) − 1 pϕ(Y)+ Y p 392 C Regular Polygons where ε(Y) and ϕ(Y) are polynomials with coefficients in N. Passing to the corre- sponding polynomials modulo p (see Corollary C.4.2) we obtain

2 Y (p ) δ(Y) = = Y p(p−1). Y p This proves that all the coefficients of δ(Y)—except the first one—are equal to zero modulo p, thus are multiples of p. Therefore

δ(Y) = Y p(p−1) + pψ(Y) where ψ(Y)is a polynomial with coefficients in N. Observe further that the constant term of − δ(Y) = (1 + Y)p(p 1) +···+(1 + Y)p + 1 is − 1p(p 1) +···+1p + 1 = 1 +···+1 + 1 = p. So Eisenstein’s criterion (Theorem A.2.2) applies and δ(Y)—thus also γ(X)—is irreducible. 

Corollary C.6.3 If p is an odd prime number, then the regular p2-gon is not con- structible with ruler and compass.

− Proof By Lemma C.6.2, the minimal polynomial of ζp2 has degree p(p 1), which is not a power of 2, since p is odd. The result follows by Corollary C.2.2. 

We are now ready to conclude:

Theorem C.6.4 (GaussÐWantzel) The regular n-gon (n ≥ 3) is constructible with ruler and compass if and only if n is of the form

k n = 2 p1 ···ps,k∈ N,s∈ N where the pi are distinct Fermat primes.

Proof If s = 0—that is, n = 2k—then k ≥ 2 because n ≥ 3. The square is con- structible with ruler and compass (see Proposition 3.1.43); and the result follows by an iterative application of Proposition C.1.3. If s = 0, for each Fermat prime pi in the decomposition of n, the regular pi -gon is constructible with ruler and compass, by Proposition C.6.1. Since these various pi ’s are distinct, each of them is relatively prime with every product of some (or all) of the others. By an iterative application of Proposition C.1.2, the regular (p1 ···ps)- gon is constructible. This proves the sufficiency of the condition on n. C.6 The GaussÐWantzel Theorem 393

To prove the necessity of the condition, consider the decomposition of n into prime factors (see Corollary 3.7.9)

= k m1 ··· ms ∈ N n 2 p1 ps , k,s,mi where p1,...,ps are distinct odd prime numbers. If mi ≥ 2forsomei, then the 2 regular pi -gon is constructible by Proposition C.1.1; this is impossible, as attested by Corollary C.6.3. Thus all the exponents mi are necessarily equal to 1. Again by Proposition C.1.1,ifsomepi appears in the decomposition of n, then the regular pi -gon is constructible and therefore, by Corollary C.3.2, pi is a Fermat prime.  References and Further Reading

1. S.K. Adhikari, Babylonian mathematics. Indian J. Hist. Sci. 33, 1 (1998) 2. E. Artin, Algèbre Géométrique (Gauthiers-Villars, Paris, 1978) 3. F. Ayres, Projective Geometry. Schaum’s Outline Series (McGraw-Hill, New York, 1967) 4. F. Borceux, Invitation à la Géométrie (CIACO, Louvain-la-Neuve, 1986) 5. F. Borceux, An Algebraic Approach to Geometry. Geometric Trilogy II (Springer, Berlin, 2014) 6. F. Borceux, A Differential Approach to Geometry. Geometric Trilogy III (Springer, Berlin, 2014) 7. C.B. Boyer, A History of Mathematics (Wiley, New York, 1968) 8. G. Choquet, L’Enseignement de la Géométrie (Hermann, Paris, 1964) 9. J.P. Collette, Histoire des Mathématiques, vol. 2 (Editions du Renouveau pédagogique, Mon- tréal, 1973Ð1979) 10. H. Coxeter, Non-Euclidean Geometry (University of Toronto Press, Toronto, 1942) 11. H. Coxeter, Introduction to Geometry (Wiley, New York, 1961) 12. H. Coxeter, The Real Projective Plane (Cambridge University Press, Cambridge, 1955) 13. H. De Sloover, Cours de Géométrie Descriptive; Méthode des Plans Côtés (De Boeck, Paris, 1969) 14. H. Davenport, The Higher Arithmetic: An Introduction to the Theory of Numbers, 7th edn. (Cambridge University Press, Cambridge, 1999) 15. J. Dieudonné, Abrégé d’Histoire des Mathématiques, vol. 2 (Hermann, Paris, 1978) 16. F. Enriques, Fragen der Elementargeometrie II (Teubner, Leipzig, 1907) 17. Euclid, Elements of Geometry. Euclidis Elementa (Teubner, Leipzig, 1885). Greek text of J.L. Heiberg, English translation by R. Fitzpatrick 18. D. Gans, An Introduction to Non-Euclidean Geometry (Academic Press, San Diego, 1973) 19. M.C. Gemignani, Axiomatic Geometry (Addison-Wesley, Reading, 1971) 20. K. Gödel, Über formal Unentscheidbare Sätze der Principia Mathematica und Verwandter Systeme, I. Monatshefte Math. Phys. 38, 173Ð198 (1931) 21. M.J. Greenberg, Euclidean and Non-Euclidean Geometries (Freeman, New York, 1974) 22. R. Hartshorne, Foundations of Projective Geometry (Benjamin, Elmsford, 1967) 23. R. Hartshorne, Geometry: Euclid and Beyond (Springer, Berlin, 1997) 24. A. Heyting, Axiomatic Projective Geometry (North-Holland, Amsterdam, 1963) 25. D. Hilbert, Foundations of Geometry (Open Court, La Salle, 1971) 26. H. Jacobs, Geometry (Freeman, New York, 1974) 27. A.P. Juschkewitsch, Geschichte der Mathematik Im Mittelalter (Teubner, Leipzig, 1964) 28. F. Karteszi, Introduction to Finite Geometries (North Holland, Amsterdam, 1976) 29. B. Kerekjarto, Les Fondements de la Géométrie. Géométrie Projective, Tome 2 (Gauthiers- Villars, Paris, 1966)

F. Borceux, An Axiomatic Approach to Geometry, DOI 10.1007/978-3-319-01730-3, 395 © Springer International Publishing Switzerland 2014 396 References and Further Reading

30. M. Kline, Mathematical Thought from Ancient to Modern Times (Oxford University Press, London, 1972) 31. D. Lehmann, R. Bkouche, Initiation à la Géométrie (Presses Universitaires France, Paris, 1988) 32. B. Leighton Wellman, Technical Descriptive Geometry (McGraw-Hill, New York, 1948) 33. M.S. Mahoney, The Mathematical Career of Pierre de Fermat (Princeton University Press, Princeton, 1973) 34. G.E. Martin, The Foundations of Geometry and the Non-Euclidean Plane. Undergraduate Texts in Math. (Springer, Berlin, 1975) 35. D. Pedoe, An Introduction to Projective Geometry (Pergamon, Elmsford, 1963) 36. A. Seidenberg, Lectures in Projective Geometry (Van Nostrand, Princeton, 1962) 37. I. Stewart, Galois Theory (Chapman & Hall, London, 1973) 38. J.P. Tignol, Galois’ Theory of Algebraic Equations (World Scientific, Singapore, 2001) 39. A. Tuller, A Modern Introduction to Geometries (Van Nostrand, Princeton, 1967) 40. B. Vitrac, Structure et Genèse des Eléments d’Euclide. CNRS, UMR 8567, Centre Louis Ger- net, Paris 41. R. Yates, The Trisection Problem (The National Council of Teachers of Mathematics, Reston, 1971) Index

Symbols projective, 207 δ(ABC), 261 sign, 205 π, 4, 103, 118, 168, 371 Apollonius, 130 Δ(ABC), 274 Ð problem, 148 Archimedes, 42, 112 A axiom, 31, 337 abstract spiral, 127 plane, 306 Archytas, 24 projective plane, 211 area acute angle, 338 circle, 112 Ahmes, 3 function of Ð, 274 Alberti, 197, 198 parabola, 125 algebraic triangle, 150 element, 356 Arguesian plane, 214, 223 extension, 356 Aristotle, 10, 39 altitude, 109, 172 axiom Anaxagoras, 18 Archimedes, 31, 337 angle, 44, 317 Arguesian plane, 214 acute, 338 continuity, 335 at the center, 70 Eudoxus, 31, 78, 336 bisector, 74, 324 incidence, 306 characteristic, 252 order, 307 exterior, 318 Pappian plane, 215 external, 51, 328 parallelism, 351 inscribed, 70 projective plane, 211 interior, 318 internal, 51, 328 obtuse, 338 B of parallelism, 268 Babylon, 5 right, 44, 332 barycentre, 171 solid, 96 basic geometric configuration, 362 trisection, 17, 369 Beltrami, 245, 281 vertex, 317 Ð-Klein disk, 281 anharmonic ratio, 158, 205 between relation, 286, 307 conic, 240 Bezout theorem, 87 of four lines, 206 bisector, 74, 174, 324

F. Borceux, An Axiomatic Approach to Geometry, DOI 10.1007/978-3-319-01730-3, 397 © Springer International Publishing Switzerland 2014 398 Index

BK-disk, 281 conic, 25, 131 angle, 291 anharmonic ratio, 240 congruent Ðs, 291 center, 138 distance, 287 conjugate directions, 138 perpendicularity, 293 diameter, 138 segment equation, 133 congruent Ðs, 290 focus, 147, 177 Bolyai, 244, 250 tangent, 139, 141 Brianchon, 198 conjugate directions, 138 broken consistency, 280 line, 54 constructible projective plane, 218 number, 362 point, 361 C continuity axiom, 335 Carnot, 198 criterion center Eisenstein, 359 of a circle, 44, 341 cube, 13, 106 of a conic, 138 duplication, 23, 369 central projection, 158, 207 cylinder, 24, 103 Ceva, 173 theorem, 173 D + characteristic angle, 252 dPA, 312 Chasles, 198, 205 Dandelin, 177 chord, 68 Decartes, 198 circle, 44, 341 Dedekind angle cut, 32, 336 at the center, 70 plane, 335 inscribed, 70 defect of a triangle, 261 area, 112 Democrates, 39 center, 44, 341 Desargues, 197, 208 chord, 68 theorem, 208 circumference, 44 diameter, 68 diameter, 68 of a circle, 68 exterior, 341 of a conic, 138 interior, 341 difference power of a point, 72, 73 of two segments, 328 radius, 341 dimension tangent, 69 field extension, 356 circular Dinostrates, 21 cone, 131 directed plane, 307 continuity principle, 346 directrix, 147 circumference, 44, 113 disk Common notion, 45 Beltrami-Klein, 281 cone, 24, 105 BK-disk, 281 base, 131 P-disk, 281 circular, 131 Poincaré, 281 oblique, 131 distance right, 131 BK-disk, 287 ruling, 131 P-disk, 287 vertex, 131 division congruence Euclidean Ð, 85 plane, 319 divisor, 85 relation, 319 greatest common Ð, 86 Index 399 dodecahedron, 13, 106 fourth proportional, 83 duality principle, 212 function duplication of the cube, 23, 369 of area, 274 Dürer, 197 G E Gauss, 172, 244, 379 Egypt, 3 ÐWantzel theorem, 392 Eisenstein criterion, 359 geometric ellipse, 29, 148 equation mean, 26 of a conic, 133 progression, 90 equilateral triangle, 44 geometric configuration, 360 equivalent polygons, 274 basic, 362 Euclid elementary extension, 361 Ðean division, 85 finite extension, 361 –’s lemma, 87 in standard position, 361 –’s postulate, 45 node, 360 fifth postulate, 245Ð248, 250, 253 geometry Euclidean Euclidean, 43, 351 geometry, 351 hyperbolic, 351 plane, 351 non-Euclidean, 258, 351 Eudoxus, 31, 35 projective, 198 axiom, 31, 78, 336 Gödel, 280 Euler, 384 incompleteness theorem, 280 exhaustion Golden section, 13 method, 35 greatest common divisor, 86 theorem, 35, 91 extension algebraic, 356 H elementary, 361 half field– line, 312 dimension, 356 plane, 312 finite, 361 harmonic quadruple, 145, 158, 208 exterior heptadecagon, 380 of a circle, 341 Herodotus, 2 of a triangle, 314 Heron, 149 of an angle, 318 Ð formula, 150 external angle, 51, 328 Hessenberg theorem, 216 F hexagon, 77 Fano, 211 hexahedron, 106 projective plane, 211 Hilbert, 198, 210, 222 Fermat, 198, 383 theorem, 223, 239 Little Ð Theorem, 386 Hipparchus, 151, 186 prime, 383 Hippasus, 29 Fibonacci, 170 numbers, 171 Hippias, 17 field trisectrix, 18 Ðextension Hippocrates, 19, 25 dimension, 356 homothety, 181 of coordinates, 363 hyperbola, 29 projective plane over a Ð, 219 hyperbolic finiteprojectiveplane,213 geometry, 351 focus, 147, 177 plane, 351 400 Index

I median, 109, 171 icosahedron, 13, 106 perpendicular, 75 incidence Menaechmus, 25 axioms of Ð, 306 Menelaus, 151 incident, 211 theorem, 152 incommensurable, 29 Mersenne prime, 91 incompleteness theorem, 280 middle point, 331 infinity minimal polynomial, 357 line at, 201 model, 280 point at, 200, 201 Mohr, 190 inscribed angle, 70 Monge, 198 interior moon, 20 of a circle, 341 Morley, 174 of a triangle, 314 multiple, 85 of an angle, 318 internal angle, 51, 328 N inversion, 180, 184 Naraniengar, 174 pole, 180 node, 360 power, 180 non-Euclidean geometry, 258, 351 involution, 180 number isosceles triangle, 44, 321 π, 103 constructible, 362 K line, 85 Kepler, 195 perfect, 91 Klein, 245, 281 plane, 85 Beltrami-Ð disk, 281 prime, 85 Mersenne, 91 L rational, 12 Lambert, 244, 251 real, 32 Legendre, 248 relatively prime Ðs, 85 Leonardo de Pisa, 170 solid, 85 limit parallels, 268 limiting parallel, 264 line, 44, 306 O at infinity, 201 obtuse angle, 338 broken, 54 octahedron, 13, 106 half, 312 order incident, 211 axioms, 307 number, 85 of an element, 386 orientation, 45 total, 327 parallel Ðs, 45, 93, 307 total strict, 311 perpendicular Ðs, 332 orientation, 45 polar Ð, 146 projective, 201 P straight, 44 P-disk, 281 Lobachevski, 244 angle, 291 congruent, 291 M distance, 287 Mascheroni, 190 segment mean congruent Ðs, 290 double proportional, 26 Pappian plane, 215, 239 geometric, 26 Pappus, 157 proportional, 25, 83 theorem, 161, 202, 209, 240 Index 401 parabola, 26 projective, 201 area, 125 vanishing, 199 directrix, 147 polar line, 146 Paradox pole, 146 Zeno, 38 of inversion, 180 parallel, 45, 307 polygon limit, 268 circumscribed, 74 limiting, 264 equivalent Ðs, 274 lines, 93 inscribed, 74 planes, 93 regular, 34 postulate, 45 similar Ðs, 34, 100 parallelepiped, 99 polyhedron, 106 parallelism cube, 13 angle, 268 dodecahedron, 13 parallelism axioms, 351 icosahedron, 13 parallelogram, 60 octahedron, 13 Pascal, 197 regular, 13, 106 Pasch axiom, 250 tetrahedron, 13 Peano, 210 polynomial pentagon, 13 minimal, 357 star, 13 symmetric, 371 perfect number, 91 Poncelet, 180, 190, 194, 198 Posidonius, 246 Peri, 198, 210 position perimeter, 313 standard, 361 perpendicular, 44, 93, 332 postulate, 45 median, 75 Euclid’s fifth Ð, 45 plane power abstract, 306 of a point, 72, 73 congruence, 319 of an inversion, 180 Dedekind, 335 prehistory, 1 directed, 307 prime Euclidean, 351 number, 85 half, 312 relatively Ð, 85 hyperbolic, 351 prism, 103 number, 85 progression parallel Ðs, 93 geometric, 90 projective, 201 projection abstract, 211 central, 158, 207 Arguesian, 214, 223 stereographic, 187 broken, 218 projective Fano, 211 abstract Ð plane, 211 finite, 213 line, 201 over a field, 219 plane, 201 Pappian, 215, 239 Arguesian, 214, 223 trigonometry, 151 broken, 218 Plato, 39 Fano, 211 Poincaré, 245, 281 over a field, 219 disk, 281 Pappian, 215, 239 universe, 303 point, 201 point, 44, 306 proportional at infinity, 200, 201 double Ð mean, 26 incident, 211 fourth Ð, 83 middle Ð, 331 mean, 25, 83 402 Index

Ptolemy, 41, 154, 186, 247 spherical theorem, 154 triangle, 151 pyramid, 103 trigonometry, 151 Pythagoras, 13 spiral theorem, 2, 6, 13, 63 Archimedean Ð, 127 tangent, 128 Q square, 62 quadrilateral, 154, 160 Ð root, 68 Saccheri Ð, 252 standard position, 361 quadruple star pentagon, 13 harmonic, 145, 158 Steiner, 194 Quetelet, 177 stereographic projection, 187 straight line, 44 R sum of two segments, 328 radius of a circle, 341 symmetric polynomial, 371 ratio, 77 anharmonic, 158 T same Ð, 77 tangent, 128 smaller Ð, 78 to a circle, 69 rational number, 12 to a conic, 139, 141 real number, 32 tetrahedron, 13, 106 rectangle, 62 Thales, 10 regular intercept theorem, 10 polygon, 16, 34 theorem, 10, 79 polyhedron, 13, 106 theorem relation Bezout, 87 between, 286, 307 Ceva, 173 congruence, 319 Desargues, 208 relatively prime, 85 exhaustion, 91 right Fermat’s Little Ð, 386 angle, 44, 332 angled triangle, 45, 332 GaussÐWantzel, 392 cone, 131 Hessenberg, 216 root Hilbert, 223, 239 square Ð, 68 incompleteness, 280 ruling, 131 intercept, 10 Menelaus, 152 S Pappus, 161, 202, 209, 240 Saccheri, 244, 251 Ptolemy, 154 quadrilateral, 252 Pythagoras, 2, 6, 13, 63 Satyanarayana, 174 Thales, 10, 79 segment, 308 torus, 24 difference, 328 total sum, 328 order, 327 side, 308 strict order, 311 similar, 80, 100 transcendental polygons, 34 element, 358 triangles, 4, 12 triangle, 308 solid altitude, 109, 172 angle, 96 area, 150 number, 85 barycentre, 171 sphere, 105 bisector, 174 volume, 121 congruent Ðs, 321 Index 403

defect, 261 V equilateral, 44 vanishing point, 199 exterior, 314 vertex interior, 314 of a cone, 131 isosceles, 44, 321 of a triangle, 308 median, 109, 171 of an angle, 317 perimeter, 313 volume sphere, 121 right angled, 45, 332 von Staudt, 198, 210 side, 308 similar Ðs, 4, 12, 80 W spherical, 151 Wallis, 247 trisector, 174 Wantzel, 379 vertex, 308 GaussÐtheorem, 392 trigonometry, 151 trisection, 17, 369 Z trisector, 174 Zeno, 38 trisectrix, 18, 21 paradox, 38