Appendix A Straightedge and Compass Constructions The ancient Greeks solved many beautiful problems about geometric constructions using straightedge and compass, and they raised some questions that became widely known because of numerous unsuccessful attempts to solve them. These problems remained open for many centuries. The ancient Greeks constructed regular n-gons for n D 2k3, 2k4, 2k5,and2k15,wherek is any nonnegative integer. It was unknown how to construct a regular n-gon for any other value of n until Gauss constructed the regular 17-gon and gave a characterization of all numbers n for which such a construction is possible. Gauss obtained this remarkable result before Galois theory had been discovered. His amazing result had a huge impact on the development of several branches of mathematics. Below, we discuss this problem and other problems on constructibility. Problems about constructions using straightedge and compass are the oldest problems on “solvability in finite terms.” We treat them just as we have treated other problems of this type. We distinguish three classes of constructions. In problems of the first, simplest, class, only three operations are allowed: constructing a line through two given points, a circle of given radius with a given center, and points of intersection of given lines and circles. In the third, hardest, class we allow all these constructions, and in addition, we also allow ourselves to choose an arbitrary point, but we require the result to be independent of the choice of arbitrary points. In the second, intermediate, class we do not allow arbitrary choice, but we allow two additional constructions: construction of the center of a given circle and construction of the line orthogonal to a given line passing through a given point not lying on the given line. Logically, constructions in the third class are different from the constructions in other classes and from everything we saw before in problems on solvability in finite terms: intermediate objects that we get in the process can be dependent on arbitrary This appendix was published originally as A. Khovanskii [57]. © Springer-Verlag Berlin Heidelberg 2014 239 A. Khovanskii, Topological Galois Theory, Springer Monographs in Mathematics, DOI 10.1007/978-3-642-38871-2 240 A Straightedge and Compass Constructions choices we have made. In this case, we do not consider them to be constructions made using straightedge and compass. We try to avoid, when possible, the operation of choice of an arbitrary point. We prove that in most problems, this operation does not give anything new (everything that is constructible using this operation can be constructed without it). Some of the classical problems can be rather satisfyingly formulated and proved inside the first class of constructions. The problem of trisection of an angle requires the operation of choice of an arbitrary point: given two lines passing through a given point, nothing new can be constructed using operations of the first class. We show that if we add to these given lines an arbitrary point on one of them, then using operations of the first class only, we can construct everything that can be constructed from two given lines using operations from the third class. Operations of the second class allow us to extend initial data. For example, operations from the first class of constructions do not allow us to construct anything if the initial datum consists of several nonintersecting circles. But if we mark the centers of the circles, then operations of the first class can be used to construct anything constructible from the initial datum using operations from the third class (except when all the given circles are concentric). In the first section, we discuss problems of solvability of algebraic equations using square roots that are later needed for problems of constructibility. We do it in greater generality than we need (we do not assume that the base field is perfect and that it is of characteristic not equal to 2). This problem is interesting in its own right, and additional generality does not cause much trouble. The second section is devoted to problems of constructibility using straightedge and compass. A.1 Solvability of Equations by Square Roots In this section, we discuss the following question about solvability of equations in finite terms: when is an irreducible algebraic equation over a field K solvable using arithmetic operations and the operation of taking the square root? Galois theory answers this question when the field K is perfect and its characteristic is not equal to 2. But some simple additional observations help to get rid of any assumptions on the structure of the field K. The material is structured as follows: In Sect. A.1.1 we give necessary back- ground material. In Sect. A.1.2 we prove necessary and sufficient conditions for an equation to be solvable using square roots if the characteristic of the field K is not equal to 2. In Sect. A.1.3, the same question is solved when the characteristic of the field is equal to 2. In Sects. A.1.4 and A.1.5, Gauss’s results on roots of unity of degree n are presented (we use them in the problem of constructibility of the regular n-gon). A.1 Solvability of Equations by Square Roots 241 A.1.1 Background Material We begin by recalling some elementary facts about fields. If the field K is a subfield of the field F ,thenF is a vector space over K.IfdimK F<1,thenF is said to be a finite extension of the field K.Thedegree of the extension is defined to be dimK F and is denoted by ŒF W K. Theorem A.1 If K F and F M are finite field extensions, then: 1. K M is a finite extension. 2. ŒM W K D ŒM W F ŒF W K. Proof Let u1;:::;un beabasisofF over K and let v1;:::;vm be a basis of M over F . It is easily shown that elements ui vj with 1 Ä i Ä n, 1 Ä j Ä m form a basis of M over K. ut Proposition A.2 1. If ŒF W K D n and a 2 F , then there exists a polynomial Q over the field K of degree not greater than n such that Q.a/ D 0. 2. If Q is an irreducible polynomial over the field K and Q.a/ D 0,thenŒK.a/ W K D deg Q. n Proof 1. Since the dimK F D n elements 1;a;:::;a are linearly dependent, there n n1 exist i 2 K such that na C n1a CC0 D 0 and for some i>0,the coefficient i is nonzero. 2. The field K.a/ is isomorphic to the field KŒx=I,whereI is an ideal generated by the polynomial Q of degree n. ut Proposition A.3 If the degree of an irreducible polynomial over a field of charac- teristic p>0is not divisible by p, then it does not have multiple roots. Proof A multiple root of the polynomial Q is also a root of the derivative Q0.An irreducible polynomial Q cannot have a common root with a nonzero polynomial of smaller degree. Thus if Q has a multiple root, then Q0 Á 0, i.e., Q.x/ D R.xp/ for some polynomial R, in which case, deg Q D p deg R, and so deg Q is divisible by p. ut A.1.2 Extensions by 2-Radicals We now return to the question of solvability of equations by square roots. Definition A.4 An extension K F is said to be an extension by 2-radicals if there exists a tower of fields K D F0 F1 Fn such that F Fn and for 2 1 Ä i Ä n, Fi D Fi1.ai / with ai 2 Fi1 and ai … Fi1. 242 A Straightedge and Compass Constructions Theorem A.5 If K F is a 2-radical extension, then ŒF W K D 2k. Proof For K D F0 F1 Fn,wehaveŒFn W K D ŒFn W Fn1 ŒF1 W n n F0 D 2 .IfK F Fn,thenŒF W K ŒFn W F D 2 . Therefore, ŒF W K is a power of 2. ut Corollary A.6 If a polynomial P is irreducible over the field K and has a root in some extension of K by 2-radicals, then deg P D 2k. Proof If P.a/ D 0,thenŒK.a/ W K D deg P . ut Corollary A.7 Let K be a field of characteristic not equal to 2. A cubic equation P D 0 over the field K is solvable in square roots if and only if one of its roots lies in K. Proof If a 2 K and P.a/ D 0,thenP D .xa/Q for some Q 2 KŒx. A quadratic equation Q D 0 is solvable in square roots, because the characteristic of the field is not equal to 2. If the cubic polynomial does not have a root in K,thenitis irreducible, and we can use Corollary A.6. ut Remark A.8 For K D É, Corollary A.7 has an effective form: the rational roots of a rational polynomial P 2 ÉŒx can be explicitly found (in particular, we can check whether there are none). If a root a is found, then the quadratic equation 0 D Q.x/ D P=.x a/ is explicitly solvable. Let EP denote the splitting field of the polynomial P over the field K. Corollary A.9 If a polynomial P is irreducible over K and has a root in some m extension of the field K by 2-radicals, then ŒEP W K D 2 .