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25 Radical Extensions This Section Contains Two Classic Applications of Galois Theory

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25 Radical Extensions This Section Contains Two Classic Applications of Galois Theory 25 Radical extensions This section contains two classic applications of Galois theory. Thefirst application, the problem of the constructibility of points in the plane raised by Greek mathematicians, shows that the underlying questions do not need to refer tofield extensions or auto- morphism groups in any way. The second application, solving polynomial equations by extracting roots, is the problem that, in a way, gave rise to Galois theory. � Construction problems In Greek mathematics,13 people used to constructfigures with a straightedge and com- pass. Here one repeatedly enlarges a given setX of at least two points in the plane by adding to it those points that can be constructed as intersections of lines and circles defined in terms of the given points. The question is, for givenX, whether certain points in the plane can be obtained fromX infinitely many construction steps. More formally, a construction step starting out from a subsetX of the plane consists of replacing the setX with the set (X) of all pointsP obtained by applying F the following algorithm: 1. Pick points a, b, c, d X witha=b andc=d. ∈ � � 2. Let� ab be either the line througha andb or the circle througha with centerb. Likewise, let� cd be either the line throughc andd or the circle throughc with centerd. 3. If� and� do not coincide, pickP � � . ab cd ∈ ab ∩ cd b a c a b d b c c d a d In order to perform step 1 of the algorithm,X needs to contain at least two points. In this case, pickingc=a andd=b shows that we haveX (X). As the permissible ⊂F intersections� � consist of 0, 1, or 2 points, (X) isfinite ifX isfinite. ab ∩ cd F 25.1. Definition. LetX=X 0 be a subset of the plane containing at least two points, and define, recursively fori Z 1, the sets ∈ ≥ Xi = (X i 1) X i 1. F − ⊃ − Then (X)= X is called the set of constructible points starting fromX. C i=0 i In addition to� constructing points in the plane, we can also speak of constructing lines and circles in the plane. A line is called constructible if it is possible to construct two 63 Algebra III– 25 § distinct points on it, and a circle is called constructible if it is possible to construct its center and a point on it. As early as thefifth century B.C., Greek mathematics gave rise to several construc- tion problems that the Greeks could not solve and which, for more than two thousand years, defeated many mathematicians, “professional” or not.14 25.2. Squaring the circle. Construct a square whose area is equal to that of a circle with given radius. Hippocrates of Chios, who lived around 430 B.C., showed that 25.2 is solvable if instead of the circle, we take certainfigures bounded by circular arcs, the so-called lunes of Hippocrates15 (see Exercise 14). 25.3. Doubling the cube. Construct a line segment that is √3 2 times as long as a given line segment. This problem is also known as the Delian problem, after the legend in which, through his oracle on the island Delos, the god Apollo decreed that the plague-ridden Athenians should “double” their cubic altar to Apollo. 25.4. Trisecting the angle. Use a straightedge and compass to divide a given angle into three equal parts. For a few angles, such as the right angle, this problem is easy to solve. In most other cases, angle trisection does not seem possible. 25.5. Constructing then-gon. Forn 3, construct a regularn-gon in a given ≥ circle. Strictly speaking, this problem does not belong to the classical “corpus” of the three unsolved Greek problems. But, as it corresponds to the division of the full angle of 2π radials inton equal parts, it is closely related. Since bisecting an angle using a straightedge and compass is easy, the interesting question in 25.5 is for which oddn the problem is solvable. The Greek found solutions forn = 3 andn = 5 but not, for example, forn = 7 orn = 9. To formulate construction problems in terms offield extensions, we identify the plane in the usual way with thefieldC of complex numbers. Using 25.1, the problems mentioned above can easily be reformulated in terms of numbers constructible from a subsetX C. After scaling, we may assume thatX contains the two points 0 and 1. ⊂ ForX equal to 0,1 , the set = (X) is simply called the set of constructible { } C C numbers. Problems 25.2, 25.3, and 25.5 then correspond to the questions of whether the numbers √π and √3 2 and the primitiventh root of unityζ =e 2πi/n C are con- n ∈ structible. The question in 25.4 is whether forα C with α = 1, the set ( 0,1,α ) ∈ | | C { } contains a third root √3 α. 25.6. Proposition. LetX C be a set that contains0 and1. Then (X) is a subfield ⊂ C ofC that containsX. It is closed under complex conjugation and under extracting square roots. 64 Algebra III– 25 § Proof. The proof is an exercise in carrying out elementary constructions. We assume the standard constructions from Exercise 13 known—an exercise we recommend for anyone who has never carried out a construction. It is clear thatX, and in particular 0 and 1, are contained in (X). It therefore suffices to show that for x, y (X), the C ∈C differencex y, the product xy, the inverse 1/x, the complex conjugate x, and the − square roots √x are constructible fromX. ± Forx y, we mark off the distance x y on the line through 0 parallel to the − | − | line throughx andy. For x, wefirst intersect the circle throughx with center 0 with the line through 0 and 1—this gives x —and then intersect it with the circle through | | x with center x . Forx/ R, the point x /x is the intersection of the circle through | | ∈ | | 1 with center 0 with the line through 0 and x; the line through 1 parallel to the line through x /x and x now cuts the line through 0 and x in 1/x. A similarfigure shows | | | | how to multiplyx by a real number y . The productx y is then rotated over the | | | | angle∠y01 to obtain xy. Exercise 1. How should the constructions be adjusted forx R? ∈ y x x y x | | x 0 1 0 1/x x 0 1 y x /x | | | | | | x y x − Taking the square root corresponds to constructing √x forx R . After all, for ∈ >0 non-realx, we simply mark off x on the bisector of the angle x01. Forx R , | | ∠ ∈ >0 we intersect the perpendicular in 0 to the line through 0 and 1 with the circle that has � the line segment from x to 1 as diameter. Lets be an intersection point. − s=i √x u x 0 1 − The angle ( x)s1 is a right angle (Thales’s theorem), so the triangles ( x)s0 and ∠ − − s10 are similar. The equality of the ratiosx:u=u : 1 shows thats is equal toi √x, and we are done. Exercise 2. (For whoever did not know this yet . ) Formulate and prove Thales’s theorem. The main result on constructible numbers is that 25.6 in fact characterizes thefield (X): it is the smallestfield that satisfies the conditions of 25.6. To see this, we must C show that the construction steps can only lead to quadraticfield extensions. 65 Algebra III– 25 § 25.7. Proposition. LetX C be given, and letK=Q(X, X) be the subfield ofC ⊂ generated by the elements inX and their complex conjugates. Then every pointz C ∈ that can be obtained fromX through a construction step is algebraic overK of degree [K(z):K] 2. ≤ Proof. Suppose that there exist points a, b, c, d X witha=b andc=d. We must ∈ � � show that an intersection point of the line or circle determined by ab and the line or circle determined by cd has degree at most 2 overK. We distinguish between the three possibilities given by thefigures before 25.1. The line through the pointsa andb consists of the pointsz C for which ∈ (z a)/(a b) is real. Expanding the identity (z a)/(a b) = ( z a)/(a b) gives − − − − − − the following equation for this line: � :(a b)z (a b) z= ab a b. ab − − − − Intersecting the lines� ab and� cd amounts to solving two linear equations inz and z with coefficients inQ(a, b, c, d, a, b, c, d) K. If� and� are not parallel, this system ⊂ ab cd has a unique solutionz K, and we haveK(z)=K. If the lines are parallel, then ∈ because� ab and� cd do not coincide, the system is inconsistent and has no solution. For the circle throughc with centerd, the equation is z d = c d ; we can | − | | − | rewrite this as (z d)( z d) = (c d)( c d) or − − − − zz dz d z=c c c d cd. − − − − For a pointz on the line� ab that lies on this circle, we can use the equation for� ab to write z as a linear expression inz with coefficients inK. Substituting this in the equation for the circle gives a quadratic relation with coefficients inK satisfied byz. Wefind [K(z):K] 2. ≤ In the event thatz is a point that lies both on the circle througha with centerb and on the circle throughc with centerd, taking the difference of the equations for the two circles gives a linear relation betweenz and z with coefficients inK.
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