# CONSTRUCTIBLE NUMBERS AND GALOIS THEORY

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1 CONSTRUCTIBLE NUMBERS AND GALOIS THEORY SVANTE JANSON Abstract. We correct some errors in Grillet [2], Section V Introduction The purpose of this note is to correct some errors in Grillet [2], Section V.9 (in particular Theorem 9.3 and Lemma 9.4). See also [1]. As in Grillet [2], we define a constructible number to be a complex number such that the corresponding point in the Euclidean plane is constructible from 0 and 1 by ruler and compass (a.k.a. straightedge and compass). Let K be the set of constructible numbers. Then, as shown in [2, Proposition 9.1 and Lemma 9.2], K is a subfield of C, which is closed under taking square roots (i.e., if z K, then ± z K); moreover, K is the smallest such subfield of C. Remark. A complex number is constructible if and only if its real and imaginary parts are constructible [2, Lemma 9.2], so it suffices to study real constructible numbers. However, for the present purpose it is simpler to allow complex numbers. 2. Main result Theorem 2.1. The following are equivalent, for a complex algebraic number z: (i) z is constructible (z K). (ii) There is a chain of field extensions Q = F 0 F 1 F m with z F m and F i = F i 1 (z i ) with z 2 i F i 1 for every i m. (iii) There is a chain of field extensions Q = F 0 F 1 F m with z F m and [F i : F i 1 ] 2 for every i m. (iv) There is a chain of field extensions Q = F 0 F 1 F m with z F m and [F i : F i 1 ] = 2 for every i m. (v) z L, where L is some normal field extension of Q of degree 2 k for some k 0. (vi) The splitting field of the irreducible polynomial Irr(z : Q) has degree 2 k for some k 0. (vii) The Galois group Gal(f : Q) of the irreducible polynomial f := Irr(z : Q) has degree 2 k for some k 0. Date: 8 December, 2009; reference added 30 January

2 2 SVANTE JANSON Proof. (i) (ii): By Grillet [2]. (ii) = (iii): For every i m, z i is a root of f i (X) = X 2 b i with b i := zi 2 F i 1 and thus X 2 b i F i 1 [X]. Either z i F i 1, and then F i = F i 1, or f i is irreducible over F i 1, and then [F i : F i 1 ] = [F i 1 (z i ) : F i 1 ] = deg(f i ) = 2. (iii) = (iv): Eliminate all repetitions in the sequence F 0,..., F m. (iv) = (ii): Let i m, and take any α F i \ F i 1. Then F i 1 F i 1 (α) F i, and since [F i : F i 1 ] = 2, F i 1 (α) = F i. Furthermore, α 2 F i, and thus (again using [F i : F i 1 ] = 2), α 2 = aα + b for some a, b F i 1. Consequently, α = a/2 ± a 2 /4 + b. Define z i := a 2 /4 + b. Then α = a/2 ± z i, and thus F i = F i 1 (α) = F i 1 (z i ), and zi 2 = a2 /4 + b F i 1. (ii) = (v): We prove by induction on i that every field F i has an extension L i such that L i is a normal extension of Q whose degree [L i : Q] is a power of 2. We then take L = L m and observe that z F m L m = L. To prove this claim, note first that it is trivial for i = 0. For the induction step, we fix i m and assume that F i 1 L i 1 where L i 1 is a normal extension of Q of degree 2 k for some k 0. Let w i := zi 2 F i 1 and let {z ij } J j=1 be the set of conjugates of z i over Q. For each j, z ij = σ(z i ) for some K-automorphism σ of the algebraic closure Q, and thus zij 2 = σ(z2 i ) = σ(w i) is conjugate to w i F i 1 L i 1. Since L i 1 is a normal extension of Q, zij 2 L i 1. Define L ij := L i 1 (z i1,..., z ij ), and L i := L ij. Thus L i0 = L i 1. For 1 j J, we have L ij = L i,j 1 (z ij ) and zij 2 L i 1 L i,j 1 ; hence [L ij : L i,j 1 ] = 1 or 2 (as in the proof of (ii) = (iii) above). Consequently, J [L i : L i 1 ] = [L ij : L i0 ] = [L ij : L i,j 1 ] = 2 l j=1 for some l, and [L i : Q] = [L i : L i 1 ][L i 1 : Q] = 2 l+k. This proves the induction step, and thus the claim. (v) = (vi): Let {z j } J j=1 be the set of conjugates of z. Since L is normal, each z j L. Define E := Q(z 1,..., z J ), the splitting field of Irr(z : Q). Then Q E L, and thus [L : E] [E : Q] = [L : Q] = 2 k ; hence [E : Q] is a divisor of 2 k, and thus a power of 2. (vi) (vii): A splitting field is a Galois extension (in characteristic 0, at least), and the degree of a Galois extension equals the order of its Galois group. (vii) = (iv): Let E be the splitting field of f, and let G be the Galois group Gal(f : Q) = Gal(E : Q). By Sylow s first theorem (and induction), there is a chain of subgroups G = H 0 H 1 H k = {1} with H i = 2 k i. The fundamental theorem of Galois theory shows that the corresponding sequence of fixed fields F i := Fix E (H i ) satisfies F 0 F 1 F k and [E : F i ] = H i = 2 k i, which yields [F i : Q] = 2 i and [F i : F i 1 ] = 2. Clearly z E = F k.

3 CONSTRUCTIBLE NUMBERS AND GALOIS THEORY 3 Recall that the degree deg K (α) of an algebraic element α in an extension of a field K is the degree of its irreducible polynomial. Furthermore, deg K (α) = [K(α) : K]; see [2, Section V.2]. Corollary 2.2. If z is constructible, then z is algebraic and its degree deg Q (z) = [Q(z) : Q] is a power of 2. However, the converse of Corollary 2.2 is not true, as shown by the following example. Example 2.3. Let α be a root of an irreducible polynomial f(x) Q[X] of degree 4, such that the Galois group Gal(f : Q) = S 4 or A 4. Since f = Irr(α : Q), and the order of the Galois group Gal(f : Q) is 24 or 12, and thus not a power of 2, Theorem 2.1 shows that α is not constructible. On the other hand, [Q(α) : Q] = deg(f) = 4 = 2 2. (Such polynomials f exist. By [2, Proposition V.5.6], any polynomial of degree 4 such that its cubic resolvent is irreducible will do. X 4 + 2X 2 is an explicit example.) Remark 2.4. It is easily seen that the following properties are equivalent, for a complex number z. (i) The degree of z over Q is a power of 2. (ii) The degree [Q(z) : Q] of Q(z) over Q is a power of 2. (iii) z lies in an extension E of Q whose degree [E : Q] is a power of 2. Corollary 2.2 thus says that these conditions are necessary for z to be constructible, but Example 2.3 shows that they are not sufficient. (Cf. Theorem 2.1(v), where the extension is assumed to be normal.) Furthermore, the set S of z C that satisfy (i) (or (ii) or (iii)) is not a field. For example, let α be as in Example 2.3, and let its conjugates be α 1 = α, α 2, α 3, α 4. Then the splitting field of f is E := Q(α 1, α 2, α 3, α 4 ). For every j, deg(α j ) = deg(α) = 4 and thus α j S. If S were a field, thus E = Q(α 1, α 2, α 3, α 4 ) S. However, E is a finite and separable extension of Q, and thus it is a simple extension: E = Q(β) for some β E. Since E, being a splitting field, is a Galois extension of Q, deg Q (β) = [Q(β) : β] = [E : Q] = Gal(E : Q) = 12 or 24, which is not a power of 2; hence β / S, and thus E S. This contradiction shows that S is not a field. Remark 2.5. Example 2.3 gives an example of a field extension E = Q(α) Q of degree 4 such that there is no intermediate field Q(α) F Q. (If there were such an F, then [Q(α) : F ] = [F : Q] = 2, and α would satisfy Theorem 2.1(iv) and thus be constructible, a contradiction.) 3. Some classical applications The results above are easily applied to the classical problems of construction by ruler and compass. (See also [2, Section V.9].)

4 4 SVANTE JANSON 3.1. Squaring the circle. In modern terms, the problem is to construct π by ruler and compass. Since π, and therefore also π, is transcendental and not algebraic, this is impossible by Corollary 2.2. (The transcendence of π is not so easy to prove; see e.g. [3, Section 11.14].) 3.2. Doubling the cube. The problem is to construct 3 2. This is a root of the irreducible polynomial X 3 2 = 0, so its degree is 3 and Corollary 2.2 shows that 3 2 is not constructible Trisecting the angle. The problem is to construct e iθ/3 from e iθ. We choose θ = 60 = π/3; then e iθ = (1 + i 3)/2 is constructible, so the task is equivalent to constructing α := e iθ/3 = e iπ/9 = e 2πi/18 from scratch. However, α is a primitive 18th root of unity, so its irreducible polynomial is the cyclotomic polynomial Φ 18 (X), and thus deg Q (α) = deg(φ 18 ) = ϕ(18) = 6. By Corollary 2.2, α is not constructible. Hence trisecting an angle by ruler and compass is in general not possible. (In particular, it is not possible for a 60 angle.) 3.4. Constructing a regular n-gon. This is equivalent to constructing the n:th root of unity ω n := e 2πi/n. The irreducible polynomial of ω n is the cyclotomic polynomial Φ n (X), which has degree ϕ(n). The splitting field of Φ n (X) is Q(ω n ) (since all other roots are powers of ω n, and thus belong to Q(ω n )). Consequently, Theorem 2.1(vi) shows that ω n is constructible if and only if the degree ϕ(n) of Φ n (X) is a power of 2. By simple number theory, see e.g. [3, Section 5.5], if n has the prime factorization n = k pa k k, for some distinct primes p k and exponents a k 1, then ϕ(n) = k pa k 1 k (p k 1), which is a power of 2 if and only if a k = 1 for every k with p k 2, and p k 1 is a power of 2 for every i. This implies that each p k is either 2 or a Fermat number F j = 2 2j + 1 for some j 0, see e.g. [2, Section V.9] or [3, Section 2.5]. Consequently: Theorem 3.1. A regular n-gon is constructible by ruler and compass if and only if n is a product of a power of 2 and distinct Fermat primes. The first 5 Fermat numbers F 0 = 3, F 1 = 5, F 2 = 17, F 3 = 257 and F 4 = are primes, but no others are known, and it seems likely (but unproven) that these are the only Fermat primes. We see that, for example, a regular n-gon is constructible for n = 3, 4, 5, 6, 8, 10, 12, 15, 16, 17, 20, but not for n = 7, 9, 11, 13, 14, 18, 19. References [1] D. A. Cox, Galois Theory. Wiley-Interscience, Hoboken, NJ, [2] P. A. Grillet, Abstract Algebra. 2nd ed., Springer, New York, [3] G. H. Hardy & E. M. Wright, An Introduction to the Theory of Numbers. 4th ed., Oxford Univ. Press, Oxford, 1960.

5 CONSTRUCTIBLE NUMBERS AND GALOIS THEORY 5 Department of Mathematics, Uppsala University, PO Box 480, SE Uppsala, Sweden address: URL: svante/

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