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Math 594, HW7- Solutions Gilad Pagi, [email protected] March 18, 2015 1 a). Since the set of constructible numbers, denoted C, is a subset of R, containing 1 and 0, suffices to show that C is closed under addition, multiplication, and inverse, all of which were proved in class. We shall give a proof for the inverse of a > 0, given that a is constructible. I. Given length a on the line AB, |AB| = a. II. Construct the point C s.t. |BC| = 1. III. Construct a perpendicular line to AB at point B. IV. Construct the point D s.t. |BD| = 1. V. Construct a similar triangle to ABD on with the edge BC correspond to AB. VI. Get |CE| = 1/a. D • • 1 • A a B • 1 • E C b). We shall construct the coordinate system as follows: Given the (initial) points, denoted as A and B, construct the line AB and a perpendicular line to AB through A. Let the A point denote the origin, and B be (1, 0), so the corner starting at A, going through B, is the positive x-axis. Any length a on AB, starting from A, will be treated as (a, 0) or (−a, 0) depending if the corresponding end point ( (C) in the figure) is on the positive x-axis or not. The positive y-axis is set according the the ”right hand rule”. Given constructible length p0 , p1 can position the p0 on the positive x-axis, and p1 on the positive y-axis s.t. the length from A Correspond. By constructing two perpendicular lines to 1 the axes we get the point P = (p0 , p1 ) of constructible coordinates. Conversely, given P = (p0 , p1 ), we construct the perpendicular line to the axes through P and obtain the coordinates. This proves the statement. y p1 • • • • (A) (B) • (P ) • (C) p0 x c). Given A = (ca , da ), B = (cb , db ) the line equation is (cb −ca )(y−da ) = (db −da )(x−ca ) , and the circle equation is (WLOG - the center is at A) (x−ca )2 +(y−da )2 = (cb −ca )2 +(db −da )2 d). Intersection (x0 , y0 ) of 0 = ax + by + c and 0 = dx + ey + f , if exists, is the solution of a linear system, which is in the field k since it consist of addition, subtraction, multiplication and division of the coefficients. For a line in the form y = ax + b, and a circle (WLOG centered at the origin) x2 + y 2 = r2 we get x20 + (ax0 + b)2 = r2 . So the solution for x0 is either in k or in a quadratic extension of k (See HW6, Q1). Being on the same k-line, x0 and y0 are in the same extension. For a vertical line x = a, the statement is obvious. e). For two circles: (x−a)2 +(y −b)2 = r2 and (x−c)2 +(y −d)2 = s2 , notice that subtracting one from the other eliminates the term x2 , y 2 so this problem is reduced to the previous one. Thus, the solution, if exists, is in k or a quadratic extension of k. f). From previous parts, and since any additional constructible point is either an intersection of 2 lines, 2 circles or a line and a circle in k, the statement is true - any additional constructible point lies in either k or a quadratic extension of k. g). Suppose that θ is constructed by a finite process of constructing P0 = 0, P1 = 1, P2 , P3 , ..., Pt = θ. From previous parts, starting from P0 = 0 and P1 = 1 in Q, P2 is either in Q or in quadratic extension of Q, denoted k2 . P3 is either in k2 or a quadratic extension of k2 , denoted k3 . There for, Pt = θ is in kt ⊃ kt−1 ⊃ ... ⊃ k2 ⊃ Q, each extension is of degree 2 2 or 1. Thus [kt : Q] = 2m for some m. Specifically, Q(θ) ⊃ Q is finite. √ h). Q(π) ⊃ Q is infinite and therefor π is not constructible. Likewise, nor is π. The side length of any constructed square is constructible. But if we could construct √ a square whose area is the same a circle with unit radius, then its side length would be π. i). Q(21/3 ) ⊃ Q is an extension of degree 3 6= 2m , so 21/3 is not constructible. Any attempt to construct a cube with volume 2 will result an edge of length 21/3 which is not constructible. j). The matrix for rotating though θ is (notice the used abbreviation) cos θ − sin θ c −s A= = sin θ cos θ s c Rotation through 3θ is the composition of this linear transformation with itself three times; so its matrix is 2 cos 3θ − sin 3θ c −s c − s2 −2cs c(c2 − s2 ) − 2cs2 ∗ 3 = =A = sin 3θ cos 3θ s c 2cs c 2 − s2 ∗ ∗ So cos(3θ) = c3 − 3cs2 = c3 − 3c(1 − c2 ) = 4c3 − 3c as desired. k). Denote cos(π/9) as c. Since constructing an angle and constructing the cosine of the angle is equivalent, we shall prove that c is not constructible. Notice that from the previous part, c satisfies cos(π/3) = 4c3 − 3c i.e. 0.5 = 4c3 − 3c i.e. c is a root of the polynomial p(x) = 8x3 − 6c − 1 ∈ Q[x]. Want to show that this is irreducible over Q, then the extension is of degree 3 and thus c is not constructible. By Gauss’s Lemma, p(x) is irreducible over Q if and only if it is irreducible over Z. But, if it factors over Z, then it has a degree one factor, it would have a root which would necessarily be one of the following {± 11 , ± 12 , ± 14 , ± 18 }, none of which is a root (by inspection). Therefor p(x) = 8x3 − 6c − 1 ∈ Q[x] is irreducible. √ Note that, for example, we can trisect π (cos(π/3) = 0.5) and π/2 (cos(π/6) = 3/2). 2 a). Suppose we have some polynomials g1 , ..., gt ∈ R s.t. g1 f1 (xf1 ) + ... + gt ft (xft ) = 1 in R. Let L be the algebraic extension of F obtained by adjoining roots α1 , . . . , αt of the polynomials f1 (x), . . . , ft (x), say, a splitting field for the product f1 f2 . . . ft . There is an obvious ring homomorphism from R to a free L-algebra S given by sending xf1 7→ α1 , ..., xft 7→ αt , and the remaining xf to indeterminates. This map applied to the relation above gives 0 = 1 in S, a contradiction. b). First note (using Zorn’s Lemma) that the proper ideal I is contained in a maximal ideal M. The maps F ,→ R → R/M = K1 produce a field extension F ,→ K1 of F . Now 3 every (monic, non constant) polynomial f (y) ∈ F [y] can be considered a polynomial in K1 [y]. But the image of xf in R/M, call it αf , is a root of f (y) in K1 , since f (xf ) ∈ M means f (αf ) = 0 in R/M. c). Continue the previous procedure, creating a Kn+1 , an extension of Kn (and of F as well) in which for every polynomial f (y) ∈ Kn [y] there’s at least one root in Kn+1 [y]. Denote K = ∪∞ 1 Kn . Every polynomial f [y] ∈ K[y] has finitely many coefficient so exist N , large enough, s.t. f (y) ∈ KN [y], therefore f has a root in KN +1 ⊂ K. Thus, K is algebraically closed. d). In the last problem set, we proved that addition/multiplication/division of algebraic numbers is algebraic (we used that when we reasoned why the ”algebraic-ness” of an extension is determined by its generators). So, the set of all algebraic numbers over F forms a field that contains F . Therefore, the set of all algebraic numbers over F in the above K form a field, denoted as F . We claim that this is an algebraic closure of F : for every f (y) ∈ F [y] there is a root a in K, but such root is algebraic so it must be in F . Divide (in F [x]) f by (x − a) to get g(x). Since K is algebraically closed, g has a root b in K. So F (b)/F is algebraic, and F /F is algebraic, so F (b)/F is algebraic(HW6Q4a) ⇒ b ∈ F . Now repeat. Since f is of finite degree the procedure ends and we show that all the roots of f are in F . Note here that we proved that for any F ⊂ K ⊂ F , and g(x) ∈ K[x], then g(x) splits completely in F . e). Suppose that F has two algebraic closures F , Fe. Observe that each contains (an isomorphic copy of) the splitting field of every polynomial in F [x]. Indeed, both F , Fe can be viewed as the union (or directed limit) of all splitting fields they contain. In class, we proved the uniqueness of a spitting field over F . As a result, for any splitting field K1 over F f1 in Fe. From the same result, the splitting field contained in F , exists an isomorphic copyK f2 in Fe which extends K f1 . These of K2 of a polynomial g ∈ K1 [x] is isomorphic to some K isomorphisms induce an isomorphism of the direct limits, and hence an F -isomorphism F → Fe. The image of F in Fe is F -isomorphic to F , hence is an algebraic closure of F . But if the inclusion is proper, this would contradict the claim that Fe is an algebraic closure of F , since it shouldn’t contain a proper subfield in which all polynomials have a root. If you prefer not to use the direct limit, you can also argue using Zorn’s lemma: Define a collection A of all subfields of F which have an isomorphic (over F ) copy in Fe. Note that A contains all the splitting fields. Ordinary inclusion of sets is a partial order on A. Given a chain K1 ⊂ K2 ⊂ ... in A, observe that the chain has an upper bound K = ∪∞ 1 Ki e which also has an isomorphic copy in K, namely the union of a corresponding chain of e i ’s in Fe. By Zorn’s lemma, the set A has a maximal element, call it K. So K is the K largest field between F and F which is F -isomorphic to some subfield of Fe. We will be done if we can show that K = F . Suppose a ∈ F but a ∈ / K. But a is an element of the splitting field of its mimimial polynomial, which is a subfield in A. Hence a ∈ K, and the proof is complete. 4 3 a). The complex numbers ζnk = (e2πi/n )k where k = 0, 1, . . . , n − 1, are n distinct numbers (situated at equally spaced intervals around the unit circle in C), all of which satisfy the polynomial xn − 1. This means that xn − 1 factors as (x − ζn0 )(x − ζn1 ) · ... · (x − ζnn−1 ), so that Q(e2πi/n ) is the unique splitting of xn − 1 contained in C. The set of roots is µn = {ζn0 , ζn1 , ..., ζnn−1 }, which is cyclic of order n, generated (for example) by ζn . b). Clear from the identification ζn = e2πi/n . c). Since xn − 1 = (x − ζn0 )(x − ζn1 ) · ... · (x − ζnn−1 ), the point is to show that µn is the disjoint union of all the primitive roots in µd s.t. d | n. But ζnk is a primitive generator for µd if and only if its order (in C, hence in µn ) is d. Now, the number of members of order d in n µn is φ(d) (known results coming from the lemma: xa ∈< x >∼ . = Cn has an order gcd(a,n) P The partition above give us the following partition for n: n = d|d φ(d). d). x8 − 1 = Φ1 · Φ2 · Φ4 · Φ8 = (x − 1) · (x + 1) · (x − i)(x + i) · (x − eiπ/4 )(x − e3iπ/4 )(x − e5iπ/4 )(x − e7iπ/4 ) where Φ1 = (x − 1), Φ2 = (x + 1), Φ4 = x2 + 1, Φ8 = x4 + 1. e). For n = 1, 2 see above. Assume that the statement is true for 1, 2, ..., n − 1. By previous Q n part x − 1 = Φn d|n,d<n Φd . By the induction assumption, xn − 1 = Φn · g(x) where both xn − 1, g(x) ∈ Z[x], which imply that Φn ∈ Q[x]. Gauss’s Lemma not implies that any factorization in Q[x] induces a factorization in Z[x] simply by multiplying by constants, however all the polynomials are monic. Therefore Φn ∈ Z[x]. f). f (x) irreducible ⇐⇒ f (x + 1) is irreducible is clear, since h(x)g(x) = f (x) ⇐⇒ p p −1 h(x + 1)g(x + 1) = f (x + 1). Now, Φp (x) = xx−1 , so Φp (x + 1) = (x+1)x −1 . Using i−1 Pp p the binomial formula we get the following polynomial: . We get a monic i=1 pi x polynomial where all the rest of the coefficient are a multiple of p, but the constant coefficient is not a multiple of p2 . By Eisenstein, the polynomial is irreducible. 4 NOTE: we use the notation from the previous question. √ √ 1 √ 0 p p p The roots of the polynomial (in C) are: aζ , aζ , ..., aζpp−1 . Denote K as the splitting p p √ √ p p field. Clearly, K ⊂ Q( a, ζp ). For the opposite including observe ( aζp )p = aζp ⇒ ζp ∈ √ p K ⇒ a ∈ K. Without loss of generality, we may assume that a is an integer, since if √ √ √ √ p p √ p √ p−1 p−1 p−1 p √ p p (bc ) b c b a = b/c, then p a = √ = b cc = where c, b ∈ Z. This means that pc = c c 5 the splitting field of xp − a is the same as the splitting field of xp − d where d = bcp−1 . Also d is not a p-th power since b/c is not. Note that if d has a prime factor q which divides d exactly once, then the irreducibility of xp − d follows from Eisenstein and Gauss’s lemma. If every prime factor of d appears with multiplicity two or more, we can not directly apply Eisenstein’s criterion; we need to replace the polynomial xp − d by another polynomial xp − d0 which has the same splitting field but in which some prime factor of d0 appears with multiplicity exactly one. Since d is not a p-th power, there is a prime factor q which appears in d with multiplicity k, where (k, p) = 1. Write Say d = aq k . Since k is a generator for the cyclic group Z/pZ, there is an m such p that km = 1pmodulo p. Note that dm = am q 1+pr = (am q)(q r )p for some r. Observe that Q( p (d)) = Q( p (d0 )) where d0 = (am q). So we can work with the splitting field of xp − d0 instead. √ 0 √ 0 In order to find the degree [Q( p d , ζp ) : Q], notice that [Q( p d ) : Q] = p as simple extension with minimal irreducible polynomial xp − d0 and [Q(ζp ) : Q] = p − 1 as a simple extension with minimal irreducible polynomial Φp . Thus, these intermediate fields have degrees p and √ 0 p − 1 relatively over Q, showing that p and p − 1 both divide [Q( p d , ζp ) : Q]. Since p and √ 0 p − 1 are relatively prime, we see that p(p − 1) divides [Q( p d , ζp ) : Q]. On the other hand, √ 0 √ 0 clearly p(p − 1) is also an upper bound for [Q( p d , ζp ) : Q], since adjoining p d to Q(ζp ) is a simple extension of degree at most p, since its minimal polynomial must divide xp − d0 . For QR problems, see the Math department website, where answers are posted under Past Exams. 6