DOCSLIB.ORG

Math 154. Algebraic Number Theory 11

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
Math 154. Algebraic Number Theory 11 MATH 154. ALGEBRAIC NUMBER THEORY LECTURES BY BRIAN CONRAD, NOTES BY AARON LANDESMAN CONTENTS 1. Fermat’s factorization method 2 2. Quadratic norms 8 3. Quadratic factorization 14 4. Integrality 20 5. Finiteness properties of OK 26 6. Irreducible elements and prime ideals 31 7. Primes in OK 37 8. Discriminants of number fields 41 9. Some monogenic integer rings 48 10. Prime-power cyclotomic rings 54 11. General cyclotomic integer rings 59 12. Noetherian rings and modules 64 13. Dedekind domains 69 14. Prime ideal factorization 74 15. Norms of ideals 79 16. Factoring pOK: the quadratic case 85 17. Factoring pOK: the general case 88 18. Ramification 93 19. Relative factorization and rings of fractions 96 20. Localization and prime ideals 101 21. Applications of localization 105 22. Discriminant ideals 110 23. Decomposition and inertia groups 116 24. Class groups and units 122 25. Computing some class groups 128 26. Norms and volumes 134 27. Volume calculations 138 28. Minkowski’s theorem and applications 144 29. The Unit Theorem 148 References 153 Thanks to Nitya Mani for note-taking on two days when Aaron Landesman was away. 1 2 BRIAN CONRAD AND AARON LANDESMAN 1. FERMAT’S FACTORIZATION METHOD Today we will discuss the proof of the following result, and highlight some of its main ideas that will be important themes in the course: Theorem 1.1 (Fermat). For x, y 2 Z, the only solutions of y2 = x3 − 2 are (3, ±5). Remark 1.2. This is an elliptic curve (a notion we shall not define, essentially the set of solutions to a certain type of cubic equation in two variables) with infinitely many Q-points, a contrast with finiteness of its set of Z-points. Fermat’s equation can be rearranged into the form x3 = y2 + 2. Lemma 1.3. For any Z-solution (x, y) to x3 = y2 + 2, the value of y must be odd. Proof. Indeed, if y is even then x is even, so x3 is divisible by 8. But y2 + 2 = 4k + 2 is not divisible by 8. Fermat’s first great idea is to introduce considerations in the ring p p Z[ −2] := fm + n −2 j m, n 2 Zg. The point is that over this ring, the equation x3 = y2 + 2 can be expressed as p p x3 = (y + −2)(y − −2). To find solutions to this, we ask the following two questions: p p Question 1.4. Do y + −2 and y − −2 have “gcd = 1” (meaning no non- trivial factor in common, where “nontrivial” means “not a unit”)? Question 1.5. If we knowp the answer to thep above question is “yes,” can we conclude that both y ± −2 are cubes in Z[ −2]? Recall the definition of unit: Definition 1.6. For R a commutative ring, an element u 2 R is a unit if there exists u0 2 R, so that uu0 = 1. We let R× := f units in Rg; this is an abelian group. Example 1.7. (1) Z× = f±1g. (2) For F a field, F× = F − f0g by the definition of a field. MATH 154. ALGEBRAIC NUMBER THEORY 3 p p (3) We have that 1 + 2 is a unit in Z[ 2] because p p (1 + 2)(−1 + 2) = 1. p p Thus, likewise (1 + 2)n 2 Z[ 2]× for all n 2 Z. Note that if u 2 R× then for any x 2 R we have x = (xu−1)u. Hence, for any consideration of “unique” factorization we must allow for adjust- ing factors by unit multiples (absorbing the inverse unit elsewhere in the factorization). Definition 1.8. A domain (sometimes also called an integral domain) is a nonzero commutative ring R such that if ab = 0 with a, b 2 R then either a = 0 or b = 0. For a domain R, if a, b 2 R − f0g and a j b and b j a then a = bu for u 2 R×. (Indeed, if a = bs and b = at then a = a(ts), so ts = 1 since a 6= 0 and hence s, t 2 R×.) Away from Z+,“ajb, bja ) a = b” is very rare. Example 1.9. Later, we’ll see p p Z[ 2]× = f±1g × (1 + 2)Z. This is non-obvious! p Question 1.10. What is Z[ −2]×? In fact, we’ll answer this more generally: Lemma 1.11. Let d be any non-square integer > 1. We have p Z[ −d]× = f±1g. Proof. Consider p a := u + v −d for u, v 2 Z. Let p a := u − v −d, so 2 2 aa = u + dv 2 Z≥0. Since ab = ab, if ab = 1 then ab = 1 = 1, which implies (aa)(bb) = 1, 4 BRIAN CONRAD AND AARON LANDESMAN so aa = 1 2 2 2 2 because aa = u + dv 2 Z≥0. The equality u + dv = 1 with d > 1 forces v = 0 and then u = ±1. p Remark 1.12. If we try the same argument for Z[ d], we get u2 − dv2 = ±1 with u, v 2 Z (and v 6= 0 for units distinct from ±1). The question of finding such non-trivial units then becomes Pell’s equation, which is u2 − dv2 = 1 and its variant with −1 on the right side. For the case d = 2, Pell’s equation has infinitely many Z-solutions with u, v > 0 (as we will recover later in the course from a more sophisticated point of view), beginning with: (3, 2), (17, 12),... p p by considering powers (1 + 2)2n = (3 + 2 2)n for n > 0. 2 2 The variant equation u − 2v =p−1 also has infinitely many Z-solutions with u, v > 0 by considering (1 + 2)2n+1 with n ≥ 0, such as: (1, 1), (7, 5),... 3 2 We’ll now resume ourp goal of finding the Z-solutions to x = y + 2, first addressing if y ± −2 have a non-unit common factor. The answer is negative: p Lemma 1.13. If d 2 Z[ −2] satisfies p p d j (y + −2) and d j (y − −2) then d is a unit. Proof. First, recall from Lemma 1.3 that y is odd. Observe that p p d j (y + −2) − (y − −2) p = 2 −2 p = ( −2)3. Then, we claim the following sublemma: p p Lemma 1.14. The element −2 2 Z[ −2] is irreducible (i.e., it is a nonzero non-unit such that any factorization as ab must have a or b a unit). p p Proof. Assuming −2 = ab, we have − −2 = ab by conjugating both sides. Multiplying these relations, 2 = (aa)(bb) 2 Z≥0, so aa = 1 or bb = 1. This implies either a or b is a unit, as desired. MATH 154. ALGEBRAIC NUMBER THEORY 5 p In HW1 it will be shown that Z[ −2] is a UFD, so the irreducibility of p p e p −2 forces d = u −2 for somep 0 ≤ e ≤ 3 and some unit u 2 Z[ −2]. Thus, if d is not a unit then −2 j d. Hence,p to getp a contradictionp (and conclude d is a unit) it is enough to show −2 - (y + −2) in Z[ −2]. Suppose for some u, v 2 Z that p p p y + −2 = −2(u + v −2) p = 2v + u −2. This forces y = 2v to be even, but y is odd by Lemma 1.3. This concludes the proof that d is a unit. p We conclude that in the UFD Z[ −2] we have p p gcd(y + −2, y − −2) = 1. This completes the preparations for: Proof of Theorem 1.1. In any UFD, any nonzero element is a finite product of irreducibles, and by lumping together any irreducibles that agree up to unit multiple we can rewrite such a product as ei u · ∏ pi i with pi pairwise non-associate irreducible elements (non-associate means one is not a unit times another; by irreducibility of the pi’s, this amounts to saying pi - pj for any i 6= j). Since p p (y + −2)(y − −2) = x3 with p p gcd(y + −2, y − −2) = 1, it follows from expressing each of p y ± −2 as a unit multiple of a productp of pairwise non-associate irreducibles that all irreduciblep factors of y ± −2 occur with multiplicity divisible by 3. Therefore, y ± −2 is the productp of a unit and a cube. But now a miracle occurs: we know the units of Z[ −2] by Lemma 1.11, and from this we see that all units are themselves cubes (as −1 = (−1)3)! Hence, p p y + −2 = (a + b −2)3 for some a, b 2 Z. 6 BRIAN CONRAD AND AARON LANDESMAN Therefore, p p y + −2 = (a + b −2)3 p = a(a2 − 6b2) + b(3a2 − 2b2) −2, so 1 = b(3a2 − 2b2), from which we see that b = ±1, so b2 = 1. This implies ±1 = 3a2 − 2 so 3a2 = 2 ± 1. Then 3a2 equals either 3 or 1. The latter is impossible because we cannot have 3a2 = 1 with a 2 Z. Thus, 3a2 = 3, so a = ±1. This implies b = 1, so y = a(a2 − 6b2) = ±(1 − 6) = ±5 and x3 = y2 + 2 = 27, so x = 3, concluding the proof. The above considerations yield the following lessons: (1) For studying Z-solutions to polynomial equations, it’s useful to con- sider arithmetic in larger numberp systems. For example, in this case, it was useful to work in Z[ −2].
Load more

Recommended publications
  • Algorithmic Factorization of Polynomials Over Number Fields
    Rose-Hulman Institute of Technology Rose-Hulman Scholar Mathematical Sciences Technical Reports (MSTR) Mathematics 5-18-2017 Algorithmic Factorization of Polynomials over Number Fields Christian Schulz Rose-Hulman Institute of Technology Follow this and additional works at: https://scholar.rose-hulman.edu/math_mstr Part of the Number Theory Commons, and the Theory and Algorithms Commons Recommended Citation Schulz, Christian, "Algorithmic Factorization of Polynomials over Number Fields" (2017). Mathematical Sciences Technical Reports (MSTR). 163. https://scholar.rose-hulman.edu/math_mstr/163 This Dissertation is brought to you for free and open access by the Mathematics at Rose-Hulman Scholar. It has been accepted for inclusion in Mathematical Sciences Technical Reports (MSTR) by an authorized administrator of Rose-Hulman Scholar. For more information, please contact [email protected]. Algorithmic Factorization of Polynomials over Number Fields Christian Schulz May 18, 2017 Abstract The problem of exact polynomial factorization, in other words expressing a poly- nomial as a product of irreducible polynomials over some field, has applications in algebraic number theory. Although some algorithms for factorization over algebraic number fields are known, few are taught such general algorithms, as their use is mainly as part of the code of various computer algebra systems. This thesis provides a summary of one such algorithm, which the author has also fully implemented at https://github.com/Whirligig231/number-field-factorization, along with an analysis of the runtime of this algorithm. Let k be the product of the degrees of the adjoined elements used to form the algebraic number field in question, let s be the sum of the squares of these degrees, and let d be the degree of the polynomial to be factored; then the runtime of this algorithm is found to be O(d4sk2 + 2dd3).
    [Show full text]
  • Algebraic Number Theory
    Algebraic Number Theory William B. Hart Warwick Mathematics Institute Abstract. We give a short introduction to algebraic number theory. Algebraic number theory is the study of extension fields Q(α1; α2; : : : ; αn) of the rational numbers, known as algebraic number fields (sometimes number fields for short), in which each of the adjoined complex numbers αi is algebraic, i.e. the root of a polynomial with rational coefficients. Throughout this set of notes we use the notation Z[α1; α2; : : : ; αn] to denote the ring generated by the values αi. It is the smallest ring containing the integers Z and each of the αi. It can be described as the ring of all polynomial expressions in the αi with integer coefficients, i.e. the ring of all expressions built up from elements of Z and the complex numbers αi by finitely many applications of the arithmetic operations of addition and multiplication. The notation Q(α1; α2; : : : ; αn) denotes the field of all quotients of elements of Z[α1; α2; : : : ; αn] with nonzero denominator, i.e. the field of rational functions in the αi, with rational coefficients. It is the smallest field containing the rational numbers Q and all of the αi. It can be thought of as the field of all expressions built up from elements of Z and the numbers αi by finitely many applications of the arithmetic operations of addition, multiplication and division (excepting of course, divide by zero). 1 Algebraic numbers and integers A number α 2 C is called algebraic if it is the root of a monic polynomial n n−1 n−2 f(x) = x + an−1x + an−2x + ::: + a1x + a0 = 0 with rational coefficients ai.
    [Show full text]
  • On the Rational Approximations to the Powers of an Algebraic Number: Solution of Two Problems of Mahler and Mend S France
    Acta Math., 193 (2004), 175 191 (~) 2004 by Institut Mittag-Leffier. All rights reserved On the rational approximations to the powers of an algebraic number: Solution of two problems of Mahler and Mend s France by PIETRO CORVAJA and UMBERTO ZANNIER Universith di Udine Scuola Normale Superiore Udine, Italy Pisa, Italy 1. Introduction About fifty years ago Mahler [Ma] proved that if ~> 1 is rational but not an integer and if 0<l<l, then the fractional part of (~n is larger than l n except for a finite set of integers n depending on ~ and I. His proof used a p-adic version of Roth's theorem, as in previous work by Mahler and especially by Ridout. At the end of that paper Mahler pointed out that the conclusion does not hold if c~ is a suitable algebraic number, as e.g. 1 (1 + x/~ ) ; of course, a counterexample is provided by any Pisot number, i.e. a real algebraic integer c~>l all of whose conjugates different from cr have absolute value less than 1 (note that rational integers larger than 1 are Pisot numbers according to our definition). Mahler also added that "It would be of some interest to know which algebraic numbers have the same property as [the rationals in the theorem]". Now, it seems that even replacing Ridout's theorem with the modern versions of Roth's theorem, valid for several valuations and approximations in any given number field, the method of Mahler does not lead to a complete solution to his question. One of the objects of the present paper is to answer Mahler's question completely; our methods will involve a suitable version of the Schmidt subspace theorem, which may be considered as a multi-dimensional extension of the results mentioned by Roth, Mahler and Ridout.
    [Show full text]
  • Subclass Discriminant Nonnegative Matrix Factorization for Facial Image Analysis
    Pattern Recognition 45 (2012) 4080–4091 Contents lists available at SciVerse ScienceDirect Pattern Recognition journal homepage: www.elsevier.com/locate/pr Subclass discriminant Nonnegative Matrix Factorization for facial image analysis Symeon Nikitidis b,a, Anastasios Tefas b, Nikos Nikolaidis b,a, Ioannis Pitas b,a,n a Informatics and Telematics Institute, Center for Research and Technology, Hellas, Greece b Department of Informatics, Aristotle University of Thessaloniki, Greece article info abstract Article history: Nonnegative Matrix Factorization (NMF) is among the most popular subspace methods, widely used in Received 4 October 2011 a variety of image processing problems. Recently, a discriminant NMF method that incorporates Linear Received in revised form Discriminant Analysis inspired criteria has been proposed, which achieves an efficient decomposition of 21 March 2012 the provided data to its discriminant parts, thus enhancing classification performance. However, this Accepted 26 April 2012 approach possesses certain limitations, since it assumes that the underlying data distribution is Available online 16 May 2012 unimodal, which is often unrealistic. To remedy this limitation, we regard that data inside each class Keywords: have a multimodal distribution, thus forming clusters and use criteria inspired by Clustering based Nonnegative Matrix Factorization Discriminant Analysis. The proposed method incorporates appropriate discriminant constraints in the Subclass discriminant analysis NMF decomposition cost function in order to address the problem of finding discriminant projections Multiplicative updates that enhance class separability in the reduced dimensional projection space, while taking into account Facial expression recognition Face recognition subclass information. The developed algorithm has been applied for both facial expression and face recognition on three popular databases.
    [Show full text]
  • Finite Fields: Further Properties
    Chapter 4 Finite fields: further properties 8 Roots of unity in finite fields In this section, we will generalize the concept of roots of unity (well-known for complex numbers) to the finite field setting, by considering the splitting field of the polynomial xn − 1. This has links with irreducible polynomials, and provides an effective way of obtaining primitive elements and hence representing finite fields. Definition 8.1 Let n ∈ N. The splitting field of xn − 1 over a field K is called the nth cyclotomic field over K and denoted by K(n). The roots of xn − 1 in K(n) are called the nth roots of unity over K and the set of all these roots is denoted by E(n). The following result, concerning the properties of E(n), holds for an arbitrary (not just a finite!) field K. Theorem 8.2 Let n ∈ N and K a field of characteristic p (where p may take the value 0 in this theorem). Then (i) If p ∤ n, then E(n) is a cyclic group of order n with respect to multiplication in K(n). (ii) If p | n, write n = mpe with positive integers m and e and p ∤ m. Then K(n) = K(m), E(n) = E(m) and the roots of xn − 1 are the m elements of E(m), each occurring with multiplicity pe. Proof. (i) The n = 1 case is trivial. For n ≥ 2, observe that xn − 1 and its derivative nxn−1 have no common roots; thus xn −1 cannot have multiple roots and hence E(n) has n elements.
    [Show full text]
  • Prime Factorization
    Prime Factorization Prime Number ­ A number with only two factors: ____ and itself Circle the prime numbers listed below 25 30 2 5 1 9 14 61 Composite Number ­ A number that has more than 2 factors List five examples of composite numbers What kind of number is 0? What kind of number is 1? Every human has a unique fingerprint. Similarly, every COMPOSITE number has a unique "factorprint" called __________________________ Prime Factorization ­ the factorization of a composite number into ____________ factors You can use a _________________ to find the prime factorization of any composite number. Ask yourself, "what Factorization two whole numbers 24 could I multiply together to equal the given number?" If the number is prime, do not put 1 x the number. Once you have all prime numbers, you are finished. Write your answer in exponential form. 24 Expanded Form (written as a multiplication of prime numbers) _______________________ Exponential Form (written with exponents) ________________________ Prime Factorization Ask yourself, "what two 36 numbers could I multiply together to equal the given number?" If the number is prime, do not put 1 x the number. Once you have all prime numbers, you are finished. Write your answer in both expanded and exponential forms. Prime Factorization Ask yourself, "what two 68 numbers could I multiply together to equal the given number?" If the number is prime, do not put 1 x the number. Once you have all prime numbers, you are finished. Write your answer in both expanded and exponential forms. Prime Factorization Ask yourself, "what two 120 numbers could I multiply together to equal the given number?" If the number is prime, do not put 1 x the number.
    [Show full text]
  • Performance and Difficulties of Students in Formulating and Solving Quadratic Equations with One Unknown* Makbule Gozde Didisa Gaziosmanpasa University
    ISSN 1303-0485 • eISSN 2148-7561 DOI 10.12738/estp.2015.4.2743 Received | December 1, 2014 Copyright © 2015 EDAM • http://www.estp.com.tr Accepted | April 17, 2015 Educational Sciences: Theory & Practice • 2015 August • 15(4) • 1137-1150 OnlineFirst | August 24, 2015 Performance and Difficulties of Students in Formulating and Solving Quadratic Equations with One Unknown* Makbule Gozde Didisa Gaziosmanpasa University Ayhan Kursat Erbasb Middle East Technical University Abstract This study attempts to investigate the performance of tenth-grade students in solving quadratic equations with one unknown, using symbolic equation and word-problem representations. The participants were 217 tenth-grade students, from three different public high schools. Data was collected through an open-ended questionnaire comprising eight symbolic equations and four word problems; furthermore, semi-structured interviews were conducted with sixteen of the students. In the data analysis, the percentage of the students’ correct, incorrect, blank, and incomplete responses was determined to obtain an overview of student performance in solving symbolic equations and word problems. In addition, the students’ written responses and interview data were qualitatively analyzed to determine the nature of the students’ difficulties in formulating and solving quadratic equations. The findings revealed that although students have difficulties in solving both symbolic quadratic equations and quadratic word problems, they performed better in the context of symbolic equations compared with word problems. Student difficulties in solving symbolic problems were mainly associated with arithmetic and algebraic manipulation errors. In the word problems, however, students had difficulties comprehending the context and were therefore unable to formulate the equation to be solved.
    [Show full text]
  • Algebraic Number Theory Summary of Notes
    Algebraic Number Theory summary of notes Robin Chapman 3 May 2000, revised 28 March 2004, corrected 4 January 2005 This is a summary of the 1999–2000 course on algebraic number the- ory. Proofs will generally be sketched rather than presented in detail. Also, examples will be very thin on the ground. I first learnt algebraic number theory from Stewart and Tall’s textbook Algebraic Number Theory (Chapman & Hall, 1979) (latest edition retitled Algebraic Number Theory and Fermat’s Last Theorem (A. K. Peters, 2002)) and these notes owe much to this book. I am indebted to Artur Costa Steiner for pointing out an error in an earlier version. 1 Algebraic numbers and integers We say that α ∈ C is an algebraic number if f(α) = 0 for some monic polynomial f ∈ Q[X]. We say that β ∈ C is an algebraic integer if g(α) = 0 for some monic polynomial g ∈ Z[X]. We let A and B denote the sets of algebraic numbers and algebraic integers respectively. Clearly B ⊆ A, Z ⊆ B and Q ⊆ A. Lemma 1.1 Let α ∈ A. Then there is β ∈ B and a nonzero m ∈ Z with α = β/m. Proof There is a monic polynomial f ∈ Q[X] with f(α) = 0. Let m be the product of the denominators of the coefficients of f. Then g = mf ∈ Z[X]. Pn j Write g = j=0 ajX . Then an = m. Now n n−1 X n−1+j j h(X) = m g(X/m) = m ajX j=0 1 is monic with integer coefficients (the only slightly problematical coefficient n −1 n−1 is that of X which equals m Am = 1).
    [Show full text]
  • Approximation to Real Numbers by Algebraic Numbers of Bounded Degree
    Approximation to real numbers by algebraic numbers of bounded degree (Review of existing results) Vladislav Frank University Bordeaux 1 ALGANT Master program May,2007 Contents 1 Approximation by rational numbers. 2 2 Wirsing conjecture and Wirsing theorem 4 3 Mahler and Koksma functions and original Wirsing idea 6 4 Davenport-Schmidt method for the case d = 2 8 5 Linear forms and the subspace theorem 10 6 Hopeless approach and Schmidt counterexample 12 7 Modern approach to Wirsing conjecture 12 8 Integral approximation 18 9 Extremal numbers due to Damien Roy 22 10 Exactness of Schmidt result 27 1 1 Approximation by rational numbers. It seems, that the problem of approximation of given number by numbers of given class was firstly stated by Dirichlet. So, we may call his theorem as ”the beginning of diophantine approximation”. Theorem 1.1. (Dirichlet, 1842) For every irrational number ζ there are in- p finetely many rational numbers q , such that p 1 0 < ζ − < . q q2 Proof. Take a natural number N and consider numbers {qζ} for all q, 1 ≤ q ≤ N. They all are in the interval (0, 1), hence, there are two of them with distance 1 not exceeding q . Denote the corresponding q’s as q1 and q2. So, we know, that 1 there are integers p1, p2 ≤ N such that |(q2ζ − p2) − (q1ζ − p1)| < N . Hence, 1 for q = q2 − q1 and p = p2 − p1 we have |qζ − p| < . Division by q gives N ζ − p < 1 ≤ 1 . So, for every N we have an approximation with precision q qN q2 1 1 qN < N .
    [Show full text]
  • DISCRIMINANTS in TOWERS Let a Be a Dedekind Domain with Fraction
    DISCRIMINANTS IN TOWERS JOSEPH RABINOFF Let A be a Dedekind domain with fraction field F, let K=F be a finite separable ex- tension field, and let B be the integral closure of A in K. In this note, we will define the discriminant ideal B=A and the relative ideal norm NB=A(b). The goal is to prove the formula D [L:K] C=A = NB=A C=B B=A , D D ·D where C is the integral closure of B in a finite separable extension field L=K. See Theo- rem 6.1. The main tool we will use is localizations, and in some sense the main purpose of this note is to demonstrate the utility of localizations in algebraic number theory via the discriminants in towers formula. Our treatment is self-contained in that it only uses results from Samuel’s Algebraic Theory of Numbers, cited as [Samuel]. Remark. All finite extensions of a perfect field are separable, so one can replace “Let K=F be a separable extension” by “suppose F is perfect” here and throughout. Note that Samuel generally assumes the base has characteristic zero when it suffices to assume that an extension is separable. We will use the more general fact, while quoting [Samuel] for the proof. 1. Notation and review. Here we fix some notations and recall some facts proved in [Samuel]. Let K=F be a finite field extension of degree n, and let x1,..., xn K. We define 2 n D x1,..., xn det TrK=F xi x j .
    [Show full text]
  • Mop 2018: Algebraic Conjugates and Number Theory (06/15, Bk)
    MOP 2018: ALGEBRAIC CONJUGATES AND NUMBER THEORY (06/15, BK) VICTOR WANG 1. Arithmetic properties of polynomials Problem 1.1. If P; Q 2 Z[x] share no (complex) roots, show that there exists a finite set of primes S such that p - gcd(P (n);Q(n)) holds for all primes p2 = S and integers n 2 Z. Problem 1.2. If P; Q 2 C[t; x] share no common factors, show that there exists a finite set S ⊂ C such that (P (t; z);Q(t; z)) 6= (0; 0) holds for all complex numbers t2 = S and z 2 C. Problem 1.3 (USA TST 2010/1). Let P 2 Z[x] be such that P (0) = 0 and gcd(P (0);P (1);P (2);::: ) = 1: Show there are infinitely many n such that gcd(P (n) − P (0);P (n + 1) − P (1);P (n + 2) − P (2);::: ) = n: Problem 1.4 (Calvin Deng). Is R[x]=(x2 + 1)2 isomorphic to C[y]=y2 as an R-algebra? Problem 1.5 (ELMO 2013, Andre Arslan, one-dimensional version). For what polynomials P 2 Z[x] can a positive integer be assigned to every integer so that for every integer n ≥ 1, the sum of the n1 integers assigned to any n consecutive integers is divisible by P (n)? 2. Algebraic conjugates and symmetry 2.1. General theory. Let Q denote the set of algebraic numbers (over Q), i.e. roots of polynomials in Q[x]. Proposition-Definition 2.1. If α 2 Q, then there is a unique monic polynomial M 2 Q[x] of lowest degree with M(α) = 0, called the minimal polynomial of α over Q.
    [Show full text]
  • Algebraic Numbers, Algebraic Integers
    Algebraic Numbers, Algebraic Integers October 7, 2012 1 Reading Assignment: 1. Read [I-R]Chapter 13, section 1. 2. Suggested reading: Davenport IV sections 1,2. 2 Homework set due October 11 1. Let p be a prime number. Show that the polynomial f(X) = Xp−1 + Xp−2 + Xp−3 + ::: + 1 is irreducible over Q. Hint: Use Eisenstein's Criterion that you have proved in the Homework set due today; and find a change of variables, making just the right substitution for X to turn f(X) into a polynomial for which you can apply Eisenstein' criterion. 2. Show that any unit not equal to ±1 in the ring of integers of a real quadratic field is of infinite order in the group of units of that ring. 3. [I-R] Page 201, Exercises 4,5,7,10. 4. Solve Problems 1,2 in these notes. 3 Recall: Algebraic integers; rings of algebraic integers Definition 1 An algebraic integer is a root of a monic polynomial with rational integer coefficients. Proposition 1 Let θ 2 C be an algebraic integer. The minimal1 monic polynomial f(X) 2 Q[X] having θ as a root, has integral coefficients; that is, f(X) 2 Z[X] ⊂ Q[X]. 1i.e., minimal degree 1 Proof: This follows from the fact that a product of primitive polynomials is again primitive. Discuss. Define content. Note that this means that there is no ambiguity in the meaning of, say, quadratic algebraic integer. It means, equivalently, a quadratic number that is an algebraic integer, or number that satisfies a quadratic monic polynomial relation with integral coefficients, or: Corollary 1 A quadratic number is a quadratic integer if and only if its trace and norm are integers.
    [Show full text]