12 the Different and the Discriminant
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APPLICATIONS of GALOIS THEORY 1. Finite Fields Let F Be a Finite Field
CHAPTER IX APPLICATIONS OF GALOIS THEORY 1. Finite Fields Let F be a finite field. It is necessarily of nonzero characteristic p and its prime field is the field with p r elements Fp.SinceFis a vector space over Fp,itmusthaveq=p elements where r =[F :Fp]. More generally, if E ⊇ F are both finite, then E has qd elements where d =[E:F]. As we mentioned earlier, the multiplicative group F ∗ of F is cyclic (because it is a finite subgroup of the multiplicative group of a field), and clearly its order is q − 1. Hence each non-zero element of F is a root of the polynomial Xq−1 − 1. Since 0 is the only root of the polynomial X, it follows that the q elements of F are roots of the polynomial Xq − X = X(Xq−1 − 1). Hence, that polynomial is separable and F consists of the set of its roots. (You can also see that it must be separable by finding its derivative which is −1.) We q may now conclude that the finite field F is the splitting field over Fp of the separable polynomial X − X where q = |F |. In particular, it is unique up to isomorphism. We have proved the first part of the following result. Proposition. Let p be a prime. For each q = pr, there is a unique (up to isomorphism) finite field F with |F | = q. Proof. We have already proved the uniqueness. Suppose q = pr, and consider the polynomial Xq − X ∈ Fp[X]. As mentioned above Df(X)=−1sof(X) cannot have any repeated roots in any extension, i.e. -
Super-Multiplicativity of Ideal Norms in Number Fields
Super-multiplicativity of ideal norms in number fields Stefano Marseglia Abstract In this article we study inequalities of ideal norms. We prove that in a subring R of a number field every ideal can be generated by at most 3 elements if and only if the ideal norm satisfies N(IJ) ≥ N(I)N(J) for every pair of non-zero ideals I and J of every ring extension of R contained in the normalization R˜. 1 Introduction When we are studying a number ring R, that is a subring of a number field K, it can be useful to understand the size of its ideals compared to the whole ring. The main tool for this purpose is the norm map which associates to every non-zero ideal I of R its index as an abelian subgroup N(I) = [R : I]. If R is the maximal order, or ring of integers, of K then this map is multiplicative, that is, for every pair of non-zero ideals I,J ⊆ R we have N(I)N(J) = N(IJ). If the number ring is not the maximal order this equality may not hold for some pair of non-zero ideals. For example, arXiv:1810.02238v2 [math.NT] 1 Apr 2020 if we consider the quadratic order Z[2i] and the ideal I = (2, 2i), then we have that N(I)=2 and N(I2)=8, so we have the inequality N(I2) > N(I)2. Observe that if every maximal ideal p of a number ring R satisfies N(p2) ≤ N(p)2, then we can conclude that R is the maximal order of K (see Corollary 2.8). -
1 Spring 2002 – Galois Theory
1 Spring 2002 – Galois Theory Problem 1.1. Let F7 be the field with 7 elements and let L be the splitting field of the 171 polynomial X − 1 = 0 over F7. Determine the degree of L over F7, explaining carefully the principles underlying your computation. Solution: Note that 73 = 49 · 7 = 343, so × 342 [x ∈ (F73 ) ] =⇒ [x − 1 = 0], × since (F73 ) is a multiplicative group of order 342. Also, F73 contains all the roots of x342 − 1 = 0 since the number of roots of a polynomial cannot exceed its degree (by the Division Algorithm). Next note that 171 · 2 = 342, so [x171 − 1 = 0] ⇒ [x171 = 1] ⇒ [x342 − 1 = 0]. 171 This implies that all the roots of X − 1 are contained in F73 and so L ⊂ F73 since L can 171 be obtained from F7 by adjoining all the roots of X − 1. We therefore have F73 ——L——F7 | {z } 3 and so L = F73 or L = F7 since 3 = [F73 : F7] = [F73 : L][L : F7] is prime. Next if × 2 171 × α ∈ (F73 ) , then α is a root of X − 1. Also, (F73 ) is cyclic and hence isomorphic to 2 Z73 = {0, 1, 2,..., 342}, so the map α 7→ α on F73 has an image of size bigger than 7: 2α = 2β ⇔ 2(α − β) = 0 in Z73 ⇔ α − β = 171. 171 We therefore conclude that X − 1 has more than 7 distinct roots and hence L = F73 . • Splitting Field: A splitting field of a polynomial f ∈ K[x](K a field) is an extension L of K such that f decomposes into linear factors in L[x] and L is generated over K by the roots of f. -
11. Counting Automorphisms Definition 11.1. Let L/K Be a Field
11. Counting Automorphisms Definition 11.1. Let L=K be a field extension. An automorphism of L=K is simply an automorphism of L which fixes K. Here, when we say that φ fixes K, we mean that the restriction of φ to K is the identity, that is, φ extends the identity; in other words we require that φ fixes every point of K and not just the whole subset. Definition-Lemma 11.2. Let L=K be a field extension. The Galois group of L=K, denoted Gal(L=K), is the subgroup of the set of all functions from L to L, which are automorphisms over K. Proof. The only thing to prove is that the composition and inverse of an automorphism over K is an automorphism, which is left as an easy exercise to the reader. The key issue is to establish that the Galois group has enough ele- ments. Proposition 11.3. Let L=K be a finite normal extension and let M be an intermediary field. TFAE (1) M=K is normal. (2) For every automorphism φ of L=K, φ(M) ⊂ M. (3) For every automorphism φ of L=K, φ(M) = M. Proof. Suppose (1) holds. Let φ be any automorphism of L=K. Pick α 2 M and set φ(α) = β. Then β is a root of the minimum polynomial m of α. As M=K is normal, and α is a root of m(x), m(x) splits in M. In particular β 2 M. Thus (1) implies (2). -
Class Group Computations in Number Fields and Applications to Cryptology Alexandre Gélin
Class group computations in number fields and applications to cryptology Alexandre Gélin To cite this version: Alexandre Gélin. Class group computations in number fields and applications to cryptology. Data Structures and Algorithms [cs.DS]. Université Pierre et Marie Curie - Paris VI, 2017. English. NNT : 2017PA066398. tel-01696470v2 HAL Id: tel-01696470 https://tel.archives-ouvertes.fr/tel-01696470v2 Submitted on 29 Mar 2018 HAL is a multi-disciplinary open access L’archive ouverte pluridisciplinaire HAL, est archive for the deposit and dissemination of sci- destinée au dépôt et à la diffusion de documents entific research documents, whether they are pub- scientifiques de niveau recherche, publiés ou non, lished or not. The documents may come from émanant des établissements d’enseignement et de teaching and research institutions in France or recherche français ou étrangers, des laboratoires abroad, or from public or private research centers. publics ou privés. THÈSE DE DOCTORAT DE L’UNIVERSITÉ PIERRE ET MARIE CURIE Spécialité Informatique École Doctorale Informatique, Télécommunications et Électronique (Paris) Présentée par Alexandre GÉLIN Pour obtenir le grade de DOCTEUR de l’UNIVERSITÉ PIERRE ET MARIE CURIE Calcul de Groupes de Classes d’un Corps de Nombres et Applications à la Cryptologie Thèse dirigée par Antoine JOUX et Arjen LENSTRA soutenue le vendredi 22 septembre 2017 après avis des rapporteurs : M. Andreas ENGE Directeur de Recherche, Inria Bordeaux-Sud-Ouest & IMB M. Claus FIEKER Professeur, Université de Kaiserslautern devant le jury composé de : M. Karim BELABAS Professeur, Université de Bordeaux M. Andreas ENGE Directeur de Recherche, Inria Bordeaux-Sud-Ouest & IMB M. Claus FIEKER Professeur, Université de Kaiserslautern M. -
FIELD EXTENSION REVIEW SHEET 1. Polynomials and Roots Suppose
FIELD EXTENSION REVIEW SHEET MATH 435 SPRING 2011 1. Polynomials and roots Suppose that k is a field. Then for any element x (possibly in some field extension, possibly an indeterminate), we use • k[x] to denote the smallest ring containing both k and x. • k(x) to denote the smallest field containing both k and x. Given finitely many elements, x1; : : : ; xn, we can also construct k[x1; : : : ; xn] or k(x1; : : : ; xn) analogously. Likewise, we can perform similar constructions for infinite collections of ele- ments (which we denote similarly). Notice that sometimes Q[x] = Q(x) depending on what x is. For example: Exercise 1.1. Prove that Q[i] = Q(i). Now, suppose K is a field, and p(x) 2 K[x] is an irreducible polynomial. Then p(x) is also prime (since K[x] is a PID) and so K[x]=hp(x)i is automatically an integral domain. Exercise 1.2. Prove that K[x]=hp(x)i is a field by proving that hp(x)i is maximal (use the fact that K[x] is a PID). Definition 1.3. An extension field of k is another field K such that k ⊆ K. Given an irreducible p(x) 2 K[x] we view K[x]=hp(x)i as an extension field of k. In particular, one always has an injection k ! K[x]=hp(x)i which sends a 7! a + hp(x)i. We then identify k with its image in K[x]=hp(x)i. Exercise 1.4. Suppose that k ⊆ E is a field extension and α 2 E is a root of an irreducible polynomial p(x) 2 k[x]. -
6 Ideal Norms and the Dedekind-Kummer Theorem
18.785 Number theory I Fall 2017 Lecture #6 09/25/2017 6 Ideal norms and the Dedekind-Kummer theorem In order to better understand how ideals split in Dedekind extensions we want to extend our definition of the norm map to ideals. Recall that for a ring extension B=A in which B is a free A-module of finite rank, we defined the norm map NB=A : B ! A as ×b NB=A(b) := det(B −! B); the determinant of the multiplication-by-b map with respect to an A-basis for B. If B is a free A-module we could define the norm of a B-ideal to be the A-ideal generated by the norms of its elements, but in the case we are most interested in (our \AKLB" setup) B is typically not a free A-module (even though it is finitely generated as an A-module). To get around this limitation, we introduce the notion of the module index, which we will use to define the norm of an ideal. In the special case where B is a free A-module, the norm of a B-ideal will be equal to the A-ideal generated by the norms of elements. 6.1 The module index Our strategy is to define the norm of a B-ideal as the intersection of the norms of its localizations at maximal ideals of A (note that B is an A-module, so we can view any ideal of B as an A-module). Recall that by Proposition 2.6 any A-module M in a K-vector space is equal to the intersection of its localizations at primes of A; this applies, in particular, to ideals (and fractional ideals) of A and B. -
Math 200B Winter 2021 Final –With Selected Solutions
MATH 200B WINTER 2021 FINAL {WITH SELECTED SOLUTIONS 1. Let F be a field. Let A 2 Mn(F ). Let f = minpolyF (A) 2 F [x]. (a). Let F be the algebraic closure of F . Show that minpolyF (A) = f. (b). Show that A is diagonalizable over F (that is, A is similar in Mn(F ) to a diagonal matrix) if and only if f is a separable polynomial. 2 3 1 1 1 6 7 (c). Let A = 41 1 15 2 M3(F ). Is A diagonalizable over F ? The answer may depend 1 1 1 on the properties of F . Most students did well on this problem. For those that missed part (a), the key is to use that the minpoly is the largest invariant factor, and that invariant factors are unchanged by base field extension (since the rational canonical form must be the same regardless of the field). There is no obvious way to part (a) directly. 2. Let F ⊆ K be a field extension with [K : F ] < 1. In this problem, if you find any results from homework problems helpful you can quote them here rather than redoing them. Note that a commutative ring R is called reduced if R has no nonzero nilpotent elements. (a). Suppose that K=F is separable. Prove that the K-algebra K ⊗F K is reduced, but is not a domain unless K = F . (b). Suppose that K=F is inseparable. Show that the K-algebra K ⊗F K is not reduced. Proof. (a). Since K=F is separable, K = F (α) for some α 2 K, by the theorem of the primitive element. -
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 . -
Super-Multiplicativity of Ideal Norms in Number Fields
Universiteit Leiden Mathematisch Instituut Master Thesis Super-multiplicativity of ideal norms in number fields Academic year 2012-2013 Candidate: Advisor: Stefano Marseglia Prof. Bart de Smit Contents 1 Preliminaries 1 2 Quadratic and quartic case 10 3 Main theorem: first implication 14 4 Main theorem: second implication 19 Introduction When we are studying a number ring R, that is a subring of a number field K, it can be useful to understand \how big" its ideals are compared to the whole ring. The main tool for this purpose is the norm map: N : I(R) −! Z>0 I 7−! #R=I where I(R) is the set of non-zero ideals of R. It is well known that this map is multiplicative if R is the maximal order, or ring of integers of the number field. This means that for every pair of ideals I;J ⊆ R we have: N(I)N(J) = N(IJ): For an arbitrary number ring in general this equality fails. For example, if we consider the quadratic order Z[2i] and the ideal I = (2; 2i), then we have that N(I) = 2 and N(I2) = 8, so we have the inequality N(I2) > N(I)2. In the first chapter we will recall some theorems and useful techniques of commutative algebra and algebraic number theory that will help us to understand the behaviour of the ideal norm. In chapter 2 we will see that the inequality of the previous example is not a coincidence. More precisely we will prove that in any quadratic order, for every pair of ideals I;J we have that N(IJ) ≥ N(I)N(J). -
Algebraic Number Theory Course Notes (Fall 2006) Math 8803, Georgia Tech
Algebraic Number Theory Course Notes (Fall 2006) Math 8803, Georgia Tech Matthew Baker E-mail address: [email protected] School of Mathematics, Georgia Institute of Technol- ogy, Atlanta, GA 30332-0140, USA Contents Preface v Chapter 1. Unique Factorization (and lack thereof) in number rings 1 1. UFD’s, Euclidean Domains, Gaussian Integers, and Fermat’s Last Theorem 1 2. Rings of integers, Dedekind domains, and unique factorization of ideals 7 3. The ideal class group 20 4. Exercises for Chapter 1 30 Chapter 2. Examples and Computational Methods 33 1. Computing the ring of integers in a number field 33 2. Kummer’s theorem on factoring ideals 38 3. The splitting of primes 41 4. Cyclotomic Fields 46 5. Exercises for Chapter 2 54 Chapter 3. Geometry of numbers and applications 57 1. Minkowski’s geometry of numbers 57 2. Dirichlet’s Unit Theorem 69 3. Exercises for Chapter 3 83 Chapter 4. Relative extensions 87 1. Localization 87 2. Galois theory and prime decomposition 96 3. Exercises for Chapter 4 113 Chapter 5. Introduction to completions 115 1. The field Qp 115 2. Absolute values 119 3. Hensel’s Lemma 126 4. Introductory p-adic analysis 128 5. Applications to Diophantine equations 132 Appendix A. Some background results from abstract algebra 139 iii iv CONTENTS 1. Euclidean Domains are UFD’s 139 2. The theorem of the primitive element and embeddings into algebraic closures 141 3. Free modules over a PID 143 Preface These are the lecture notes from a graduate-level Algebraic Number Theory course taught at the Georgia Institute of Technology in Fall 2006. -
NORM GROUPS with TAME RAMIFICATION 11 Is a Homomorphism, Which Is Equivalent to the first Equation of Bimultiplicativity (The Other Follows by Symmetry)
LECTURE 3 Norm Groups with Tame Ramification Let K be a field with char(K) 6= 2. Then × × 2 K =(K ) ' fcontinuous homomorphisms Gal(K) ! Z=2Zg ' fdegree 2 étale algebras over Kg which is dual to our original statement in Claim 1.8 (this result is a baby instance of Kummer theory). Npo te that an étale algebra over K is either K × K or a quadratic extension K( d)=K; the former corresponds to the trivial coset of squares in K×=(K×)2, and the latter to the coset defined by d 2 K× . If K is local, then lcft says that Galab(K) ' Kd× canonically. Combined (as such homomorphisms certainly factor through Galab(K)), we obtain that K×=(K×)2 is finite and canonically self-dual. This is equivalent to asserting that there exists a “suÿciently nice” pairing (·; ·): K×=(K×)2 × K×=(K×)2 ! f1; −1g; that is, one which is bimultiplicative, satisfying (a; bc) = (a; b)(a; c); (ab; c) = (a; c)(b; c); and non-degenerate, satisfying the condition if (a; b) = 1 for all b; then a 2 (K×)2 : We were able to give an easy definition of this pairing, namely, (a; b) = 1 () ax2 + by2 = 1 has a solution in K: Note that it is clear from this definition that (a; b) = (b; a), but unfortunately neither bimultiplicativity nor non-degeneracy is obvious, though we will prove that they hold in this lecture in many cases. We have shown in Claim 2.11 that a less symmetric definition of the Hilbert symbol holds, namely that for all a, p (a; b) = 1 () b is a norm in K( a)=K = K[t]=(t2 − a); which if a is a square, is simply isomorphic to K × K and everything is a norm.