Simple Algebras and Involutions
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Nilpotent Elements Control the Structure of a Module
Nilpotent elements control the structure of a module David Ssevviiri Department of Mathematics Makerere University, P.O BOX 7062, Kampala Uganda E-mail: [email protected], [email protected] Abstract A relationship between nilpotency and primeness in a module is investigated. Reduced modules are expressed as sums of prime modules. It is shown that presence of nilpotent module elements inhibits a module from possessing good structural properties. A general form is given of an example used in literature to distinguish: 1) completely prime modules from prime modules, 2) classical prime modules from classical completely prime modules, and 3) a module which satisfies the complete radical formula from one which is neither 2-primal nor satisfies the radical formula. Keywords: Semisimple module; Reduced module; Nil module; K¨othe conjecture; Com- pletely prime module; Prime module; and Reduced ring. MSC 2010 Mathematics Subject Classification: 16D70, 16D60, 16S90 1 Introduction Primeness and nilpotency are closely related and well studied notions for rings. We give instances that highlight this relationship. In a commutative ring, the set of all nilpotent elements coincides with the intersection of all its prime ideals - henceforth called the prime radical. A popular class of rings, called 2-primal rings (first defined in [8] and later studied in [23, 26, 27, 28] among others), is defined by requiring that in a not necessarily commutative ring, the set of all nilpotent elements coincides with the prime radical. In an arbitrary ring, Levitzki showed that the set of all strongly nilpotent elements coincides arXiv:1812.04320v1 [math.RA] 11 Dec 2018 with the intersection of all prime ideals, [29, Theorem 2.6]. -
Classifying the Representation Type of Infinitesimal Blocks of Category
Classifying the Representation Type of Infinitesimal Blocks of Category OS by Kenyon J. Platt (Under the Direction of Brian D. Boe) Abstract Let g be a simple Lie algebra over the field C of complex numbers, with root system Φ relative to a fixed maximal toral subalgebra h. Let S be a subset of the simple roots ∆ of Φ, which determines a standard parabolic subalgebra of g. Fix an integral weight ∗ µ µ ∈ h , with singular set J ⊆ ∆. We determine when an infinitesimal block O(g,S,J) := OS of parabolic category OS is nonzero using the theory of nilpotent orbits. We extend work of Futorny-Nakano-Pollack, Br¨ustle-K¨onig-Mazorchuk, and Boe-Nakano toward classifying the representation type of the nonzero infinitesimal blocks of category OS by considering arbitrary sets S and J, and observe a strong connection between the theory of nilpotent orbits and the representation type of the infinitesimal blocks. We classify certain infinitesimal blocks of category OS including all the semisimple infinitesimal blocks in type An, and all of the infinitesimal blocks for F4 and G2. Index words: Category O; Representation type; Verma modules Classifying the Representation Type of Infinitesimal Blocks of Category OS by Kenyon J. Platt B.A., Utah State University, 1999 M.S., Utah State University, 2001 M.A, The University of Georgia, 2006 A Thesis Submitted to the Graduate Faculty of The University of Georgia in Partial Fulfillment of the Requirements for the Degree Doctor of Philosophy Athens, Georgia 2008 c 2008 Kenyon J. Platt All Rights Reserved Classifying the Representation Type of Infinitesimal Blocks of Category OS by Kenyon J. -
Generic Splitting Fields of Central Simple Algebras: Galois Cohomology and Non-Excellence
GENERIC SPLITTING FIELDS OF CENTRAL SIMPLE ALGEBRAS: GALOIS COHOMOLOGY AND NON-EXCELLENCE OLEG T. IZHBOLDIN AND NIKITA A. KARPENKO Abstract. A field extension L=F is called excellent, if for any quadratic form ϕ over F the anisotropic part (ϕL)an of ϕ over L is defined over F ; L=F is called universally excellent, if L · E=E is excellent for any field ex- tension E=F . We study the excellence property for a generic splitting field of a central simple F -algebra. In particular, we show that it is universally excellent if and only if the Schur index of the algebra is not divisible by 4. We begin by studying the torsion in the second Chow group of products of Severi-Brauer varieties and its relationship with the relative Galois co- homology group H3(L=F ) for a generic (common) splitting field L of the corresponding central simple F -algebras. Let F be a field and let A1;:::;An be central simple F -algebras. A splitting field of this collection of algebras is a field extension L of F such that all the def algebras (Ai)L = Ai ⊗F L (i = 1; : : : ; n) are split. A splitting field L=F of 0 A1;:::;An is called generic (cf. [41, Def. C9]), if for any splitting field L =F 0 of A1;:::;An, there exists an F -place of L to L . Clearly, generic splitting field of A1;:::;An is not unique, however it is uniquely defined up to equivalence over F in the sense of [24, x3]. Therefore, working on problems (Q1) and (Q2) discussed below, we may choose any par- ticular generic splitting field L of the algebras A1;:::;An (cf. -
Hyperbolicity of Hermitian Forms Over Biquaternion Algebras
HYPERBOLICITY OF HERMITIAN FORMS OVER BIQUATERNION ALGEBRAS NIKITA A. KARPENKO Abstract. We show that a non-hyperbolic hermitian form over a biquaternion algebra over a field of characteristic 6= 2 remains non-hyperbolic over a generic splitting field of the algebra. Contents 1. Introduction 1 2. Notation 2 3. Krull-Schmidt principle 3 4. Splitting off a motivic summand 5 5. Rost correspondences 7 6. Motivic decompositions of some isotropic varieties 12 7. Proof of Main Theorem 14 References 16 1. Introduction Throughout this note (besides of x3 and x4) F is a field of characteristic 6= 2. The basic reference for the staff related to involutions on central simple algebras is [12].p The degree deg A of a (finite dimensional) central simple F -algebra A is the integer dimF A; the index ind A of A is the degree of a central division algebra Brauer-equivalent to A. Conjecture 1.1. Let A be a central simple F -algebra endowed with an orthogonal invo- lution σ. If σ becomes hyperbolic over the function field of the Severi-Brauer variety of A, then σ is hyperbolic (over F ). In a stronger version of Conjecture 1.1, each of two words \hyperbolic" is replaced by \isotropic", cf. [10, Conjecture 5.2]. Here is the complete list of indices ind A and coindices coind A = deg A= ind A of A for which Conjecture 1.1 is known (over an arbitrary field of characteristic 6= 2), given in the chronological order: • ind A = 1 | trivial; Date: January 2008. Key words and phrases. -
Commutative Algebra
Commutative Algebra Andrew Kobin Spring 2016 / 2019 Contents Contents Contents 1 Preliminaries 1 1.1 Radicals . .1 1.2 Nakayama's Lemma and Consequences . .4 1.3 Localization . .5 1.4 Transcendence Degree . 10 2 Integral Dependence 14 2.1 Integral Extensions of Rings . 14 2.2 Integrality and Field Extensions . 18 2.3 Integrality, Ideals and Localization . 21 2.4 Normalization . 28 2.5 Valuation Rings . 32 2.6 Dimension and Transcendence Degree . 33 3 Noetherian and Artinian Rings 37 3.1 Ascending and Descending Chains . 37 3.2 Composition Series . 40 3.3 Noetherian Rings . 42 3.4 Primary Decomposition . 46 3.5 Artinian Rings . 53 3.6 Associated Primes . 56 4 Discrete Valuations and Dedekind Domains 60 4.1 Discrete Valuation Rings . 60 4.2 Dedekind Domains . 64 4.3 Fractional and Invertible Ideals . 65 4.4 The Class Group . 70 4.5 Dedekind Domains in Extensions . 72 5 Completion and Filtration 76 5.1 Topological Abelian Groups and Completion . 76 5.2 Inverse Limits . 78 5.3 Topological Rings and Module Filtrations . 82 5.4 Graded Rings and Modules . 84 6 Dimension Theory 89 6.1 Hilbert Functions . 89 6.2 Local Noetherian Rings . 94 6.3 Complete Local Rings . 98 7 Singularities 106 7.1 Derived Functors . 106 7.2 Regular Sequences and the Koszul Complex . 109 7.3 Projective Dimension . 114 i Contents Contents 7.4 Depth and Cohen-Macauley Rings . 118 7.5 Gorenstein Rings . 127 8 Algebraic Geometry 133 8.1 Affine Algebraic Varieties . 133 8.2 Morphisms of Affine Varieties . 142 8.3 Sheaves of Functions . -
Nilpotent Ideals in Polynomial and Power Series Rings 1609
PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 138, Number 5, May 2010, Pages 1607–1619 S 0002-9939(10)10252-4 Article electronically published on January 13, 2010 NILPOTENT IDEALS IN POLYNOMIAL AND POWER SERIES RINGS VICTOR CAMILLO, CHAN YONG HONG, NAM KYUN KIM, YANG LEE, AND PACE P. NIELSEN (Communicated by Birge Huisgen-Zimmermann) Abstract. Given a ring R and polynomials f(x),g(x) ∈ R[x] satisfying f(x)Rg(x) = 0, we prove that the ideal generated by products of the coef- ficients of f(x)andg(x) is nilpotent. This result is generalized, and many well known facts, along with new ones, concerning nilpotent polynomials and power series are obtained. We also classify which of the standard nilpotence properties on ideals pass to polynomial rings or from ideals in polynomial rings to ideals of coefficients in base rings. In particular, we prove that if I ≤ R[x]is aleftT -nilpotent ideal, then the ideal formed by the coefficients of polynomials in I is also left T -nilpotent. 1. Introduction Throughout this paper, all rings are associative rings with 1. It is well known that a polynomial f(x) over a commutative ring is nilpotent if and only if each coefficient of f(x) is nilpotent. But this result is not true in general for noncommutative rings. For example, let R = Mn(k), the n × n full matrix ring over some ring k =0. 2 Consider the polynomial f(x)=e12 +(e11 − e22)x − e21x ,wheretheeij’s are 2 the matrix units. In this case f(x) = 0, but e11 − e22 is not nilpotent. -
Problem 1. an Element a of a Ring R Is Called Nilpotent If a M = 0 for Some M > 0. A) Prove That in a Commutative Ring R
Problem 1. An element a of a ring R is called nilpotent if am = 0 for some m> 0. a) Prove that in a commutative ring R the set N of all nilpotent elements of R is an ideal. This ideal is called the nilradical of R. Prove that 0 is the only nilpotent element of R/N. b) Let R be a commutative ring and let a1, ..., an R be nilpotent. Set I for the ∈ ideal < a1, ..., an > generated by a1, ..., an. Prove that there is a positive integer N N such that x1x2...xN = 0 for any x1, ..., xN in I (i.e. that I = 0). c) Prove that the set of all nilpotent elements in the ring M2(R) is not an ideal. d) Prove that if p is a prime and m > 0 then every element of Z/pmZ is either nilpotent or invertible. e) Find the nilradical of Z/36Z (by correspondence theorem, it is equal to nZ/36Z for some n). Solution:a) Suppose that a, b N and r R. Thus an = 0 and bm = 0 for ∈ ∈ some positive integers m, n. By the Newton’s binomial formula we have m+n m + n (a b)m+n = ai( b)m+n−i. (1) − i − i=0 X Note that, for every i 1, 2, ..., m+n , either i n or m+n i m and therefore ∈{ } ≥ − ≥ either ai =0or( b)m+n−i = 0. It follows that every summand in the sum (1) is 0, − so (a b)m+n = 0. Thus a b N. -
Semisimple Cyclic Elements in Semisimple Lie Algebras
Semisimple cyclic elements in semisimple Lie algebras A. G. Elashvili1 , M. Jibladze1 , and V. G. Kac2 1 Razmadze Mathematical Institute, TSU, Tbilisi 0186, Georgia 2 Department of Mathematics, MIT, Cambridge MA 02139, USA Abstract This paper is a continuation of the theory of cyclic elements in semisimple Lie algebras, developed by Elashvili, Kac and Vinberg. Its main result is the classification of semisimple cyclic elements in semisimple Lie algebras. The importance of this classification stems from the fact that each such element gives rise to an integrable hierarchy of Hamiltonian PDE of Drinfeld-Sokolov type. 1 Introduction Let g be a semisimple finite-dimensional Lie algebra over an algebraically closed field F of characteristic 0 and let e be a non-zero nilpotent element of g. By the Morozov- Jacobson theorem, the element e can be included in an sl(2)-triple s = fe; h; fg (unique, up to conjugacy [Ko]), so that [e; f] = h,[h; e] = 2e,[h; f] = −2f. Then the eigenspace decomposition of g with respect to ad h is a Z-grading of g: d M (1.1) g = gj; where g±d 6= 0. j=−d The positive integer d is called the depth of the nilpotent element e. An element of g of the form e + F , where F is a non-zero element of g−d, is called a cyclic element, associated to e. In [Ko] Kostant proved that any cyclic element, associated to a principal (= regular) nilpotent element e, is regular semisimple, and in [Sp] Springer proved that any cyclic element, associated to a subregular nilpotent element of a simple exceptional Lie algebra, is regular semisimple as well, and, moreover, found two more distinguished nilpotent elements in E8 with the same property. -
Are Octonions Necessary to the Standard Model?
Vigier IOP Publishing IOP Conf. Series: Journal of Physics: Conf. Series 1251 (2019) 012044 doi:10.1088/1742-6596/1251/1/012044 Are octonions necessary to the Standard Model? Peter Rowlands, Sydney Rowlands Physics Department, University of Liverpool, Oliver Lodge Laboratory, Oxford St, Liverpool. L69 7ZE, UK [email protected] Abstract. There have been a number of claims, going back to the 1970s, that the Standard Model of particle physics, based on fermions and antifermions, might be derived from an octonion algebra. The emergence of SU(3), SU(2) and U(1) groups in octonion-based structures is suggestive of the symmetries of the Standard Model, but octonions themselves are an unsatisfactory model for physical application because they are antiassociative and consequently not a group. Instead, the ‘octonion’ models have to be based on adjoint algebras, such as left- or right-multiplied octonions, which can be seen to have group-like properties. The most promising of these candidates is the complexified left-multiplied octonion algebra, because it reduces, in effect, to Cl(6), which has been identified by one of us (PR) in a number of previous publications as the basic structure for the entire foundation of physics, as well as the algebra required for the Standard Model and the Dirac equation. Though this algebra has long been shown by PR as equivalent to using a complexified left-multiplied or ‘broken’ octonion, it doesn’t need to be derived in this way, as its real origins are in the respective real, complex, quaternion and complexified quaternion algebras of the fundamental parameters of mass, time, charge and space. -
The Descent of Biquaternion Algebras in Characteristic
The descent of biquaternion algebras in characteristic two Demba Barrya,b, Adam Chapmanc, Ahmed Laghribid aFacult´edes Sciences et Techniques de Bamako, BP: E3206 Bamako, Mali bDepartement Wiskunde–Informatica, Universiteit Antwerpen, Belgium cDepartment of Computer Science, Tel-Hai College, Upper Galilee, 12208 Israel dUniversit´ed’Artois, Facult´edes Sciences Jean Perrin, Laboratoire de math´ematiques de Lens EA 2462, rue Jean Souvraz - SP18, 62307 Lens, France Abstract In this paper we associate an invariant to a biquaternion algebra B over a field K with a subfield F such that K/F is a quadratic separable extension and char(F) = 2. We show that this invariant is trivial exactly when B B0 ⊗ K for some biquaternion algebra B0 over F. We also study the behavior of this invariant under certain field extensions and provide several interesting examples. Keywords: Kato-Milne Cohomology, Cohomological invariants, Algebras with involution, Biquaternion algebras 2010 MSC: primary 11E81; secondary 11E04, 16K20, 19D45 1. Introduction Given a central simple algebra C over a field K with a subfield F, we say that C has a descent to F if there exists a central simple algebra C0 over F such that C C0 ⊗ K. When [K : F] = exp(C), one necessary condition for C to have a descent to F is that corK/F(C) ∼Br F, because if C C0⊗K ⊗[K:F] then corK/F(C) = corK/F(C0 ⊗ K) = C0 ∼Br F, where ∼Br denotes the Brauer equivalence. If K/F is a separable quadratic extension and Q is a quaternion algebra over K, then corK/F(Q) ∼Br F is a necessary and sufficient condition for Q to have a descent to F by [1, Chapter X, Theorem 21]. -
Honors Algebra 4, MATH 371 Winter 2010 Assignment 4 Due Wednesday, February 17 at 08:35
Honors Algebra 4, MATH 371 Winter 2010 Assignment 4 Due Wednesday, February 17 at 08:35 1. Let R be a commutative ring with 1 6= 0. (a) Prove that the nilradical of R is equal to the intersection of the prime ideals of R. Hint: it’s easy to show using the definition of prime that the nilradical is contained in every prime ideal. Conversely, suppose that f is not nilpotent and consider the set S of ideals I of R with the property that “n > 0 =⇒ f n 6∈ I.” Show that S has maximal elements and that any such maximal element must be a prime ideal. Solution: Suppose that f ∈ R is not nilpotent and let S be the set S of ideals I of R with the property that “n > 0 =⇒ f n 6∈ I.” Ordering S by inclusion, note that every chain is bounded above: if I1 ⊆ I2 ⊆ · · · is a chain, then I = ∪Ii is an upper bound which clearly lies in S. By Zorn’s lemma, S has a maximal element, say M, which we claim is prime. Indeed, suppose that uv ∈ M but that u 6∈ M and v 6∈ M. Then the ideals M + (u) and M + (v) strictly contain M so do not belong to S by maximality of M. Thus, there exist m and n such that f n ∈ M + (u) and f m ∈ M + (v). It follows that f m+n ∈ M + (uv) = M and hence that M is not in S, a contradiction. Thus, either u or v lies in M and M is prime. -
Homework 4 Solutions
Homework 4 Solutions p Problem 5.1.4. LetpR = fm + n 2 j m; n 2 Zg. 2 2 (a) Show that m +pn 2 is a unit in R if and only if m − 2n = ±1. (b) Show that 1 + 2 has infinite order in R×. (c) Show that 1 and −1 are the only units that have finite order in R×. p Proof. (a) Define a map N : R ! Z where N(r) = m2 − 2n2 where r = m + n 2. One can verify that N is multiplicative, i.e. N(rs) = N(r)N(s) for r; s 2 R. Thus if r 2 R× then rs = q for some s 2 R. This implies 2 2 that N(r)N(s) = N(rs) = N(1) = 1 andp hence N(rp);N(s) 2 Zpimplies that m − 2n = N(r) = ±p1. 2 2 2 2 × For the converse observep that (m + n 2)(m +pn 2)(m − n 2) = (m − 2np) = 1, i.e. m + n 2 2 R . (b) Note that 1 + 2 > 1 and therefore (1 + 2)k > 1k = 1. Therefore 1 + 2 has infinite order. (c) Since R is a subring of R and the only elements in R with finite order in R× are ±1 it follows that the only elements in R× with finite order are ±1. Problem 5.1.5. Let R be a subset of an integral domain D. Prove that if R 6= f0g is a ring under the operations of D, then R is a subring of D. Proof.