Exercises in Classical Ring TheorySpringer Science & Business Media, 9 mai 2006 - 364 pages This useful book, which grew out of the author's lectures at Berkeley, presents some 400 exercises of varying degrees of difficulty in classical ring theory, together with complete solutions, background information, historical commentary, bibliographic details, and indications of possible improvements or generalizations. The book should be especially helpful to graduate students as a model of the problem-solving process and an illustration of the applications of different theorems in ring theory. The author also discusses "the folklore of the subject: the `tricks of the trade' in ring theory, which are well known to the experts in the field but may not be familiar to others, and for which there is usually no good reference". The problems are from the following areas: the Wedderburn-Artin theory of semisimple rings, the Jacobson radical, representation theory of groups and algebras, (semi)prime rings, (semi)primitive rings, division rings, ordered rings, (semi)local rings, the theory of idempotents, and (semi)perfect rings. Problems in the areas of module theory, category theory, and rings of quotients are not included, since they will appear in a later book. T. W. Hungerford, Mathematical Reviews |
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... integers integers modulo n (or Z/nZ) field of rational numbers field of real numbers field of complex numbers finite field with q elements M set of n × n matrices with entries from S S n symmetric group on {1,2,...,n} A n alternating ...
... integers integers modulo n (or Z/nZ) field of rational numbers field of real numbers field of complex numbers finite field with q elements M set of n × n matrices with entries from S S n symmetric group on {1,2,...,n} A n alternating ...
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... integer coefficients. Solution. (1) Let R = k (a, v) and A = a, R. For re R, we have re IR(A) iff r , a, e a R. Writing r = ro + r1 where ro is the constant term of r, we see that r . a = roa; -- r1a e a R iff r1 e acR. This shows that ...
... integer coefficients. Solution. (1) Let R = k (a, v) and A = a, R. For re R, we have re IR(A) iff r , a, e a R. Writing r = ro + r1 where ro is the constant term of r, we see that r . a = roa; -- r1a e a R iff r1 e acR. This shows that ...
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... integer or a rational number. Since End(S(R/A)) is a subring of End(Z(R/A)), we have (7) =⇒ (1) =⇒ (2). Comment ... integers, etc. Therefore, if R/A is any one of these groups, A must be an ideal of R. On the other hand, if R/A ∼= (Zm) ...
... integer or a rational number. Since End(S(R/A)) is a subring of End(Z(R/A)), we have (7) =⇒ (1) =⇒ (2). Comment ... integers, etc. Therefore, if R/A is any one of these groups, A must be an ideal of R. On the other hand, if R/A ∼= (Zm) ...
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... integer such that agg # 0. From O = f; g = (aga' + · · · + ao), g, we have as b = 0, so deg(ajg) < d. But I (ajg) C I g = 0, which contradicts the choice of g. Thus, we must have d = 0 and g e R. For the inductive step, write A = Ba:n) ...
... integer such that agg # 0. From O = f; g = (aga' + · · · + ao), g, we have as b = 0, so deg(ajg) < d. But I (ajg) C I g = 0, which contradicts the choice of g. Thus, we must have d = 0 and g e R. For the inductive step, write A = Ba:n) ...
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... integers. Show that (1 + a)" = 1. Solution. The proof is by induction on r. The case r = 0 being clear, we assume r > 0. Since ma = 0, the binomial theorem gives (1 + a)” = 1 + aob where b is a polynomial in a with integer coefficients ...
... integers. Show that (1 + a)" = 1. Solution. The proof is by induction on r. The case r = 0 being clear, we assume r > 0. Since ma = 0, the binomial theorem gives (1 + a)” = 1 + aob where b is a polynomial in a with integer coefficients ...
Table des matières
Jacobson Radical Theory | 49 |
Introduction to Representation Theory | 99 |
Ordered Structures in Rings 247 | 246 |
Perfect and Semiperfect Rings 325 | 324 |
Name Index | 349 |
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Expressions et termes fréquents
0-divisor 2-primal abelian algebra artinian ring assume automorphism commutative ring conjugate constructed contradiction cyclic Dedekind-finite defined direct product direct summand division ring domain element endomorphism equation Exercise exists fact field finite group finite-dimensional follows group G hence homomorphism hopfian idempotent identity implies indecomposable induction infinite integer inverse irreducible isomorphism J-semisimple Jacobson radical k-algebra kG-module left ideal left primitive Lemma Let G local ring Math maximal ideal maximal left ideal Mn(k Mn(R module multiplication Neumann regular ring nil ideal Nilº noetherian ring noncommutative nonzero polynomial prime ideal primitive rings proof prove R-module R/rad rad kG representation resp right ideal right R-module ring theory semilocal ring semiprime semisimple ring show that rad simple ring Solution stable range subdirect product subgroup submodule subring suffices to show surjective Theorem unit-regular von Neumann regular zero
Informations bibliographiques
| Titre | Exercises in Classical Ring Theory Problem Books in Mathematics |
| Auteur | T.Y. Lam |
| Édition | 2, illustrée |
| Éditeur | Springer Science & Business Media, 2006 |
| ISBN | 0387217711, 9780387217710 |
| Longueur | 364 pages |
| Exporter la citation | BiBTeX EndNote RefMan |