2004
DOI: 10.1017/cbo9780511841699
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An Introduction to Noncommutative Noetherian Rings

Abstract: This 2004 introduction to noncommutative noetherian rings is intended to be accessible to anyone with a basic background in abstract algebra. It can be used as a second-year graduate text, or as a self-contained reference. Extensive explanatory discussion is given, and exercises are integrated throughout. Various important settings, such as group algebras, Lie algebras, and quantum groups, are sketched at the outset to describe typical problems and provide motivation. The text then develops and illustrates the… Show more

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Cited by 553 publications
(664 citation statements)
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“…For the commutative cases i = 1, 2, D i is the familiar field of rational functions, but the non-commutative D 3 is likely less familiar (see Ref. [37] for general discussion). Note that the inverse in D 3 of an element r of R 3 can be expressed in a similar way as for quaternions, as an element of R 3 divided by a real polynomial, that is, by an element |r| 2 of R 2 .…”
Section: B Syzygy Theorem and Relation With K0(r)mentioning
confidence: 99%
See 1 more Smart Citation
“…For the commutative cases i = 1, 2, D i is the familiar field of rational functions, but the non-commutative D 3 is likely less familiar (see Ref. [37] for general discussion). Note that the inverse in D 3 of an element r of R 3 can be expressed in a similar way as for quaternions, as an element of R 3 divided by a real polynomial, that is, by an element |r| 2 of R 2 .…”
Section: B Syzygy Theorem and Relation With K0(r)mentioning
confidence: 99%
“…The Gaussian elimination algorithm can be carried out in D i , and the resulting solutions form a set of n 0 − m n 0 -component vectors with entries in D i , and are linearly independent over D i . Finally, we can multiply each n 0 -component vector by a common denominator of its entries [37] to obtain vectors with entries in R i . These must be linearly independent over R i , because if not then the original vectors in D n0 i would be linearly dependent over both R i and D i .…”
Section: B Syzygy Theorem and Relation With K0(r)mentioning
confidence: 99%
“…Then R is clearly a domain (hence Armendariz) and J(R) = pR. But R/J(R) is isomorphic to Mat 2 (Z p ) by the argument in [11,Exercise 2A] Considering Lemma 1.1 (3,5), one may naturally ask whether the converse of Lemma 1.1(5) also holds. But the answer is negative as can be seen by R = U n (A) (n ≥ 2) over a reduced ring A.…”
Section: On Radicals When Factor Rings Are Armendarizmentioning
confidence: 99%
“…Then R is a domain and thus nil-semicommutative. However, for any odd prime integer q, there exists a ring isomorphism R/qR ∼ = Mat 2 (Z q ) by the argument in [8,Exercise 2A]. But Mat 2 (Z q ) is not nil-semicommutative by Example 1.7, and thus R/qR cannot be nil-semicommutative.…”
Section: Introductionmentioning
confidence: 99%