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Results tagged with noncommutative-algebra
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user 160378
Non-commutative rings and algebras, non-associative algebras. Can be used in combination with ra.rings-and-algebras
6
votes
2
answers
436
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Survey of recent developments of the Gelfand-Kirillov dimension
It is almost two decades since the now classical books by McConnell and Robinson's
[ Noncommutative Noetherian rings. With the cooperation of L. W. Small. Revised edition. Graduate Studies in Mathema …
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1
vote
Survey of recent developments of the Gelfand-Kirillov dimension
Complementing Manuel Norman's excelent answer, recently I've found a very nice survey about the Gelfand-Kirillov dimension, from 2015, by Jason Bell, called Growth Functions.
This survey discusses man …
- 3,318
8
votes
0
answers
218
views
Differential birational equivalence
Suppose the base field algebraically closed and of zero characteristic.
There are two fascinating questions in the intersection of ring theory and algebraic geometry (for which an excellent discussion …
- 3,318
1
vote
1
answer
152
views
Polynomial identities satisfied by the Weyl algebra in prime characteristic
The rank $n$ Weyl $A_n(\mathsf{k})$ algebra over a field $\mathsf{k}$ of zero characteristic does not satisfies any polinomial identity. If it were a PI-algebra, Kaplansky theorem would apply (since t …
- 398
4
votes
0
answers
80
views
On noncommutative transcendence degrees
The original transcendence degree for (noncommutative) division algebras is the Gelfand-Kirillov transcendence degree, due to I. Gelfand and K. Kirillov ([ Sur les corps li´es aux algèbres envoloppant …
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3
votes
0
answers
231
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On the Gelfand-Kirillov Conjecture
The base field $k$ is of zero characteristic.
Notation: $A_{n,s}(k):= A_n(k(x_1,\ldots,x_s))$, the Weyl agebra over a purely transcedental extension of the base field; $F_{n,s}(k)$, the Weyl field, is …
- 3,318
8
votes
non-associative but commutative algebra
The class of Jordan algebras are the most important class of algebras in this direction.
They are defined by the two identities,
(commutativity): $xy=yx$,
(Jordan identity): $(xy)(xx)=x(y(xx))$.
They …
- 3,318