Search Results
| Search type | Search syntax |
|---|---|
| Tags | [tag] |
| Exact | "words here" |
| Author |
user:1234 user:me (yours) |
| Score |
score:3 (3+) score:0 (none) |
| Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
| Views | views:250 |
| Code | code:"if (foo != bar)" |
| Sections |
title:apples body:"apples oranges" |
| URL | url:"*.example.com" |
| Saves | in:saves |
| Status |
closed:yes duplicate:no migrated:no wiki:no |
| Types |
is:question is:answer |
| Exclude |
-[tag] -apples |
| For more details on advanced search visit our help page | |
Results tagged with noncommutative-algebra
Search options not deleted
user 4008
Non-commutative rings and algebras, non-associative algebras. Can be used in combination with ra.rings-and-algebras
6
votes
Homological dimension of a graded ring which is like polynomial ring
I do not know the answer to your first question. As for the next two the answer
is positive; one need only slightly modify standard proofs for the usual
polynomial ring:
If $R$ is a graded ring then …
- 22.6k
3
votes
Accepted
The Jacobson radical of an infinite dimensional algebra
As there seem to be some differing opinions in the comments as to whether all
irreducible representations are finite-dimensional let me give the argument I
had in mind. A module over the path algebra …
- 22.6k
3
votes
Accepted
Morphisms of a simple sheaf over an algebra to its double dual
Any $R$-homomorphism (in fact any $\mathcal O_S$-homomorphism) $M \to M^{**}$ extends to a morphism $M^{**}\to M^{**}$ (as $M$ is locally free in codimension $1$ and $M^{**}$ is the maximal extension …
- 22.6k
23
votes
Why does the Grothendieck group $K_0(R)$ of a ring not depend on our choice of using left mo...
Here is an alternative to Andreas proof (which if you unfold it is not so different): We have a functor $M\mapsto \mathrm{Hom}_R(M,R)=:M^\ast$ which gives both a contravariant functor from left $R$-mo …
- 22.6k