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 ra.rings-and-algebras
Search options not deleted
user 134942
Non-commutative rings and algebras, non-associative algebras, universal algebra and lattice theory, linear algebra, semigroups. For questions specific to commutative algebra (that is, rings that are assumed both associative and commutative), rather use the tag ac.commutative-algebra.
2
votes
1
answer
64
views
Finistic dimensions under scalar extensions
Let $A$ be a finite dimensional algebra over some field $K$. Denote the finistic dimension of A by fin($A$), that is, the supremum of the projective dimensions of finite generated modules whose projec …
- 775
5
votes
1
answer
97
views
Tensor decomposition under derived equivalence
Let $K$ be a field. Suppose $A$ and $B$ are $K$-algebra and there is a derived equivalence $F:D^b(A)\cong D^b(B)$ between their bounded derived categories. If we assume that $A$ has a tensor decomposi …
- 775
5
votes
2
answers
349
views
Is this quiver with relations of finite representation type
Let $Q=(Q_0,Q_1)$ be the following quiver, $Q_0$ consist of 2 vertices, denoted by 1,2. $Q_1$ consist a loop at 1 called $\gamma$, an arrow $\alpha$ from 1 to 2 and an arrow $\beta$ from 2 to 1. The r …
- 775
8
votes
1
answer
193
views
Is there always a simple module whose Green correspondent is a simple module under some cond...
Let $G$ be a finite group and $KG$ its group algebra over some field $K$ with $\mathrm{char}\ K$ dividing the order of $G$. It's well-known that the Green correspondence is compatible with the Brauer …
- 775