Skip to main content
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
Search options not deleted user 35416

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.

7 votes
Accepted

Wedderburn decomposition of special linear groups

$\DeclareMathOperator\M{M}\DeclareMathOperator\Gal{Gal}\DeclareMathOperator\End{End}$As has been mentioned in the comments, the question for algebraically closed fields of characteristic $0$ is equiva …
Alex B.'s user avatar
  • 13k
14 votes

Algebra with a certain abelian group as the multiplicative group

I am going to assume that by "algebra" you simply mean a ring. The answer is "no", in general. For example $\mathbb{Z}/5\mathbb{Z}$ is not the unit group of a ring. Indeed, suppose it was the unit gro …
Alex B.'s user avatar
  • 13k
5 votes

Structure theorem for finitely generated Z[G] modules

The answer to your question is 'no'. Even if you limit yourself to modules that are free over $\mathbb{Z}$, there is no classification known. Indeed, if your abelian group is not cyclic or its order i …
Alex B.'s user avatar
  • 13k