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 |
The study of algebraic structures and properties applying to large classes of such structures. For example, ideas from group theory and ring theory are extended and considered for structures with other signatures (systems of basic or fundamental operations).
3
votes
Generalization of results from specific algebraic theories to Universal Algebra
A Google search immediately brings a few results including Master thesis of Junpei Sekino from 1969 where a generalisation of the Jordan-Hölder theorem for sufficiently arbitrary algebraic structures …
9
votes
Do non-associative objects have a natural notion of representation?
Let me concentrate on your first question (frankly speaking, the way you formulate your second question slightly lacks motivation).
The case where there is a reasonable suggestion, assumes that you wo …
10
votes
Accepted
Reference for an old result of P. M. Cohn
This is in P. M. Cohn, "On the Embedding of Rings in Skew Fields", Proceedings of the London Mathematical Society, Volume s3-11, Issue 1 (1961), Pages 511-530. I do not think that the zero characteris …
17
votes
Accepted
Is the Amitsur-Levitzki identity essentially unique?
The answer to your question is "no", as explained by Anton Klyachko in his answer. Let me refer you to a remarkable statement of Razmyslov and Procesi that describes all identities. They proved (indep …
1
vote
Accepted
Polynomial identities of supercommutative-gradable algebras
I believe that the identity $(xy-yx)z-z(xy-yz)$ generates everything (in characteristic 0, at least). To show that no further identities are needed, it is enough to exhibit one algebra that has no fur …
3
votes
Algebraic axiomatization for AB+BA^T operation on matrices
Including the operation $A\mapsto A^T$ can be viewed, in the language of operads, in many different closely related ways: via adjoining a new unary operation $J$ that satisfies $J^2=\operatorname{id}$ …