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Results tagged with ra.rings-and-algebras
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user 1306
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.
1
vote
Accepted
characterization of strong nilpotent elements
I believe that the counterexample given in the accepted answer to the question you link (strong nilpotent elements) shows that there are strongly nilpotent elements for which $RxR$ is not nilpotent. …
- 16.9k
0
votes
Universal enveloping ring–symmetric algebra isomorphism for Lie rings
It is known if $g$ is free as a $\mathbb{Z}$-module, and fails otherwise, see e.g. P.M.Cohn, A Remark on the Birkhoff-Witt Theorem J. London Math. Soc. (1963) s1-38(1): 197-203
- 16.9k
4
votes
What is growth of ass. algebra with 3 generators and relation a1a2a3 + a2a3a1 +a3a1a2 - a1a...
(edited a bit to cover a few questions about the notation)
A slight simplification of David Speyer's argument: his argument using Groebner bases explains that is we degenerate the relation into $a_1a …
- 16.9k
3
votes
Basis in a factor algebra of free associative algebra by one homogeneous relation
What are the conditions on $P$ such that monomials non-containing the given monomial $m$ from $P$ will form a basis in $A/P$?
In general, there might be some coincidences here (so I am not giving any …
- 16.9k
4
votes
Perspective on the diamond lemma in ring theory
What I believe is really important is that you can use Diamond Lemma to make homological conclusions, and not just prove results on the size of your algebra. For instance, a homogeneous algebra with a …
- 16.9k
2
votes
Perspective on the diamond lemma in ring theory
What I totally did not have in mind when writing my first answer to this in 2012 is that the confluence condition of the Diamond Lemma can be interpreted quite nicely in terms of the Maurer-Cartan equ …
- 16.9k
2
votes
Accepted
Specialization of PBW-algebras over rational function field
I fail to figure out where exactly you see potential problems.
Diamond lemma indeed says that each ambiguity is resolvable, that is for each "common multiple" of some $W_\sigma$ and $W_\tau$, WLOG $ …
- 16.9k
5
votes
Two (other) rings...are they isomorphic?
This is a bit too long for a comment.
Let me denote by $L_k=\sum_i x_i^{k+1}\frac{\partial\phantom{x_i}}{\partial x_i}$ the operators via which vector fields on a line act on polynomials in several …
- 16.9k
5
votes
Where should I search for resolutions?
There is a way to approach it going back to one of the most efficient general computational methods for associative algebras. Namely, if you know a "Groebner basis" of your algebra (a confluent presen …
- 16.9k
1
vote
Vector "product" diagonalization
The answer to your first question is - generally, it is not possible. For instance, let $V=\mathbb{C}$ viewed as a 2-dimensional vector space over $\mathbb{R}$, with the usual product of complex numbe …
- 16.9k
4
votes
Terminology: Algebras where long strings of products are 0?
Just to confirm the accepted answer, I link here relevant definitions from Springer Online Encyclopaedia of Mathematics (definitely more reliable source than MathWorld and even PlanetMath...):
Nilpote …
- 16.9k
2
votes
Accepted
polarization/linearization as in jordan forms
Alternatively, you can polarise right away as follows: if $p(x)$ is homogeneous of degree $n$ (here $x$ may be a variable with values in $\mathbb{R}^k$, e.g. $p(x)=\mathop{\mathrm{tr}}(x^4)$, where $x …
- 16.9k
8
votes
How to recognize a Hopf algebra?
A homological condition that might be useful: in the Hopf case, the Yoneda algebra $Ext_A^\bullet(k,k)$ embeds into the Hochschild cohomology $HH^\bullet(A,A)$, moreover, there is a Gerstenhaber algeb …
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3
votes
Accepted
Are there examples of brace algebras that are not operads?
To record my answer in comments properly: brace algebras coming from operads satisfy one obvious constraint: for every $x$ and sufficiently large $n$ we have $x\{x_1,\ldots,x_n\}=0$, since we cannot p …
- 16.9k
2
votes
Analogy of Gerstenhaber algebra
One possible answer is contained in the paper of Victor Ginzburg and Travis Schedler, "Free products, cyclic homology, and Gauss-Manin connection", https://arxiv.org/abs/0803.3655. You will be in part …
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