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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.
7
votes
Accepted
Does Manin's construction of non-commutative endomorphism algebra $\mathrm{End}(A)$ produce ...
Q1: This algebra is just the Manin black product of $A$ and $A^!$ (in other words, the Koszul dual of the Segre product of $A$ and $A^!$), and hence it is Koszul. (As requested, the Segre product of t …
- 16.9k
4
votes
Accepted
Second cohomology group of the contact Lie algebra $K_3$
Yes, it is true. In fact, it is true that $H^i(K_{2n+1},F)=0$ for $0<i\le 2n$. This can be deduced from the theorem of Feigin sketched in
Feigin, B.L. Cohomology of contact Lie algebras.
(Russian) C. …
- 16.9k
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
7
votes
Accepted
Subalgebras of quadratic algebras that are not quadratic
Here is an example that shows that you can expect things to get as bad as it goes (I learned about this algebra from the wonderful article The Non-Commutative Gröbner Freaks by Green, Mora, and Ufnaro …
- 16.9k
1
vote
Non-associative deformation quantization
I figured out that in full generality this problem has no chance of leading to a different algebraic structure for which the given one is a quasi-classical limit (like it is for associative/Poisson): …
- 16.9k
5
votes
Is there a $3$-commutative algebra?
I have just thought of another interesting example, which is a bit peculiar, in that the 3-commutativity property arises for a natural subspace of an algebra which itself does not satisfy any identity …
- 16.9k
5
votes
Is there a $3$-commutative algebra?
This variety of algebras is well studied. In particular, the description of the underlying vector space of the free algebra in this variety follows immediately from
"A note on the T-ideal generated by …
- 16.9k
2
votes
Accepted
Weak associativity
Let me assume that the characteristic of the ground field is different from two. Let me start by replacing your identity by something where the existing symmetries are a bit more apparent. I claim tha …
- 16.9k
8
votes
0
answers
111
views
Identity for the associator involving a third root of unity
This is a reference request. I came across the class of nonassociative algebras satisfying the following identity:
$$
(a,b,c)+\omega(b,c,a)+\omega^2(c,a,b)=0.
$$
Here:
by an "algebra" I mean a vect …
- 16.9k
1
vote
Is there a name for a noncommutative generalization of Poisson algebra?
This seems to first have been considered by Dirac under the name "quantum Poisson bracket" - an easy accessible reference is Fock's "Fundamentals of Quantum Mechanics", discussion around formula (2.10 …
- 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
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 …
- 16.9k
7
votes
Accepted
Curious anti-commutative ring
I noticed this now, and I want to remark that the underlying abelian group can in fact be described very precisely. To do that, note that:
(1) the defining relations easily imply that the abelian gro …
- 16.9k
24
votes
Accepted
Does any derivation of commutative algebra preserve its nil-radical?
Suppose $x\in N$, so that $x^n=0$ for some $n$. Then using the product rule for derivations many times, we see that
$$
0=D^n(x^n)=n! D(x)^n+Y,
$$
where $Y$ is divisible by $x$. Therefore,
$D(x)^{n …
- 16.9k
3
votes
Accepted
A commutative variant of the exterior algebra
For $k=\infty$, this algebra appears when studying integrable representations of level 1 of the Lie algebra $\widehat{\mathfrak{sl}}_2$, see, for example, discussion in Section 2 of
A. V. Stoyanovs …
- 16.9k