12
votes
Accepted
Faithfully injective projective modules
There are certainly finite dimensional algebras for which there are no projective modules that are also injective. An example is the $D_4$ quiver algebra with two incoming and two outgoing arrows to ...
- 16.2k
10
votes
Accepted
Tensor product of irreducible representations of an algebra
By a famous theorem of R. Steinberg, STEINBERG, R. Complete sets of representations of algebras. PROC. Am. Math. Soc. 13 (1962), 746-747, if $V$ is a faithful representation for a monoid $M$, then $T(...
- 38.6k
9
votes
Injection of the Universal enveloping algebra
This question is known as the Isomorphism Problem for enveloping algebras and, in general, the answer is negative. The following counterexample is due to Mikhalev, Umbirbaev and Zolotykh. (See A.A. ...
- 5,878
7
votes
Accepted
Are polynomial algebras over fields (that are not algebraically closed) tame?
The definition of tame representation type usually assumes that you're dealing with a finite dimensional algebra $A$ over a field $k$. In this case, the definition of a one parameter family of modules ...
- 16.2k
7
votes
Tensor product of irreducible representations of an algebra
Here is an example in which $A$ is freely generated by two elements $x,y$. Let $W$ be $\mathbb C^2$ with representation given by
$$
x(a_1,a_2)=(0,a_1)
$$$$
y(a_1,a_2)=(a_2,0).
$$
This is irreducible. ...
- 55.9k
7
votes
Accepted
Given a representation-infinite algebra, when is every AR component infinite?
If $A$ is connected and has infinite representation type, then every component of its Auslander-Reiten quiver is infinite. See, for example, Theorem 5.4 in Assem, Simson and Skowronski’s Elements of ...
- 35.2k
6
votes
Accepted
What's an illustrative example of a tame algebra?
I think the following is an example of a tame algebra where there is more than one component to a moduli space of fixed dimension. I don't know any examples where there are dimension vectors with ...
- 156k
6
votes
Accepted
Enveloping algebra of affine Lie algebra is (not) noetherian
It is conjectured that over fields $k$ of $0$ characteristic, the universal enveloping algebra of infinite dimensional Lie algebra is never left or right Noetherian.
However, only a few cases of this ...
- 3,318
5
votes
Accepted
Trying to understand "a refinement of the Peter–Weyl theorem" by Lusztig
For a complex reductive algebraic group $ G $, the Peter-Weyl theorem gives an isomorphism of $ G \times G $ representations
$$
\mathbb C[G] \cong \oplus_{\lambda} V(\lambda)^* \otimes V(\lambda)
$$
...
- 4,612
5
votes
What's an illustrative example of a tame algebra?
For the Kronecker quiver (two vertices, two arrows in the same direction) and dimension vector (1,1), over an algebraically closed ground field, the indecomposables are naturally parameterized by ...
- 6,282
5
votes
Classification of finite-dimensional (nilpotent) associative algebras
In
Belitskii, Genrich; Lipyanski, Ruvim; Sergeichuk, Vladimir V., Problems of classifying associative or Lie algebras and triples of symmetric or skew-symmetric matrices are wild, Linear Algebra Appl....
- 35.2k
4
votes
The inner product of a Clifford Algebra
$
\newcommand\lcontr{\,\lrcorner\,}
\newcommand\rcontr{\,\llcorner\,}
\newcommand\lcontrr{{\rfloor}}
\newcommand\rcontrr{{\lfloor}}
\newcommand\form[1]{\langle#1\rangle}
\newcommand\Ext\bigwedge
\...
4
votes
Accepted
Operation of a p'-group on a set of p-power order and fix points
I think the guess in your comment is correct: There is the following result of Glauberman:
Theorem (Glauberman). Let the finite group $G$ act on the finite group $N$ by automorphisms, where $(\...
- 6,919
3
votes
Classification of finite-dimensional (nilpotent) associative algebras
There are already infinitely many isomorphism classes of finite-dimensional commutative associative (nilpotent) $\mathbb{C}$-algebras of rank $\geq 6$. See for example Suprunenko and Tueshkevich's ...
- 7,127
3
votes
What's an illustrative example of a tame algebra?
Any quiver whose graph is affine Dynkin graph $\tilde{D}_4$, $I=0$. If all arrows look at the center, this is related to the 4-subspace problem, which is tame.
- 12.3k
3
votes
Accepted
Literature on the polynomials and equations, in structures with zero-divisors
For associative algebras, as your required, see Plotkin, Algebras with the same (algebraic) geometry, Israel J. Math., 96 (2) (1996), 511–522.
This is, being more precise, part of this nice relatively ...
- 3,318
3
votes
Elementary proof that a central simple algebra over a field having a maximal subfield is a cyclic algebra
The proof uses as an essential ingredient Proposition 2.2.8, which itself relies on Lemma 2.2.9 telling you that the k-algebras in $M_n(k)$ isomorphic to $k^n$ are conjugate to the subalgebra of ...
- 1,017
3
votes
Accepted
Orthogonality in Hilbert algebras and congruence
No way. Take $A = V = \mathbb{C}^2$, giving $A$ the coordinatewise product. Let $e_1$ and $e_2$ be the standard basis vectors of $\mathbb{C}^2$ and take $x = (e_1 - e_2)\otimes e_1$ and $y = (e_1 - ...
- 42.8k
2
votes
Accepted
Jordan-Hölder series of $k$-subalgebras?
Since $A$ is an $n$-dimensional commutative associative unital $\Bbb R$-albebra, it has finite length as a module over itself.
Claim: The length $\ell_A(A)$ as a module over itself equals $n$.
...
- 3,079
2
votes
Accepted
A weak Schur's lemma for non-semisimple finite dimensional algebras
No, not necessarily.
Consider the case $B=kH$, $C=kG$ of finite group algebras over a field $k$, where $H\leq G$. I'll write $\downarrow$ and $\uparrow$ for restriction and induction. This case has a ...
- 35.2k
2
votes
Number of rings with additive group $(\mathbb{Z}_{16})^2$. A341547(16) in OEIS
It is not an answer but rather a method, outlined in https://arxiv.org/abs/1903.01623v1 . We adapt it to the ring $Z_{16}$.
We say an algebra is curled if every element $x$ of order 16 is linearly ...
- 12.3k
2
votes
Accepted
Non-rigid modules and Auslander-Reiten quiver
Let $Y$ be the module in the second row between $N$ and $A$. Using the mesh relations, the composition
$$\tau^{-1}X\to N\to C\to M\to X$$
can be rewritten (up to a sign) as
$$\tau^{-1}X\to N\to Y\to A\...
- 35.2k
2
votes
A problem about extensions of division rings
I find that the answer of this problem is affirmative and obvious by a trick of tensors:
$D\otimes_{E} \bar{F} \cong (D\otimes_{F}F)\otimes_{E} \bar{F}\cong D\otimes_{F}(F\otimes_{E} \bar{F})\cong D\...
- 331
2
votes
Accepted
Infinite-dimensional, non-unital Frobenius algebras
It is studied under the name "nearly Frobenius algebras" in this paper. In Example 3.3, the algebra $\mathbb{C}[[x,x^{-1}]]$ of Laurent series can be endowed with (countably) many unital but ...
- 985
2
votes
Enveloping algebra of affine Lie algebra is (not) noetherian
It is known that enveloping algebras of Kac–Moody algebras are not noetherian. In the affine case, any affine Kac–Moody algebra contains an infinite-dimensional abelian Lie subalgebra, which is enough ...
- 21
1
vote
Enveloping algebra of affine Lie algebra is (not) noetherian
The free Lie algebra in two letters L satisfies the conjecture since U(L) is the tensor algebra of a 2-dimensional vector space.
If A is a matrix of indefinite type in Kac classification, then the ...
- 11
1
vote
Accepted
How to compute the associated reduced ring for this finitely generated algebra?
In $R/xR$, all $xy^l, l<m$ are nilpotent. To see this, notice that $(xy^l)^m= x^{m-l}(xy^m)^l$. Thus, modulo nilpotents, $R/xR$ is generated by the single element $xy^m$ over $k$. The rest is clear....
- 6,262
1
vote
Zhu's algebra for the Virasoro VOA
Since the question got at least one upvote, perhaps I should post my own answer instead of deleting the question.
It can be proved by induction that $(L_0 + L_{-1})b \in O'(M_c)$, i.e., $(L_0+L_{-1})b$...
- 323
1
vote
Is there a short proof for the permutation invariance of this combinatorial map?
One conceptual proof is to observe that it comes from the associativity of the character ring of $\mathrm{PSL}(2,q)$, interpolated (the ring) to every integer $q$. See Corollary 4.6 in this paper. ...
Only top scored, non community-wiki answers of a minimum length are eligible