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Jan
20
comment Homomorphisms of modules over a corner ring
Another dual way to look at this is to use the other pair $(-\otimes_R Re_1, Hom_S(Re_1,-))$ of adjunction (you are using the adjunction $(-\otimes_S e_1 R , Hom_R(e_1R , -))$, and $Hom_R(e_1 R,-)\cong \otimes_R Re_1$). Then you obtain the condition which require you have to have $Hom_S(Re_1,e_j R)\cong e_j R$, which is only possible if $Re_1 \cong S = e_1 Re_1$.
Jan
20
comment Homomorphisms of modules over a corner ring
It follows from you edit that you want $Re_1\otimes_{S} e_1 R \cong R$ as bimodule. But $Re_1 \otimes_S e_1 R \to Re_1 R$ given by the multiplication $re_1\otimes e_1 r' \mapsto re_1 r'$ is a surjective $R$-$R$-bimodule map, which embeds strictly into $R$. In case $e_1 = 1_R$, then you need also $(Re_1 R)$ to be projective as left $R$-module and projective as right $R$-module in order to have bimodule isomorphism.
Dec
10
comment Reference request: bounded derived categories and their Auslander-Reiten quivers
Have you read Happel's book "Triangulated categories in the representation theory of finite dimensional algebras"? Then the next step is probably Keller's paper "Orbit triangulated categories".
Nov
29
answered A question about the representation theory of finite dimensional algebra
Nov
29
revised A question about the representation theory of finite dimensional algebra
Applied LaTeX and explained notations, spellings.
Nov
29
suggested approved edit on A question about the representation theory of finite dimensional algebra
Nov
29
comment For any positive integer $n$, is there an irreducible representation of degree $n$?
Is your $G$ the matrix algebra $M_n(\mathbb{C})$? You can check easily that $M_n(\mathbb{C})$ has an irreducible representation of degree $n$ given by the natural action of $n$-by-$n$-matrix on the $n$-dimensional vector space, essentially by what you wrote. Also I don't understand why your argument proves $Z(M_n(\mathbb{C}))=\mathbb{C}I_n$.
Nov
17
comment Verification of Ext groups and projective resolution for S3 over F3
How do you map $P_1$ to $T$? If you map the generator $-s-1$ to the generator $v$ of $T$, then we have $s(-s-1)=-1-s$ (hence maps to itself), but $s(v)=0$; how does this give a $\mathbb{F}_3S_3$-module homomorphism? Or did I misunderstood something you wrote?
Nov
17
answered $e_{i}Ae_{j} \neq 0$ implies that $e_{i}rad(A)e_{j} \neq 0$.
Nov
14
comment $e_{i}Ae_{j} \neq 0$ implies that $e_{i}rad(A)e_{j} \neq 0$.
Nakayama lemma would be an overkill. You started with $Hom_A(Ae_i,Ae_j) = e_i Ae_j \neq 0$, and assume the subspace $Hom_A(Ae_i, rad(Ae_j)) = Hom_A(Ae_i, rad(A) e_j) = e_i rad(A) e_j = 0$, then clearly the quotient (vector space) $Hom_A(Ae_i, (A/rad(A))e_j) = e_i (A/rad(A)) e_j \neq 0$.
Nov
13
comment $e_{i}Ae_{j} \neq 0$ implies that $e_{i}rad(A)e_{j} \neq 0$.
By Wedderburn theorem $B=A/rad(A)$ is just direct sum of matrix algebras, since $e_i$'s are orthogonal, we can assume $B$ is direct sum of the underlying field, where $e_i$ is the projection onto the $i$-th copy. If $e_i rad(A) e_j = 0$, then we have $e_i B e_j\neq 0$, but $e_i Be_j = Hom_B(Be_i,Be_j)$, then Schur lemma tells you this is zero, a contradiction.
Nov
11
comment Difference between $\mathbb{Z} G$-module and $G$-module
When people say $G$-module, they probably have already fixed a ground ring/field $R$ beforehand. Then $RG$-cohomology is just the tenosring of $\mathbb{Z}G$-cohomology with $R$ over $\mathbb{Z}$.
Nov
11
comment $e_{i}Ae_{j} \neq 0$ implies that $e_{i}rad(A)e_{j} \neq 0$.
Follows from Schur lemma; also radical-square zero is not necessary.
Nov
9
comment An opposite to the process of forming extensions
I am not sure how one can answer this question, surely there will be rings for which one can do this classification; I think a more appropriate question is "how?". Also, to make this problem more "fancy", this question is the same as ("up to an appropriate kind of equivalence") understanding the comultiplication of the Hall algebra of your original algebra.
Oct
14
comment Definition of Regular Representation
And the regular representation is just to take $X=G$. I don't understand what you mean by $x\in K$, and I am guessing you are confused by some notations. Here $\bigoplus_{h\in G} K$ means the $|G|$-dimensional $K$-vectorspace with basis given by elements of $G$.
Oct
14
comment Definition of Regular Representation
Why the map should be in GL? Well, this is really by definition. If you have a finite set $X=\{x_1,\ldots,x_n\}$, then you can extend it linearly to a vector space $V = K$-span$\{x_1,\ldots,x_n\}$ with basis $X$. Since $G$-action on $X$ means that $g(x)\in X$ for all $x\in X$, so for any $v\in V$, we have $g(v) = g(\sum_{i=1}^n \lambda_i x_i)$ which is by definition $\sum_{i=1}^n\lambda_i g(x_i)$ and hence is back in $V$. Since group has inverse, and a representation is a homomorphism, so the transforamtion represented by $g$ is invertible, hence it is in GL.
Aug
31
comment Quiver algebra as a wreath product?
It seems to me the first important issue I can clarify for you is that the wreath product of an algebra $A$ is given as follows. Take the $n$-fold tensor product $A^{\otimes n}$ of $A$ over the underlying field (or some nice ring), then the symmetric group of rank $n$ acts on $A^{\otimes n}$ by permuting the coordinates. So now you can define the semi-direct product of $A^{\otimes n}$ with $k\mathfrak{S}_n$, which is what we call the wreath product $A\wr\mathfrak{S}_n$. For more details, you can refer to J.Chuang-K.M.Tan's Reprsentations of wreath product algebras
Aug
31
comment Quiver algebra as a wreath product?
I am having trouble understanding your terminologies and notation (as well as the question). First, it seems to me you are asking about quiver Hecke algebras, not quiver algebra; they are two completely different thing. A quiver algebra is the algebra with basis given by compositions of arrows in a quiver. On the other hand, a quiver Hecke algebra, is an algebra defined by Hecke-like relation determined the structure of the quiver. Second, what is $k[x]^{\Omega}$? Third, what do you mean by $k[x]$ indexed by $\Omega^n$ or $\Omega$?
Jul
14
comment Question concerning a self-injective algebra and a faithful module
Faithfulness of $M$ means we have ses $0\to A\to M^r \to C\to 0$ (where $C$ is the cokernel of the injection). Self-injectivity of $A$ means that this ses splits and in particular $add(M)=add(A)$. The final question follows from the theorem (Double centraliser property of $A$ and $B$)
May
1
comment How much information about $R-\mathrm{Mod}$ can be extracted from $\underline{R-\mathrm{Mod}}$ and $K_0(R)$?
In that case, I think it is quite difficult to just use $K_0(R)$ to reconstruct $R$ - describing $R$ is the same as asking describing morphisms (in the ordinary module cat.) between the indec. projectives. $K_0(R)$ effectively forgets morphisms between indec projectives, how can we use information from $\underline{\mathrm{mod}}-R$ to get back the morphisms between them? The problem when we try to reconstruct $R$ (or any stably equiv. ring) from $\underline{\mathrm{mod}}-R$ is that many maps got killed in the stable category, and you don't know which one got killed.