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 Apr 28 comment Computing (the ring structure of) $\mathrm{Ext}^\bullet_R(k,k)$ for $R=k[x]/(x^2)$ It really is just maps of chains; otherwise, the "cohomology" gives something much stronger than just chain homotopy. If you have guessed the differential map (of the endomorphism complex) correctly, then you should see the cocycles are chain maps, and cohomologies are chain maps up to homotopy equivalence. If your differential map does not give you this, try again. Apr 27 comment Computing (the ring structure of) $\mathrm{Ext}^\bullet_R(k,k)$ for $R=k[x]/(x^2)$ Well, that is a short exact sequence of vector spaces, the first term has dimension 1 and the last term has dimension 1; you know that it does not split, what is the only natural choice you can get? For longer ones, look at your projective resolution. Apr 27 comment Computing (the ring structure of) $\mathrm{Ext}^\bullet_R(k,k)$ for $R=k[x]/(x^2)$ Oh, your example is the smallest non-trivial example of Koszul algebra; but may be you should read up on that after you are finished with this. The isomorphism is evident as groups/graded vector space; you just need to write everything out explicitly to see why it is true.... The elements of the endomorphism complex are maps between chain complexes but not necessarily chain maps, but then when you take cohomology, you will be effectively considering the chain maps. The $k$-th degree component is given by maps of degree $k$; the differential is the obvious choice... Apr 27 comment Computing (the ring structure of) $\mathrm{Ext}^\bullet_R(k,k)$ for $R=k[x]/(x^2)$ For ring structure, use the fact that $\mathrm{Ext}_R^\bullet(k,k) \cong H^\bullet( \mathrm{End}_{Ch(R)}(P_k^\bullet))$, where $P_k^\bullet$ is the projective resolution of $k$; the multiplication on the left hand side is the same as the (homology of) the composition of the right. Alternatively, use Koszul theory (but I guess the point you started this computation is to understand Koszul theory with more hands-on experience...?). Apr 11 awarded Yearling Mar 10 comment Intuition/Motivation behind Algebras (R-Algebras, Q-Algebras, etc.) To add on/clarify David's comment, basically by saying something is an $k$-algebra where $k$ is a field, you should always remember it is a ring AND a $k$-vector space. In particular, the natural place to do representation theory is over $k$ (and any extension of $k$), i.e. you want to consider $k$-algebra hom (i.e. ring hom and vector space hom) $\rho: A \to GL_n(k)$, or equivalently $A$-modules which are also $k$-vector spaces. 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.