Timeline for Classification of certain selfinjective algebras
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 11, 2017 at 15:38 | comment | added | Jeremy Rickard | @Mare Yes, sorry. In the last sentence of my first comment, both instances of "projective" should be "non-projective". | |
| Jan 11, 2017 at 15:37 | comment | added | Mare | $Hom_K(M,M)$ is projective iff $M$ is projective for general hopf algebras. I wonder if such a characterisation exists for $Hom_K(M,N)$. | |
| Jan 11, 2017 at 15:36 | comment | added | Mare | We need $Hom_k(X,Y)$ nonprojective but you said at the end of your 2. comment that it is projective. I think you mean "non-projective" instead. The argument should be taht $M$ and $N$ contain $k$ as a direct summand as representations over $Z$ and thus $Hom_K(M,N)$ has also $K$ as a direct summand and is not projective. | |
| Jan 11, 2017 at 15:33 | comment | added | Jeremy Rickard | @Mare And in answer to your second comment, $\text{Hom}_K(M,N)$ can be projective for the group algebra of any $p$-group of $p$-rank greater than one (i.e., containing a subgroup $C_p\times C_p$). This follows from the theory of varieties for modules. | |
| Jan 11, 2017 at 15:31 | comment | added | Jeremy Rickard | @Mare Your second comment makes the answer to your first comment easier! It follows from Chouinard's Theorem (a module for a finite group algebra $KG$ over a field of characteristic $p$ is projective iff its restriction to every elementary abelian $p$-subgroup of $G$ is projective). The only non-trivial elementary abelian $2$-subgroup of $Q_8$ is its cyclic centre $Z\cong C_2$. So if $M$, $N$ are non-projective, then their restrictions to $Z$ are non-projective, and so $\text{Hom}_K(M,N)$ is projective on restriction to $Z$, and hence projective. | |
| Jan 11, 2017 at 15:28 | comment | added | Mare | For a local (nonsemisimple) Hopf algebra with nonprojective indecomposable modules X,Y one has $Ext^{1}(X,Y)=Ext^{1}(k \otimes_k X,Y) =Ext^{1}(k, Hom_k(X,Y))$ which is always nonzero if $Hom_k(X,Y)$ is not projective. So can $Hom_k(X,Y)$ ever be projective in case the hopf algebra is local? | |
| Jan 11, 2017 at 15:18 | comment | added | Mare | Do you have a reference or quick argument for that? | |
| Jan 11, 2017 at 15:17 | history | answered | Jeremy Rickard | CC BY-SA 3.0 |