Timeline for Permutation covering of a $G$-lattice
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Jun 26, 2015 at 18:34 | comment | added | YCor | Oh, I misread the question, namely that the group acting on the tensor product was the product... | |
| Jun 26, 2015 at 17:43 | comment | added | David Hill | Ahhh! I have been using the diagonal action of $C_p$ on $L$ to make computations. | |
| Jun 26, 2015 at 17:12 | vote | accept | Mikhail Borovoi | ||
| Jun 26, 2015 at 15:58 | comment | added | Frieder Ladisch | @MikhailBorovoi: A reference is Lam, A First Course in Noncommutative Rings, Proposition (19.11). The basic idea in characteristic p is that the augmentation ideal $J$ is nilpotent, namely $J^{|G|}=0$. (The augmentation ideal of $kG$ is the kernel of the map $kG \to k$ sending every group element to $1$.) | |
| Jun 26, 2015 at 15:30 | comment | added | Mikhail Borovoi | @FriederLadisch: You write: "It is, however, well known that the group ring of a $p$-group over a field of characteristic $p$ or over a local ring with residue field of characteristic $p$ is local." Could you please give a reference? Many thanks in advance, | |
| Jun 26, 2015 at 15:03 | comment | added | Frieder Ladisch | @DavidHill: Are you sure there are two one-dimensional submodules for $p=3$? I think for $p=3$ we have two submodules of dimension $2$? | |
| Jun 26, 2015 at 11:27 | history | edited | Frieder Ladisch | CC BY-SA 3.0 |
Proof of indecomposability rewritten
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| Jun 26, 2015 at 9:53 | comment | added | Frieder Ladisch | @YCor: The problem is not only about the decomposition of $J(1,p-1) \otimes J(1,p-1)$, but you have two commuting operators $J(1,p-1) \otimes 1$ and $1\otimes J(1,p-1)$, and the decomposition would have to be with respect to both at the same time. | |
| Jun 26, 2015 at 9:48 | comment | added | Frieder Ladisch | @DavidHill: it is true that $L$ has $p-1$ invariant submodules, corresponding to the decomposition of $L\otimes_{\mathbf{Z}} \mathbf{Q}$ into $p-1$ different irreducibles, but their sum is not all of $L$. Of course, the factor module of $L$ mod this sum is a finite torsion module. | |
| Jun 26, 2015 at 6:18 | comment | added | YCor | PS I did the computation and it seems that in any characteristic the rank of $J(1,k)\otimes J(1,k)$ is $k^2-k$, so the dimension of its kernel is $k$, so its number of Jordan blocks is $k$. So even in characteristic $p$ $J(1,p-1)\otimes J(1,p-1)$ has $p-1$ Jordan blocks, which means that tensoring with characteristic $p$ will not provide any obstruction. | |
| Jun 26, 2015 at 5:31 | comment | added | YCor | It's clear that the invariant subspaces of all nontrivial $C_p$-modules in char. $p$ are nontrivial, because the action is given by a unipotent operator (in $P_i$ the invariants are 1-dimensional, generated by the class of the vector $(0,1,\dots,p-1)$. So the problem in general is about the decomposition of $K=J(1,p-1)\otimes J(1,p-1)$ as Jordan blocks ($J(1,k)$ being the unipotent Jordan block of size $k$); it's enough here to know the number of blocks, that is the dimension of the kernel of $K-1$. | |
| Jun 25, 2015 at 23:53 | comment | added | David Hill | I am having a bit of trouble verifying your answer. In particular, I don't see why $L$ has no invariant subspaces. For example, when $p=3$, I calculated that $L$ has exactly two 1-dimensional submodules. This means there should be a permutation covering of rank $5$. Am I missing something here? | |
| Jun 25, 2015 at 22:14 | history | answered | Frieder Ladisch | CC BY-SA 3.0 |