Timeline for Conjugacy for p-adic matrices of finite order II
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 12, 2011 at 9:03 | comment | added | Alex B. | Admittedly, my solution to the first question is not terribly explicit. The other problem in trying to find such a counterexample is that the reductions might not be isomorphic for the chosen $M$ and $N$, but might be isomorphic for some other choices of lattice in the same $\mathbb{Q}_p[G]$-modules. | |
| Aug 12, 2011 at 8:20 | comment | added | user91132 | If I was looking for a counterexample, I'd cook up two explicit non-isomorphic $\mathbb{Z}_p[C_{p^3}]$-modules $M$ and $N$ (as in your answer to the first question) with $M/pM$ isomorphic to $N/pN$ and see if they become isomorphic over $\mathbb{Q}_p$. This should be doable in theory. | |
| Aug 11, 2011 at 17:49 | comment | added | Alex B. | @Geoff I think you are right. I am only proving that direct sums of lattices sitting inside the irreducible $\mathbb{Q}_p[G]$-modules cannot have conjugate reductions, unless the modules are conjugate, but that is of course much too weak. | |
| Aug 11, 2011 at 17:42 | comment | added | Geoff Robinson | @ Alex: I am not completely convinced by this argument. It hinges on what you mean by "a sum of such things". A general lattice is not necessarily a direct sum of such lattices (consider for example a projective indecompsable lattice), so I am not sure what you are intending here. | |
| Aug 11, 2011 at 17:19 | history | answered | Alex B. | CC BY-SA 3.0 |