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May 24, 2017 at 18:07 comment added Jay Taylor @LSpice Sadly not, it's still sitting on my hard drive. If you want to take a look then I can send you a copy. Just drop me an email.
May 24, 2017 at 7:26 comment added LSpice @JayTaylor, have you posted that preprint?
Jan 26, 2016 at 16:29 comment added Matthias Klupsch Thank you for your quick response. I just wanted to be sure about this and I can imagine how this is going to work on the level of root data.
Jan 25, 2016 at 12:43 comment added Jay Taylor You are right that the argument above is not correct. That is the same counterexample I came to in the end. This result is not so easy to prove correctly. A proper proof of this result using root data will appear in a forthcoming joint paper that I am working on. I was going to come back and update the answer once the preprint had appeared on the arXiv. In the meantime if you want to discuss this more you should just send me an email.
Jan 25, 2016 at 10:19 comment added Matthias Klupsch It has been some time since I asked this question, I know, but lately I came to wonder whether your argument really works, because for example $SL_p \to GL_p$ is a regular embedding but $GL_p \not\cong SL_p \times Z(GL_p)$, right? Maybe you could explain to me where my misunderstanding lies?
May 8, 2015 at 13:58 vote accept Matthias Klupsch
May 8, 2015 at 13:58 comment added Matthias Klupsch You are right, it seems my main problem was that I was not able to show that every regular embedding is isomorphic to one of the form $G \times_Z T$. Thank you for this very detailed answer.
May 8, 2015 at 12:07 history edited Jay Taylor CC BY-SA 3.0
added 82 characters in body
May 8, 2015 at 9:43 history answered Jay Taylor CC BY-SA 3.0