Timeline for How to interpret divided powers in Kostant Z Form when passing to a field of characteristic p > 0?

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Dec 5, 2020 at 22:07	history	edited	tyrese	CC BY-SA 4.0	added 2 characters in body
Dec 5, 2020 at 20:23	comment	added	Nicolas Hemelsoet		I think they are "correct" (i.e better behaved) analogous of $e_{ij}$ in positive characteristic. For example, $d_p := (\partial/ \partial x)^p$ is always zero in positive characteristic, so a better analogous of this differential operator is $d' := d_p/p!$.
Dec 5, 2020 at 18:48	comment	added	LSpice		I think the answer is the most naïve possible: you don't interpret them. That is, the whole point is that the elements $e_{i j}^r$ themselves don't exist, or at least are $0$ (unless $r < p$), but only the divided power, which should be thought of as a formal symbol, albeit denoted by a very suggestive notation. Any equality that can be performed in the $\mathbb Z$-form remains valid in any specialisation.
Dec 5, 2020 at 18:40	history	asked	tyrese	CC BY-SA 4.0