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Apr 22, 2019 at 20:53 answer added yshuai Qin timeline score: 0
Apr 22, 2019 at 1:02 comment added LSpice In your comments, you mentioned pull-backs, but I have been reminded by @anon's answer that your actual question addresses push-forwards. Accordingly, I'm not sure whether your comment about pull-backs was a separate question, or whether I missed your point.
Apr 22, 2019 at 0:32 history edited Mikhail Borovoi CC BY-SA 4.0
Typo in the title corrected
Apr 22, 2019 at 0:28 answer added anon timeline score: 2
Apr 19, 2019 at 2:04 comment added LSpice Ah, I see. If you wish to work with pullbacks in the scheme-theoretic sense (not underlying reduced schemes), then the statement is not true. Let $k = \mathbb F_2((t))$, let $G$ and $G'$ be the group schemes underlying $\ker \mathrm N_{D/k}$ and $D^\times/k^\times$ where $D/k$ is the quaternionic division algebra, and let $f : G \to G'$ be the natural projection. Then the maximal split torus in $G'$ is trivial, but its pullback to $G$ is the non-smooth scheme $Z(G) = \mu_2$.
Apr 19, 2019 at 0:51 comment added yshuai Qin @LSpice, I think the pull back (in scheme sense) is not a smooth subgroup scheme over a non perfect field $F$ in general. The argument works when F is perfect field.
Apr 18, 2019 at 23:05 comment added LSpice This is not the place to get detailed proofs of standard results. One approach (probably not optimal) is to notice that the character lattice of $f^{-1}(T_{F^{\text{alg}}})$ (which is certainly a torus) has the trivial Galois action, so that the $F$-algebra it generates is an $F$-structure for $f^{-1}(T_{F^{\text{alg}}})$.
Apr 18, 2019 at 22:11 comment added yshuai Qin @LSpice Thank you for the explanation. I only know how to show it if F is perfect field. The inverse image of a maximal torus defined over F is also maximal torus defined over F. But I don't know how to show it for reductive group over a non perfect field. Could you explain how to proved it in details.
Apr 18, 2019 at 13:50 review Close votes
Apr 30, 2019 at 3:05
Apr 18, 2019 at 13:33 comment added LSpice Because it induces an isomorphism of rational-ised character lattices (with Galois action). This question is not research level, and can be found in the part of any of the standard books dealing with rationality questions.
Apr 18, 2019 at 6:38 history edited yshuai Qin CC BY-SA 4.0
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Apr 18, 2019 at 3:10 review First posts
Apr 18, 2019 at 4:02
Apr 18, 2019 at 3:09 history asked yshuai Qin CC BY-SA 4.0