Why central isogeny of reductive group over general field F map maximal F split torus onto a maximal split F torus

let $$f$$ be a central isogeny of reductive groups over a field F, why $$f$$ map a maximal split $$F$$ torus onto a maximal split $$F$$ torus.

• 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. – LSpice Apr 18 '19 at 13:33
• @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. – yshuai Qin Apr 18 '19 at 22:11
• 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 $f^{-1}(T_{F^{\text{alg}}})$. – LSpice Apr 18 '19 at 23:05
• @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. – yshuai Qin Apr 19 '19 at 0:51
• 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 $Z(G) = \mu_2$. – LSpice Apr 19 '19 at 2:04

The image $$f(T)$$ of a maximal split torus $$T$$ is a split torus of the same dimension, which is contained in a maximal split torus $$T'$$. But the maximal split tori have the same dimension, and so $$f(T)=T'$$ (the maximal split tori are even conjugate, see, for example, Milne 2017, 25.10). [I am assuming that, as the question originally stated, $$f$$ maps a group to itself. Otherwise, you need to use that the two groups have the same split rank.]
@LSpice, I figured out how to show the pre-image of a maximal torus defined over $$F$$ is also defined over $$F$$( geometrical reduced subscheme of $$G_{F}$$). Let $$f: G\rightarrow G^{\prime}$$ be a central isogeny. By base change to $$F_{s}$$( separable closure of $$F$$), all $$U_{\alpha}$$ ( the unipotent group correspondents to the root $$\alpha$$) are defined over $$F_{s}$$. $$T(F_{s})$$ and all $$U_{\alpha}(F_s)$$ generates $$G(F_{s})$$ by Bruhat's decomposition. Central isogeny implies $$f$$ restricted to $$U_{\alpha}(F_{s})$$ is an isomorphism to $$U^{\prime}_{\alpha^{\prime}} (F_{s})$$. Therefore, two maximal torus defined over $$F_{s}$$ of $$G^{\prime}$$ are conjugate by an element in $$f(G(F_{s}))$$. This implies the preimage is defined over $$F_{s}$$. I think we essentially need it is a central isogeny. Do you think it still holds if it is only an isogeny?