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.

$\begingroup$ Because it induces an isomorphism of rationalised 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. $\endgroup$ – LSpice Apr 18 '19 at 13:33

$\begingroup$ @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. $\endgroup$ – yshuai Qin Apr 18 '19 at 22:11

$\begingroup$ 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 $f^{1}(T_{F^{\text{alg}}})$. $\endgroup$ – LSpice Apr 18 '19 at 23:05

$\begingroup$ @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. $\endgroup$ – yshuai Qin Apr 19 '19 at 0:51

$\begingroup$ Ah, I see. If you wish to work with pullbacks in the schemetheoretic 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 nonsmooth scheme $Z(G) = \mu_2$. $\endgroup$ – 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.]

$\begingroup$ Just to emphasise, it's split because the isogeny induces a Galoisequivariant isomorphism of ratinalised character lattices. $\endgroup$ – LSpice Apr 22 '19 at 12:35

1$\begingroup$ @LSpice sorry I didn't write the question clear. The central isogeny is between two different reductive groups (could be not isomorphism to each other). So we don't assume the split ranks are same. $\endgroup$ – yshuai Qin Apr 22 '19 at 19:05

$\begingroup$ @yshuaiqin, ah, now I see the relevance of pullbacks to your question! This answer explains why the image of a split torus is split. As you argue, the preimage of a torus need not be a torus, but the reduced scheme underlying the preimage is a torus, with the same image, and the rest of the argument goes through. $\endgroup$ – LSpice Apr 22 '19 at 20:12
@LSpice, I figured out how to show the preimage 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?