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.
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$\begingroup$ 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. $\endgroup$– LSpiceCommented Apr 18, 2019 at 13:33
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$\begingroup$ @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 QinCommented Apr 18, 2019 at 22:11
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$\begingroup$ 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 $f^{-1}(T_{F^{\text{alg}}})$. $\endgroup$– LSpiceCommented Apr 18, 2019 at 23:05
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$\begingroup$ @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 QinCommented Apr 19, 2019 at 0:51
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$\begingroup$ 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 $Z(G) = \mu_2$. $\endgroup$– LSpiceCommented Apr 19, 2019 at 2:04
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2 Answers
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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.]
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$\begingroup$ Just to emphasise, it's split because the isogeny induces a Galois-equivariant isomorphism of ratinalised character lattices. $\endgroup$– LSpiceCommented Apr 22, 2019 at 12:35
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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. $\endgroup$ Commented Apr 22, 2019 at 19:05
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$\begingroup$ @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 pre-image of a torus need not be a torus, but the reduced scheme underlying the pre-image is a torus, with the same image, and the rest of the argument goes through. $\endgroup$– LSpiceCommented Apr 22, 2019 at 20:12
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@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?