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Let $G$ be a reductive algebraic group and denote by $G^{\operatorname{der}}$ its derived group (as defined in [1], 6.15). In [1], 17.28, it is stated that the quotient of $G$ by $G^{\operatorname{der}}$, let's denote it by $T$, is a torus, such that there is an exact sequence $$ 1 \rightarrow G^{\operatorname{der}} \rightarrow G \rightarrow T \rightarrow 1. $$ It is not clear to me why this is the case.

It holds in the isogeny category by [2], 22.125, but I do not understand why it should hold "on the nose."

[1] Milne, J.S.: Reductive Groups. https://www.jmilne.org/math/CourseNotes/RG.pdf
[2] Milne, J.S.: Algebraic Groups. https://www.jmilne.org/math/CourseNotes/iAG200.pdf

fklein
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  • As the quotient of a smooth connected algebraic group, $T$ is smooth and connected. It is also commutative and reductive, so it contains no connected unipotent subgroups, even after extension of the base field. Now, it comes down to the study of smooth connected commutative groups (not trivial). –  May 31 '21 at 13:07
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    We can assume the base field is algebraically closed, and embed $T$ in $GL_{V}$. There exists a basis for $V$ for which $T$ is contained in the upper triangular matrices (linear algebra). Because it is reductive, it is actually contained in the diagonal matrices. Now being smooth and connected it must be a torus. –  May 31 '21 at 13:16

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By the exposition you've cited, there is a torus $T'$ in $G$ such that the multiplication map $T'\times G^{der} \to G$ is an isogeny. The map $T' \to T$ has the same kernel as the map $T'\to G$ and is therefore also an isogeny.

We're reduced to showing that a connected algebraic group which is isogenous to a torus is isomorphic to a torus. By definition of a torus, this is equivalent to showing that a connected algebraic group over an algebraically closed field which is isogenous to $\mathbb{G}_m^n$ is isomorphic to $\mathbb{G}_m^n$. This is true by the classification (see https://mathoverflow.net/questions/102746/structure-of-abelian-connected-complex-linear-algebraic-groups for references); there may also be an easy proof!

hunter
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  • Thanks for the answer. While this makes sense in the case that the base field is algebraically closed, Milne states the result in higher generality, without any assumption on the field. – fklein May 30 '21 at 20:18
  • @Durr let me spell out the reduction of the general case to the algebraically closed case in a bit more detail. You have a map $T'\to T$ of varieties over $k$, where $T$ is a torus, with finite kernel. Base change the map to $\overline{k}$. Now by definition of a torus, $T'{\overline{k}}$ is isomorphic to $\mathbb{G}_m^n$ for some $n$. Once we prove this is also true for $T{\overline{k}}$ then (by definition of a torus) we will know $T$ was a torus. – hunter May 30 '21 at 20:33
  • of course. I've been staring at this problem for too long. Thank you for clarifying. – fklein May 30 '21 at 20:43
  • @Durr You're welcome, althouh after stepping away I suspect this fact might be needed to prove the classification of connected abelian algebraic groups over an alg. closed field, so appealing to the classification is likely circular. There's probably a direct invariant-theory type argument that an algebraic group which is a quotient of $G_m^n$ by a finite group has to be $G_m^n$ without appealing to big theorems. – hunter May 30 '21 at 21:37
  • @hunter Any quotient $Q$ of a torus $T$ is a torus. Since quotients commute with base change we can assume our base field $F$ is algebraically closed. If $R_u(Q)(F)$ is trivial, then so is $R_u(Q)$. But, since $T(F)\to Q(F)$ is surjective, and every element of $T(F)$ is semisimple, so is every element of $Q(F)$ and thus of $R_u(Q)(F)$. But, $R_u(Q)(F)$ consists entirely of unipotent elements. So, $Q$ is reductive, and connected. It's also abelian, and so it suffices to show that an abelian reductive connected group is a torus. But, it's well known that if $T$ is a maximal torus in $Q$ then – Alex Youcis Jun 01 '21 at 20:01
  • $Z_Q(T)=T$, but since $Q$ is abelian we know that $Z_Q(T)=Q$. QED. IDK if you consider any of those results high-powered. You can also just show, by hand, that $Q$ is isomorphic to the torus with cocharacter lattice $X_\ast(T)/X_\ast(K)$ where $K=\ker(T\to Q)$. – Alex Youcis Jun 01 '21 at 20:02