For the past few weeks I've been trying to get myself acquainted with the language and basic theory of linear algebraic group schemes. In an attempt to see whether I have learned enough to read a paper that I once told myself that I would read once I know enough about linear algebraic groups, I have been stumped by the following definition. In the paper, they let $H$ be a linear algebraic group (not nec. smooth nor connected), and they defined $H^{mult}$ to be the "largest quotient group of $H$ which is a group of multiplicative type". (Grammatical nitpicking: They meant "that" instead of "which" of course, otherwise $H^{mult}$ would be trivial.)


Why does this definition make sense? I didn't find an explanation for this phenomenon in any of the texts I've been reading. Is there a standard way for constructing $H^{mult}$?


I'm sure this question is trivial for experts, but unfortunately I'm very far from being an expert at this point...



Why does the expression "the largest quotient of a linear algebraic group that is multiplicative type" make sense?
This is a subtle question, since the formation of that quotient need *not* commute with extension of the ground field (and it is not said if the ground field is algebraically closed, or separably closed, or what).


Let's take up the story at the start; I assume you are working over a field. Let $k$ be a field, and let $G$ be an affine $k$-group scheme of finite type. A $k$-group scheme $M$ is of *multiplicative type* if any of the following equivalent conditions hold: $M$ is a closed $k$-subgroup of a $k$-torus, $M\_K$ is a closed $K$-subgroup of a $K$-torus for some separably closed extension $K/k$, or $M$ is "${\rm{GL}}\_1$-dual" to a discrete ${\rm{Gal}}(k\_s/k)$-module $M$ finitely generated over $\mathbf{Z}$. (In SGA3 a more general notion is allowed, but it is irrelevant for practical purposes or even theoretical developments of practical interest, so let's ignore it.) There are other characterizations as well, but in lieu of a rationale let's not get into that.


The key thing is that a closed $k$-subgroup of a multiplicative type $k$-group $M$ is again of multiplicative type (and likewise for quotients by closed $k$-subgroup schemes), so if $f:G \rightarrow M$ is a $k$-homomorphism then $G/(\ker f)$ is of multiplicative type since $G/(\ker f) \rightarrow M$ is a closed immersion (as for any $k$-homomorphism with trivial kernel between affine $k$-group schemes of finite type). 


Thus, to give an initial $k$-homomorphism from $G$ to multiplicative type $k$-group schemes amounts to finding a closed normal $k$-subgroup $H \subset G$ such that (i) $G/H$ is of multiplicative type and (ii) any closed normal $k$-subgroup $H' \subset G$ such that $G/H'$ is of multiplicative type contains $H$. That is, we seek such an $H$ that is "as small as possible" (in the sense of being contained in all others; it is clear that any closed $k$-subgroup of $G$ containing $H$ is normal and yields a multiplicative type quotient since $G/H$ is commutative and every quotient of a multiplicative type $k$-group modulo a closed $k$-subgroup scheme is again of multiplicative type). 


Now we use Marty's suggestion, expressed in the form of noetherian induction. Namely, if $H\_1, H\_2 \subset G$ are closed normal $k$-subgroups such that each $G/H\_j$ is of multiplicative type then so is $G/(H\_1 \cap H\_2)$ since it is naturally a closed $k$-subgroup scheme of $(G/H\_1) \times (G/H\_2)$. Thus, if $\{H\_i\}$ is the collection of closed normal $k$-subgroups of $G$ such that $G/H\_i$ is of multiplicative type then any finite intersection among the $H\_i$'s is also in the collection. Since $G$ is a noetherian scheme, it follows that by intersecting a sufficiently large finite collection of such $H\_i$'s we get one that is contained in all others. The associated quotient is $G^{\rm{mult}}$, the "maximal multiplicative type quotient" over $k$.


By Galois descent, the formation of $G^{\rm{mult}}$ is compatible with scalar extension to $k\_s$, and so with any separable algebraic extension on $k$. A finer argument using smoothness and specialization shows that the formation of $G^{\rm{mult}}$ commutes with any separable extension on $k$ (not necessarily algebraic, such as the map from a global field to a completion). 




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But if $k$ is not perfect then there is a "problem": the formation of $G^{\rm{mult}}$ need not commute with non-separable extension on $k$. For example, if $k$ is any imperfect field of characteristic $p > 0$ and if $k'/k$ is a nontrivial finite purely inseparable extension of degree $p$ then let $G$ be the Weil restriction ${\rm{R}}\_{k'/k}({\rm{GL}}\_1)$ (informally, this is "${k'}^{\times}$ as a $k$-group"). This $k$-group is smooth, commutative, and connected of dimension $p$, and it contains as its maximal $k$-torus $T$ an evident copy of ${\rm{GL}}\_1$. The quotient $G/T$ is killed by $p$ as we may check on $k\_s$-points since $({k'\_s}^{\times})^p \subset k\_s^{\times}$, so $G/T$ is unipotent. 


Now over $\overline{k}$ we have
$$G\_{\overline{k}} = T\_{\overline{k}} \times \mathcal{R}\_u(G\_{\overline{k}})$$
as for any smooth connected commutative $\overline{k}$-group, so in other words
there is a geometric character $G\_{\overline{k}} \rightarrow {\rm{GL}}\_1$ that restricts to an isomorphism on $T\_{\overline{k}}$. This is visibly the maximal multiplicative type quotient over $\overline{k}$. But as a quotient it does *not* descend to $k$, or even to $k\_s$. Indeed, any such descent would have kernel that descends the geometric unipotent radical, so it suffices to show that for a separable algebraic extension $K/k$ and $K' := k' \otimes\_k K$ (a field since $K/k$ is separable!) the $K$-group $G\_K = {\rm{R}}\_{K'/K}({\rm{GL}}\_1)$ contains *no* nontrivial smooth connected unipotent $K$-subgroup. 


Let $j:U \hookrightarrow G\_K$ be the inclusion of a smooth connected unipotent $K$-subgroup. By the universal property of Weil restriction, this "corresponds" to a $K'$-homomorphism $U\_{K'} \rightarrow {\rm{GL}}\_1$. This latter homomorphism must be trivial, as it maps a smooth connected unipotent $K'$-group to a $K'$-torus (or for a zillion other reasons). Hence, $j$ is trivial, yet it is an inclusion, so $U=1$ as claimed.


Note that in this example $G^{\rm{mult}}$ is nonetheless nontrivial. Indeed, the $p$-power endomorphism of $G$ vanishes on $G/T$, so it is valued in $T$, so that provides a nontrivial $k$-homomorphism $f:G \rightarrow T = {\rm{GL}}\_1$. It is an instructive exercise to check that this quotient map $f$ is $G^{\rm{mult}}$ over $k$. (Hint: consider ${\rm{Hom}}\_k(G, {\rm{GL}}\_1)$ as a subgroup of ${\rm{Hom}}\_{\overline{k}}(G\_{\overline{k}}, {\rm{GL}}\_1)$.)




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As another example of failure to commute with non-separable extension of the ground field, using the same notation as in the preceding example, consider the $k$-group scheme $N = {\rm{R}}\_{k'/k}(\mu\_p)$. This is the $p$-torsion in the preceding example, so $N\_{\overline{k}} = \mu\_p \times \mathbf{G}\_a^{p-1}$. Hence, $(N\_{\overline{k}})^{\rm{mult}} = \mu\_p$ via the evident quotient map. This does not descend to a quotient of $N$ over $k$ (or over any separable extension of $k$) by a similar argument to the preceding example (using that there is no nontrivial homomorphism over a field from a unipotent group to a multiplicative type group), so $N^{\rm{mult}}=1$.


