9
$\begingroup$

Let $S$ be a fixed base scheme and $G, H$ be group schemes over $S$. Since I am mainly interested in commutative group schemes over fields, we may assume that $G,H$ are commutative and $S$ is a field if this helps.

(1) Let $f:G\to H$ be a morphism of group schemes. To define the cokernel of this map, we need to choose which topology to work with. Some people use the fppf topology (as in van der Geer & Moonen's book) and other people use the fpqc topology (as in Cornell-Silverman). My question is: what is the difference of those two topologies in terms of group schemes? Is fppf quotient and fpqc quotient of group schemes different? Which topology do people prefer when they are working with group schemes?

(2) Let $H$ be a (normal) closed subgroup scheme of $G$. I think there are at least three plausible definitions of the quotient $G/H$:

  1. Categorical quotient: Since $H$ naturally acts on $G$, we can think categorical quotient $G/H$ of the action $H\times G\to G$.

  2. Fppf/fpqc quotient: $G/H$ represents the quotient of $H\to G$ in the category of fppf/fpqc sheaves.

  3. Naive quotient: A group scheme $G/H$ with a surjective (wrt fppf/fpqc topology) map $p:G\to G/H$ such that kernel of $p$ is the inclusion $H\to G$

Are they equivalent in some good situations? In van der Geer & Moonen's book, it is proved that a fppf quotient is also a categorical quotient. But I cannot find proof nor prove other directions.

context of the question (2): Let $f:A\to B$ be an isogeny of abelian varieties with kernel $\ker f$. Then we have the dual exact sequence $0\to \widehat{B}\to \widehat{A}\to \widehat{\ker f}\to 0$. In Milne's book on abelian variety, to prove the dual exact sequence, consider $0\to \ker f\to A\to B\to 0$ as an exact sequence in the category of commutative group schemes over a field and use a long exact sequence with $\text{Hom}(-, \mathbb{G}_m)$. To use the long exact sequence, we need to prove $B$ is $A/\ker f$ as a fppf/fpqc quotient (In fact I don't know which topology to work with. This is why I ask the question (1)...). However, I only know that $B$ is the `naive quotient (3)' $A/\ker f$.

(3) Is the category of commutative group schemes over a field an abelian category? This statement is in Milne's book on abelian variety, but I cannot find proof. The main point is existence of cokernel, i.e. representability of fppf/fpqc quotient. However, I only know the following theorem in Cornell & Silverman,

Theorem. Let $G$ be a finite type $S$-group scheme and let $H$ be a closed subgroup scheme of $G$. If $H$ is proper and flat over $S$ and if $G$ is quasi-projective over $S$, then the quotient sheaf $G/H$ is representable.

and this is too weak to prove our statement.

Also one more quick question: do you know any good reference dealing with sufficiently general group schemes? I know Shatz's paper in Cornell-Silverman, Tate's paper in Cornell-Silvermann-Stevens, and Stix's lecture note, but they focus on finite flat group schemes. Also, I know some other articles & books which mainly focus on affine algebraic groups. Are there some more general references?

Thank you for reading my stupid questions.

$\endgroup$
2
  • $\begingroup$ Have you looked at Milne's book, Algebraic Groups, CUP 2017. That should answer all your questions over fields. $\endgroup$
    – anon
    Commented May 30, 2020 at 20:18
  • $\begingroup$ @anon I take a look at that book, and I think this book contains essentially everything I want. Thank you for your recommendation! Also, I found answers to my question in other books, and I posted it as an answer. $\endgroup$ Commented May 31, 2020 at 8:07

2 Answers 2

3
$\begingroup$

Let me self-respond to my question. First of all, about reference: I found answers to these questions in `Rational Points on Varieties' by Bjorn Poonen. This book contains an excellent summary of essential facts on algebraic groups. Proof of these facts are contained in, of course, SGA 3-1. I still wonder why I cannot think of SGA while searching for reference.

(3) Consider the following theorems:

Theorem. [Theorem 5.2.5 of Poonen's book, Theorem 3.2 of Expose VI$_A$ of SGA 3-1] Let $H$ be a closed normal subgroup scheme of a group scheme of finite type $G$ over an Artinian ring $A$. Then the fppf quotient $G/H$ exists as a group scheme. Also, the quotient map $p:G\to G/H$ is faithfully flat.

Theorem [Theorem 5.2.9 of Poonen's book, Corollary 7.4 of FGA] Let $f:G \to H$ be a homomorphism between algebraic groups over a field. Then $f$ is factored into homomorphism $G\to G/\ker f\to H$, where $G/\ker f\to H$ is a closed immersion.

By combining these two theorems, we can show that the cokernel of a map always exists.

(2) By these two theorems, we know that the fppf quotient always exists. As I mentioned in the question, fppf quotient is also a categorical quotient. Since the categorical quotient is determined by its universal property, the categorical quotient must be the fppf quotient. Hence these two notions of quotient coincide. Equivalence of the fppf quotient and the naive quotient can be shown in a similar way. (I prove the fact in this way because I use the existence of fppf quotient as a Blackbox, but I think this argument is redundant, because in my understanding, what SGA proved is that the categorical quotient is the fppf quotient.)

(1) Since Poonen's book and SGA both use the fppf topology, I think the fppf topology is a better choice. By the above theorems, at least in the commutative algebraic group case, a map between commutative algebraic group schemes is surjective in the category of algebraic group schemes if and only if it is surjective as fppf sheaves. However, I'm not sure whether we can do this with fpqc topology. If we are working with algebraic groups, then everything is finitely presented, so it is hard to imagine that fpqc topology makes any difference. If someone know something about algebraic group schemes over fpqc topology, then please let me know.

$\endgroup$
1
  • 2
    $\begingroup$ As long as the group schemes are all finitely presented, it should make no difference whether one uses the fppf or fpqc topology. However, if you allow non-finitely presented group schemes -- they often occur in the Tannakian formalism -- you should use the fpqc topology; in general, for the correct meaning of $G/H$, the map $G\to G/H$ may fail to be an fppf cover. $\endgroup$ Commented Mar 17, 2021 at 12:31
2
$\begingroup$

So I think for this sort of question (quotients of flat finitely presented group schemes) the best is to use the theory of algebraic stacks and spaces. I am not an expert, so if someone could double check this that would be great.

Let $G$ be an fppf group scheme over a scheme $S$, and $H$ an fppf subgroup scheme of $G$. Let $\mathcal{X}=[G/H]$ be the stack quotient. Since $G \times H \to G \times G$ is an fppf groupoid, it is algebraic and $G \to \mathcal{X}$ is an fppf presentation of $\mathcal{X}$. Since the inertia is $H$, it is fppf, so $\mathcal{X}$ is a gerbe over the fppf sheaf quotient $G/H$ (which is an algebraic space), and so $\mathcal{X} \to G/H$ is smooth. So $G \to G/H$ is fppf, where $G/H$ is the quotient in algebraic spaces (or in fppf sheafs).

Now if $G/H$ is a nice space, for instance qs (this is always the case in practice, for instance it is if $H \to G$ is qc), then it contains an open subscheme. If the base $S$ is a field, then since $G$ acts transitively in $G/H$ by acting on this subscheme we get that $G/H$ is a scheme (this is the same trick as for proving that a group algebraic space over a field is a group scheme. In fact we also have that an abelian algebraic space over a base $S$ is always an abelian scheme but this is harder to prove).

Remark: if $H \to G$ is proper, then $[G/H]$ is separated.

$\endgroup$
2
  • $\begingroup$ fppf is familiar, and I guess qc is 'quasi-compact', but what is qs? $\endgroup$
    – LSpice
    Commented Sep 8, 2020 at 20:25
  • 2
    $\begingroup$ fppf='faithfully flat locally of finite presentation', qc is indeed 'quasi-compact' and qs is 'quasi-separated' $\endgroup$ Commented Sep 9, 2020 at 7:49

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .