# Is the category of affine fppf groups closed under normal quotients?

Let $S$ be a scheme and let $N$, $G$ be affine flat group schemes of finite presentation over $S$. If we assume that $N$ is a closed normal subgroup of $G$, we may form the fppf quotient sheaf $G/N$, which is a sheaf of groups. By using descent and Artin's representability result for algebraic spaces, it follows that $G/N$ is an algebraic space which is flat, separated and of finite presentation over $S$. By a classic result in the case where $S$ is the spectrum of a field, it also follows formally that $G/N$ has affine fibres. My question is:

Is $G/N$ always affine over $S$?

Phrased in another way: Is the category of affine fppf group schemes over $S$ a semi-abelian subcategory of the semi-abelian category of fppf group sheaves over $S$?

Another related question is if a flat, affine, finitely presented group scheme always may be embedded in the automorphism group of a locally free coherent sheaf on the base. There is a remark about this in SGAIII_1 EXPOSE VI_B 11.11.1, where it is stated without proof that this is supposed to be true for some base schemes under certain regularity conditions.

EDIT: A third related question is if someone knows of an example of a separated, flat, finitely presented group algebraic space with affine fibres which is not affine.

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In general, the quotient $G/N$ is not representable. Lemma X.14 of Raynaud's book "Faisceaux amples sur les schémas en groupes et les espaces homogènes" gives a counter-example with $S=\mathbb{A}^2_k$ (which is regular 2-dimensional), $G=(\mathbb{G}_{a,S})^2$ and $N\subset G$ étale over $S$.

However if $S$ is locally noetherian of dimension at most 1, then $G/N$ is representable by a scheme : this is theorem 4.C of Anantharaman's thesis "Schémas en groupes, espaces homogènes et espaces algébriques sur une base de dimension 1". If you assume moreover that $S$ has finite normalization (e.g. $S$ excellent), then $G/N$ is affine if $G$ is. Indeed by Serre's cohomological criterion, affineness is not affected by nilpotents hence you may base-change by $S_{red}\to S$ and hence assume that $S$ is reduced (note that the formation of $G/N$ commutes with base change). Then let $S'\to S$ be the normalization, a finite morphism, and let $G',N'$ be the pullbacks to $S'$. The restriction of $(G/N)'=G'/N'$ to the generic points of $S'$ is affine (since we are then over a field) and because $S'$ is Dedekind, it follows that $(G/N)'$ is affine (Anantharaman prop. 2.3.1). Finally $(G/N)'\to G/N$ is finite surjective hence $G/N$ is affine by Chevalley's theorem.

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Thanks for a very nice answer! It was pointed out to me that the online TeX version of SGA 3 contains an extra section treating my second question. There is a proof (Prop 13.2) for the case when the base is regular noetherian of codimension 2. As for my third question, I obviously didn't think that through properly. Thanks Angelo for giving the simple counterexample. –  Daniel Bergh Jun 8 '11 at 14:42
Thanks for pointing out this new section in SGA3. By the way, since you had assumptions of finite presentation for $G$ and $N$, we can remove the locally noetherian assumption for $S$ in my argument. Indeed things are local on $S$ so we may assume $S$ affine and then everything comes by base change from the spectrum of a finitely generated $\mathbb{Z}$-algebra, which is noetherian and excellent. –  Matthieu Romagny Jun 9 '11 at 12:06

Yes, $G/N$ is always an affine group scheme. This is explained, for example, in one of the last chapters of Waterhouse's book on affine group schemes (an excellent reference for these questions).

 As Matthieu points out, this is only correct over a field.

As to your second question, consider the product $(\mathbb Z/2 \mathbb Z) \times \mathbb A^2_k$ (here $k$ is a field), which is an affine group scheme over $\mathbb A^2_k$, and delete the closed point $(1, 0)$.

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Angelo, you seem to assume that $S$ is the spectrum of a field but Daniel is asking about what's happening over a general $S$. In that case, as you certainly know, the quotient $G/N$ is not representable in general (hence not affine). For a counter-example of Raynaud with $S=\mathbb{A}^2_k$, $G=(\mathbb{G}_{a,S})^2$ and $N$ etale over $S$, see Lemma X.14 in Faisceaux amples sur les schemas en groupes et les espaces homogenes. –  Matthieu Romagny Jun 6 '11 at 20:35
Matthieu, you are absolutely right. You should probably post your comment as an answer. –  Angelo Jun 7 '11 at 6:00