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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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  • $\begingroup$ 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. $\endgroup$
    – Daniel Bergh
    Commented Jun 8, 2011 at 14:42
  • $\begingroup$ 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. $\endgroup$
    – Matthieu Romagny
    Commented Jun 9, 2011 at 12:06
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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).

[Edit] 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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    $\begingroup$ 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. $\endgroup$
    – Matthieu Romagny
    Commented Jun 6, 2011 at 20:35
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    $\begingroup$ Matthieu, you are absolutely right. You should probably post your comment as an answer. $\endgroup$
    – Angelo
    Commented Jun 7, 2011 at 6:00

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