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Suppose G is a group object in the category of algebraic spaces (over a field, if you like, or even over ℂ if you really want). Is G necessarily a scheme?

My feeling is that the answer is "yes" because an algebraic space group which is not a scheme would be too awesome. Any group homomorphism from such a G to an algebraic group (a scheme group) would have to have infinite kernel since an algebraic space which is quasi-finite over a scheme is itself a scheme. In particular, G would have no faithful representations or faithful actions on projective varieties (probably). There can't be a surjective group homomorphism from an algebraic group H to G, since that would identify G with H/K, which is a scheme. In particular, you cannot put a group structure on any étale cover of G.

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Wow, I was thinking about this back in 2006, and asked a few experts, but got nowhere. Great question. –  David Roberts Nov 15 '10 at 4:51

6 Answers 6

up vote 13 down vote accepted

I think the answer is yes. If S is a noetherian scheme and G is a relative algebraic group space over S, then there is a stratification of S such that over each stratum, G is a group scheme (see K. Behrend, Derived \ell-adic categories for Algebraic Stacks, 5.1.1). When S is Spec k there is no non-trivial stratification.

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Behrend assumes right before 5.1.1 that all his group algebraic spaces are of finite type, so it is not a priori clear that this answers the question. –  Kestutis Cesnavicius Sep 7 '13 at 20:25

Wikipedia says, "Group objects in the category of algebraic spaces over a field are schemes." There is no citation, but in an arXiv paper, arXiv:0907.3880, the claim is cited to the original paper of Artin, "Algebraization of formal moduli".


Looking over the other answers provided, I gather that the question is analogous to asking, "Is a group object in the category of orbifolds a Lie group?" Yes, because it can't have a stratification. It is apparently similar but more abstract in the setting of stacks (say). Anton's other question about a coset space $G/H$ is presumably a similar phenomenon. But I would guess that a double coset space $H\backslash G/K$ can be a stack, since in the Lie group setting it can be an orbifold.

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When I first saw your post, I thought you meant Emil Artin, and I thought, "Wow, that sucks, he's going to have to read it in german?" –  Harry Gindi Dec 14 '09 at 23:34

Let me say first that I am not an algebraic geometer. Nevertheless, in trying to understand the answers to this very question I asked the following question. After hearing the ensuing answers it seems that the answer to your question depends on whether you consider an algebraic space to be quasi-separated or not. The standard definition, as I am told, is that an algebraic space is required to be quasi-separated. In this case the answer to your question is yes, as the answers here show. However, if algebraic spaces are not necessarily quasi-separated, then the theorems quoted here fail, and it seems the answer is no. The example in this question appears to give a counter example.

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The answer is yes. Every algebraic space $X$ has an (non-empty) open subspace $U$ such that i) $U$ is a scheme, and ii) a point $p \rightarrow X$ is in $U$ if and only if that point is scheme-like. A scheme-like point is one where there is an affine scheme $V$ and an open immersion $V \rightarrow X$ factoring the point. Further, $X$ is a scheme if and only if all its points are scheme-like. Now, if $X$ is a group algebraic space, the group action is transitive on points. Using this, there can be no non-scheme like points and so $X$ is a scheme.

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This argument requires some care, since such kind of translation requires using rational points (I assume the base is meant to be a field k here, and the algebraic space group locally of finite type), so one has to give an argument to justify passing to an alg. closed base field. (In the non-smooth case there may be no nontrivial k_s-points as well.) Artin's article omits discussion of this point. –  BCnrd Feb 10 '10 at 16:18
    
Ah thanks Brian. Is it true that we can reduce to the alg. closed case? –  mdeland Feb 10 '10 at 18:18
    
Yes. I scribbled a proof in the margin of my copy of Artin's paper, so I suppose with a bit of time you should be able to cook up an argument too. –  BCnrd Feb 10 '10 at 20:26

Over a field, the reference is indeed "Algebraization of formal moduli, I".

Over a base more general than a field, the answer may be no. For example, Artin proved in "Algebraization of formal moduli, I" that under some natural conditions the Picard functor is representable by an algebraic space (it is a group algebraic space of course). There are several sufficient conditions for it to be a scheme, due to Artin and Mumford. You will find a discussion, if not the proof, of them in Bosch-Lutkebohmert-Raynaud's "Neron models". But in general case is not known, I think.

On the other hand, abelian algebraic spaces over a scheme S (i.e. smooth proper with irreducible geometric fibers) is an abelian scheme over S. See p.3 of Faltings-Chai's "Degenerations of abelian varieties", where this is attributed to Raynaud and Deligne.

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That's relative Picard functor $Pic_{X/S}$, just to be sure. –  VA. Dec 19 '09 at 1:09

Over a general base the answer is no, as pointed out to me by Matthieu Romagny in response to this question. He gives a reference to Lemma X.14 of Raynaud's book "Faisceaux amples sur les schémas en groupes et les espaces homogènes".

There is an example given where $N$ and $G$ are smooth affine group schemes over the base $S = \mathbb{A}^2_k$, with $N$, which is étale over $S$, a normal closed subgroup of $G$. It is shown that the quotient $G/N$ is not a scheme. But it is certainly an algebraic space, since the quotient map $G \to G/N$ is an étale atlas for $G/N$. Moreover, $G/N$ is both separated (as $N$ is closed in $G$) and smooth (since $G$ is smooth).

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