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Is the category of commutative group algebraic spaces (commutative group objects in algebraic spaces) locally of finite type over a field, an abelian category?

I would benefit from a reference

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    $\begingroup$ If you replace "locally" of finite type by "finite type", then every algebraic group space is a group scheme (because over a noetherian scheme, there's a stratification of the base such that the pullback over each stratum is a gp scheme, and a point has only the trivial stratification), and group schemes of finite type form an abelian category, eg by arguments in SGA3, Exp. VI_A, around Thm. 3.2. But I am assuming you want to know if this is true for algebraic group spaces locally of finite type. $\endgroup$ – user87684 Jan 15 '18 at 2:15
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    $\begingroup$ There is a commutative group algebraic $K$-space $F$ that is a disjoint union of countably many copies of $\text{Spec}(K)$ and whose associated group of $K$-points, $F(\text{Spec}(K)),$ is isomorphic to the group $\mathbb{Z}$. For every group algebraic $K$-space $G$, for every element $g\in G(\text{Spec}(K))$, there is a unique morphism of group algebraic $K$-spaces $F\to G$ mapping a specified generator of $F(\text{Spec}(K))$ to $g$. What would be the cokernel of this morphism? $\endgroup$ – Jason Starr Jan 15 '18 at 2:17
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    $\begingroup$ @JasonStarr: it would be a rather non-quasi-separated quotient of $G$ in general. :) $\endgroup$ – nfdc23 Jan 15 '18 at 7:49
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The intervention of algebraic spaces is a red herring because algebraic space groups $G$ locally of finite type (and quasi-separated!) over a field $k$ are necessarily schemes. (I am assuming you are only interested in the quasi-separated case, though no motivating context for the question was given.) This is Lemma 4.2 in Artin's paper "Algebraization of Formal Moduli I" (where all algebraic spaces are taken to be quasi-separated, pervading the foundations such as for a reasonable theory of irreducible components); the proof is given near the end of section 4 uses results about biratonal group laws. So let's give a variant on that proof which avoids birational group laws (as that tool feels too heavy for the situation at hand, for which the real technical point is that only over the typically infinite-degree algebraic closure do we have enough rational points to make translation arguments).

Remark: The quasi-separatedness hypothesis is essential, even if we only consider algebraic space groups of finite type. For instance, if $k$ has characteristic 0 then the quotient $\mathbf{A}^1_k/\mathbf{Z}$ of the affine line modulo the constant $k$-group $\mathbf{Z}$ is an algebraic space group of finite type over $k$ that is not quasi-separated and is certainly not a scheme.

Coming back to the proof of Artin's Lemma 4.2, first note that $G^0$ is of finite type (same proof as for schemes, as Artin sketches). Moreover, it remains connected after any finite extension on $k$ since it is connected with a rational point (same proof as for schemes, using that $(G^0)_{k'} \to G^0$ is finite flat surjection with fiber ${\rm{Spec}}(k')$ over the identity $k$-point of $G^0$). In fact, as will be useful later, even $(G^0)_{\overline{k}}$ is connected (since any idempotent on this algebraic space of finite type descends to one on some $(G^0)_{k'}$ for some finite extension $k'/k$ inside $\overline{k}$, yet any such $(G^0)_{k'}$ is connected, so the descended idempotent is 0 or 1).

Similarly, for any connected component $E$ of $G$, $E_{k'}$ is closed and open in $G_{k'}$ with $E_{k'} \to E$ a finite flat surjection, so by picking $k'/k$ big enough to split the residue field over $k$ at some closed point of $E$ we get that all connected components of $E_{k'}$ have a $k'$-point and thus are translated copies of $(G^0)_{k'}$.

We claim each $E$ is a quasi-projective $k$-scheme (so in particular $G$ is a scheme). Grant this for $G^0$ for a moment. Fix $E$ and pick finite $k'/k$ so $E_{k'}$ has a $k'$-point in all connected components. Thus, as noted above, $E_{k'}$ is a disjoint union of finitely many copies of $(G^0)_{k'}$, so it is a quasi-projective scheme. But $E_{k'}$ as a $k$-scheme is equipped with evident descent data relative to the finite flat cover ${\rm{Spec}}(k') \to {\rm{Spec}}(k)$, and this is effective by quasi-projectivity with the descent also a quasi-projective over $k$ (see SGA1, VIII, 7.6). A further argument with descent for morphisms of algebraic spaces (left as an exercise) shows that this descent is identified with $E$, so $E$ is a scheme (granting the same for $G^0$).

Now it remains to show that $G^0$ is a quasi-projective scheme. This is the step for which Artin appeals to the atom bomb of birational group laws, and we'll now make a bit more use of descent to provide an alternative procedure without nuclear weapons. It suffices to show $(G^0)_{k'}$ is a quasi-projective scheme for some finite extension $k'/k$, as then we can use a descent argument as above. By "chasing coefficients", it suffices to show $(G^0)_{\overline{k}}$ is a quasi-projective scheme (as we write $\overline{k}$ as the direct limit of finite subextensions over $k$ to descend an immersion $(G^0)_{\overline{k}} \hookrightarrow \mathbf{P}^n_{\overline{k}}$ to a map $(G^0)_{k'} \to \mathbf{P}^n_{k'}$ that is moreover an immersion for some finite extension $k'/k$; here we are using that $G^0$ is "finitely presented" over $k$, not just finite type, precisely because $G^0$ inherits quasi-separatedness from $G$). But $(G^0)_{\overline{k}}$ is connected (as we saw above!), so this finally allows us to replace $k$ with $\overline{k}$ so that $k$ is algebraically closed. We want to show over such $k$ that any locally finite type (and quasi-separated!) algebraic space group $H$ over $k$ is a scheme.

Every non-empty closed subspace of $H$ has a $k$-point since $k$ is algebraically closed, so it suffices to make an open subspace $U \subset H$ that is a scheme and contains all $k$-points (as then $(H-U)(k)$ is empty, forcing $H=U$ to be a scheme). It suffices to find a non-empty open subspace $V$ in $H$ that is a scheme, since that must have a $k$-point (as $k = \overline{k}$) and so can be translated to get an open subscheme around any $k$-point, the union of which (over all $k$-points of $H$) is a $U$ as desired.

The desired $V$ exists because (i) any quasi-separated algebraic space contains a dense open subspace that is a scheme (see II, 6.13 in D. Knutson's book Algebraic Spaces) and (ii) we are assuming that $H$ is quasi-separated.

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    $\begingroup$ This is a great answer. I just wanted to point out that the argument towards the end is similar to the proof of Lemma 6.2 in arxiv.org/pdf/1505.02249.pdf . However, quite unfortunately, the authors of that paper forgot to explicitly mention that the algebraic space in Lemma 6.2 is (assumed to be) finitely presented over $k$. $\endgroup$ – Ariyan Javanpeykar Jan 17 '18 at 19:48
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    $\begingroup$ @AriyanJavanpeykar: The authors of that paper didn't forget anything in their hypotheses (just perhaps in a step in their proof), since the hypotheses there force the algebraic space to be quasi-separated (and even separated). Indeed, the diagonal map becomes quasi-compact (even a closed immersion) after an extension of the ground field and therefore had to be quasi-compact (even a closed immersion) in the first place by elementary arguments. $\endgroup$ – nfdc23 Jan 18 '18 at 3:02
  • $\begingroup$ Thank you for your comment. You are right. What probably confused me when I wrote my comment is that the quasi-separatedness of $G$ in your answer is used (crucially) to guarantee that $G_{\overline{k}}$ is quasi-projective. On the other hand, if one were to know $G_{\overline{k}}$ is quasi-projective (over $\overline{k}$) (as in Lemma 6.2 of that paper), then quasi-separatedness of $G$ over $k$ is automatic. Thank you for clarifying. $\endgroup$ – Ariyan Javanpeykar Jan 19 '18 at 18:14

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