## quotients of varieties as non-noetherian schemes?

Let $X$ be a variety (i.e. a reduced scheme of finite type over a field) and let $G$ be an abstract group, finitely generated, acting of $X$ algebraically freely. The example I have in mind is $\mathbb Z$ acting by shifts on $\mathbb A^1$. The quotient in this example clearly does not exist as a noetherian scheme since the fibres are discrete infinite. Is it possible to still make sense of the quotient algebraically? I guess the best way to put it formally is:

Is it possible to put the category of varieties over the field k into a bigger "algebraic" category such that the functor of points of the quotient (i.e. orbits of the values of the usual functior of points) is always representable? What I mean by "algebraic" is that I am definitely not interested in a quotient in the sense of complex geometry.

Is it possible to achieve this goal by considering a suitable category of non-Noetherian schemes?

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 If $k$ has characteristic $0$, then $\mathbb{A}^1_k / \mathbb{Z}$ exists and is just $\mathrm{Spec}(k)$. – Martin Brandenburg Nov 26 at 2:16 is it in the sense I've mentioned? it doesn't represent the functor $F(T)$={$\mathbb Z$-orbits of T-points of $\mathbb A^1$}. by the way in characteristic $p>0$ the quotient exits and is just $\mathbb A^1$ – Dima Sustretov Nov 26 at 3:26 The functor of points of the quotient is not "points invariant under the action," it's "orbits under the G-action". Of course this cannot be representable in a "geometric" setting, since it's not a sheaf in e.g. the etale topology. Its etale sheafification is representable by a (non-separated) algebraic space, however. – Daniel Litt Nov 26 at 3:54 Daniel, yes, my mistake, I should have said "orbits". What is this algebraic space? I.e. if we regard an algebraic space as a quotient of a scheme by an étale equivalence relation, what the scheme and the equivalence relation it would be? – Dima Sustretov Nov 26 at 8:52 @ Dima: I don't think you really mean "transitively"! – Laurent Moret-Bailly Nov 26 at 20:18
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The quotient exists as an algebraic space (if we ignore the literature that defines algebraic spaces to be separated). The equivalence relation is given by the standard action groupoid. This is étale, since both the projection and the action map from $X \times G$ to $X$ are locally finitely presented and formally étale (indeed, they are locally isomorphisms). I'm not sure what you are looking for by passing to non-Notherian schemes, beyond allowing the "morphism space" $X \times G$ to be used.
 shouldn't the equivalenc relation be a closed subscheme of $X \times X$? or is it something that we relax when we do not require the algebraic space to be separated? – Dima Sustretov Nov 26 at 17:27 The equivalence relation is a monomorphism of schemes $R\to X\times X$. In your case, $R$ would be $G\times X$. Separated algebraic spaces correspond to the case where $R\to X\times X$ is a closed immersion. In your case, the morphism $R\to X\times X$ is not even quasicompact (as soon as $G$ is infinite), so the quotient is not quasiseparated and therefore does not qualify as an "algebraic space" for many authors. I am not quite sure this is justified, but on the other hand such spaces have not really proved useful yet. – Laurent Moret-Bailly Nov 26 at 20:30 The reason I am asking this question is beause I have been told that sometimes passing to formal schemes allows taking quotients that would otherwise not exist (as a scheme), the exemplary construction being the Tate curve. But perhaps this is a wrong analogy. – Dima Sustretov Nov 26 at 21:17 Formal schemes allow you to consider uniformization of curves that lie over a formal neighborhood of the maximally degenerate locus (after Mumford), but they form a category that is different from non-Noetherian schemes, since coordinate rings become topologized. If you are looking for a world where you can take sheaf quotients of arbitrary free actions of group schemes or group sheaves, you could just consider the category of sheaves itself. – S. Carnahan♦ Nov 26 at 22:11