A standard example of an ind-scheme over a field $\mathrm{k}$ which is not a $\mathrm{k}$-scheme is $\mathrm{k}((\varepsilon))$. My question is how to prove that rigorously? To put it more precisely, let $$\mathrm{k}((\varepsilon)) = \{ a \in \prod_{-\infty}^{\infty}\mathrm{k}: a_i =0, i \ll 0 \}$$ An ind-scheme is an injective limit of schemes. So here, $$\mathrm{k}((\varepsilon)) = \lim_{i \rightarrow -\infty}\varepsilon^i\mathrm{k}[[\varepsilon]]$$ But why isn't it an algebraic subset of $\prod_{-\infty}^{\infty}\mathrm{k}$?

EDIT: I seem to have mixed up some notions, and have asked two different questions at once (or maybe even three) so I'll try to make myself clear. My motivation for the question was to be able to justify the following "$k((\epsilon))$ is not an algebraic subset of $\prod_{-\infty}^{\infty}k$ so we define it as $k$-points of an ind-scheme". So the original question is: why $k((\epsilon)) \subseteq \prod_{-\infty}^{\infty}k$ isn't algebraic and it is answered by Jason Starr (though I'm not sure if I understand the answer). We can also define $k((\epsilon))$ as and ind-scheme by $$k((\epsilon)) = colim_n Spec(k[x_{-n},x_{-n+1}, \ldots])$$ Now, one can ask, why isn't the ind-scheme we've constructed a scheme after all (btw. wouldn't it contradict Jason's argument?), and this question is answered by Scott Carnahan below. Finally, there is a question: if the co-limit exists in the category of schemes which Scott Carnahan addresses below as well ...

  • $\begingroup$ Do you want a prove that this limit does not exist in the category of schemes? In this case you may find mathoverflow.net/questions/65506/… helpful. Often people just "argue" in this case that there does not seem to be any natural model of this limit, thus it does not exist ... perhaps motivated by some sort of "good limits" in the same spirit as GIT uses good quotients. $\endgroup$ – Martin Brandenburg Jul 11 '11 at 21:03
  • $\begingroup$ Yes, that's what I'm trying to do. Just hoping for an "elementary" proof, since this seems to be the easiest example (maybe alongside with infinite dimensional vector space). $\endgroup$ – Michal Zydor Jul 11 '11 at 22:03
  • $\begingroup$ I think the easiest example is the completion of $\mathbb{A}^1_k$ at a $k$-point. As a locally ringed space, it is a topological point whose ring of functions is $k[[t]]$ - obviously not a scheme. $\endgroup$ – S. Carnahan Jul 12 '11 at 5:31
  • $\begingroup$ Do you really mean $k[[\epsilon]]$? If you are writing $k$-valued points above, this does not give the correct algebraic set. Perhaps you mean $k((\epsilon)) := \text{colim}_n \text{Spec}(k[x_{-n},...,x_0,x_1,...])$? $\endgroup$ – Martin Brandenburg Jul 12 '11 at 6:19

I think you can just consider the decreasing sequence of ideals of the increasing sequence of algebraic subsets $\epsilon^i k[[\epsilon ]]$ of $\prod_{-\infty}^{\infty}k$. For each $i$, the ideal is $( a_j | j\leq i-1 )$. The intersection of this sequence of ideals is $\{ 0 \}$. The corresponding algebraic subset of $\prod_{-\infty}^{\infty} k$ is the whose scheme. Since $k((\epsilon))$ is not the whole scheme, it is not an algebraic subset.

  • $\begingroup$ Thanks, I was sure I had understood, but now I have doubts though. Doesn't it only show that $k((\epsilon))$ is not affine? But, anyway, you made me realize that I was missing one thing - $\epsilon^ik[[\epsilon]]$ are closed in $k((\epsilon))$. $\endgroup$ – Michal Zydor Jul 11 '11 at 22:00
  • $\begingroup$ Yes this proof is incomplete. $\endgroup$ – Martin Brandenburg Jul 12 '11 at 6:27
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    $\begingroup$ It depends on what is being asked. In the post, it is asked why this subset of $\prod_{k=-\infty}^{\infty} k$ is not an algebraic subset. That is the question I answered. The separate question, is there some way of making this set a scheme, is not the one answered. $\endgroup$ – Jason Starr Jul 12 '11 at 12:47
  • $\begingroup$ I got two ideas confused for a while, now I understand your answer, thanks. $\endgroup$ – Michal Zydor Jul 12 '11 at 13:54

Based on the comments, it looks like you have two questions mixed up. There is the question of whether the colimit exists in the category of schemes, and there is the question of whether the ind-scheme described by the colimit in the category of set-valued functors on schemes (or the category of Zariski sheaves of sets, or fpqc sheaves, etc.) is represented by a scheme. The two questions are quite different. Since you mentioned ind-schemes, I'll answer the (easier) question about ind-schemes.

Following Martin's comment, I'm assuming by $k((\epsilon))$, you are referring to $\operatorname{colim}_n \operatorname{Spec}(k[x_{-n},x_{-n+1},\ldots])$ where the colimit is taken in set-valued functors on schemes. It is an ind-scheme in the sense that it is a colimit over a directed system of closed immersions of schemes. This particular ind-scheme is ind-affine, so it can be written as the formal spectrum of the topological ring $A = \varprojlim_n k[x_{-n},x_{-n+1},\ldots]$ (see Beilinson and Drinfeld's Quantization of Hitchin’s integrable system and Hecke eigensheaves section 7.11.2). Since $A$ has a non-discrete localization at zero, and all local rings of schemes at points are discrete, the locally topologically ringed space $\operatorname{Spf} A$ is not isomorphic to a scheme.

I think the directed system $\{ \operatorname{Spec}(k[x_{-n},x_{-n+1},\ldots]) \}_{n \geq 0}$ has a colimit in schemes, given by $\operatorname{Spec} A$, where $A$ is given the discrete topology. $\operatorname{Spec} A$ is certainly the colimit in affine schemes, by the anti-equivalence between affine schemes and commutative rings. Surely there must be a theorem somewhere ...?

  • $\begingroup$ This still dies not answer why the ind-scheme is not a scheme, right? $\endgroup$ – Martin Brandenburg Jul 12 '11 at 11:51
  • $\begingroup$ I think it does. The category of schemes embeds into the category of locally topologically ringed spaces, which in turn embeds into the category of functors from schemes to sets. The colimit described in the question is represented as a locally topologically ringed space that is not in the essential image of the first embedding. When you say "why", are you asking a more philosophical question? $\endgroup$ – S. Carnahan Jul 12 '11 at 16:33

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