$\let\opn=\operatorname$For my BA thesis I have to describe formal groups from the functorial point of view. I am hence reading Strickland - Formal Schemes and Formal Groups, which is apparently the only article that deals with this topic in that way.

He defines (4.1) an formal scheme as a functor $X: \opn{CRings}\to \opn{Set}$ that is a small filtered colimit of affine schemes i.e., $X(R)=\lim\limits_{\rightarrow i}X_i(R)$.

The first example (4.2) is given by the functor $\widehat{\mathbb {A}}^{1} $ defined as $\widehat{\mathbb {A}}^{1}(R)\mathrel{:=}\opn{Nil}(R)$.

I don't understand why this functor is the colimit over $N$ of the functors $\opn{spec}(\mathbb{Z}[x]/x^{N+1})\mathrel{:=}\opn{Hom}_{\opn{CRing}}(\mathbb{Z}[x]/x^{N+1},\_)$.

I would appreciate it if someone could explain it in general and kindly give an illustrating example. Other simple examples of formal schemes are also highly welcome. Many thanks!

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    $\begingroup$ You seem to be asking a lot of questions about basic examples and definitions from this paper. In order that your BA thesis represent your own work, it is probably better to work these out yourself, or at least to seek help from your advisor or classmates. $\endgroup$ – LSpice Jul 14 '19 at 19:01

It might be illuminating to first work the example of (ordinary) affine space $\mathbb{A}^1_\mathbb{Z}$ over the integers.

As a functor, $\mathbb{A}^1_\mathbb{Z}$ is the forgetful functor $\mathit{Rings}^\mathrm{op}\rightarrow\mathit{Sets}$, sending a ring $R$ to its underlying set. It is representable by $\mathbb{Z}[t]$, as can be seen by the isomorphism $$\mathrm{Hom}_\mathit{Rings}(\mathbb{Z}[t],R)\cong R.$$ (As a homomorphism $f\colon\mathbb{Z}[t]\rightarrow R$ is determined by its value $f(t)$ at $t$, we can define an isomorphism $\mathrm{Hom}_\mathit{Rings}(\mathbb{Z}[t],R)\rightarrow R$ by $f\mapsto f(t)$ for all $f\in\mathrm{Hom}_\mathit{Rings}(\mathbb{Z}[t],R)$.)

$\widehat{\mathbb{A}}^1_\mathbb{Z}$ is similar: there is an isomorphism $$\mathrm{Hom}_\mathit{Rings}(\mathbb{Z}[t]/(t^n),R)\cong\mathrm{Nil}_n(R),$$ where $\mathrm{Nil}_n(R)$ denotes the set of nilpotent elements $r$ of $R$ of order $n$ (i.e. $r^n=0$). (We define an isomorphism as before, but now the element $f(t)$ that we send $f$ to must be nilpotent in $R$, for it to preserve the ring structure: $f(t)^n=f(t^n)=f(0)=0$.)

The answer from here own was changed following Dmitri Pavlov's comment:

As (co)limits of presheaves are computed objectwise, we see that $\widehat{\mathbb{A}}^1_\mathbb{Z}$ sends $R$ to the colimit $\mathrm{colim}(\mathrm{Nil}_n(R))=\bigcup_{n\geq0}\mathrm{Nil}_n(R)=\mathrm{Nil}(R)$.

  • $\begingroup$ (Sorry for the probably very overcomplicated proof.) $\endgroup$ – Théo Jul 14 '19 at 17:44
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    $\begingroup$ Why do you need EGA here? Once you observed that Nil_n computes nilpotents of order n, you can immediately conclude the proof by observing that colimits of functors CRing→Set are computed objectwise and Nil(R)=⋃_{n≥0}Nil_n(R) because any nilpotent element of R has finite order. $\endgroup$ – Dmitri Pavlov Jul 14 '19 at 18:31
  • $\begingroup$ @DmitriPavlov That's precisely the fact I was missing. Thank you! $\endgroup$ – Théo Jul 14 '19 at 19:08
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    $\begingroup$ @DmitriPavlov Continuing the discussion of the Nil functor, one can show that Nil is not an affine scheme and hence that the category of affine schemes does not admit all colimits. Trying to construct the cases for which coproducts does exist, Strickland remarks (Remark 2.21) that, for affine schemes X, Y, (X \coprod Y)(R) is NOT equal to X(R) \coprod Y(R). Doesn't this statement contradict the theorem of objectwise computation of colimits? $\endgroup$ – sagirot Aug 4 '19 at 11:12
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    $\begingroup$ @sagirot: The category of affine schemes admits all small limits and colimits, since it is the opposite category of commutative rings, and the latter is locally presentable, being the category of algebras over a small algebraic theory. Observe, however, that Nil was not defined as a colimit in the category of affine schemes, but rather as a colimit in the category of presheaves on affine schemes. $\endgroup$ – Dmitri Pavlov Aug 4 '19 at 14:49

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