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$letopn=operatorname$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 $X: opnCRingsto opnSet$ that is a small filtered colimit of affine schemes i.e., $X(R)=limlimits_rightarrow iX_i(R)$.

The first example (4.2) is given by the functor $widehatmathbb A^1 $ defined as $widehatmathbb A^1(R)mathrel:=opnNil(R)$.

I don't understand why this functor is the colimit over $N$ of the functors $opnspec(mathbbZ[x]/x^N+1)mathrel:=opnHom_opnCRing(mathbbZ[x]/x^N+1,_)$.

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!

It might be illuminating to first work the example of (ordinary) affine space $mathbbA^1_mathbbZ$ over the integers.

As a functor, $mathbbA^1_mathbbZ$ is the forgetful functor $mathitRings^mathrmoprightarrowmathitSets$, sending a ring $R$ to its underlying set. It is representable by $mathbbZ[t]$, as can be seen by the isomorphism $$mathrmHom_mathitRings(mathbbZ[t],R)cong R.$$
(As a homomorphism $fcolonmathbbZ[t]rightarrow R$ is determined by its value $f(t)$ at $t$, we can define an isomorphism $mathrmHom_mathitRings(mathbbZ[t],R)rightarrow R$ by $fmapsto f(t)$ for all $finmathrmHom_mathitRings(mathbbZ[t],R)$.)

$widehatmathbbA^1_mathbbZ$ is similar: there is an isomorphism
$$mathrmHom_mathitRings(mathbbZ[t]/(t^n),R)congmathrmNil_n(R),$$
where $mathrmNil_n(R)$ 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 $widehatmathbbA^1_mathbbZ$ sends $R$ to the colimit $mathrmcolim(mathrmNil_n(R))=bigcup_ngeq0mathrmNil_n(R)=mathrmNil(R)$.