Let $k$ be a field. Then consider the rings $k[T] / (T^n)$ with $n \in \mathbb{N}$. The inverse limit of these is given by $k[[T]]$. Passing on to the category of schemes, one concludes that the direct limit of $\text{Spec}(k[T] / (T^n)) $ is given by $\text{Spec}(k[[T]]) $. Now to the question. $\text{Spec}(k[T] / (T^n)) $ can be viewed as one point on the affine line with an infinitisimal neighborhood that remembers some derivatives of functions, right? As the direct limit, $\text{Spec}(k[[T]]) $ is one point on the affine line, that remembers all derivatives of functions. But this scheme has two points, one being a specialisation of the other. Are there any intuitive geometric explanations for this phenomena?
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1$\begingroup$ The generic point $(0)$ is what you described, a point which remembers all derivatives. The specialization $(T)$ is the underlying closed point of which your scheme may be regarded as a thickening. $\endgroup$– Tabes BridgesCommented Aug 24, 2017 at 7:03
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$\begingroup$ Thank you! But then I wonder why the other schemes have only one point, maybe they aren't thick "enough"? $\endgroup$– Manuel HoffCommented Aug 24, 2017 at 8:30
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6$\begingroup$ I would consider $\mathrm{Spec}(k[[T]])$ as a non-integrable curve though. The object which gives the infinitesimal neighborhood of one point on the affine line should be $\mathrm{Spf}(k[[T]])$. $\endgroup$– Wille LiuCommented Aug 24, 2017 at 10:34
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6$\begingroup$ I second @WilleLiou; Spec does not commute with limits in the category of rings. So, Spec $k[[t]]$ is different than the colimit of Spec $k[t]/t^n$ (e.g. using functor of points in the category of sheaves of sets). $\endgroup$– math no moreCommented Aug 24, 2017 at 13:41
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I'm not sure if you''re looking for something deeper than this (and therefore would have preferred to make this a comment if I could have squeezed it in), but:
$Spec(A)$ has to be just rich enough so that every map from $A$ to a field shows up as a function on $Spec(A)$.
Now the issue is that, for a field $F$, the functor $Hom_{\bf Rings}(-,F)$ does not preserve inverse limits. So if $A$ is the inverse limit of rings $A_n$, you can't in general understand maps $A\rightarrow F$ as limits of maps $A_n\rightarrow F$, which means you shouldn't expect to understand the points of $Spec(A)$ as limits of points of $Spec(A_n)$.
In particular, put $A_n=Spec\Big(k[[t]]/(t^n)\Big)$ and $A=k[[t]]=\lim_\leftarrow A_n$. Put $F=A[t^{-1}]$. Then there is a ring map $A\rightarrow F$ (namely the obvious inclusion) that does not arise from maps $A_n\rightarrow F.$ This necessitates a point in $Spec(A)$ that does not arise from points in the various $Spec(A_n)$.
answered Aug 24, 2017 at 13:27