TITLE: Question concerning $\text{Spec}(k[[T]]) $
QUESTION [7 upvotes]: 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?

REPLY [11 votes]: 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 $Spec(A_n)$.