The Disk and the Punctured Disk Can you explane me why
$$D =  \operatorname{Spec}\mathbb{C}[[t]]$$
is the disk and
$$D^{\times} =  \operatorname{Spec}\mathbb{C}((t))$$
is the punctured disk? Or give me some links on intelligible books, lectures, etc...
Thanks a lot!
 A: Yes, $\operatorname{Spec}\mathbb C[[t]]$ only has two points, one is open and the other is closed. The closed point correspond to the maximal ideal generated by $t$ which corresponds to the origin: if we consider $\operatorname{Spec}\mathbb C[[t]]\rightarrow \operatorname{Spec}\mathbb C[t]$ given by the inclusion $\mathbb C[t]\to \mathbb C[[t]]$ then zero corresponds to the ideal $(t)$ which corresponds to the origin. The open point is a neigborghood of zero since contains it $t$. 
If to $D$ you take out a point, algebraicly it means that you have to localize the ring $\mathbb C[[t]]$ with respect to the ideal of the point. If your point is the open point, its ideal is $(0)$ and so to get the open disk you localize $\mathbb C[[t]]$ at $(0)$ which is just taking the fraction field of $\mathbb C[[t]]$ which is $\mathbb C((t))$. So the open punctured disk is $\operatorname{Spec}\mathbb C((t))$.
$$D-\{0\} = \operatorname{Spec}\mathbb C[[t]] - V(t) = \operatorname{Spec}\mathbb C((t))$$
A: There's a canonical projection from $\mathbb{C}[[t]]$ down onto the residue field $\mathbb{C} \cong \mathbb{C}[[t]]/t$, and so dually, there is a distinguished geometric point:
$$O: \operatorname{Spec} \mathbb{C} \to \operatorname{Spec}\mathbb{C}[[t]]$$
This morphism doesn't lift (canonically) to $\operatorname{Spec}\mathbb{C}((t))$ though, i.e. there is not a canonical morphism from $\operatorname{Spec}\mathbb{C}$ to $\operatorname{Spec}\mathbb{C}((t))$, and so we call $\operatorname{Spec}\mathbb{C}((t))$ the punctured formal disc.
