Following the first definition on page 3 of 1, the blow-up is defined as a closure of the image set in the corresponding product space. I have seen a few examples of this, for instance, the blow-up of the origin of the nodal curve.
In this case (correct me if I'm wrong), what happens is that the origin is sent to a line, where each point corresponds to the slope of a line in the projective space $\mathbb{P}^1$ in the new variables, say $[s,t]$ (isomorphic to $\mathbb{P}^1$ with variables $[x,y]$ where the original curve is defined). As I understand, the blow-up is considered to be the black curve above the plane (if we introduce this new projective axis orthogonal to the plane). In this case, this curve is closed in $\mathbb{R}^2\times \mathbb{P}^1$ (the blue sheet), and therefore the closure is itself. My questions are:
- I would like to see an example (graphical if possible) where the blow-up is not its closure and see then why this requirement is added.
- When we consider this change of coordinates, we usually set one of the variables in projective space equal to one, as explained in this lecture lecture in min 5, which corresponds to one chart of the smooth structure of $\mathbb{P}^1$. So for instance in the nodal curve (also explained in the lecture), one sets $s=1$ in the plane $(s,t)$, this corresponds to the chart $s\neq 0$ of $\mathbb{P}^1$ in the product $\mathbb{R}^2\times\mathbb{P}^1$. So in the axis to which the origin blew up, we can't consider the infinity slope (corresponding to $s=0$ in the $(s,t)$ plane). Would this only affect in the sense that the blow-up is not defined in that limit (the projective axis approaching infinity)?
Answers to these questions aren't found in the standard textbooks I have read, so any help is appreciated.