Why do we blow-up only closed points on curves I was reading Cutkosky's Resolution of Singularities and at chapter 3 we endeavor into the resolution of embedded curves through blowing-up. The situation is simple: we have a regular surface $S$ and on it we have a possibly singular curve $C$. The strategy we have to solve its singularities is to blow up the closed points $p$ of $S$ which are singular on $C$ and then take the strict transform of the Zariski closure of the preimage of $C-\{p\}$: it is its strict transform.
I do understand that the map we get from the blow-up $\pi : \widetilde{S}\rightarrow S$ is birationnal and proper, but my question is the following: since all other points other than $p$ are preserved on $C$ when taking its strict transform, how can we be sure that its non-closed points aren't singular after a certain amount of blow-ups?
Edit: Thanks to AreaMan's response I think I have found an answer to my question. Tell me what you think of it.
The main point is that there are just two types of points in an algebraic curve: closed points and generic points of irreducible components. However a generic point $\eta$ is regular since $\mathcal{O}_{C,\eta}$ is a field which is alway regular.
Now in Cutkosky's book, he supposes tht a variety is equidimensional: each irreducible component has the same dimension, the dimension of our variety. So if I take a point $x\in C$ on our curve (variety of dimension 1) there is an irreducible component of $C$, let's say $Z$ that contains $x$ andis of dimension 1. Let's set $X=\overline{\{x\}}\subset Z$ and let's show that $X$ contains a closed point. I shall use Lemma 5 in Makoto's answer here Is the set of closed points of a $k$-scheme of finite type dense?
I take an open affine $U=$Spec$(A)$ containing $x$, where he is represented by a prime ideal $P$. By Krull's theorem, there is a maximal ideal $M$ in $A$ such that $P\subset M$. But since $\overline{\{x\}}\cap U$ is the closure of $x$ in $U$ we have
$$\overline{\{x\}}\cap U = V(P)=\{[P']\ | P' \text{ prime ideal }P'\supseteq P\}$$
So if $z$ is the point in $C$ represented by $M$ in $U$ then we have $z\in\overline{\{x\}}\cap U$, but by the aforementioned lemma, $z$ is a closed point in $C$. So we have the following inclusions of closed irreducible sets:
$$\{z\}\subset\overline{\{x\}}\subset Z$$
And since dim $Z=1$ we can't have two strict inclusions, thus two cases arise:
-$\{z\}=\overline{\{x\}}$ in which case $z=x$ is closed;
-$\overline{\{x\}}=Z$ in which case, $x$ i generic.
 A: The non-closed points of $S$ are contained in $S \setminus \{p\}$, so the blow up map is an isomorphism on them. (The blow up at $p$ is not just birational, the inverse is defined on the complement of $p$, i.e. the open subscheme $S \setminus \{p\}$.)
As far as the non-closed points of the blow up of the curve go - away from the excpetional divisor the blow down is an isomorphism. As the exception divisor $E$ is a line (so has only one non-closed point), and when we $E$ (and take the closure) to produce the strict transform, we have removed this non-closed point. (Since the intersection of the strict transform with the excpetion divisor is a finite set of points.)
After all, irreducible curves (over fields) have just one non-closed point, which is the generic point. (I think...) The generic point of the blow up maps down to the generic point of the original curve $C$, and since it does so in the complement of $p$, the induced map on local rings is an isomorphism.
I think this answers your question? (I'm certainly no AG expert - meaning that I encourage you to please check my suggestions carefully - but the internet is a fun place to blab on about stuff, so here you are.)
