TITLE: Very ample sheaf on a blowup
QUESTION [5 upvotes]: Suppose one wants to prove that a $1$-dimensional integral proper scheme over an algebraically closed field is projective. This is a step in how Hartshorne has you prove that any $1$-dimensional proper scheme (over an algebraically closed field) is projective.
The method is that first note the normalization is non-singular and hence projective. Consider $f: \tilde{X}\to X$, and let $\mathcal{L}$ be a very ample sheaf on $\tilde{X}$. The goal is to prove that there is an effective divisor $D=\sum P_i$ such that $\mathcal{L}(D)\simeq \mathcal{L}$ and such that each $f(P_i)$ are non-singular points.
The rest of the proof follows from the fact that there merely exists some very ample sheaf with that property, but the exercise seems to imply that any very ample sheaf satisfies this property.
I think I'm missing something really obvious, but take a proper curve over $k$, say $C$, that has a single singularity. Blow-up that singularity, $\pi: \tilde{C}\to C$, this is the normalization. Take a point $P\in \tilde{C}$ such that $\pi(P)$ is the singularity. It seems to me that $D=P$ is an effective divisor, so $\mathcal{L}(D)$ is an invertible sheaf. If the above is correct, then $\mathcal{L}(D)$ cannot be very ample. Does anyone have a simple reason for this?

REPLY [4 votes]: Notice that in your 2nd paragraph you say the goal is to prove that «there is an effective divisor $D=\sum P_i$ such that $\mathcal{L}(D)\simeq \mathcal{L}$ and such that each $f(P_i)$ are non-singular points», but in your 4th paragraph you are picking a non singular point and asking if there is an effective divisor with a specific support. 
The two things have quantifiers in different places. In particular, the claim in your 2nd paragraph does not say that you can pick the support of the divisor as you want!