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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-dim 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?

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A very ample invertible sheaf $\mathcal L(D)$ is characterized by there being an embedding $\tilde{C} \hookrightarrow \mathbb P^n$ so that $D$ is the intersection of $\tilde{C}$ with a hyperplane $H$ in $\mathbb P^n$. If a hyperplane intersects $\tilde{C}$ in the point $P$ with multiplicity one, then $\tilde{C}$ has to be a degree one curve in $\mathbb P^n$, which is to say, a line. Thus $\tilde{C}$ must have genus zero. Thus all points on $\tilde{C}$ must be linearly equivalent, and so in this case $\mathcal L(P) = \mathcal L(P')$ for any point $P'$ on $\tilde{C}$ (just to say the same thing a different way, we have that $\tilde{C} = \mathbb P^1$, and that $\mathcal L(D) = \mathcal O(1)$ on this $\mathbb P^1$). In particular, we can take $P'$ to be a point whose image in $C$ is non-singular.

Another way to phrase the argument, which avoids this precise intersection theoretic analysis, is to say that if $\mathcal L$ is very ample on $\tilde{C}$, then it comes from an embedding $\tilde{C} \hookrightarrow \mathbb P^n$ by pulling back $\mathcal O(1)$, and so is of the form $\mathcal L(D)$ for any intersection of $\tilde{C}$ with a hyperplane in $\mathbb P^n$. But we can always choose this hyperplane so that its intersection with $\tilde{C}$ misses any given finite set of points. So we can always choose this hyperplane so that $D$ consists only of points whose images in $C$ are non-singular, which is what we want.

In general, when thinking about what very ample means, you want to relate it to the corresponding projective embedding, and then use geometric facts about hyperplane intersections. (Primarily: that there are a lot of hyperplanes (!), so you can hope to get a lot of control over the $D := \tilde{C} \cap H$ for which your very ample invertible sheaf equals $\mathcal L(D)$. Although it isn't relevant here, because you are just working with curves, a typical tool that you might apply to understand the possible such $D$ is Bertini's theorem.)

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Thanks! Both these answers are great, and point out the fundamental flaw in my head. How embarrassing, it wasn't clear from my question, but the thing I was overlooking was the fact that given any $D'$ linearly equivalent to $D$, then $\mathcal{L}(D')\simeq \mathcal{L}(D)$ and all we care about is up to isomorphism. Anyway, great explanation more generally too. – Matt Sep 28 '10 at 17:59

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!

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