I'm a bit confused by some notation in Hartshorne's chapter II (theorems II.5.17, II.7.6, and some discussion on ample invertible sheaves in section 7).

Theorem 5.17 (Serre). Let $X$ be a projective scheme over a noetherian ring $A$, let $\mathcal{O}(1)$ be a very ample invertible sheaf on $X$, and let $\mathcal{F}$ be a coherent $\mathcal{O}_X$-module. Then there is an integer $n_0$ such that for all $n \geq n_0$, the sheaf $\mathcal{F}(n)$ can be generated by a finite number of global sections.

Proof. Let $i: X \rightarrow \mathbb{P}_A^r$ be a closed immersion of $X$ into a projective space $A$, such that $i^*(\mathcal{O}(1)) = \mathcal{O}_X(1)$... [The remainder of the paragraph is devoted to reducing to the case where $X$ is Proj of a polynomial ring.]

First, what is meant by the hypothesis "$\mathcal{O}(1)$ be a very ample invertible sheaf on $X$", since, a priori, $X$ is not projective space or Proj of a graded ring? I see that one can appeal to Corollary 5.16 to see that $X$ is $\mathrm{Proj}(S)$ for some graded ring $S$, finitely generated by $S_1$ as an $S_0 = A$-algebra. Then one can take $\mathcal{O}(1) = \widetilde{S(1)}$. Then this is very ample since we can use the description of $S$ to construct a closed immersion $i: X \rightarrow \mathbb{P}_A^r$ with $i^*(\mathcal{O}_{\mathbb{P}_A^r}(1)) = \mathcal{O}_X(1)$ (by Proposition 5.12(c)), which is just $\mathcal{O}(1)$. Is this the correct interpretation?

Second, on p. 153, this result is later quoted as having established: for $\mathcal{L}$ any very ample invertible sheaf on a projective scheme $X$ over $A$, for any coherent $\mathcal{F}$, there is an $n_0$ such that $n \geq n_0$ implies $\mathcal{F} \otimes \mathcal{L}^n$ is generated by global sections, which appears slightly different than the above result, since we worked with the particular very ample invertible sheaf $\mathcal{O}_X(1)$. Based on the above argument, it seems every very ample invertible sheaf over a projective scheme (at least those coming from closed immersions coming from surjective ring maps, by proposition 5.12(c)) will be isomorphic to $\mathcal{O}_X(1)$. Is this true? If it's not, in what sense was the quoted result proven by theorem 5.17? Also, is this rephrasing merely meant to suggest the definition of ample provided on p. 153?

Finally, I am confused about the notation found in the proof of theorem 7.6, in chapter II. The setting is $X$ a scheme of finite type over a noetherian ring $A$, and $\mathcal{L}$ an invertible sheaf on $X$. One direction of the theorem is that if $\mathcal{L}^m$ is very ample implies $\mathcal{L}$ is ample. The proof begins with $i: X \rightarrow \mathbb{P}_A^n$, an immersion with $\mathcal{L}^m \simeq i^*(\mathcal{O}(1))$. Then given any coherent $\mathcal{F}$, consider the closure $\overline{X}$ in projective space and $\overline{\mathcal{F}}$ the coherent sheaf which is the extension of $\mathcal{F}$ to $\overline{X}$. Then it is noted that if $\overline{\mathcal{F}} \otimes \mathcal{O}_{\overline{X}}(l)$ is generated by global sections, hence $\mathcal{F} \otimes \mathcal{O}_X(l)$ is also. My question is, what is the meaning of $\mathcal{O}_X(l)$? The map $i$ is just an immersion, so we don't know that $X$ is Proj of some graded ring. Is $\mathcal{O}_X(l)$ just a short-hand for $\mathcal{L}^l$?


The notation $\mathcal{O}_X(1)$ in the statement of the theorem is just a name for an invertible sheaf which by hypothesis is very ample. The motivation of such a name is exactly the fact that a very ample invertible sheaf $L$, plus a choice of a basis of $\Gamma(X, L)$, where $X$ is a variety over some algebraically closed field, is the same as giving a closed immersion $i:X\hookrightarrow \mathbb{P}^n$ plus an isomorphism $\phi: L \simeq i^*\mathcal{O}_{\mathbb{P}^n}(1)$. As you were suggesting, $\mathcal{O}_X(l)$ is a shorthand for $L^l$.

The isomorphism $\phi$ above shows that every very ample line bundle is isomorphic to $i^*\mathcal{O}(1)$, or $\mathcal{O}_X(1)$ if you prefer this notation. Let me remark that there is not a canonically defined invertible sheaf $\mathcal{O}_X(1)$ on $X$ as in the case of $\mathbb{P}^n$. In other terms, there is not a canonically defined very ample invertible sheaf $L$ on $X$. This is the same as saying that you can embed $X$ in $\mathbb{P}^n$ in different ways, and it can be that the associated very ample line bundles $\mathcal{O}_X(1)$ are one different from each other.

For example, think of a non hyperelliptic curve of genus 3. You have different embeddings given by $K$ and $2K$, where $K$ is the canonical line bundle. By embedding the curve using these two line bundles, you have that $\mathcal{O}_X(1)$ will be isomorphic to $K$ in the first case and to $2K$ in the second one. You can also see this by looking at $X=\mathbb{P}^1$ : this is isomorphic both to $Proj(k[x,y])$ and $Proj(k[u,v,w]/(uw-v^2))$ , so that the $\mathcal{O}_X(1)$ in the first case is the sheafification of the $k[x,y]$-module $k[x,y](1)$ , in the other case is the sheafification of $k[x,y](2)$. Clearly, in the case of the projective line you do have a canonical $\mathcal{O}(1)$ but this is a special case, and that's because you have a canonical embedding, if you want, of the projective line into a projective space, namely the identity (you can also see this by looking at the functorial definition of the projective line). But in general, you do not have such a canonical embedding.

  • $\begingroup$ I'm still confused why on p. 153, theorem 5.17 is referred to as having shown for any very ample sheaf $L$ on $X$ a projective scheme over a noetherian ring $A$, $L$ is ample. I guess, more succinctly, I am asking, is every very ample sheaf on a projective scheme $X$ over a noetherian ring isomorphic to $\mathcal{O}_X(1)$? $\endgroup$
    – notes
    Aug 13 '17 at 6:21
  • $\begingroup$ I edited the answer. $\endgroup$
    – Andrea
    Aug 13 '17 at 8:27
  • $\begingroup$ I added an example in the end, hoping that it clarifies a little bit more. $\endgroup$
    – Andrea
    Aug 13 '17 at 17:06
  • $\begingroup$ Ah, yes, I guess I was confused on the point that $\mathcal{O}_X(1)$ does not always mean the same thing, and depends on the embedding into projective space. Thank you! $\endgroup$
    – notes
    Aug 13 '17 at 20:46
  • 2
    $\begingroup$ @Andrea This might be a long shot since the question is six months old, but I just wanted to clarify something. In your answer you say "The notation $\mathcal{O}_{X}(1)$ in the statement of the theorem is just a name for an invertible sheaf..." That would make sense, except that the notation $\mathcal{O}_{X}(1)$ is not used in the theorem's statement. He uses $\mathcal{O}(1)$ in the statement to denote a very ample sheaf on $X$. But in the proof says that $i^{*}(\mathcal{O}(1)) = \mathcal{O}_{X}(1)$. Is it a typo in the statement and $\mathcal{O}(1)$ is meant to be $\mathcal{O}_{X}(1)$? $\endgroup$
    – Luke
    Jan 5 '18 at 6:22

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