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I wonder if the definition of very ample sheaves of Hartshorne coincides with the definition of Stack projects.

Let $f: X \to S$ be a morphism of schemes. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module. Assume that $X$ is noetherian so that Hartshorne's definition of "immersion" coincides with projects'.

Definition of Stack projects:

We say $\mathcal{L}$ is relatively very ample if there exists a quasi-coherent $\mathcal{O}_ S$-module $\mathcal{E}$ and an immersion $i : X \to \mathbf{P}(\mathcal{E})$ over $S$ such that $\mathcal{L} \cong i^*\mathcal{O}_{\mathbf{P}(\mathcal{E})}(1)$.

Definition of Hartshorne:

We say $\mathcal{L}$ is very ample relative to $S$ if there is an immersion $i: X \to \mathbf{P}_S^r$ for some $r$, such that $\mathcal{L} \cong i^*(\mathcal{O}(1))$.

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    $\begingroup$ Hartshorne's definition is the special case of the stacks project's definition when $\mathcal{E}$ is a free sheaf. As such, the stacks project definition is much more general. In fact, you can find a locally free sheaf $\mathcal{E}$ on a scheme $X$ so that $\mathcal{O}_{\mathbb{P}(\mathcal{E})}(1)$ is not very ample over $X$ in the sense of Hartshorne. $\endgroup$
    – Daniel
    Commented May 14, 2023 at 15:49
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    $\begingroup$ @Daniel that looks like an answer to me - would you care to record it as such below? $\endgroup$
    – KReiser
    Commented May 14, 2023 at 18:29

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If $\mathcal{E} = \mathcal{O}_Y^{\oplus r}$, then $$\mathbb{P}_Y(\mathcal{E}) = \operatorname{Proj} \mathcal{O}_Y[x_0, \dots, x_{r - 1}] = \mathbb{P}^{r - 1}_Y.$$

As such, Hartshorne's definition is a specialization of the stacks project's definition to the case that $\mathcal{E}$ is a free sheaf of modules.

Even when $\mathcal{E}$ is taken to be locally free, we still don't have an equivalence of the definitions, since according to Hartshorne's exercise II.7.14(a), there exists a noetherian scheme $Y$ and a locally free sheaf $\mathcal{E}$ on $Y$ so that $\mathcal{O}_{\mathbb{P}(\mathcal{E})}(1)$ is not (Hartshorne) very ample relative to $Y$.

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