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Let $\mathcal{L},\mathcal{U}$ be invertible sheaves over a noetherian scheme $X$, where $X$ is of finite type over a noetherian ring $A$. If $\mathcal{L}$ is very ample, and $\mathcal{U}$ is generated by global sections, then $\mathcal{L} \otimes \mathcal{U}$ is very ample.

Since $\mathcal{L}$ is very ample, there exists $n$, s.t. $i: X\mapsto \mathbb{P}^n$ is an immersion with $\mathcal{L}= i^*\mathcal{O}(1)$, and since $\mathcal{U}$ is generated by global sections, one can construct $j:X \to \mathbb{P}^m$ with $j^*\mathcal{O}(1) = \mathcal{U}$. From this I can construct the following morphism:

$$ h: X \xrightarrow{\Delta} X\times X \xrightarrow{i\times j} \mathbb{P}^n \times \mathbb{P}^m \xrightarrow{ \operatorname{segre \ embedding}} \mathbb{P}^N $$

I can prove $\mathcal{L}\otimes \mathcal{U } \cong h^*\mathcal{O}(1)$, and the segre embedding is a closed immersion. But I don't know whether the map $(i\times j) \circ \Delta$ is an immersion, which is suspicious to be such, especially for the $\Delta$.

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This is Hartshorne's Exercise II.7.5 (d), and his Exercise II.7.4 (a) shows that $\Delta$ is a closed immersion. – darij grinberg Nov 28 '11 at 4:30
Yes, you are right. But is $i \times j$ also an immersion? I can prove $X \times \mathbb{P}^m \to \mathbb{P}^n \times \mathbb{P}^m$ is an immersion, but not for the whole morphism. – Li Zhan Nov 29 '11 at 0:43
$i\times j=\left(i\times \mathrm{id}\right)\circ \left(\mathrm{id}\times j\right)$, and the composition of two immersions is an immersion (this follows from the fact that a closed subscheme of an open subscheme is always an open subscheme of a closed subscheme if everything is Noetherian; this is Exercise 9.3.C in Vakil's notes ). – darij grinberg Nov 29 '11 at 3:34
Note that Vakil comments that calling these things "immersion" here is a bad idea. – darij grinberg Nov 29 '11 at 3:35
Oh, Liu has the proof (more or less): Let $\pi$ be the projection $\mathbb P^m\to \mathrm{Spec}A$. Then, $\left(\mathrm{id}\times \pi\right)\circ \left(i\times j\right)\circ \Delta = i$ is an immersion, while $\mathrm{id}\times \pi$ is separated (by Hartshorne's Corollary II.4.6 (c)), so that Hartshorne's Exercise II.4.8 (applied to $\mathcal P = \text{"being an immersion"}$) yields that $\left(i\times j\right)\circ \Delta$ is an immersion. Let's hope this isn't wrong again... – darij grinberg Nov 29 '11 at 4:29

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