[Math] $\mathcal{L}$ is very ample, $\mathcal{U}$ is generated by global sections $\Rightarrow$ $\mathcal{L} \otimes \mathcal{U}$ is very ample

algebraic-geometry

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
$\Delta$.

Best Answer

From the comments, to clear this from the unanswered queue:

Let $\pi:\Bbb P^m\to \operatorname{Spec} A$ be the projection. Then, $(\text{id}\times\pi)\circ (i\times j) \circ \Delta = i$ is an immersion, while $\text{id}\times \pi$ 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 $(i\times j)\circ \Delta$ is an immersion.

This argument is sourced to Liu's Algebraic Geometry and Arithmetic Curves, Chapter 5, Excercise 1.28. This comment was originally written by darij grinberg (2011-11-29), and at the time of this posting was comment-upvoted at +7 as well as confirmed by the original asker in a subsequent comment.

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