Liu's "Algebraic Geometry and Arithmetic Curves" - Proposition 4.4. (Ch. 7) and Exercises 5.1.29 and 5.1.30 In Liu's book, Chapter 7, Proposition 4.4, the question is about closed embeddings (or: closed immersions) coming from very ample divisors. The theorems are quite well-known when the ground field $k$ is algebraically closed (e.g., Hartshorne's Proposition 7.3 and Sec. 7, Chap. 2 there, in general). But Liu does not assume $k$ to be algebraically closed because of the Exercises 5.1.29 and 5.1.30. and not assuming $k$ to be algebraically closed is also a major difference to various other texbooks!
Here it is said that the statements in the exercises could be incorrect. Can one prove them (at least) in special cases?
Ex. 5.1.29: Let $\pi: X'\rightarrow X$ be a faithfully flat morphism of schemes, $F$ quasi-coherent sheaf on $X$.
(a) There is a morphism $\phi: \mathcal{O}_{X}^{(I)} \rightarrow F$ such that $\phi(X)$ is surjective. Further, $F$ is generated by global sections iff $\phi$ is surjective.
(b) $F$ generated by global sections iff $\pi^{\ast}F$ generated by global sections.
(c) Now $X,X'$ quasi-compact, $L$ invertible sheaf on $X$. If $\pi^{\ast}L$ is ample, then $L$ is ample.
Ex. 5.1.30 (light version): Let $X$ be aproper scheme over $\mathbb{R}$, $L$ invertible sheaf on $X$, $\pi: X_{\mathbb{C}} \rightarrow X$ projection (to base change $\mathbb{R}$ to $\mathbb{C}$) and let $\pi^{\ast}L$ be very ample.
(a) Show that $L$ is ample and generated by some global sections $s_0, \ldots , s_n$. Let $f: X \rightarrow Y:=\mathbb{P}^{n}_\mathbb{R}$ be the morphism associated to $L$ and the $s_i$. Show that $f_\mathbb{C}: X_\mathbb{C} \rightarrow Y_\mathbb{C} $ is a closed immersion.
(b) Let $y \in Y$ and let $y' \in Y_\mathbb{C}$ be a point lying over $y$. Show that the fiber of $X_\mathbb{C}$ over $y'$ is isomorphic to $X_y \times_{Spec(k(y))} Spec(k(y')) $. Deduce from this that $f^{-1}(y)$ consists of at most one point, and that $f$ is a topological closed immersion.
(c) Show that $\mathcal{O}_Y \rightarrow f_{\ast}\mathcal{O}_X$ is surjective and that $f$ is a closed immersion.
 A: Exercise 5.1.29(b) and (c) are just plain wrong. Mohan's counterexamples are completely correct, and some further exposition of the general case of a cyclic cover of a curve can be found here. The key issue is that there is no good way to go from a surjective map $\mathcal{O}_{X'}^I\to \pi^*\mathcal{F}$ to a surjective map $\mathcal{O}_{X}^I\to \mathcal{F}$.
On the other hand, when we are in the situation of exercise 5.1.30, everything works out. The big idea that lets us bypass the issue from exercise 5.1.30 is that $H^0(X,\mathcal{L})\otimes_A B \cong H^0(X_B,\mathcal{L}_B)$ by flat base change (ref Stacks 02KH, for instance). So pick a generating set $\{s_i\}_{i\in I}$ for $H^0(X,\mathcal{L})$, construct the obvious morphism $g:\mathcal{O}_X^I\to \mathcal{L}$, and then note that $\{s_i\otimes 1\}_{i\in I}$ is a generating set for $H^0(X,\mathcal{L})\otimes_A B = H^0(X_B,\mathcal{L}_B)$. Thus $\pi^*g:\pi^*\mathcal{O}_X^I\to\pi^*\mathcal{L}$ is a map $\mathcal{O}_{X_B}^I\to\mathcal{L}_B$ which is surjective on global sections. So $\mathcal{L}_B$ is globally generated iff $\pi^*g$ is surjective, which happens if and only if $g:\mathcal{O}_X^I\to\mathcal{L}$ is surjective by the argument in Niven's post addressing 5.1.29(b). This fixes all of the issues in Niven's outline of a solution to exercise 5.1.30.
A: $\newcommand\msO{\mathscr O}\newcommand\pull[1]{#1^*}\newcommand\C{\mathbb C}\newcommand\by\times$
There are a lot of parts, so these are more hints/sketches than full arguments.
For 5.1.29(a), use that a morphism $\msO_X\to F$ is the same thing as a choice of global section $s\in F(X)$.
For 5.1.29(b), you want to show $\msO_X^I\to F$ is surjective its pullback $\msO^I_{X'}\to\pi^*F$ is. Surjectivity can be checked on stalks. Use that $\msO_{X,x}\to\msO_{X',\pi(x)}$ is a faithfully flat ring extension so a map $A\to B$ of $\msO_{X,x}$-modules is surjective iff the map $A\otimes_{\msO_{X,x}}\msO_{X',\pi(x)}\to B\otimes_{\msO_{X,x}}\msO_{X',\pi(x)}$ is surjective.
For 5.1.29(c), use part (b). If $F$ is a f.g. qcoh sheaf on $X$, then $\pull\pi(F\otimes L^n)=\pull\pi F\otimes(\pull\pi L)^n$ is generated by global sections for $n\gg0$ since $\pull\pi L$ ample, so $F\otimes L^n$ is generated by global sections for $n\gg0$, so $L$ is ample.
For 5.1.30(a), faithful flatness is preserved by base change, so $X_\C\to X$ is faithfully flat, so $L$ is ample by 5.1.29(c) and globally generated by 5.1.29(b). Finally, $f_\C$ is a closed immersion since it is the morphism associated to the very ample line bundle $\pull\pi L$ (gives immersion) and since $X_\C$ is proper (gives closed).
For 5.1.30(b), check/convince yourself that the square
$$\require{AMScd}\begin{CD}
X_\C @>>> X\\
@VVV @VVV\\
Y_\C @>>> Y
\end{CD}$$
is Cartesian. Then it is a formal fact that the fiber above some $y'\in Y_\C$ is the same as the fiber above the corresponding $y\in Y$. That $\#f^{-1}(y)\le1$ follows from $X_\C\to Y_\C$ being topologically injective. This gives that $X\to Y$ is a continuous injection with closed image. I don't know an easy way off the top of my head to see that it is an open map as well (so that $f$ will be a topological closed embedding). One possible method is to appeal to the fact that if $S'\to S$ is faithfully flat and quasi-compact, then the topology on $S$ coincides with the quotient topology induced by this map. Since $Y_\C\to Y,X_\C\to X$ are faithfully flat and quasi-compact, this would let you deduce openness of $X\to Y$ from openness of $X_\C\to Y_\C$.
For 5.1.30(c), use the same strategy as in 5.1.29(b). The point is that surjectivity can be checked after faithful flat base change where we already know it to hold. Knowing that $\msO_Y\to f_*\msO_X$ is surjective and that $f$ is a topological closed embedding says that $f$ is a closed immersion.
Edit: As KReiser points out in the comments, my suggestion for 5.1.29(b) is incomplete. Hence also are my arguments for 5.1.29(c) and 5.1.30(a). KReiser also explains how to recover a correct argument for 5.1.30 in their posted answer, so anyone looking at this question should read their answer as well.
