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This proposition on page 370 is to prove any ruled surface over a nonsingular curve $C$ is $\bf{P}(\mathscr E)$, the projective space bundle of a locally free sheaf $\mathscr E$ of rank $2$ on $C$ and vice versa. In the direction that for a given rule surface $\pi:\, X \to C$, $X \cong \bf P(\mathscr E)$ over $C$, the proof uses the fact that $X$ admits a section $\sigma$. So define the divisor $D = \sigma(C)$ on $X$ and define $\mathscr E = \pi_*(\mathscr L(D))$, where $\mathscr L (D)$ is the sheaf on $X$ corresponding to the divisor $D$. Then $\mathscr E $ is locally free of rank $2$ on $C$ from Grauert's theorem and clearly there is a natural map $$\pi^*(\mathscr E) \to \pi^*(\pi_*(\mathscr L (D))) \to \mathscr L (D).$$ But then there are some parts I do not fully understand (here both $X$ and $C$ are defined over some algebraically closed field):

  1. To prove this map $\pi^*(\mathscr E) \to \mathscr L(D)$ is surjective, it says ''by Nakayama's lemma, it is enough to check this on any fiber $X_y$''. Here $y$ is any point on $C$. But I can not see how to use Nakayama's lemma in this case to prove the surjectivity provided the surjectivity is proved on any fiber $X_y$.

  2. After proving the surjectivity of the above morphism, we can say there is a morphism $g:\, X \to \bf P(\mathscr E)$ over $C$ with $\mathscr L(D) \cong g^*(\mathcal O_{\bf{P}(\mathscr E)}(1))$. I see $g$ is an isomorphism on any fibre because basically in this case $g$ induces a morphism $g_y:\, \bf P^1 \to \bf P^1$ over any point $y$ in $C$ and since $\mathscr L(D)$ is very ample on any fiber, $g_y$ is a closed embedding hence an isomorphism. But I can not see why $g$ itself is an isomorphism. Maybe it is trivial.

Thank you for anyone who can give me clues or details.

user41541
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    For $1$, I think it follows from the following fact. Let $A$ be a ring and $M \to N$ a homomorphism of f.g. $A$-modules. Then the cokernel of this homomorphism, call it $K$ is itself a $f.g.$ $A$-module. Then $K = 0$ if and only if $K_p = 0$ for any prime ideal of $A$. By Nakayama, $K_p = 0$ if and only if $K_p \otimes_{A_p} k(p) = 0$. – David Benjamin Lim May 04 '15 at 16:37

1 Answers1

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For question 1, see "How can Nakayama's Lemma be used to extend surjectivity over a point to surjectivity over an open subset?"

For question 2, the condition that $g_y$ is an isomorphism for any point $y\in C$ implies that $g$ is unramified and bijective. By miracle flatness, $g$ is flat, so $g$ is bijective and étale. According to this answer, $g$ is universally injective. A universally injective étale morphism is an open immersion, so $g$ is an isomorphism.

notime
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