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Theorem 7.1 (a) says that- If $\phi$: $X \rightarrow \mathbb P_A^n $ is an $A$- morphism, then $\phi^*(\mathcal O(1)) $ is an invertible sheaf on $X$, which is generated by the global sections $s_i=\phi^*(x_i) $, $i$=0,1,...,n.

I do not know how to prove that the global sections $s_i$ generate $ \phi ^*(\mathcal O(1)) $.

Also I have one more question in the proof of 7.1 (b)- While giving the ring homomorphism $A[y_0,...,y_n]$ $\rightarrow$ $\Gamma$($X_i$, $\mathcal O_X{_i} $), I do not undrstand what is $ s_i / s_j $ . Here $s_i $ and $s_j$ are two global sections of an $ \mathcal O_X $- module. How to undersatnd that their quotient(?) is an element of $\Gamma$ ($X_i$, $\mathcal O_X{_i}$).

Can anyone please explain these things.

Suhas
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  • I strongly recommend taking a few toy examples and writing everything out explicitly. – RghtHndSd Feb 01 '14 at 14:13
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    Since the $x_i$ generate $\mathcal{O}(1)$, we have a surjective homomorphism $\mathcal{O}\mathbb{P}^{n+1} \to \mathcal{O}(1)$. The pullback functor is right exact and preserves the structure sheaves, thus we get a surjective homomorphism $\mathcal{O}_X^{n+1} \cong \phi^* \mathcal{O}\mathbb{P}^{n+1} \to \phi^* \mathcal{O}(1)$. – Martin Brandenburg Feb 01 '14 at 22:49
  • http://therisingsea.org/notes/Section2.7-ProjectiveMorphisms.pdf This pdf might be helpful. – WWK Apr 17 '15 at 01:01

1 Answers1

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The sections $s_i$ and $s_j$ are sections of an invertible sheaf (not just any old $\mathcal O_X$-module). By definition $s_i$ is nowhere-zero on $X_i$, and so then $s_i$ forms a basis for the invertible sheaf over $X_i$, i.e. provides a trivialization of the invertible sheaf over $X_i$. We may thus write, on $X_i$, the $s_j = f s_i$ for some section $f$ of $\mathcal O_X$ over $X_i$. For obvious reasons, one denotes $f$ by $s_j/s_i$.

As for your first question, check that $x_0, \ldots, x_n$ generate $\mathcal O(1)$ on $\mathbb P^n$. Then, check that this property is preserved under pull-back.

Matt E
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