Question on a projective bundle associated to a vector bundle

Following is quoted from -Nakayama, On Weierstrass models-,

" Let $S$ be a complex surface, and $L$ a line bundle on it. consider $P=\mathbb{P}(\mathcal{O}_S\oplus L^2 \oplus L^3)$. Let $a$ and $b$ be arbitrary sections of $L^{-4}$ and $L^{-6}$ and let $(x,y,z)$ be the canonical sections of $\mathcal{O}_P(1)\otimes L^{-2}$, $\mathcal{O}_P(1)\otimes L^{-3}$ and $\mathcal{O}_P(1)$ respectively which correspond to the natural injections of $L^2$, $L^3$ and $\mathcal{O}_S$ into $\mathcal{O}_S\oplus L^2 \oplus L^3$. Then the Weierstrass model is given by equation $y^2z=x^3+axz^2+bz^3$ in $P$ and is an elliptic fibration over $S$ ..."

My question: When I do calculations for my self, I see that the embeddings given above should correspond to canonical sections of $\mathcal{O}_P(1)\otimes L^{2}$, $\mathcal{O}_P(1)\otimes L^{3}$ and $\mathcal{O}_P(1)$, not the one he is saying and therefore we should consider $(a,b)$ as sections in dual of what he has said and at the end we should get an equation in $\mathcal{O}(3)\otimes L^6$ not in $\mathcal{O}(3)\otimes L^{-6}$ ? How does he get $x$ (and similarly $y$) from embedding $L^2 \rightarrow \mathcal{O}_S \oplus L^2 \oplus L^3$??

My calculation: We have the exact sequnce $$0\rightarrow \mathcal{O}_P(-1) \rightarrow \mathcal{O}\oplus L^2 \oplus L^3 \rightarrow Q \rightarrow 0$$

over $P$, from which we get

$$0 \rightarrow Hom(\mathcal{O}(-1),\mathcal{O}(-1)) \rightarrow Hom(\mathcal{O}(-1),\mathcal{O}\oplus L^{2} \oplus L^{3}) \rightarrow T_{W/S} \rightarrow 0$$

Here the last object is the relative tangent bundle and the first map is given by three sections $(z,x,y) \in Hom(\mathcal{O}(-1),\mathcal{O}\oplus L^{2} \oplus L^{3})\cong \Gamma(\mathcal{O}(1)) \oplus \Gamma(\mathcal{O}(1)\otimes L^{2}) \oplus \Gamma(\mathcal{O}(1)\otimes L^{3})$ as above. These are analogue of $(x,y,z)$ coordinate on projective space.

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1 Answer

There are two competing definitions for $\mathbb P(\mathcal E)$, one classifies subbundles of $\mathcal E$ of rank 1, the other classifies quotients. With the latter (as used in EGA or Hartshorne), you have a canonical quotient $p^*\mathcal E\to\mathcal O(1)$ instead of a canonical subbundle $\mathcal O(-1)\to p^*\mathcal E$. As long as $\mathcal E$ is locally free of finite rank, there is no big difference, you can just dualize $\mathcal E$ to pass from one to the other.

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OK, so he is considering the quotient. What is the benefit of that? – Mohammad F. Tehrani Aug 11 '11 at 12:23
Tehrani -- Have you ever looked at EGA? The benefit of working with quotients is that the functor of flat quotients is compatible with arbitrary base change, and thus very often representable, e.g., whenever $\mc{E}$ is quasi-coherent. The functor of flat subsheaves is not compatible with arbitrary base change, since injections are not compatible with arbitrary base change. – Jason Starr Aug 11 '11 at 12:26
No, I have never looked at EGA. what you said was beyond my knowledge. – Mohammad F. Tehrani Aug 11 '11 at 18:53