Lemma 50.11.1. There exists a short exact sequence

$0 \to \Omega \to \mathcal{O}(-1)^{\oplus n + 1} \to \mathcal{O} \to 0$

Proof. To explain this, we recall that $\mathbf{P}^ n_ A = \text{Proj}(A[T_0, \ldots , T_ n])$, and we write symbolically

$\mathcal{O}(-1)^{\oplus n + 1} = \bigoplus \nolimits _{j = 0, \ldots , n} \mathcal{O}(-1) \text{d}T_ j$

The first arrow

$\Omega \to \bigoplus \nolimits _{j = 0, \ldots , n} \mathcal{O}(-1) \text{d}T_ j$

in the short exact sequence above is given on each of the standard opens $D_+(T_ i) = \mathop{\mathrm{Spec}}(A[T_0/T_ i, \ldots , T_ n/T_ i])$ mentioned above by the rule

$\sum \nolimits _{j \not= i} g_ j \text{d}(T_ j/T_ i) \longmapsto \sum \nolimits _{j \not= i} g_ j/T_ i \text{d}T_ j - (\sum \nolimits _{j \not= i} g_ jT_ j/T_ i^2) \text{d}T_ i$

This makes sense because $1/T_ i$ is a section of $\mathcal{O}(-1)$ over $D_+(T_ i)$. The map

$\bigoplus \nolimits _{j = 0, \ldots , n} \mathcal{O}(-1) \text{d}T_ j \to \mathcal{O}$

is given by sending $\text{d}T_ j$ to $T_ j$, more precisely, on $D_+(T_ i)$ we send the section $\sum g_ j \text{d}T_ j$ to $\sum T_ jg_ j$. We omit the verification that this produces a short exact sequence. $\square$

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