The Stacks project

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. $\square$

Comments (0)

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0FMH. Beware of the difference between the letter 'O' and the digit '0'.