Let $A$ be an integral domain, let $\mathbb{P}^n_A=\operatorname{Proj}A[x_0,x_1,...,x_n]$, in this case, the Weil, Cartier divisor, and invertible sheaves can be used interchangeably. I have seen some heuristic argument that $\mathcal{O}(1)$ corresponding to Weil divisor $[\mathbb{P}^{n-1}_A]$, but unfortunately, I cannot prove this rigorously. Does anyone know a rigorous way to see this? That will be very helpful for me!


1 Answer 1


Well, I'm not an expert in Algebraic Geometry, so I don't know whether this works for abritrary $A$, but maybe the case $A = \mathbb{C}$ could help you.

If D is a Cartier divisor on $\mathbb{P}^n$, given by meromorphic (=rational) functions $f_i \in K^*(U_i)$ such that $f_i f_j^{-1} \in \mathcal{O}^*(U_i \cap U_j)$ then $\mathcal{O}(D)$ is the line bundle (=invertible sheaf) given by the cocycles $f_i f_j^{-1}$ on $U_i \cap U_j$.

Now, take $H \simeq \mathbb{P}^{n-1}$ given by the equation $z_0 = 0$. Let $U_i = \lbrace z_i \neq 0 \rbrace$ the standard open covering. Then, on $U_0$, $H$ is given by the zero set of $1 \in \mathcal{O}^*(U_0)$ and, on $U_j, j>0$, $H$ is given by the zero set of $z_0/z_j \in \mathcal{O}(U_j)$. So the line bundle $\mathcal{O}(H)$ is given by the cocycles $z_j/z_i$ on $U_i \cap U_j$, which are the cocycles for $\mathcal{O}(1)$.

  • $\begingroup$ The last paragraph works mutatis mutandis for any commutative ring $A$. $\endgroup$
    – user18119
    Commented Oct 26, 2011 at 19:36
  • $\begingroup$ @Lucas Kaufmann Yes, it helps, thank you! $\endgroup$
    – Li Zhan
    Commented Oct 27, 2011 at 19:26

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