Cartier Divisor for Hyperplane

Question

Let $H$ be the hyperplane in $\mathbb P^n$ defined by $x_0=0$

What is the Cartier Divisor $(U_i,f_i)$ corresponding to H?

I was thinking that the $U_i$ should be the distinguished open sets $D_+(x_i)$. But I'm unsure what $f_i$ should be.

Answer 1

concerning your remark in the comments: you_re right, the defining $f_i$ shouldn't be identically zero, but they can have zeros or poles.

And don't confuse $V(x_i)$, the closed subset of $\mathbf{P}^n$ where $x_i$ vanishes with the principal open $D(x_i)$ where $x_i$ is nonzero, hence invertible. Note that giving Cartier divisor is the same as giving a global section of the sheaf $\mathcal{K}^*/\mathcal{O}_X^*$, where $\mathcal{K}$ is the sheaf of meromorphic functions on $X$. Such a section is a collection of functions $\{f_i\}$ for an open cover $U_i$ of $X$, satisfying on intersections $\frac{f_i}{f_j} \in \mathcal{O}_{U_i\cap U_j}^*$.

In this case, $x_0/x_i$ gets identified with $x_0/x_j$ on the overlaps, by multiplying with $x_i/x_j$. The fact that these transition functions are units means they don't change the order of zeros or poles of $\{f_i\}$ when going from one patch of the cover to another. So $f_i$ certainly can have zeros (and poles). If it has only zeros and no poles, then it defines, like in this case, an effective divisor $\sum ord(f_i)V_i$, so $ord(f_i)\geq0$ where we run over closed $V_i \subset X$ intersecting $U_i$. In our case, the effective divisor is just $H$, because $x_0$ vanishes only there and with multiplicity 1.