# Cartier divisor example in Harthsorne

This question is about example 6.11.4 in Chapter II of Hartshorne. The example is about computing the Cartier divisor class group of the cuspidal cubic curve $$y^2z = x^3$$ in $$\mathbb{P}_{k}^{2}$$. He begins by saying "note that any Cartier divisor is linearly equivalent to one whose local function is invertible in some neighbourhood of the singular point $$Z = (0,0,1)$$".

I am not sure what this statement means at all. A Cartier divisor is, by definition, represented by a family $$\{(U_{i}, f_{i}) \}$$ where the $$U_{i}$$ cover $$X$$ and the $$f_{i} \in \mathcal{K}^{*}(U_{i})$$. But the elements of $$\mathcal{K}^{*}(U_{i})$$ are invertible by definition. So every local function is invertible, right? So what does Hartshorne mean by hsi statement?

I agree he did not phrase it very well. Let $$U$$ be an open chart containing the cusp. By removing the cuspidal point from the other charts, we may assume it is the only one containing the cuspidal point. Let $$f$$ be the rational function on $$U$$. What he means, is that we may assume that $$f$$ is invertible in $$\mathcal{O}$$ (and not in $$\mathcal{K}$$, which would be obvious as $$\mathcal{K}$$ is a field). So, he is assuming that $$f \in \mathcal{O}^*(V)$$, where $$V \subset U$$ is an open set containing the cusp.
The way to achieve it is the following. Notice that $$f^{-1}$$ is an element of $$\mathcal{K}^*$$, as so is $$f$$. Thus, $$f^{-1}$$ gives a principal Cartier divisor. So, the divisor $$\lbrace (U_i,f_i) \rbrace$$ is the same as $$\lbrace (U_i,f_i \cdot f^{-1}) \rbrace$$. Since $$U = U_{i_0}$$ for some index $${i_0}$$, this shows that we may assume that on that chart the divisor is represented by $$(U,1)$$.