Page 142, Example 6.11.4.
I've been trying to go through the details of the sentence
The proof of (6.10) shows that if $f \in K$ is invertible at $Z$, then the principal divisor $(f)$ on $X - Z$ has degree $0$.
$X$ is the singular curve given by the homogeneous equation $y^2z = x^3$ in $\mathbb P^2$ and $Z$ is the singular point $[0,0,1]$. $K$ is the field of rational functions of $X$. (6.10) refers to the result that a principal divisor on a complete nonsingular curve has degree $0$.
So first of all $X$ is not nonsingular, so I guess the idea is to relax all the statements involved and still get the proof. My main issue is that the morphism $X - Z \to \mathbb P^1$ is not finite (it is not proper and finite morphisms are). Everything in the section was about finite morphisms of curves with nonsingularity assumptions almost everywhere, so this is a pain. I understand that the morphism $\varphi : X \to \mathbb P^1$ obtained from the field extension $k(f) \subseteq K(X)$ will be finite and $X$ is complete, but then the map $\varphi^* : \mathrm{Div}(\mathbb P^1) \to \mathrm{Div}(X)$ assumes $X$ is nonsingular, and I don't see how to correctly get rid of the issue of the singular point.
I thought about using the projection from a point $\mathbb P^3 \backslash \{*\} \to \mathbb P^2$ which sends a twisted cubic $C$ to this singular curve $X$, and since the induced morphism $C \to X$ is a bijection of sets, the induced morphism of complete nonsingular curves $C \to X \to \mathbb P^1$ would give me a Weil divisor on $C$, which I could pushforward to $X - Z$ using the fact that if $P \in C$ is the point which maps to $Z$, then the map $C \to X$ restricts to an isomorphism $C - P \to X - Z$, so I can push my Weil divisor of $C - P$ to $X - Z$. This seems to be a correct approach, but if it was the only one, it would have deserved a comment from Hartshorne...
Added : to show that principal Cartier divisors of $X$ have a Weil divisor of degree $0$ on $X - Z$, I encounter similar issues.
So my question is : does anyone know exactly what was the intended meaning of the sentence I quoted above?
Corollary I.6.10 states that every nonsingular quasi-projective curve is isomorphic to an open subset of a nonsingular projective curve. Thus we have $X - Z \subset W$ where $W$ is a projective nonsingular curve. Now the morphism $X - Z \rightarrow \mathbb{P}^1$ extends uniquely to $W$, and we can apply Theorem 6.10.