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In this example one determines the Cartier divisor class group of the cuspidal cubic curve $y^2z=x^3$, let's call it $X$. Let $Z$ be the singular point $(0,0,1)$.

To each closed point of $X−Z$, one associates a Cartier divisor and the goal is to prove (by contradiction) that this map is injective.

At some point in this proof Hartshorne deduces that there exists an $f\in K^*$, which is invertible at $Z$, and such that $(f)=P−Q$ on $X−Z$ for distinct points $P$ and $Q$ of $X−Z$.

He continues: "Then $f$ gives a morphism of $X$ to $\mathbb{P}^1$, which must be birational."

So my first question is: why is it a well-defined morphism from $X$ to $\mathbb{P}^1$? I can see $f$ gives one from $X-Z$ to $\mathbb{P}^1$ but that doesn't seem to be extendable to the entire $X$, since $X$ is non-singular to begin with. In fact, can one give an explicit construction to show why $f$ gives a morphism of $X$ to $\mathbb{P}^1$?

EDIT: I don't see $f$ gives a morphism from $X$ to $\mathbb P^1$ (it's indeed a rational map from $X$ to $\mathbb P^1$). Does Hartshorne really mean it's a morphism from $\mathbb P^1$ to $X$?

My second question is: why is it birational then? In this post someone remarks "It's birational because it has one zero and one pole, i.e., it's generically one-to-one." But I don't understand what does it mean to be generically one-to-one, and why $f$ has one zero and one pole.

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If you pay close attention to the definitions of the objects at play, you will see that $f$ is by construction a regular function on a neighborhood of $Z$, not an extension of some other regular function. To see this, recall first that a Cartier divisor is nothing but a collection of rational maps $f_i$ on a covering $(U_i,f_i)$, such that on the overlaps, the $f_i$'s share the same zeroes and poles (i.e. $f_i/f_j$ is a regular map on $U_i \cap U_j$). On top of that, there is an additional equivalence relation: two Cartier divisors $(U_i,f_i)$ and $(V_j,g_j)$ are considered equal (as divisors, not even divisor classes yet) if the functions $f_i$ and $g_j$ differ (multiplicatively) by some nonvanishing regular map on $U_i \cap V_j$. This is yet another way to ensure that we only care about the locations (and multiplicities) of zeroes and poles.

Now, Hartshorne's map is as follows: if $P$ is a point of $X - Z$, he defines a Cartier divisor $D_P$ as:

  • the constant function 1 on the open subset $X \setminus \{P,P_0\}$ ($P_0 = (0,1,0)$ is fixed)
  • whatever set of rational maps on $X \setminus Z$ which represents the Weil divisor $P-P_0$ (this exists, as Weil divisors are well defined on $X \setminus Z$, and always come from Cartier divisors on that same variety).

Whatever functions were used for that second point, they do not vanish on the subset $X \setminus \{Z,P,P_0\}$ (the intersection of the covering we picked). So this does indeed fulfill the requirements to define a Cartier divisor.

Now if we assume that for two distinct points $P$ and $Q$, $D_P \sim D_Q$ (i.e. the map to Cartier divisor classes is not injective), then, BY DEFINITION of linear equivalence, there is a rational map $f$ on $X$ which satisfies $$D_P = D_Q + \operatorname{div}(f)$$ which is an equality at the level of Cartier divisors. In particular, if we look at the corresponding equality of rational maps on the open subset $X \setminus \{P,Q,P_0\}$, this tells us that the rational map $f$ coincides with the constant map 1, up to some nonvanishing regular map (see the end of my first paragraph). In other words, $f$ is indeed a regular function on $X \setminus \{P,Q,P_0\}$.

Now it remains to show that $f$ defines a regular map $X \rightarrow \mathbb{P}^1$. What we have shown so far is that $f$ defines a regular map $X \setminus P_0 \rightarrow \mathbb{A}^1$, and we want to show that it extends to a regular map $X \rightarrow \mathbb{P}^1$. Unless there is a simpler way I am unaware of, we prove this via an application of Hartshorne's chapter I.6. This chapter states that any rational map between two complete nonsingular curves $C$ and $C'$ can be extended to a regular map $C \rightarrow C'$. In our case, we would like to pick $C = X$ and $C' = \mathbb{P}^1$ (in which case $f$ does define a rational map between the two), but there is one issue: $X$ is singular.

Luckily, $X$ is nonsingular on the open subset $U = X - P_0$ on which we want to extend $f$. Let $\tilde{C}$ be the nonsingular completion of $U$, which is different from $X$, but have isomorphic open subsets via $U$. we know that $f$ can be extended into a regular map $\tilde{C} \rightarrow \mathbb{P}^1$, which we can then restrict to the open subset $U$. We have now extended $f$ to a regular map $U \rightarrow \mathbb{P}^1$, which is what we wanted to do.

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  • $\begingroup$ Thank you for your response. The thing I'm confused about is, $f$ is indeed a rational map from so defined, but how is it a morphism from the entire $X$ to $\mathbb P^1$? Rational maps were well-defined on the open subset, as you pointed out $X\setminus\{P, P_0, Q\}$, but how does that give a morphism on the entire $X$: i.e. how could we extend this $f$ to the entire $X$ when $X$ is nonsingular. $\endgroup$
    – Jasper
    Commented Jul 10 at 14:30
  • $\begingroup$ We know that if $X$ is regular then such extension is possible, but here $X$ is obviously not regular at $Z$ unless I'm missing anything. $\endgroup$
    – Jasper
    Commented Jul 10 at 14:35
  • $\begingroup$ Alright, I will add that to my answer. $\endgroup$ Commented Jul 10 at 16:15
  • $\begingroup$ You were already quite close: $f$ only needs to be extended on an open subset which is nonsingular. The trick is now to restrict to this nonsingular open, complete it into a nonsingular complete curve, extend $f$, and restrict again to the open subset of interest. $\endgroup$ Commented Jul 10 at 16:37

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