# Hartshorne chapter II Example 6.11.4: induced morphism is birational: proof and explicit form

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.

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.

• 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. Commented Jul 10 at 14:30
• 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. Commented Jul 10 at 14:35
• Alright, I will add that to my answer. Commented Jul 10 at 16:15
• 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. Commented Jul 10 at 16:37