# Finite morphism induced by section of a line bundle

Assume $$X$$ is smooth, irreducible curve (curve = $$1$$-dimensional proper scheme over a fixed base field $$k$$) and $$x \in X$$ a closed point. Since $$\{x\}$$ is a divisor of $$X$$, we obtain a line bundle $$\mathcal{L}:= O_X(\{x\})$$ a.k.a. invertible sheaf on $$X$$. We assume that $$\dim_k H^0(\mathcal{L}, X) \ge 2$$. Therefore $$H^0(\mathcal{L}, X)$$ contains a non constant section $$f \in H^0(\mathcal{L}, X)$$. $$f$$ we can also see as an element of fraction field $$K(X)$$ of $$X$$ and therefore we can talk about $$(f),(1/f)$$ as $$O_X$$-modules.

Let $$U_f:=X \backslash \text{Supp}(f)$$ and $$U_{1/f}:=X \backslash \text{Supp}(1/f)$$. We obtain a cover $$X=U_f \cup U_{1/f}$$ of $$X$$ and can define a well defined map $$f: X \to \mathbb{P}^1 _k= \text{Proj}(k[T_0,T_1])$$ as follows on affine pieces:

$$U_f \to D_+(T_0)$$ corresponds to $$T_1/T_0 \mapsto f$$ and $$U_{1/f} \to D_+(T_1)$$ corresponds to $$T_0/T_1 \mapsto 1/f$$. It's easy exercise to verify that these maps glue to a map $$f: X \to \mathbb{P}^1 _k$$. By construction $$f$$ is affine and $$f^{-1}(\infty)=x$$, because $$f$$ is non constant and $$\mathcal{L}= O_X(\{x\})$$ (recall, $$\infty =(0:1) \in \mathbb{P}^1$$).

My question is why is $$f$$ a finite morphism? Since it's a local property and $$f$$ is defined on affine pieces symmetrically, that suffice to understand why is $$k[T_1/T_0] \to O_X(U_f), T_1/T_0 \mapsto f$$ is a finite map or equivalently why is $$O_X(U_f)$$ a finite $$k[T_1/T_0]$$-module?

• I'd say there is finite type ($O_X(U)$ is finitely generated $k$-algebra) in the definition of proper $k$-scheme, irreducible means $O_X(U)$ is an integral domain (since otherwise looking at its minimal prime ideals would give reducibility), so $Frac(O_X(U))$ is a finite extension of $k(t_1,\ldots,t_n)$, 1-dimensional means $n=1$ so $Frac(O_X(U))$ is a finite extension of $k(f)$ for any $f$ non-constant (non-algebraic over $k$) Commented Oct 17, 2019 at 2:54
• yes, this looks nice. we consider the finite injection of fields $k(T_1/t_0) \to Frac(O_X(U))$. thus we can identify $k(T_1/T_0)$ with it's image $k(f)$ and $Frac(O_X(U))$ is finite over $k(f)$. the last problem is to conclude that this implies $O_X(U)$ is finite over $k[f] \cong k[T_1/T_0]$. or generally is that true that if $\phi: R \to A$ is a injective ring morphism and the induced map $\bar{\phi}:Frac(R) \to Frac(A)$ is finite, then $\phi$ is finite?
– user705174
Commented Oct 17, 2019 at 21:05
• @katalaveino I think the more general statement does not hold: $\varphi: \mathbb Z\rightarrow \mathbb Q$ is not finite, but it induces a finite map $\mathbb Q\rightarrow \mathbb Q$. Commented Oct 18, 2019 at 3:06

This was intended as a comment, but it is too long to fit in.

I think in this case $$\mathcal L$$ is an ample line bundle of $$X$$, and the morphism $$f$$ in question is the canonical morphism (stacks project Lemma 01PZ).

Since the canonical morphism is an open immersion (stacks project Lemma 01Q1), the morphism $$k[T_1/T_0]\rightarrow\mathcal O_X(U_f)$$ should be a localization. Since being finite is a local property (stacks project Lemma 01WI), we may assume it is the localization at one element.

Since $$X$$ is a curve over $$k$$, that element should be finite over $$k$$ (as it is proper over $$k$$), i.e. satisfy a polynomial equation with coefficients in $$k$$. This shows that the localization in question is finite.

Perhaps there is a simpler way to show this, but I hope this could still be of some use.

Please point it out if some errors occur or if there are any inappropriate points, thanks.

EDIT:

$$(1):$$

Consider a ring extension $$A\subseteq B$$ and $$x\in B$$ which is finite over $$A$$. Then $$A_x$$ is finite over $$A$$. This is because $$x$$ being finite over $$A$$ implies $$x$$ is integral over $$A$$, and hence $$x$$ satisfies a monic equation $$x^n+\sum_{i=0}^{n-1}a_ix^i=0$$. Dividing by $$x^n$$, we have $$1+\sum_{i=0}^{n-1}a_ix^{i-n}=0.$$ As a consequence $$A_x/A$$ is generated by $$\{1/x^k\mid k = 0, \cdots, n-1\}$$ as a module.

As to why $$x$$ is finite, see this MO question, or this blog post.

$$(2):$$

Not any localization is the localization at one element: For example, consider the localization $$\mathbb Z_{(2)}$$ of $$\mathbb Z$$ at the prime ideal $$(2)$$: it consists of elements of the form $$\frac ab$$, where $$a,b\in\mathbb Z$$ and $$2\not\mid b$$. There is no element $$x$$ such that $$\mathbb Z_{(2)}=\mathbb Z_x$$ (otherwise $$x$$ has infinitely many prime divisors).

On the other hand, I think my original argument involving PID is wrong, so I edited the above argument as well. Thanks for pointing out the flaw.

• why the fact, that the single element $s$, which gives the localisation $k[T_1/T_0]\rightarrow\mathcal O_X(U_f)=k[T_1/T_0]_s$, is proper (over $k$), implies that it is also finite?
– user705174
Commented Oct 17, 2019 at 21:10
• a second remark: you say "ince the canonical morphism is an open immersion , the morphism $k[T_1/T_0]\rightarrow\mathcal O_X(U_f)$ should be a localization (at one element, since $k[T_1/T_0]$ is a P.I.D.). I don't understand the interplay with P.I.D. property. isn't that true that every ring map $A \to B$, that is a localization, can be realized by localisation at a single element? i.e. $B \cong A_s$ with $s \in A$. in other words: is the PID condition neccessary?
– user705174
Commented Oct 17, 2019 at 21:35
• I made an edit, trying to explain the questions. If it is not clear or if you have more questions, feel free to ask. :) Commented Oct 18, 2019 at 3:03
• do you maybe know if the claim could also be proved as a 'corollar' from Noether normalisation theorem? namely NNT tells me in our case that if $A$ is integral $k$-algebra of dimension $1$, then there exist a finite injection $K[X] \to A$. can from NNT instantly derived that if we take arbitrary non constant $f \in A$, i.e. $f \not \in k$, then after replacing $X$ by $f$, $A$ stays finite over $k[f]$. your proof above is indeed fine, but looking through the statement of NNT it looks so alluring that the claim of my post can also be easily dereved from that.
– user705174
Commented Oct 18, 2019 at 10:08
• Indeed NNT tells us that there is a finite injection $K[x]\hookrightarrow A$. But I do not know how to show that the image of $x$ can be chosen to be any $f\not\in K$. Commented Oct 18, 2019 at 10:41