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Consider the following setting: $X$ an irreducible smooth proj. curve and $f$ a morphism $X \rightarrow \mathbb{P}^1(k)$, with $k$ algebraically closed. Call $C$ the complement of $f^{-1}\{(1:0)\}$ and assume it non-empty. I am trying to prove that $f|_C$ seen as a map to $\mathbb{A}^1$ (which we can do because we took a point away from $\mathbb{P}^1$) defines an element $f^{\vee}$ of $K(X)$ (rational functions on X) and actually $f \mapsto f^{\vee}$ establishes a bijection between the set of morphisms $X \rightarrow \mathbb{P}^1$ whose image is not $(1:0)$ and $K(X)$.

The proof that there is this bijection will probably involve divisors and Riemann-Roch but my problem is actually that I don't see how exactly the restriction to $C$ defines a rational function $f^{\vee}$. Thanks for your attention.

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    $\begingroup$ What is your definition of the function field $K(X)$? As a direct limit, as in Ch. I of Hartshorne? $\endgroup$ Dec 7, 2018 at 21:29
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    $\begingroup$ It doesn't really mention direct limits but I think it's the same concept in disguise. A rational function would be an equivalence class of couples of the type $(U,f)$ where $U$ is open and $f$ is a regular function on $U$ and two such couples $(U,f)$,$(V,g)$ coincide if there exists $W \subset U \cap V$ s.t. $f=g$ on $W$. $\endgroup$
    – Dalamar
    Dec 7, 2018 at 23:56

1 Answer 1

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$f|_C$ gives a map $C\to \newcommand\AA{\mathbb{A}}\AA^1_k=\newcommand\Spec{\operatorname{Spec}}\Spec k[x]$. This corresponds to a morphism $k[x]\to \newcommand\calO{\mathcal{O}}\calO_X(C)$. Then $f^\vee$ is the pair consisting of $C$ and the image of $x$ in $\calO_X(C)$.

This can be thought of as being analagous to how meromorphic functions $f$ on $\Bbb{C}$ define holomorphic functions $F:\Bbb{C}\to\Bbb{P}^1_{\Bbb{C}}$ with the set of poles of $f$ being the set of points $F$ sends to $\infty=(1:0)$.

Edit

It's ok if you're not aware of $\Spec$, I'll rephrase so I don't use it.

Firstly, note that $f$ is a continuous map, so $\newcommand\inv{^{-1}}f\inv(\{(1:0)\})$ is closed as a subset of $X$ (since the point $(1:0)$ is closed in the Zariski topology on $\Bbb{P}^1_k$, so its complement $C$ is open. Open subsets of varieties are usually considered to be varieties as well, so yes $C$ is a variety. That said, I don't think you need that $C$ is a variety, just that it is open.

Then since $f|_C$ gives a map from $C$ to $\Bbb{A}^1_k$, it induces a pullback map on the regular functions (I assume you think of this pullback map as composition with $f|_C$), which gives a map $f^*:k[x]\to \mathcal{O}_X(C)$, where to be clear $k[x]$ is the ring of polynomial functions, which is the coordinate ring of $\Bbb{A}^1_k$, and $\mathcal{O}_X(C)$ denotes the ring of regular functions on $C$ as an open subset of $X$. Then the image of $x$ under this map gives a particular regular function on $C$, and it is the pair of $C$ together with this regular function that is our rational function.

Some intuition for this correspondence This is not formal. Aspects of this aren't quite valid. In fact, in some ways, it's sort of circular, but I think it's helpful for intuition.

If you think of $X$ as being embedded in some suitable projective space, and the map $f:X\to \Bbb{P}^1$ as being a polynomial map, then if $x\hookrightarrow \Bbb{P}^n_k$ with the coordinates on $\Bbb{P}^n_k$ being $x_0,\ldots,x_n$, then a map from $X$ to $\Bbb{P}^1$ is locally given by a pair of homogeneous polynomials of the same degree that don't simultaneously vanish on $X$. Say these are $f_0,f_1\in k[x_0,\ldots,x_n]$. Then the image of a point $\bar{a}=(a_0:a_1:\cdots:a_n)$ in $X$ is the point $(f_0(\bar{a}):f_1(\bar{a})$.

Thus those points with image $(1:0)$ are precisely the points $\bar{a}$ with $f_1(\bar{a})=0$. This is a closed subset of $X$, so once again we see that $C$ is open. Then on $C$, $f_1$ doesn't vanish. Thus the point $(f_0(\bar{a}):f_1(\bar{a}))$ is equivalent to the point $$\left(\frac{f_0(\bar{a})}{f_1(\bar{a})}:1\right),$$ which, under the standard correspondence with affine space is simply the point $$\frac{f_0(\bar{a})}{f_1(\bar{a})}.$$ Then applying the regular function on affine space $x$ which takes points in $\Bbb{A}^1_k$ to their corresponding value in $k$, we end up with a regular function on $C$ that takes a point $\bar{a}$ to $\frac{f_0(\bar{a})}{f_1(\bar{a})}\in k$, which is what we usually think of as a rational function.

Obtaining the full bijection

To show that this correspondence is a bijection, we must show that any rational function on $X$ induces a map to projective space and that this is the inverse of the correspondence above.

Let $f$ be a rational function on $X$. You say that you're aware that $f$ has a finite set of zeros $Z$ and poles $P$ and that $f$ is defined on $X\setminus P$ and $\frac{1}{f}$ is defined on $X\setminus Z$. Then the morphism to projective space is defined by gluing the two maps $$ x \mapsto (f(x):1) $$ for $x\not\in P$ and $$x\mapsto \left(1:\frac{1}{f}(x)\right)$$ for $x\not\in Z$.

To check that these maps glue, note that when $x\not\in Z$ and $x\not\in P$, then the value of the second function is $$\left(1:\frac{1}{f}(x)\right),$$ and since $x\not\in P$, $\frac{1}{f}(x)\ne 0$, and since $x\not\in Z$, $f(x)\ne 0$, so we can multiply both homogeneous coordinates by $f(x)$ and use that for $x\not \in Z\cup P$, $f(x)\frac{1}{f}(x)=1$ to get that the value of the second function in fact equals $(f(x):1)$, which is the value of the first function at $x$. Thus the two maps glue.

You can check that the points that map to $(1:0)$ will be precisely those in $P$ (since they are the zeros of $\frac{1}{f}$).

An aside on notation. I'm using $\frac{1}{f}(x)$ which is perhaps not standard notation to emphasize that the inversion of the rational function occurs in the rational function field, rather than by inverting the values of the function.

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  • $\begingroup$ Hi and thanks for your reply. I am not aware of the concept of Spec yet (sorry for not mentioning this), but I think I know a correspondence similar to the one you are wielding it's just that it would require $C$ itself to be a variety in order to work. Can $U$ as the complement of an anti image be regarded as a variety? Further: does this correspondence already imply the bijectivity that I mention later in my question? $\endgroup$
    – Dalamar
    Dec 8, 2018 at 10:45
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    $\begingroup$ @Dalamar edited $\endgroup$
    – jgon
    Dec 8, 2018 at 15:15
  • $\begingroup$ I reformulate my comment. Starting from a rational function $f \in K(X)$, I know it will have finitely many zeroes, say a set $Z$ and poles, say a set $P$. I can then define a function $X/P \rightarrow \mathbb{A}^1$ defined through $k[x] \rightarrow \mathcal{O}_X(X/P)$ sending $x$ to $f$ and a function $X/Z \rightarrow \mathbb{A}^1$ defined through $k[y] \rightarrow \mathcal{O}_X(X/Z)$ sending $y$ to $1/f$. How do I show they glue along $X/Z \cap Z/P$ thus defining a function $X \rightarrow \mathbb{P}^1$? Do divisors help? If you think this requires another question, let me know. $\endgroup$
    – Dalamar
    Dec 9, 2018 at 12:26
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    $\begingroup$ @Dalamar I'm editing now, but in response to your second comment, it doesn't have to be. It's perfectly possible to choose it to be the image of $x-1$ or something, since there is an automorphism of $k[x]$ sending $x-1$ to $x$. But $x$ is the natural choice. $\endgroup$
    – jgon
    Dec 9, 2018 at 14:45
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    $\begingroup$ @Dalamar finished editing. $\endgroup$
    – jgon
    Dec 9, 2018 at 14:54

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