Why does non-constant $f\in \Gamma(\mathcal{O}_X(nP),X)$ mean that $f$ has a pole of order $n$?

Question

Let $X$ be a curve over an algebraically closed field $k$. Why does non-constant $f\in \Gamma(\mathcal{O}_X(nP),X)$ mean that $f$ has a pole of order $n$? I have seen this used in the proof that for every point $P$ on $X$ there exists a non-constant rational function $f$ that is regular everywhere except for $P$.

By definition, if $f\in \Gamma(\mathcal{O}_X(nP),X)$, then $(f)+nP\geq 0$, which implies that $v_P(f)\geq -n$. But this is not necessarily an equality, so why do we have such a pole? This inequality alone doesn't preclude the possibility that $v_P(f)$ is a positive number. What am I missing?

Answer 1

Suppose $X$ is a complete curve. Then for all $n\ge1$ there is an injection $\Gamma(X,\mathcal O_X((n-1)P))\hookrightarrow\Gamma(X,\mathcal O_X(nP))$, from the injection of sheaves $\mathcal O_X((n-1)P)\to\mathcal O_X(nP)$ together with the fact that $\Gamma(X,-)$ is left exact. Here $\Gamma(X,\mathcal O_X((n-1)P))$ consists of functions $f$ that are regular away from $P$ and such that $v_P(f)\ge-(n-1)$. Thus, $\Gamma(X,\mathcal O_X(nP))\setminus \Gamma(X,\mathcal O_X((n-1)P))$ consists of the functions $f$ that are regular away from $P$ and such that $v_p(f)=n$ exactly.

Thus, to have a $f$ with pole of order exactly $n$ at $P$ we need $\Gamma(X,\mathcal O_X((n-1)P))\hookrightarrow\Gamma(X,\mathcal O_X(nP))$ to be a proper injection. One criteria that assures this is $n\ge 2g$. Indeed, by Riemann-Roch we have $\dim\Gamma(X,\mathcal O_X(nP))=n-g+1$ and $\dim\Gamma(X,\mathcal O_X((n-1)P))=n-g$.

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Note that as a special case, when $X=\mathbb P^1$, Riemann-Roch tells us $\Gamma(X,\mathcal O_X(P))$ is $2$-dimensional and $\Gamma(X,\mathcal O_X)=\mathbb C$, the space of constant functions, is $1$-dimensinoal. Thus any non-constant function in $\Gamma(X,\mathcal O_X(P))$ has a pole of order exactly $1$, which may be what you were referring to.

Answer 2

You'll note that in my linked comment, I never say that there is a function $\newcommand{\L}{\mathscr{L}} f \in \L(nP)$ with a pole of order exactly $n$, only that there is a nonconstant $f \in \L(nP)$ if $n \geq g + 1$. For $n \geq g+1$ we have $$ \ell(nP) \geq 1 - g + n \geq 2 \, . $$ The constant functions contribute a $1$-dimensional subspace, so the fact that $\dim \L(nP) \geq 2$ means there is some nonconstant function $f \in \L(nP)$.

I think the part you might be missing is that every nonconstant rational function $h$ on a projective curve $X$ has a pole somewhere. Considering $h$ as a morphism $X \to \mathbb{P}^1$, since $X$ is irreducible and projective (actually, proper suffices) then the image of $h$ is irreducible and closed. For any point $P$ in the image of $h$, we have the containments of irreducible closed sets $$ \DeclareMathOperator{\img}{img} \{P\} \subsetneq \img(h) \subseteq \mathbb{P}^1 $$ and since $h$ is nonconstant, the first containment is strict. Since $\mathbb{P}^1$ has dimension $1$, this means that we must have $\img(h) = \mathbb{P}^1$. Thus $h$ is surjective, so in particular takes on the value $\infty$.

So now we've established that the nonconstant $f \in \L(nP)$ has a pole somewhere, and the fact that $\operatorname{div}(f) + nP \geq 0$ means that the only possibility is at $P$. Thus $f$ has a pole at $P$ and is regular elsewhere.

Indeed, as you say, $f \in \L(nP)$ implies $v_P(f) \geq -n$, which simply means that $f$ has a pole of order at most $n$ at $P$. There are curves $X$ and values of $n$ and $P$ such that there is no function with a pole of order exactly $n$ at $P$ and no other poles: such $n$ are called Weierstrass gaps.