Doubt on base point for a linear system

Question

I'm studying divisors on Riemann surfaces and I got stuck on a thing said by the professor at lesson. Probably I'm getting lost in a glass of water, or there is something wrong in my notes: if this is so please let me notice what is the error.

Consider a linear system $V$ on a (compact) Riemann surface $X,$ i.e. a subset of a complete linear system $$|D|=\{E\in \operatorname{Div}(X) \mid E\geq 0,\ \exists f:X\to\mathbb{C}\text{ meromorphic such that }E=\operatorname{div}(f)+D \}$$ we say that $p\in X$ is a base point for $V$ if $p$ is in the support of each divisor in $V$ (with this statement I mean that the coefficient $E(p)$ of $p$ in each divisor $E\in V$ is nonzero).

I know that $V$ is in correspondence with a linear subspace $\overline{V}$ of $\mathbb{P}(L(D)),$ where $$L(D)=\{f:X\to\mathbb{C} \mid \operatorname{div}(f)\geq -D \}$$

In the notes there is this statement: $p$ is a base point for $V$ if and only if for each $f\in\overline{V}$ we have $f(p)=0$

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My attempt: I know that the divisors in the linear system $V$ are exactly those of the form $$\operatorname{div}(f)+D$$ with $f\in\overline{V}$.

If $p$ is a base point for $V$ I have that $E(p)\neq0$ for each $E\in V:$ hence we must have $E(p)\geq 1,$ since each divisor $E\in|D|$ satisfies $E\geq0.$ If we take a function $f\in\overline{V}$ we can consider the divisor $E=\operatorname{div}(f)+D\in V:$ using the previous observation and we must have $$\operatorname{ord}_p(f)+D(p)\geq 1$$ To conclude I must prove that $\operatorname{ord}_p(f)\geq1,$ and this is certainly true if $D(p)\leq0,$ but I don't see how to continue from this point if $D(p)\geq 1.$

I don't think this is a difficult thing to prove, since this is a simple observation in the notes, but I don't see how to conclude...

Answer 1

$f(p)=0$ should be understood as $f_p\in\mathfrak{m}_p(\mathcal{O}_X(D))_p$ here, not that $f$ as a meromorphic function evaluates to $0$. Here is an example demonstrating that this is the correct interpretation: if $p$ is a point on an elliptic curve $X$, then for the divisor $D=p$ we have the global sections of $\mathcal{O}_X(D)$ are just the constants, and $|D|$ is just the point $P$ - so we can't evaluate these as meromorphic functions and hope that all our constants are zero. On the other hand, as $(\mathcal{O}_X(D))_p$ is generated by a local function which has a pole of order 1 at $p$, any constant function is locally something in the maximal ideal times this generator.

Once we understand what we're really up to, we can solve the problem. Locally, at any point $x\in X$, the stalk $\mathcal{O}_X(D)$ is isomorphic to $\mathcal{O}_{X,x}$ by sending $1\in\mathcal{O}_{X,x}$ to a meromorphic function with pole of order equal to $\nu_x(D)$, the coefficient of $x$ in $D$. Now if $f_x$ is in $\mathfrak{m}_x(\mathcal{O}_X(D))_x$, we must have that as a meromorphic function, $f$ has a pole of order less than $\nu_x(D)$, giving that $\operatorname{div}(f)+D$ has a positive coefficient on $x$. Conversely, if $f_x$ isn't in $\mathfrak{m}_x(\mathcal{O}_X(D))_x$, then as a meromorphic function, $f$ has a pole of order equal to $\nu_x(D)$ and therefore $x$ does not appear in $\operatorname{div}(f)+D$. From here we may quickly conclude that $x$ being a base point of a linear system $V$ is equivalent to $f_x\in\mathfrak{m}_x(\mathcal{O}_X(D))_x$ for all $f\in V$.

Answer 2

@KReiser I want to try to give another point of view to see if I've understood well what you said in your answer, without talking about stalks, at least not explicitly: please tell me if there is something wrong.

With the same notations in the question we have a morphism of sheaves $\phi:\mathcal{O}_X[D]\to \mathbb{C}_p$ defined in this way:

1. We choose a local coordinate $z$ centered in $p$
2. Given an open subset $U\subseteq X,$ if $p\not\in U$ we simply set $\phi_U$ to be the zero morphism; if $p\in U,$ given $f\in \mathcal{O}_X[D](U)$ we may consider the local expression for $f$ in the coordinate $z$ $$f(z)=\sum_{r=-D(p)}^{+\infty}c_rz^r$$ and we set $$\phi_U(f)=c_{-D(p)}$$

Clearly this morphism of sheaves is onto and its kernel is the sheaf $\mathcal{O}_X[D-p]$ With this definition the morphism $\phi$ depends from the local coordinate $z$ we have chosen, but the minimum $r\geq -D(p)$ such that $c_r\neq0$ doesn't depend from this coordinate, and this data is can be recovered using stalks.

We have the long exact sequence in cohomology $$0\to H^0(X,D-p)\to H^0(X,D)\overset{\text{val}_p}{\longrightarrow} \mathbb{C}\to H^1(X,D-p)\to H^1(X,D)\to 0$$ where the map $\text{val}_p$ should be considered as the map $\phi_X$

Using this language the correct way to express the condition "$f(p)=0$ for each $f\in L(D)=H^0(X,D)$" is obtained substituting $f(p)=0$ with $\text{val}_p(f)=0$ (i.e. $f\in H^0(X,D-p)=\text{ker}\ \text{val}_p$): the map $\text{val}_p$ coincides with the evaluation at $p$ only if $D(p)=0.$

Now I can prove that the two characterizations of base points I have given are equivalent, and the proof should be the same as yours expressed in a different way:

Suppose first that $p\in X$ is a point in the support of each divisor $E\in|D|:$ that means exactly $$E(p)\geq 1\qquad \forall E\in|D|$$ If we fix $f\in H^0(X,D)$ we can consider the divisor $$E=\text{div}(f)+D$$ By construction $E\in|D|,$ hence $p$ is in its support and we have $$\text{ord}_p(f)=E(p)-D(p)>-D(p)$$ and hence $f\in H^0(X,D-p).$

Conversely suppose $\text{val}_p(f)=0$ for each $f\in H^0(X,D):$ that means $\text{ord}_p(f)>-D(p)$ (this information doesn't depend, as I've written above, by the local coordinate chosen) and consider a divisor $E\in|D|:$ by definition there is a $f\in H^0(X,D)$ such that $$E=\text{div}(f)+D$$
Then $$E(p)=\text{ord}_p(f)+D(p)>0$$ which means that $p$ is in the support of the divisor $E.$