Very Ample Implies Separation of Points and Tangents: Miranda As a preface, this is not homework. I'm a professor, after all.
I'm reading through Miranda's "Algebraic Curves and Riemann Surfaces," and getting caught up on one point from Chapter VI. 
First, some background: a divisor $D$ is called very ample on compact Riemann Surface $X$ if a) its associated embedding of $X$ into projective space is a holomorphic embedding whose image is a Riemann Surface and b) its linear system $|D|$ is base point-free. It is a theorem that this happens if and only if $\dim L(D - q - p) = \dim L(D) - 2$ for all points $p, q \in X$ (including the case $p = q$), where $L(D)$ is the set of meromorphic functions on $X$ such whose principal divisors (divisors given by orders of points) are $\geq - D$. 
Claim: a very ample divisor $D$ must have the property that the field generated by $L(D)$ separates points and tangents on $X$. That is, given distinct $p, q \in X$, there is a meromorphic function on $X$ in the field generated by $L(D)$ separating them (separates points), and, given any $p \in X$, there must be a meromorphic function on $X$ in that same field whose multiplicity at $p$ is one, which means there is a meromorphic function $f$ on $X$ (in the field generated by $L(D)$) which is either a) holomorphic at $p$ with the order of $f - f(p)$ equal to one at $p$, or b) which has a simple pole at $p$ (separates tangents). 
What I know: the very ample condition implies
$$L(D - p - q) \subset L(D - p) \subset L(D)$$
is a strict chain of subspaces for any $p, q \in X$, which means, for example, there is a function $f$ in $L(D - p)$ not in $L(D - p - q)$. This function would satisfy the property that $ord_p(f) \geq -D(p) + 1$, but that there is some point $x$ in its support where $ord_x(f) < -D(x)$, but I am lost after that. 
I have begun to slowly get lost in the concepts over this past chapter, and would appreciate some help here. For reference, this claim is made on page 170, and is problem A on p 178.
EDIT: The embedding associated to a divisor $D$ is given by
$$\phi_D : X \to \mathbb P^n$$
where $p \to [f_0(p):...:f_n(p)]$, and the $f_i$ are a basis for $L(D)$. We also know this basis can be chosen so that $ord_p f_0(p) = -D(p)$ and $ord_p f_i(p) > -D(p)$ for all other $i$.
 A: My attempt at solving this :
To see it separates points first take $p,q\in X$ such that $p\neq q$, and we know that since $D$ is a very ample divisor $\phi_D$ will be injective and so there exists a function $f$ such that $f\in L(D-p)$ but not in $L(D-p-q)$ and so there exists a function $f$ such that $ord_q(f)=-D(q)$ and $ord_p(f)\geq -D(p)+1$.Now since $D$ is base-point free we know that there exists a function in $L(D)$ but not in $L(D-p)$ and so we have a function $g$ such that $ord_p(g)=-D(p)$ and $ord_q(g)\geq -D(q)$.So putting both there results together and considering the function $\frac{f}{g}$ we will have that $ord_q(\frac{f}{g})=ord_q(f)-ord_q(g)\leq 0$ and $ord_p(\frac{f}{g})=ord_p(f)-ord_p(g)\geq 1$, and so we can see that it will have a zero at the point $p$ and at $q$ it will have a pole or not a zero, hence proving that this function separates the points and notice that this function is the function field generated by the functions in $L(D)$.
And now using the criteria that $\phi_D$ has to be an embedding we show that there is a function $f_i$ that as simple zero at $p, \forall p\in X$, this is because since we have a very ample divisor it will be base point free so there will be an $f_0\in L(D)$ such that $ord_p(f_0)=-D(p)$, and since we also have that $\phi_D$ is an embedding we have there exists a function $f_1\in L(D-p)-L(D-2.p)$ and so this function will have order $-D(p)+1$ at $p$, and so with this the desired function with a pole at $p$ will be $\frac{f_0}{f_1}$, and this is the reason we are working with the field generated by the functions in $L(D)$ and not only with $L(D)$.
