In Stichtenoth’s book (Algeraic function fields and codes) I have encountered a small technical difficulty in a Lemma ($\textbf{Lemma 1.6.14}$) that proves the following:
Given a function field $F/K$ with $K$ an infinte field and with genus $g>0$, let $A,B$ Be divisors such that $l(A)>0$ and $l(B)>0$. Then we have $l(A)+l(B)\leq 1+l(A+B)$ . Here $l(A),l(B)$ denotes the dimensions of the Riemann-Roch spaces $\mathcal{L}(A),\mathcal{L}(B)$ resp.
The proof goes like this :
Since $l(A),l(B)>0$ we pick effective divisors $A_0,B_0$ which are equivalent to $A,B$ resp. (this is done by picking any nonzero element say $x\in \mathcal{L}(A)$ and setting $A_0:=A+(x)$ where $(x)$ denotes the principal divisor of $x$). Then we choose a divisor $D_0$ of minimal degree satisfying $D_0\leq A$ and $\mathcal{L}(D_0)=\mathcal{L}(A_0)$. Now if $\{P_1,...,P_r\}$ is the support of the divisor $B_0$ then by minimality of degree $\mathcal{L}(D_0-P_i)$’s are proper subspaces of $\mathcal{L}(D_0)$ and since $K$ is an infinte field one can pick $z\in \mathcal{L}(D_0)\setminus \bigcup\limits_{i=1}^r \mathcal{L}(D_0-P_i)$.
The technical difficulty is in the following statement that the author claims to be “obvious” from the above choice of $z$.
The map $x\mapsto xz (mod\, \mathcal{L}(A_0): \mathcal{L}(B_0)\to \mathcal{L}(D_0+B_0)/\mathcal{L}(A_0)$ has kernel $K$.
This is not clear to me as the author claims it should be. May be I am missing something obvious. The kernel surely contains $K$. I have to show that if $x\in \mathcal{L}(B_0)$ is transcendental (note that the book assumes that $K$ is algebraically closed in $F$) then $xz\notin \mathcal{L}(A_0)$. From the choice of $z$, it is clear to me that $v_{P_i}(z)=-v_{P_i}(D_0)$. For any divisor $Q$ outside the support of $B_0$ one has $v_Q(xz)=v_Q(x)+v_Q(z)\geq -v_Q(D_0)\geq -v_Q(A_0)$, so there must be some $i$ for which $v_{P_i}(xz)<-v_{P_i}(A_0)$ and hence $(xz)\not\geq -A_0$. How to proceed from here?
Thanks in advance for any help with this.
Refer : Algebraic Function fields and Codes by Henning Stichtenoth, Second Edition, page 36-37.
If $x$ is transcendental then it must have a pole $P\in\mathbb{P}_F$ (cf. Corollary 1.1.20). Since $x\in\mathcal{L}(B_0)$, for any divisor $Q\notin \text{supp} B_0$ we have $v_Q(x)\geq 0$. Thus $P=P_i$ for some $i\in\{1,2,\dots,r\}$.
Now, $v_P(x)<0$ and $v_P(z)=-v_P(D_0)$, hence $$v_P(xz)=v_P(x)+v_P(z)<-v_P(D_0)$$ and so $xz\notin\mathcal{L}(D_0)=\mathcal{L}(A_0)$.