TITLE: Proof that Riemann Roch Space is a Vector Space
QUESTION [5 upvotes]: I'm having some Problems with a counterexample, showing that $L(D)$ is not a vector space.
Just imagine $D=-(Q)+2(\infty)$. Then
$$L(D)=\{f\in K(C)^*: \mbox{div}(f)-(Q)+2(\infty)\geq 0\}\cup \{0\}$$
so take (in one dimension) $f=(X-P)\cdot(X-Q)$, thus $\mathrm{div}(f)= (P)+(Q)-2(\infty)\in L(D)$ and
$$-f=-(X-P)\cdot(X-Q)\in L(D)$$
since 
$$\mathrm{div}(-f)= (P)+(Q)-2(\infty). $$
Now if I take $f+ -f=0$ I get $\mathrm{div}(0)=0$, so $f + -f$ is not in $L(D)$.
Where is my mistake???

REPLY [8 votes]: It is specifically written into the definition of $L(D)$ that $0 \in L(D)$!  Intuitively, if $D = \sum a_i P_i$ with $a_i \geq 0$ (only for convenience of explanation; the general case is handled by considering orders of vanishing as well) you should think of $L(D)$ as the space of all rational functions that have poles no worse than order $a_i$ at $P_i$.  Since the zero function has no poles (and, for the general case, also vanishes everywhere to arbitrary high order), it should be included in every $L(D)$ space.  Because $\text{div}(0)$ is not well-defined (in particular, the statement $\text{div}(0) = 0$ is false), $0$ must be included in $L(D)$ by definition.