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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???

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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.

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  • $\begingroup$ thx, helped me a lot. So $\mathrm{div}(0)$ is something like the empty divisor or empty sum? ("(in particular, the statement $div(0)=0$ is false)") $\endgroup$
    – Dan
    Commented Mar 29, 2013 at 17:52
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    $\begingroup$ The divisor of $0$ is not defined. $\endgroup$
    – Zhen Lin
    Commented Mar 29, 2013 at 18:09
  • $\begingroup$ It is not defined, since the sum of zeroes, i.e. $\sum v_P P$ would be infinite, right? $\endgroup$
    – Dan
    Commented Mar 31, 2013 at 20:43

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