Riemann-Roch space ($L(D)$)

The Riemann-Roch space of a divisor $D$ is the finite-dimensional vector spaceIts elements are the rational functions whose poles are bounded by the negative coefficients of $D$.

Let $P=p_1+\cdots +p_k$ and $Q=q_1+\cdots +q_k$ be disjoint sums of distinct points on a smooth projective curve. A nonzero function with principal divisor $P-Q$ exists exactly when $L(Q-P)\ne 0$: a nonzero member makes $(f)+Q-P$ an effective divisor of degree zero, forcing equality. Evaluation $f\mapsto (f(p_1),\dots ,f(p_k))$ on $L(Q)$ has kernel of a linear map $L(Q-P)$.

If $\ell (Q)=k$, a choice of basis $f_1,\dots ,f_k$ turns the criterion into $\det (f_j(p_i))=0$. This determinant is not identically zero. Starting with $L(Q)$, repeatedly choose a point where some nonzero function in the current space does not vanish; evaluation then lowers the dimension of that space by one. After $k$ choices the joint kernel is zero. Thus a generic tuple of points gives an invertible evaluation matrix, and no function with divisor $P-Q$. On a genus one curve, Riemann-Roch theorem gives $\ell (Q)=k$, so this obstruction applies for every $k>0$.

If a divisor on an algebraic curve $D$ on a smooth projective curve has negative degree of a divisor, then its Riemann-Roch space is zero. Otherwise a nonzero $f\in L(D)$ would make $(f)+D$ an effective divisor of degree $\mathrm{deg}D<0$, since a principal divisor has degree zero. This is impossible. In genus one the Riemann-Roch theorem consequently gives $\ell (D)=\mathrm{deg}D$ for every divisor of positive degree.

For finitely many distinct points $P_i$ on a smooth projective genus-$g$ curve, choose $N\ge 2g+1$. The Riemann-Roch theorem gives $\ell (N{P}_i)-\ell ((N-1)P_i)=1$, so choose $f_i$ with a sole pole of exact order $N$ at $P_i$. Then ${\sum }_i{f}_i$ has exactly those poles with that order: all other summands are regular at each $P_i$, so no principal part can cancel. This is an explicit construction rather than an unspecified generic choice of section.