Past exam of the algebraic curves course of the University of Oxford / 2023 / 3 / e

First fix distinct $q_1,\dots ,q_k$ and let $Q=q_1+\cdots +q_k$. A negative-degree divisor has no sections: if $0\ne f\in \mathcal{L}(D)$, then $(f)+D$ is an effective divisor of degree $\mathrm{deg}D$, since a principal divisor has degree zero, so $\mathrm{deg}D$ cannot be negative. As $\mathrm{deg}K=0$ and $\mathrm{deg}(K-Q)=-k<0$, the Riemann-Roch theorem therefore gives

Now choose distinct $p_1,\dots ,p_k$ outside the support of $Q$, and put $P=p_1+\cdots +p_k$. A function with the required zeros and poles must be a nonzero element of $\mathcal{L}(Q-P)$. Conversely, if $0\ne f\in \mathcal{L}(Q-P)$, then $(f)+Q-P$ is an effective divisor of degree zero, hence is zero. It follows that $(f)=P-Q$, giving precisely the required simple zeros and simple poles, with no others. Thus the existence question is exactly whether $\mathcal{L}(Q-P)\ne 0$.

Take a basis $f_1,\dots ,f_k$ of the Riemann-Roch space $\mathcal{L}(Q)$. Evaluation at the $p_i$ defines the linear mapwhose kernel is $\mathcal{L}(Q-P)$. The evaluation criterion for a prescribed principal divisor becomesThe crucial point is that this determinant does not vanish identically; a count of unknowns and equations alone would not prove this.

To verify it, begin with $V_0=\mathcal{L}(Q)$, of dimension $k$. Whenever $V_{j-1}\ne 0$, choose a nonzero $f\in V_{j-1}$. Its zeros are finite. Choose $p_j$ outside those zeros, the support of $Q$ and the previously chosen points. Evaluation at $p_j$ is then a nonzero linear functional on $V_{j-1}$, soAfter $k$ steps, $V_k=0$. Thus there exists a tuple of distinct points with $\Delta \ne 0$.

The entries of this determinant vary holomorphically away from the $q_i$, and are rational functions on the nonsingular projective curve. Its vanishing is therefore a proper closed exceptional locus on this configuration space; its complement is open and dense in the Zariski topology. For a generic choice of the $p_i$, $E_P$ is invertible andThis holds for every fixed tuple of distinct $q_i$, which proves the assertion for generic choices of both sets of points. At $k=1$, it is especially transparent: $\mathcal{L}(q_1)=\mathbb{C}$, so no nonzero member can vanish at a distinct $p_1$.