Past exam of the mathematics course of the University of Cambridge / 2015 / ii / Paper 4 / 20F (Algebraic Geometry)

For a smooth projective algebraic curve, a linear system of divisors comes from a finite-dimensional space $V\subseteq H^0(C,\mathcal{O}(D))$. A basis of sections $s_0,\dots ,s_n$ defines $f(P)=[s_0(P):\cdots :s_n(P)]$, using any local trivialization; a common scalar change does not alter the projective point. A base-point-free linear system makes this defined everywhere, and $f^{\ast }{\mathcal{O}}_{{\mathbb{P}}^n}(1)\cong \mathcal{O}(D)$. Pulling back a hyperplane gives the zero divisor of a linear combination of the sections, an element of the original system.

A system with base divisor $B$ can instead define a map after removing $B$, but its hyperplane pullbacks have class $D-B$. Thus base-point-freeness is needed for the stated original-system property. For a closed immersion, the system must also separate distinct points and tangent directions; equivalently it is a very ample linear system. On a smooth curve the latter tangent condition is separation of the length-two subscheme at each point.

The Riemann-Roch theorem states, for a divisor $D$ on a smooth projective curve of genus $g$,where $\ell (D)=\dim H^0(C,\mathcal{O}(D))$ and $K_C$ is a canonical divisor, of degree $2g-2$. Negative-degree divisors have no nonzero global sections. The theorem's usual formulation here is over an algebraically closed field, as in the curve course.

Here $\mathrm{deg}{K}_C=2$. For $D$, $D-P$ and $D-P-Q$, including $Q=P$, the degrees are $5,4,3$, all greater than two. Thus the correction term in the Riemann-Roch theorem vanishes andThe first drop supplies a section nonzero at every chosen point, proving base-point-freeness of a linear system. For distinct $P,Q$, the second drop supplies a section vanishing at one but not the other, proving point separation. For $Q=P$, it supplies a section vanishing to order exactly one, proving tangent separation. Hence the complete linear system of divisors is very ample and $\boxed{C\hookrightarrow {\mathbb{P}}^3}$ is the induced closed immersion.