Past exam of the mathematics course of the University of Cambridge / 2011 / ii / Paper 3 / 23H (Algebraic Geometry)

For a divisor $D={\sum }_p{n}_{p}p$ on the smooth projective curve,This is a $k$-vector space of rational functions with the prescribed pole bounds. A nonzero rational function on a projective curve has divisor of degree zero. Thus if $\mathrm{deg}D<0$ there can be no nonzero element of $L(D)$.

For any point $p$ with coefficient $n_p$ in $D$, use a local parameter $t$ at $p$. The linear map sending $f\in L(D)$ to the residue-field value of $t^{n_p}f$ has kernel $L(D-p)$ and target $k$, since $k$ is algebraically closed. Therefore $\dim L(D)/L(D-p)\le 1$. If $d=\mathrm{deg}D\ge 0$, successively subtract a point $d+1$ times. The final divisor has degree $-1$ and no nonzero sections, givingEquivalently the universally valid bound is $\dim L(D)\le \max (0,\mathrm{deg}D+1)$. The PDF states the unqualified bound $\dim L(D)\le \mathrm{deg}D+1$; literally this is false when $\mathrm{deg}D\le -2$, because its right side is negative. For example $D=-2p$ has $L(D)=0$.

Let $K=k(X)$. The space of Kähler differentials ${\Omega }_{K/k}^1$ is the $K$-vector space generated by symbols $df$, $f\in K$, with relationsIts elements are rational differentials. A local parameter $t$ is a nonconstant element of $K$, so $k(t)$ is a rational function field and $K/k(t)$ is finite. In characteristic zero this extension is separable.

For any $f\in K$ let $P(t,f)=0$ be its minimal polynomial over $k(t)$. Differentiating givesSeparability ensures $P_f(t,f)\ne 0$, so $df$ is a multiple of $dt$, and hence ${\Omega }_{K/k}^1=Kdt$. To show $dt\ne 0$, the derivation $d/dt$ on $k(t)$ extends uniquely to the separable extension $K$, by the same differentiated minimal-polynomial formula; it sends $t$ to $1$, so the universal differential $dt$ cannot vanish. ThusFor a nonzero differential $\omega =fdt$, define its valuation of a rational differential at $p$ byConcretely $\omega =t^{m}udt$ with $u$ a unit in the local ring, and this valuation is $m$. For $\omega =0$ one conventionally assigns $+\infty$, not an integer.