Hartshorne's proof of Halphen's theorem

Question

Apologies if this is not quite at the level of MathOverflow, but it has already been asked at MSE and gone unresolved for several years despite a bounty.

Hartshorne states the theorem as follows:

Proposition IV.6.1. A curve $X$ of genus $g\geq 2$ has a nonspecial very ample divisor $D$ of degree $d$ if and only if $d\geq g+3$.

The necessity is shown, and then sufficiency. The idea is to show that the set $S$ of divisors $D \in X^d$ such that there exists $D' \sim D$ and points $P,Q \in X$ with $E = D'-P-Q$ an effective special divisor has dimension $\leq g+2$. Because $d\geq g+3$ this means there is some $D\notin S$ that is nonspecial and very ample of degree $d$.

Hartshorne shows that the set of divisors of the form $E+P+Q$ in $X^d$ that are nonspecial with $E$ a special effective divisor has dimension $\leq g+1$.

The part that confuses me comes afterwards. Namely, as $E$ is special the Riemann-Roch tells us that $\dim |E| \geq d-1-g$, and similarly that $\dim |E+Q+P| = d-g$. Because the difference between these two dimensions is at most 1, this somehow implies that the set of divisors $S$ as above has dimension $\leq g+2$.

I don't understand this implication. We have divisors of the form $E+Q+P$, each of which gives a linear system of dimension $d-g$, and all of which form a set of divisors of dimension $\leq g+1$. How does the difference in dimension of $E$ and $E+P+Q$ tell us the dimension of $S$?

Answer 1

I think this Hartshorne's explanation is very rough. By using Picard variety, this step can be proved more clearly as follows.

Write $\mathrm{Pic}^d(X)$ for the scheme which parametrized all line bundles of degree $d$ on $X$, $\mathrm{Div}^d(X)$ for the scheme which parametrized all effective divisors of degree $d$ on $X$, and $D_{\mathrm{univ}}\subset X\times \mathrm{Div}^d(X)$ for the universal effective divisor of degree $d$. Then, by the universality of $\mathrm{Pic}^d(X)$, the line bundle $\mathcal{O}_{X\times \mathrm{Div}^d(X)}(D_{\mathrm{univ}})$ induces a morphism $$ \varphi_d:\mathrm{Div}^d(X) \to \mathrm{Pic}^d(X). $$ This morphism can be written as $\varphi_d(D) = \mathcal{O}_X(D)$. Hence each fiber of $\varphi_d$ is linearly equivalent class. In particular, for any $L\in \mathrm{Pic}^d(X)$, it holds that $\dim(\varphi_d^{-1}(L)) = \dim H^0(X,L) - 1$.

Write $\mathrm{SpDiv}^d\subset \mathrm{Div}^d(X)$ for the closed subscheme determied by all special effective divisors. Then, by Riemann-Roch, for any $D\in \mathrm{SpDiv}^d$, it holds that $$ \dim(\varphi_d^{-1}(\mathcal{O}_X(D))) = 1+d-g+l(K-D)-1 \geq 1+d-g. $$ Since $\dim(\mathrm{SpDiv}^d) = g-1$, it holds that $$ \dim(\varphi_d(\mathrm{SpDiv}^d)) \leq (g-1)-(1+d-g) = 2g-2-d. $$

Now, let us consider the following two morphisms: \begin{align*} f:X\times X\times \mathrm{Div}^{d-2}(X)\to \mathrm{Div}^d(X), (P,Q,D)\mapsto P+Q+D, \\ g:X\times X \times \mathrm{Pic}^{d-2}(X) \to \mathrm{Pic}^d(X), (P,Q,L)\mapsto L\otimes \mathcal{O}(P+Q). \end{align*} Then we obtain the following commutative diagram: $$ \require{AMScd} \begin{CD} X\times X\times \mathrm{SpDiv}^{d-2} @>{\subset}>> X\times X\times \mathrm{Div}^{d-2}(X) @>{f}>> \mathrm{Div}^d(X) \\ @. @V{\psi}VV @VV{\varphi_d}V \\ @. X\times X\times \mathrm{Pic}^{d-2}(X) @>{g}>> \mathrm{Pic}^d(X), \end{CD} $$ where $\psi := \mathrm{id}_X\times \mathrm{id}_X \times \varphi_{d-2}$. Write $T:= g(\psi(X\times X\times \mathrm{SpDiv}^{d-2}))\subset \mathrm{Pic}^d(X)$. Then $$\dim(T) \leq 2g-2-(d-2)+2 = 2g-d+2.$$ Since $\dim(\mathrm{Div}^d)) = d$, and $\dim(\mathrm{Pic}^d) = g$, the dimension of general fibers of $\varphi_d$ are $d-g$. Hence $$\dim(\varphi_d^{-1}(T)) \leq (2g-d+2)+(d-g) = g+2,$$ (where we note that if $T \subset \varphi_d(\mathrm{SpDiv}^d)$, then the dimension of general fibers of $\varphi_d^{-1}(T) \to T$ are $>d-g$, however, in this case, since $\varphi_d^{-1}(T)\subset \mathrm{SpDiv}^d$, it holds that $\dim(\varphi_d^{-1}(T)) \leq \dim(\mathrm{SpDiv}^d) = g-1 < g+2$). Moreover, by our construction, the scheme $\varphi_d^{-1}(T)$ parametrizes all effective divisors $D\subset X$ which are linearly equivalent to $E+P+Q$, the sum of a special effective divisor $E\subset X$ and points $P,Q\in X$. This is desired conclusion.