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 $S$?


1 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.

  • $\begingroup$ Dear YJ_cat, thank you very much for your informative answer. The diagram in particular was a great help to my understanding. $\endgroup$ Dec 8, 2021 at 2:08

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