I am struggling with details of the proof of Halphen's Theorem in Hartshorne's Algebraic Geometry (chapter 4.6, Proposition 6.1, page 349). The statement of the theorem is:
A curve $X$ of genus $g\geq 2$ has a nonspecial very ample divisor $D$ of degree $d$ iff $d\geq g+3$.
By curve, I mean a complete, nonsingular curve over $k=\overline{k}$. (i.e. integral scheme of dimension 1, proper over $k$, all local rings are regular).
I struggle with the second part of the proof. Given $d\geq g+3$, Hartshorne proves the existence of a nonspecial very ample divisor of degree $d$.
In his proof he associates $X^{d}$ with divisors of the form $D=P_{1}+\ldots+P_{d}$ where the $P_{i}$ are closed points.
He then defines the set \begin{align*} S=&\{D \in X^d\; | \;\exists \textrm{ divisor } D' \textrm{and points } P,Q: \\ &D \sim D', E=D'-P-Q \textrm{ is effective and special} \} \end{align*} and discusses its dimension. So, my first question: Does he refer to the subscheme associated to $\overline{S}$ with the reduced induced structure? If not, how can the authore refer to the notion of the dimension of this set?
He further notes that any effective special divisor is a subset of an effective canonical divisor and since $\textrm{dim}_k |K|=g-1$, the set of all effective special divisors also has dimension at most $g-1$.
My problem here is that I can not make this precise. (I also struggle again with the fact that he talks about the dimension of the set). If $E$ is an effective, special divisor, then there is an effective divisor $F$ such that $K-E \sim F$ and thus $K\sim F+E$ and $F+E$ is effective. So we could define a map $E \mapsto E+F$ and its injectivity would imply the bound of the dimension. However, there might be multiple suitable $F$ (so a choice has to be made?) and the bound of the dimension only makes sense if the map respects some sort of structure.
If anybody could clarify some of my confusions, I would be very thankful. I apologize for the long post, I wasn't able to shorten it.
I would also appreciate the reference to another proof of this theorem.
I think this Hartshorne's proof is very rough. Actually, the scheme that parametrizes all special effective divisors of degree $d$ is constracted scheme-theoretically as follows:
Write $\mathrm{Div}^d = X^d/S_d$ for the scheme that parametrizes all effective divisors of degree $d$, $D_{\mathrm{univ}}\subset X\times\mathrm{Div}^d$ for the universal effective divisor of degree $d$, $p:X\times\mathrm{Div}^d \to \mathrm{Div}^d$ and $q:X\times\mathrm{Div}^d\to X$ are natural projections. Then the exact sequence $$ 0 \to \mathcal{O}_{X\times\mathrm{Div}^d} \xrightarrow{"+D_{\mathrm{univ}}"} \mathcal{O}_{X\times\mathrm{Div}^d} (D_{\mathrm{univ}}) \to \mathcal{O}_{X\times\mathrm{Div}^d} (D_{\mathrm{univ}})|_{D_{\mathrm{univ}}} \to 0 $$ induces the following exact sequence on $\mathrm{Div}^d$: $$ R^1p_* \mathcal{O}_{X\times\mathrm{Div}^d} \to R^1p_*\mathcal{O}_{X\times\mathrm{Div}^d}(D_{\mathrm{univ}}) \to R^1p_*(\mathcal{O}_{X\times\mathrm{Div}^d}(D_{\mathrm{univ}})|_{D_{\mathrm{univ}}}). $$ Since the composite $D_{\mathrm{univ}}\subset X\times\mathrm{Div}^d\to \mathrm{Div}^d$ is finite (flat and of degree $d$), it holds that $R^1p_*(\mathcal{O}_{X\times\mathrm{Div}^d}(D_{\mathrm{univ}})|_{D_{\mathrm{univ}}}) = 0$. Moreover, by flat base change and Serre duality, we obtain the following natural isomorphisms: $$ R^1p_* \mathcal{O}_{X\times\mathrm{Div}^d} \cong R^1p_* q^*\mathcal{O}_X \cong H^1(X,\mathcal{O}_X)|_{\mathrm{Div}^d} \cong H^0(X,\Omega_X)|_{\mathrm{Div}^d}^{\vee}, $$ where $(-)^{\vee}$ denotes the dual of $(-)$. Hence we obtain the following surjection of coherent sheaves on $\mathrm{Div}^d$: $$ H^0(X,\Omega_X)|_{\mathrm{Div}^d}^{\vee} \to R^1p_*\mathcal{O}_{X\times\mathrm{Div}^d}(D_{\mathrm{univ}}). $$ By taking $\mathrm{Proj}(\mathrm{Sym}(-))$ on $\mathrm{Div}^d$, we obtain the following morphisms: $$ R := \mathrm{Proj}(\mathrm{Sym}(R^1p_*\mathcal{O}_{X\times\mathrm{Div}^d}(D_{\mathrm{univ}}))) \hookrightarrow \mathrm{Div}^d\times \mathbb{P}(H^0(X,\Omega_X)^{\vee}) \xrightarrow{\text{proj}.} \mathbb{P}(H^0(X,\Omega_X)^{\vee}) = |K_X|. $$
Now let us give an interpretation of the $\mathrm{Div}^d$-scheme $R$. Let $D\subset X$ be an effective divisor of degree $d$. Then $D$ determines a point $[D]\in \mathrm{Div}^d$. To consider the fiber of $R\to \mathrm{Div}^d$ on $[D]$, write $k([D])$ for the residue field at $[D]\in \mathrm{Div}^d$. Then, by cohomology and base-change theorem (cf. [Hartshorne, theorem 3.12.11 (a)]) and Serre duality, we obtain \begin{align*} &R^1p_*\mathcal{O}_{X\times\mathrm{Div}^d}(D_{\mathrm{univ}})\otimes k([D]) \\ &\cong H^1(X\times \{[D]\}, \mathcal{O}_{X\times\mathrm{Div}^d}(D_{\mathrm{univ}})|_{X\times \{[D]\}}) \\ &\cong H^1(X,\mathcal{O}_X(D)) \\ &\cong H^0(X,\mathcal{O}_X(K_X-D))^{\vee}. \end{align*} Since the operation $\mathrm{Proj}(-)$ and $\mathrm{Sym}(-)$ commute base-change, this implies that the fiber of $R\to \mathrm{Div}^d$ at $[D]\in \mathrm{Div}^d$ is naturally isomorphic to $\mathbb{P}(H^0(X,\mathcal{O}_X(K_X-D))^{\vee}) = |K_X-D|$. Thus, we may conclude that taking a point $r\in R$ is equivalent to be takeing an effective divisor $D\subset X$ (this corresponds to the image of $r$ in $\mathrm{Div}^d$) and an effective divisor $D'\subset X$ which is linearly equivalent to $K_X-D$ (this corresponds to the point $r\in |K_X-D|$, as the fiber of $[D]$), i.e., $R$ is the scheme which parametrizes all pairs $(D,D')$ of effective divisors of $X$ such that $D+D' \sim K_X$.
In particular, the image of $R\to \mathrm{Div}^d$ parametrizes all special effective divisors of degree $d$. Moreover, since the above morphism $$ R \to |K_X| $$ comes from the morphism $\mathcal{O}_{X\times\mathrm{Div}^d} \xrightarrow{"+D_{\mathrm{univ}}"} \mathcal{O}_{X\times\mathrm{Div}^d} (D_{\mathrm{univ}})$, this morphism can be written by $(D,D')\mapsto D+D'$.
Now, as explained by Hartshorne, the morphism $R\to |K_X|$ is finite. Hence $\dim R = \dim |K_X| = g-1$. Thus the image of $R\to \mathrm{Div}^d$ has dimension at most $g-1$, i.e., the scheme that parametrizes all special effective divisors of degree $d$ has dimension at most $g-1$. This is desired conclusion.