It can be seen from the Riemann-Roch's theorem that it is not easy to compute $h^1(X,D)$ in general, and of course, we want to work with $h^0(X,D)$ more than. The Serre's duality, as we will see later, implies that there is a canonical divisor $K\in Div(X)$ such that for all $D\in Div(X)$, $h^0(X,D)-h^0(X,K-D)=1+\deg D-g$, i.e. $h^0(X,K-D)=h^1(X, D)$. To do this, we need to look at the differential 1-forms in curves.
1. Differential 1-forms. We want to study curves from local information, because it seems to be easier to approach much more than the global ones. In calculus, the derivatives are very useful, they control the change of the function at a point. We want to create the same things for curves, but the tools for now is more complicated. We need to define first the whole structure of such derivatives.
Definition 1.1. Let $X$ be a curve (affine or projective) over an algebraically closed field $k$, we denote $\Omega^1(X)$ the $k[X]$-module generated by $df$, for $f\in k[X]$ with these properties hold for all $f,g\in k[X], \alpha\in k$
i. $d(f+g) = df + dg$
ii. $d(fg) = fdg + gdf$
iii. $d(\alpha g)=\alpha d(g)$
We call $\omega\in \Omega^1(X)$ the differential 1-form in $X$, sometimes it is called regular differential form, and such $\omega$ is represented as sum of $fdg$, for some $f,g\in k[X]$.
From this, one can see that it is easy to construct such $\Omega_1(X)$, as we just take the free $k[X]$-module generated by $df$, and then take the quotient of these properties. We give some examples.
Example 1.2. Let $X:y=x^2$, then $dy = d(x^2)=2xdx$. Hence, any $\omega\in \Omega^1(X)$ is generated by $dx$.
Example 1.3. Let $X: y^2 = x^3 + x$, where $char(k)\ne 2,3$. Then it can be seen $d(y^2)=2ydy=d(x^3+x)=(3x^2+1)dx$. Let us denote $\omega=\frac{dy}{3x^2+1}=\frac{dx}{2y}$. We will prove for now $\omega\in \Omega^1(X)$. To do this, we must verify if $\omega$ is sum of $fdx+gdy$, for some $f,g\in k[X]$. It can be seen that $dy =\omega(3x^2+1),dx=\omega 2y$. And it is equivalent to prove $1 = (3x^2+1)f + 2yg$, for some suitable $f,g$. It can be seen that $X$ is non-singular, and the partial derivatives do not vanish identically in $X$, this implies, in $k[X]$, the ideal generated by $3x^2 + 1$ and $2y$ is the whole ring. And hence, there exists $f,g$ satisfying our conditions, and $\omega$ is the differential 1-form.
Example 1.4. We will prove that there exists no differential 1-form in $\mathbb{P}^1$, other than 0. We first see that $\mathbb{P}^1$, with coordinate $(X:Y)$ is covered by two affine open subset $U_1:=\{(X:1)|x_1\ne 0\}, U_2=\{(1:Y)|Y\ne 0\}$. And in $U_{12}:=U_1\cap U_2$, the projective point $(X:Y)$ is represented by $(X/Y:1)$ in $U_1$, and $(1:Y/X)$ in $U_2$, and so, $Y\in U_2$ is $1/X\in U_1$.
Let $\omega\in\Omega^1(\mathbb{P}^1)$, then its restriction to $U_1$ is of the form $f(X)dX$, and its restriction to $U_2$ is of the form $g(Y)dY$. And in $U_{12}$, we have $f(X)dX = g(Y)dY = g(1/X)d(1/X) = g(1/X)X^{-2}dX$. But it can not be the case since there is no $f,g\in K[X]$ such that $f(X) = g(1/X)X^{-2}$, other than 0.
NOTE. We can easily deduce from the definition that $d(\alpha)=0, \forall\alpha\in k$, and because $\mathcal{O}_{P^1}(P^1)=k$ (see Part 1), we can deduce that $df = 0$, for any $f\in \mathcal{O}_{P^1}(P^1)$. And now, Example 1.4 easily follows.
We now extend a little bit the definition, from now on, $X$ is a smooth, irreducible, projective curve. Define $\Omega^1_{k(X)}=\{(U,\omega)\}/\sim$, where $U\subset X$ is open affine, and $\omega$ is the differential $1$-form in $\Omega^1(U)$. The equivalent relation $\sim$ is defined $(U, \omega)\sim (V,\eta)$ if $\omega_{V\cap U}=\eta_{V\cap U}$.
Fact 1.5. $\Omega^1_{k(X)}$ is 1 dimensional $k(X)$-vector space. Also, for all $P\in X$, there exists $P\in U\subset X$ an open affine, and $t_P$ the uniformizer at $P$ such that $(U, dt_P)$ is the basis for $k(X)$-vector space $\Omega^1_{K(X)}$. Then, for any $\omega\in \Omega^1_{K(X)}$, there exists only one $g\in \mathcal{O}_X(U)$ such that $\omega=gdt_P$.
And this allows us to define $v_P(\omega)=v_P(g)$. For now, if $g$ has pole at $P$ with multiplicity $n$, we can represent $g=a_{-n}t_P^{-n}+...+a_{-1}t_P^{-1}+h$, for some $h\in K(X)$, and $h$ is regular at $P$. The coefficient $a^{-1}$ is called the residue of $\omega$ at $P$, which is denoted $res_P(\omega)$. And the residue theorem in this case is
Proposition 1.6. $\sum_{P\in X}res_P(\omega)=0$.
The divisor of $\omega$ is defined $div(\omega)=\sum_{P\in X}v_P(\omega)P$. And for any divisor $D\in Div(X)$, we define $H^0(X,\Omega^1(D))=\{\omega\in \Omega^1_{K(X)}|div(\omega)-D\ge 0\}$. We denote $H^0(X,\Omega^1(0))$ as $H^0(X,\Omega^1)$.
Example 1.7. $H^0(X,\Omega^1)$ consists of all $\omega\in \Omega^1_{K(X)}$ such that $v_P(\omega)\ge 0$ for all $P$, and hence at each affine neighborhood at $P$, such that there exists $g\in K(X)$ $\omega=gdt_P$, we have $v_P(\omega)=v_P(g)\ge 0$. And hence, $g_{|U}\in \mathcal{O}_X(U)_P$. And hence, taking the intersection for all $P$, we found $\omega_{|U}\in \Omega^1(U)$, and hence $\omega\in \Omega^1(X)$. This yields $H^0(X,\Omega^1)=\Omega^1(X)$.
Because $X$ is an irreducible, projective curve, we can cover $X$ by two affine open subsets $U_1, U_2$, with $U_{12}=U_1\cap U_2$. We now define the following important pairing, for any $D\in Div(X)$ the map $H^0(U_{12},D)\oplus H^0(X,\Omega^1(D))\to k$, which sends $(f,\omega)$ to $\sum_{P\in X-U_2}res_P(f\omega)$. And we will prove for now that this pairing factors over $H^1(X,D)$.
Let $f\in \mathcal{L}(U_1,D), g\in \mathcal{L}(U_2,D)$, so that $f_{U_{12}}-g_{U_{12}}$ is in the image of the map $\delta$ in the first part, we have $div(f|_{U_1})+D\ge 0$, and for any $\omega\in H^0(X,\Omega^1(-D))$, we have $div(\omega)-D\ge 0$. This implies, in $U_1$, $div(f\omega)=div(f)+div(\omega)\ge 0$, and hence $(f\omega)$ has no pole in $U_1$. That means the pole of $f\omega$ is only possible at $P\in X-{U_1}$. By the Proposition 1.6, we have $\sum_{P\in X-U_1}res_P(f\omega)=0$, and also, $\sum_{P\in X}res_P(f\omega)=0$. This yields $\sum_{P\in X-U_2}res_P(f\omega)=0$. Similarly, we see $\sum_{P\in X-U_2}res_P(g\omega)=0$. From this, the pairing $(f_{U_{12}}-g_{U_{12}}, \omega)=0$.
In conclusion, this yields the isomorphism between $H^1(X,D)$ and $H^0(X,\Omega^1(-D))$.
2. Serre's Duality and applications. In the last section, we have proved that for any $D$, there exists $\omega\in \Omega^1_{K(X)}$ such that there exists an isomorphism between $H^1(X,D)$ and $H^0(X,\Omega^1(-D))$. However, $\Omega^1_{K(X)}$ is of 1 dimensional $K(X)$-vector space, and hence, for any $\omega$, there exists $\omega_0\ne 0$ such that $\omega=f\omega_0$, and we know from the first part that multiplying some principal divisor does not change the dimension of $H^0$. This yields a stronger result, that for any $\omega_0\ne 0$, and for any $D\in Div(X)$, $H^1(X,D)$ and $H^0(X,\Omega^1(-D))$ is isomorphic. It is called the Serre's duality.
If we denote $K:=div(\omega_0)\in Div(X)$, which is called canonical divisor. We will prove that, in fact
Proposition 2.1. Let $K=div(\omega_0)$, then $\dim H^0(X, K-D)=\dim H^0(X,\Omega^1(-D))$.
Proof. We can see that
$$f\in H^0(X,K-D)\Leftrightarrow div(f) + K - D\le 0 \Leftrightarrow div(f\omega_0)-D\ge 0\Leftrightarrow$$
$$\Leftrightarrow div(\omega)-D\le 0\Leftrightarrow \omega\in H^0(X,\Omega^1(-D))$$
(Q.E.D)
It is a very fruitful result, and when we combine things together, we get the following result
Theorem 2.2. Let $X$ be a smooth, irreducible, projective curve, then there exists $K\in Div(X)$ such that for any $D\in Div(X)$, $h^0(X,D)-h^0(X,K-D)=1+\deg D-g$.
The applications of this theorem is a lot, and we will list some of them, note that they are essential in the proof of the Weil's conjectures for curves.
1. In Theorem 2.2, if we take $D=0$, then $h^0(X,0)=1$, and hence, $h^0(X, K)=g$, the genus of $X$.
2. In theorem 2.2, if we take $D=K$, then $h^0(X,K)=g$, by 1. Also, $h^0(X, K-D)=1$, and hence, $\deg K = 2g-2$, which is called the Euler's characteristic of $X$. This implies that all non-zero $\omega\in \Omega^1_{K(X)}$ have degree $2g-2$.
3. For any principal divisor $f\in K(X)^\times$, we already know from Part 1 that $\deg(div(f))=0$. We will give another proof for this. It can be seen that the multiplication by $f$ will move $\omega\in \Omega^1_{K(X)}$ to another $\omega'\in \Omega^1_{K(X)}$, and hence, by Part 2, $\deg(div(f))=0$. From this, one can easily seen that if $D\in Div(X)$ with $\deg(D)<0$, then $h^0(X,D)=0$.
4. For $D\in Div(X)$, with $\deg(D)>2g-2$, we can see that $h^0(X, K-D)=0$, and hence, $h^0(X,D)=1+\deg D- g$.
5. From applications of the Riemann-Roch's theorem (7.) we can see that $h^0(\mathbb{P}^1,n\infty)=n+1$ for all $n$. By 4. this implies that for $n$ is big enough, $h^0(\mathbb{P}^1,n\infty)=n+1-g$. This implies $g=0$. And that the genus of $\mathbb{P}^1$ is 0.
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