1. Basic sheaf cohomology. We will use sheaf cohomology as one of the main tool in some further notes. It is said in the book "Algebraic Curves and Riemann Surfaces" of Rick Miranda that "For many geometers, cohomology theory is applied mathematics: something to be appreciated and actively used more than studied and analyzed in its own right. Its use may be likened to driving a car: you can easily get a license to drive without being able to build an engine".
For this reason, some technical points will be omitted in this section, and the reader may want to refer to Gathmann's note on Algebraic Geometry (Version 2002), Chapter VIII for proofs of some propositions.
Let $\mathfrak{U}=\{U_i\}$ be an affine open cover of $X$, and $\mathscr{F}$ is a sheaf (of abelian group) on $X$. Then we define $C^n(\mathscr{F}, \mathfrak{U}):=\prod_{i_0,...,i_n}\mathscr{F}(U_{i_0}\cap...\cap U_{i_n})$, i.e. an element in $C^n(\mathscr{F},\mathfrak{U_i})$ is of the form $\prod_{i_0,...,i_n}\alpha_{i_0,...,i_n}$, where $\alpha_{i_0,...,i_n}\in \mathscr{F}(U_{i_0}\cap...\cap U_{i_n})$. For convenience, we always denotes $C^{-1}(\mathscr{F}, \mathfrak{U})=0$. For all $n\ge 0$, the map $d^n: C^n(\mathscr{F},\mathfrak{U})\to C^{n+1}(\mathscr{F},\mathfrak{U})$ is induced by sending each $\alpha_{i_0,...,i_n}$ to $\sum_{i=0}^{n+1}(-1)^k\alpha_{i_0,...,i_{k-1},i_k,...,i_{n+1}}|U_{i_0}\cap...\cap U_{i_{n+1}}$. When $n=-1$, the map simply sends $0$ to $0$.
Let us give explicit computation for small $n$. When $n=0$, the map $d^0$ will map $\prod_{i_0}\alpha_i$ to $\prod_{i_0,i_1}(\alpha_{i_1}-\alpha_{i_0})$. When $n=1$, the map $d^1$ will map $\prod_{i_0,i_1}\alpha_{i_0,i_1}$ to $\prod_{i_0,i_1,i_2}(\alpha_{i_1,i_2}-\alpha_{i_0,i_2}+\alpha_{i_1,i_2})$. And one can check easily that $d^1\circ d^0$ is the zero map from $C^0(\mathscr{F}, \mathfrak{U})\to C^2(\mathscr{F},\mathfrak{U})$. In general, one can check by direct computation that $d^{n+1}\circ d_n=0$ for all $n\ge -1$, i.e. $Im(d_n)\subset \ker(d_{n+1})$. We then obtain the complex
$$0\to C^0(\mathscr{F},\mathfrak{U})\xrightarrow{d^0}C^1(\mathscr{F},\mathfrak{U})\xrightarrow{d^1}C^2(\mathscr{F},\mathfrak{U})\xrightarrow{d^2}...$$
$$0\to C^0(\mathscr{F},\mathfrak{U})\xrightarrow{d^0}C^1(\mathscr{F},\mathfrak{U})\xrightarrow{d^1}C^2(\mathscr{F},\mathfrak{U})\xrightarrow{d^2}...$$
And one has an important
Definition 1.1. The $n$-the cohomology group of $\mathscr{F}$ is defined by $H^n(X,\mathscr{F})=\ker(d^n)/Im(d^{n-1})$.
In Section VIII.5 of Gathmann's note, he proved that the $n$-th cohomology group is well-defined and does not depend on the choice of affine open cover of $X$. We then come to an important
Proposition 1.2. $H^0(X,\mathscr{F})\cong \mathscr{F}(X)$.
Proof. One can see $H^0(X,\mathscr{F})=\ker d^0=\{\prod_{i,j}\alpha_{ij}|\alpha_i=\alpha_j\text{ on } U_i\cap U_j\}$. By sheaf axioms, we can directly have $H^0(X,\mathscr{F})=\mathscr{F}(X)$. (Q.E.D)
If $X$ is a curve, then one can see there exists two open affine cover of $X$, namely $U_0,U_1$. Then it can be seen that $\mathfrak{U}=\{U_0,U_1,\emptyset\}$ is an open cover of $X$. And then, $C^n(\mathscr{F}, \mathfrak{U})=0$, for all $n\ge 2$. Hence, we obtain the complex sequence
$$0\to C^0(\mathscr{F},\mathfrak{U})\xrightarrow{d^0}C^1(\mathscr{F},\mathfrak{U})\xrightarrow{d^1}0\xrightarrow{d^2}0...$$
And hence, for all $n\ge 2$, $H^n(X,\mathscr{F})=0$. That is why for curves, we need to care about the first cohomology group.
Now, if we have a short exact sequence of sheaves
$$0\to \mathscr{F}\to \mathscr{G}\to \mathscr{H}\to 0$$
This will induce the long exact sequence (Refer to VIII.2 of Gathamann's note)
$$0\to H^0(X,\mathscr{F})\to H^0(X, \mathscr{G})\to H^0(X,\mathscr{H})$$
$$\to H^1(X,\mathscr{F})\to H^1(X,\mathscr{G})\to H^1(X,\mathscr{H})$$
$$\to...$$
Again, for curves, all terms $H^n$ (n\ge 2) will vanish, and just $H^1$ left.
2. Locally constant sheaf. In this section, $X$ is a smooth projective curve. Let $G$ be an abelian group, and $U\subset X$ is an open subset, a locally constant function from $U$ to $G$ is a map $f: U\to G$ such that for all $p\in X$, there exists an open neighborhood $V\subset U$ of $p$ such that $f_{V}: V\to G$ is a constant function. Let us define
$$\mathscr{F}(U)=\{f: U\to G|f\text{ is locally constant on } U\}$$
It can be seen that $\mathscr{F}(U)$ has the abelian group structure induced from $G$, and it is obvious a sheaf of abelian group on $X$. That is called locally constant sheaf. If $X$ is irreducible, one can see that any two open subsets of $X$ have non-empty intersection. This will yield in fact, $\mathscr{F}$ is a constant sheaf.
We will next prove that the $n$-th cohomology groups of constant sheaf vanish for all $n\ge 1$ on curves.
Proposition 2.1. Let $\mathscr{F}$ be a constant sheaf on $X$. Then $H^n(X,\mathscr{F})$ vanishes for all $n\ge 1$.
Proof. We already see that $H^n(X,\mathscr{F})$ vanishes for all $n\ge 2$. It is sufficient for us to check that $H^1(X,\mathscr{F})$ vanish. But this is easy, choose $\mathfrak{U}=\{U_0,U_1,\emptyset\}$ as above, then because $C^2(\mathscr{F},\mathfrak{U})=0$, all $\alpha_{01}\in C^1(\mathscr{F}, \mathfrak{U})$ is mapped to 0 via $d^1$. If we let $\beta_0=0$ and $\beta_1=\alpha_{01}$, then it can be seen that $\beta_0,\beta_1$ are in $\mathscr{F}(U_0),\mathscr{F}(U_1)$, respectively and $d^0(\beta_0,\beta_1)=\beta_1-\beta_0=\alpha_{01}\in \mathscr{F}(U_0\cap U_1)$. And hence, $d^0$ is a surjective map. It then yields $H^1(X,\mathscr{F})=0$ (Q.E.D)
In particular, if $G=k(X)^*$, where $k(X)$ is the function field of $X$ (in case $X$ is irreducible), the constant sheaf $\mathscr{F}$ with respect to $G$ is denoted $\mathscr{K}$, and one can see $H^1(X,\mathscr{K})=0$. It is the fundamental result for our next parts.
3. $Pic(X)$ and $H^1(X,\mathscr{O}^*)$. We are now in a position for our main result of this note. We always fix $X$ an irreducible, smooth projective curve in this section. First, let $Div(U)=\{\text{all divisors on with finite support in} U\}$. It can be seen that $Div(U)$ is a presheaf of abelian group. We need to check the gluing axiom to prove that $Div(U)$ is actually a sheaf. If $\{U_i\}$ is an open cover of $U$, and $D_i\in Div(U_i)$ such that $D_i|_{U_i\cap U_j}=D_j|_{U_i\cap U_j}$. If we can prove that $U$ can be cover by finitely many $U_1,.,,,U_n\in \{U_i\}$, then we can define $D\in Div(U)$ by $D(p)=D_j(p)$, for some $1\le j\le n$, and then, for sure $D\in Div(U)$.
Lemma 3.1. $X$ is noetherian topological space, then any subset of $X$ is noetherian topological space, and is quasi-compact.
Proof. Exercise.
Via this lemma, one can see $Div$ is a sheaf of abelian group. And one can easily has a sheaf map $div: \mathscr{K}\to Div$ defined by $div_U(f)=div(f)|_U$, where $div(f)$ is the divisor of a function $f\in k(X)^*$.
Lemma 3.2. The sheaf map $div$ defined above is an onto sheaf map.
Proof. For any $p\in X$, any open neighborhood $U$ of $p$, and any $D\in Div(U)$, if $D(p)=0$, we can choose $V$ is an open neighborhood of $p$ by removing all support of $D$ from U. Then it can be seen that on $V$, $div(1)=D$.
Otherwise, if $D(p)=n\ne 0$, let $z$ be the local coordinate at $p$, we then choose $V$ by removing all zeros (except $p$) and poles, and all support of $D$ (except $p$) from $U$. Then one can see on $V$, $div(z^n)=n[p]=D|_V$.
Hence, $div$ is an onto sheaf map. (Q.E.D)
What is the kernel of $div_U$? Well, it exactly consists of all nowhere zero regular functions on $U$, which is denoted $\mathscr{O}^*(U)$. We then obtain an exact sequence of sheaves
$$0\to \mathscr{O}^*\to \mathscr{K}\to Div\to 0$$
It then yields the long exact sequence, by using the vanishing of $H^1(X,\mathscr{K})$, we get
$$0\to H^0(X,\mathscr{O}^*)\to H^0(X,\mathscr{K})\to H^0(X,Div)\to H^1(X,\mathscr{O}^*)\to 0$$
Using Proposition 1.2, we have $H^0(X,\mathscr{O}^*)\cong \mathscr{O}^*(X)\cong \mathbb{k}^*$, $H^0(X,\mathscr{K})\cong \mathscr{K}(X)=k(X)^*$ and $H^0(X,Div)\cong Div(X)$. This then yields the exact sequence
$$0\to k^*\to k(X)^*\to Div(X)\to H^1(X,\mathscr{O}^*)\to 0$$
Due to the ker-coker exact sequence, we have $Pic(X)\cong H^1(X,\mathscr{O}^*)$.
Definition 1.1. The $n$-the cohomology group of $\mathscr{F}$ is defined by $H^n(X,\mathscr{F})=\ker(d^n)/Im(d^{n-1})$.
In Section VIII.5 of Gathmann's note, he proved that the $n$-th cohomology group is well-defined and does not depend on the choice of affine open cover of $X$. We then come to an important
Proposition 1.2. $H^0(X,\mathscr{F})\cong \mathscr{F}(X)$.
Proof. One can see $H^0(X,\mathscr{F})=\ker d^0=\{\prod_{i,j}\alpha_{ij}|\alpha_i=\alpha_j\text{ on } U_i\cap U_j\}$. By sheaf axioms, we can directly have $H^0(X,\mathscr{F})=\mathscr{F}(X)$. (Q.E.D)
If $X$ is a curve, then one can see there exists two open affine cover of $X$, namely $U_0,U_1$. Then it can be seen that $\mathfrak{U}=\{U_0,U_1,\emptyset\}$ is an open cover of $X$. And then, $C^n(\mathscr{F}, \mathfrak{U})=0$, for all $n\ge 2$. Hence, we obtain the complex sequence
$$0\to C^0(\mathscr{F},\mathfrak{U})\xrightarrow{d^0}C^1(\mathscr{F},\mathfrak{U})\xrightarrow{d^1}0\xrightarrow{d^2}0...$$
And hence, for all $n\ge 2$, $H^n(X,\mathscr{F})=0$. That is why for curves, we need to care about the first cohomology group.
Now, if we have a short exact sequence of sheaves
$$0\to \mathscr{F}\to \mathscr{G}\to \mathscr{H}\to 0$$
This will induce the long exact sequence (Refer to VIII.2 of Gathamann's note)
$$0\to H^0(X,\mathscr{F})\to H^0(X, \mathscr{G})\to H^0(X,\mathscr{H})$$
$$\to H^1(X,\mathscr{F})\to H^1(X,\mathscr{G})\to H^1(X,\mathscr{H})$$
$$\to...$$
Again, for curves, all terms $H^n$ (n\ge 2) will vanish, and just $H^1$ left.
2. Locally constant sheaf. In this section, $X$ is a smooth projective curve. Let $G$ be an abelian group, and $U\subset X$ is an open subset, a locally constant function from $U$ to $G$ is a map $f: U\to G$ such that for all $p\in X$, there exists an open neighborhood $V\subset U$ of $p$ such that $f_{V}: V\to G$ is a constant function. Let us define
$$\mathscr{F}(U)=\{f: U\to G|f\text{ is locally constant on } U\}$$
It can be seen that $\mathscr{F}(U)$ has the abelian group structure induced from $G$, and it is obvious a sheaf of abelian group on $X$. That is called locally constant sheaf. If $X$ is irreducible, one can see that any two open subsets of $X$ have non-empty intersection. This will yield in fact, $\mathscr{F}$ is a constant sheaf.
We will next prove that the $n$-th cohomology groups of constant sheaf vanish for all $n\ge 1$ on curves.
Proposition 2.1. Let $\mathscr{F}$ be a constant sheaf on $X$. Then $H^n(X,\mathscr{F})$ vanishes for all $n\ge 1$.
Proof. We already see that $H^n(X,\mathscr{F})$ vanishes for all $n\ge 2$. It is sufficient for us to check that $H^1(X,\mathscr{F})$ vanish. But this is easy, choose $\mathfrak{U}=\{U_0,U_1,\emptyset\}$ as above, then because $C^2(\mathscr{F},\mathfrak{U})=0$, all $\alpha_{01}\in C^1(\mathscr{F}, \mathfrak{U})$ is mapped to 0 via $d^1$. If we let $\beta_0=0$ and $\beta_1=\alpha_{01}$, then it can be seen that $\beta_0,\beta_1$ are in $\mathscr{F}(U_0),\mathscr{F}(U_1)$, respectively and $d^0(\beta_0,\beta_1)=\beta_1-\beta_0=\alpha_{01}\in \mathscr{F}(U_0\cap U_1)$. And hence, $d^0$ is a surjective map. It then yields $H^1(X,\mathscr{F})=0$ (Q.E.D)
In particular, if $G=k(X)^*$, where $k(X)$ is the function field of $X$ (in case $X$ is irreducible), the constant sheaf $\mathscr{F}$ with respect to $G$ is denoted $\mathscr{K}$, and one can see $H^1(X,\mathscr{K})=0$. It is the fundamental result for our next parts.
3. $Pic(X)$ and $H^1(X,\mathscr{O}^*)$. We are now in a position for our main result of this note. We always fix $X$ an irreducible, smooth projective curve in this section. First, let $Div(U)=\{\text{all divisors on with finite support in} U\}$. It can be seen that $Div(U)$ is a presheaf of abelian group. We need to check the gluing axiom to prove that $Div(U)$ is actually a sheaf. If $\{U_i\}$ is an open cover of $U$, and $D_i\in Div(U_i)$ such that $D_i|_{U_i\cap U_j}=D_j|_{U_i\cap U_j}$. If we can prove that $U$ can be cover by finitely many $U_1,.,,,U_n\in \{U_i\}$, then we can define $D\in Div(U)$ by $D(p)=D_j(p)$, for some $1\le j\le n$, and then, for sure $D\in Div(U)$.
Lemma 3.1. $X$ is noetherian topological space, then any subset of $X$ is noetherian topological space, and is quasi-compact.
Proof. Exercise.
Via this lemma, one can see $Div$ is a sheaf of abelian group. And one can easily has a sheaf map $div: \mathscr{K}\to Div$ defined by $div_U(f)=div(f)|_U$, where $div(f)$ is the divisor of a function $f\in k(X)^*$.
Lemma 3.2. The sheaf map $div$ defined above is an onto sheaf map.
Proof. For any $p\in X$, any open neighborhood $U$ of $p$, and any $D\in Div(U)$, if $D(p)=0$, we can choose $V$ is an open neighborhood of $p$ by removing all support of $D$ from U. Then it can be seen that on $V$, $div(1)=D$.
Otherwise, if $D(p)=n\ne 0$, let $z$ be the local coordinate at $p$, we then choose $V$ by removing all zeros (except $p$) and poles, and all support of $D$ (except $p$) from $U$. Then one can see on $V$, $div(z^n)=n[p]=D|_V$.
Hence, $div$ is an onto sheaf map. (Q.E.D)
What is the kernel of $div_U$? Well, it exactly consists of all nowhere zero regular functions on $U$, which is denoted $\mathscr{O}^*(U)$. We then obtain an exact sequence of sheaves
$$0\to \mathscr{O}^*\to \mathscr{K}\to Div\to 0$$
It then yields the long exact sequence, by using the vanishing of $H^1(X,\mathscr{K})$, we get
$$0\to H^0(X,\mathscr{O}^*)\to H^0(X,\mathscr{K})\to H^0(X,Div)\to H^1(X,\mathscr{O}^*)\to 0$$
Using Proposition 1.2, we have $H^0(X,\mathscr{O}^*)\cong \mathscr{O}^*(X)\cong \mathbb{k}^*$, $H^0(X,\mathscr{K})\cong \mathscr{K}(X)=k(X)^*$ and $H^0(X,Div)\cong Div(X)$. This then yields the exact sequence
$$0\to k^*\to k(X)^*\to Div(X)\to H^1(X,\mathscr{O}^*)\to 0$$
Due to the ker-coker exact sequence, we have $Pic(X)\cong H^1(X,\mathscr{O}^*)$.
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