In this series of post, we will briefly mention about Riemann-Roch's theorem, its generalization (Serre's duality), and their interesting applications, including the proof the Weil's conjectures for smooth, irreducible, projective curve over finite fields.
Let $X$ be a smooth, irreducible, projective curve, $P\in X$, and $f\in k(X)^\times$, the function field of $X$, if there exists an open affine neighborhood $U$ of $P$, such that there exists and $f$ is non-zero in $U-\{P\}$. We define the valuation of $P$ is $v_P(f) = \dim_k\mathcal{O}_X(U)/(f_{|U})$.
We may wonder what the condition for $f$ has no zero in $U-\{P\}$ is good for? Let $M_P$ be the maximal ideal corresponds to $P$ in $U$. From the condition, one can see $(f_{|U})$ just contain at most one zero at $P$, and hence, due to the nullstellsatz, $M_P\subset\sqrt{(f_{|U})}$. Now, because $M_P$ is finitely generated, we can assume $M_P=(f_1,...,f_r)$, and there exists $a_i \in \mathbb{N} (1\le i\le r)$ such that $f_i^{a_i}\in (f_{|U})$. Now, if we choose $a=\sum_{i}a_i$, then $M_P^a\subset (f_{|U})$. This yields $\dim \mathcal{O}_X(U)/(f_{|U})\le \dim \mathcal{O}_X(U)/M_P^a$.
And one can see $\dim \mathcal{O}_X(U)/M_P^a = \sum_{i=0}^{a-1}M_P^i/M_P^{i+1}$, where $M_P^0=\mathcal{O}_X(U)$. Also, $M_P^i/M_P^{i+1}$ is finitely generated $M_P^0/M_P^1\cong k$-module, and hence, finite dimensional $k$-vector space. And this yields $v_P(f)<\infty$.
Also, it can be seen that the $v_P(f)$ is independent from the choice of $U$. And with $g$ has the same condition, i.e. $g$ does not vanish on $U-{P}$, we have $v_P(fg)=v_P(f)+v_P(g)$, due to the short exact sequence
$$0\to \mathcal{O}_X(U)/(f)\to \mathcal{O}_X(U)/(fg)\to \mathcal{O}_X(U)/(g)\to 0$$
Next, if $f\in K(X)^\times$, and $U$ is an open affine subset of $X$ containing $x$, such that $f=g/h$ on $\mathcal{O}_X(U)$, and both have non-zero in $$ we have $v_P(f) = v_P(g)-v_P(h)$. This gives us the important following
Definition 1.1. A divisor $D$ in $X$ is a formal sum of the term $D:=\sum_{P\in X}n_PP$, where $n_P\in \mathbb{Z}$, and $n_P\ne 0$ at finitely many $P$. The set of all divisors in $X$ is denoted $Div(X)$. The degree of $D$ is defined as $\sum_{P\in X}n_P$. Let $f\in K(X)^\times$, then the divisor of $f$, which is called principal divisor, defined by $div(f):=\sum_{P\in X}v_P(f)P$.
Because $X$ is a curve, i.e. of dimension $1$, so is its open affine subset. This implies $v_P(f)$ is non-zero at finitely many $P\in X$, that means $div(f)\in Div(X)$. Also, the natural addition give $X$ a group structure, and so is the set of all principal divisors, which is denoted $Prin(X)$. We denote the Picard group $Pic(X)=Div(X)/Prin(X)$. If $D=\sum_{P\in X}n_PP$, and $n_P\ge 0$ for all $P$, we will simplify denote $D\ge 0$, and it is called effective divisor. Also, for $D,D'\in Div(X)$, we denote $D\ge D'$ if $D-D'\ge 0$. It is also seen that the degree map defines a group homomorphism from $Div(X)$ to $\mathbb{Z}$.
The following definition is also essential.
Definition 1.2. We define $\mathcal{L}(U,D)=\{f\in k(X)^\times|div(f_{|U})+D\ge 0\}\cup\{0\}$. If $U=X$, we denote this $H^0(X, D)$. And if $D=0$, we sometimes denote this $\mathcal{L}(U,\mathcal{O}_X)$.
Example 1.3. We can see that $\mathcal{L}(X,0)$ is the set of 0, together with all $f\in K(X)^\times$, that has no pole, i.e. $f\in \mathcal{O}_X(X)$, the ring of regular function on $X$.
There are important facts about $\mathcal{L}(U,D)$.
Proposition 1.4. Let $X$ be a smooth, irreducible, projective curve
i. Let $U$ be an open affine subset of $X$, then $\dim_k\mathcal{L}(U,D)<\infty$.
ii. $\mathcal{L}(U,\mathcal{O}_X)=\mathcal{O}_X(U)$.
Proof. The proof of i. can be found in the book of Dino Lorenzini, and it is quite difficult. We will prove ii. instead. It can be seen that $\mathcal{L}(U,\mathcal{O}_X)=\mathcal{L}(U,0)$, which consists of all rational function $f$ in $K(X)$, which has no pole in $U$, and hence, $v_P(f)\ge 0$, for all $P\in U$. However, $X$ is smooth, and hence, the local ring at $P$ is a DVR, and all functions $f$ that has valuation at $P$, $v_P(f)\le 0$ is in $\mathcal{O}_{X,P}$-the local ring at $P$. And actually, because $U$ is affine, and $P\in U$, the $\mathcal{O}_{X,P}=\mathcal{O}_X(U)_P$. This yields $\mathcal{L}(U,\mathcal{O}_X)=\cap_{P\in U}\mathcal{O}_X(U)_P$. Now, this follows easily from commutative algebra that an integral domain is equal to the intersection of all of its localizations at all maximal ideal. Hence, $\mathcal{L}(U,\mathcal{O}_X)=\mathcal{O}_X(U)$. (Q.E.D)
The proposition above implies the interesting fact that any global regular function from a smooth, irreducible, projective curve is just constant.
Proposition 1.5. Let $X$ be a smooth, irreducible, projective curve, then $\mathcal{O}_X(X)=k$.
Proof. We can see that $\mathcal{L}(X,\mathcal{O}_X)$ is of finite dimensional $k$-vector space (Part i). And by Part (ii), $\mathcal{L}(X,\mathcal{O}_X)=\mathcal{O}_X(X)$-which is a domain since $X$ is irreducible. Let $0\ne f\in \mathcal{O}_X(X)$, the the multiplication by $f$ defines an injective $k$-linear map, and hence, bijective. This implies $\mathcal{O}_X(X)$ is a field, and a finite extension of $k$, which is algebraically closed. Hence, $\mathcal{O}_X(X)=k$. (Q.E.D)
We come for now the theorem of Riemann-Roch. Because $X$ is a curve, there exists two affine patches cover $X$, which is denoted $U_1, U_2$. We also denote $U_{12}:=U_1\cap U_2$. Let $D\in Div(X)$, we define the map $\delta: \mathcal{L}(U_1,D)\oplus\mathcal{L}(U_2, D)\to \mathcal{L}(U_{12}, D)$, that sends $(f,g)$ to $f_{U_{12}}-g_{U_{12}}$. For the kernel of $\delta$, if $f_{U_{12}}-g_{U_{12}}=0$, where $f,g\in k(X)^\times, div(f_{|U_1})+D\ge 0, div(g_{|U_2}+D\ge 0)$. We $U_1\cap U_2$ is a dense subset of $X$, and hence, $f=g\in K(X)^\times$, also, $U_1\cup U_2= X$ implies that $div(f)+D\ge 0$, and that $f\in \mathcal{L}(X,D)$. Hence, $\ker\delta =\mathcal{L}(X,D)$. By our earlier notation, it is $H^0(X,D)$. Let $H^1(X,D)$ the cokernel of $\delta$. We have the following
Theorem 1.6 (Riemann-Roch). Let $X$ be a smooth, irreducible, projective curve, there exists a natural number $g$ such that for any $D\in Div(X)$, $\dim H^0(X,D)-\dim H^1(X,D)=1+\deg(D)-g$. Such number $g$ is called the genus of $X$.
We will not prove this theorem, instead, we will go directly to some of its consequences. From now on, we always denote $X$ the smooth, irreducible, projective curve, and $\dim H^i(X,D)=h^i(X,D)$
1. Let $div(f)$ be any principal divisor of $X$, where $f\ne 0$, then $\deg(div(f))=0$. For any $D\in Div(X)$, the multiplication map by $f$ defines an $k$-linear isomorphism from $H^0(X,D)$ to $H^0(X,D+div(f))$, and also, the isomorphism between $H^1(X, D)$ and $H^1(X,D+div(f))$. By the Riemann-Roch's theorem, we have
$$h^0(X,D)-h^1(X,D)=1+\deg(D) -g=1+\deg(D+div(f))-g=$$
$$=1+\deg(D)+\deg(div(f))-g$$.
This yields $deg(div(f))=0$. That means, any principal divisor has degree 0.
2. If $\deg(D)<0$, then $H^0(X,D)=\{0\}$. In fact, for any $f\in k(X)^\times$ in $H^0(X,D)$, then $\deg(D+div(f))=\deg(D)+\deg(div(f))=\deg D<0$ (By 1). Therefore, $H^0(X,D)=0$.
3. If $\deg(D)=0$, then $h^0(X,D)\le 1$. In fact, if there exists $f\in k(X)^\times$, such that $f\in H^0(X,D)$, then it can be seen that $\deg(div(f)+D)=0$, then $div(f)=-D$, hence, for all $g\in H^0(X,D)$, we have $div(f)=div(g)$, and this implies $f=\lambda g$, where $\lambda\in k^\times$.
4. Let $P\in X$ be a point, then $X-\{P\}$ is not a projective variety. Let us denote $U:=X-\{P\}$, then it can be seen that $\mathcal{L}(U,\mathcal{O}_X)=\{f\in k(X)^\times|div(f_{|U})\ge 0\}$. From this, the only possible pole for $f$ is just at $P$, with some multiplicity $n\in \mathbb{N}$, and hence, $f\in H^0(X,nP)$. This implies
$$\mathcal{L}(U,\mathcal{O}_X)=\{f\in k(X)^\times|f \text{ has pole at P, with multiplicity } n\in \mathbb{N}\}=\cup_{n\ge 0}(H^0(X, nP))$$
And by Riemann-Roch's theorem, $H^0(X,nP)\ge 1+n-g$, for all $n\in \mathbb{N}$. This yields $\dim\mathcal{L}(U,\mathcal{O}_X)=\infty$. And due to Proposition 1.4 (ii), we can see $\mathcal{L}(U,\mathcal{O}_X)=\mathcal{O}_X(U)=\mathcal{O}_U(U)$. If $U$ is a projective variety, it is of dimension 1, and hence, curve. But it is impossible, due to Proposition 1.5, that $\dim_k(\mathcal{O}_U(U))$ must be 1 in this case.
5. We will compute explicitly the genus of $\mathbb{P}^1$ with coordinate $x,y$. It can be seen that $\mathbb{P}^1$ is covered by affine open subsets $U_1=\{(x:y)|y\ne 0\}=\{(\frac{x}{y}:1)|y\ne 0\}$, and $U_2=\{(x:y)|x\ne 0\}=\{1:\frac{y}{x}\}$. In $U_1\cap U_2=\{(\frac{x}{y}:\frac{y}{x})|x,y\ne 0\}$. Also, one has $\mathcal{O}_X(X) = k$, and $\mathcal{O}_X(U_1)=\mathcal{O}_{U_1}(U_1)=k[x/y]$, and similarly $\mathcal{O}_X(U_2)=k[y/x]$, also $\mathcal{O}_{U_{12}}(U_{12})=k[x/y,y/x]$. And $k(\mathbb{P}^1)=k(X)$, which a remark that $f\in k(\mathbb{P}^1)$ iff $f=g/h$, where $g,h$ are homogeneous polynomials of the same degree.
Let us compute $h^0, h^1$ related to the divisor $D=0$. It can be seen that $H^0(X,0)=\mathcal{O}_X(X)=k$, and hence, $h^0(X,0)=1$. Also, $\mathcal{L}(U_1,0)=\mathcal{L}(U,\mathcal{O}_X)=\mathcal{O}_X(U)=k[y/x]$, and similarly, $\mathcal{L}(U_2,0)=\mathcal{O}_X(U)=k[x/y]$, and $\mathcal{L}(U_{12}, 0)=\mathcal{O}_X(U_{12})=k[x/y,y/x]$. The map $\delta$ in this case is $k[x/y]\oplus k[y/x]\to k[x/y,y/x]$ which sends $(f,g)$ to $f-g$. It can be seen easily that the map is surjective, and hence $h^1(X,0) = \dim coker(\delta)=0$. Due to the Riemann-Roch's theorem, we have $h^0(X,0)-h^1(X,0)=1=1-g$. And hence, $g=0$, and the genus of $\mathbb{P}^1$ is 0.
6. We will prove that the Picard group of $\mathbb{P}^1$ is isomorphic to $\mathbb{Z}$. First, it can be seen that for any $P=(x_P:x_P), Q=\{x_Q:y_Q\}\in \mathbb{P}^1$, then $D:=P-Q$ is a principal divisor. It is easy to construct, i.e. $f=\frac{y_Px-x_Py}{y_Qx-x_Qy}$. And from this, if $D\in Div(X)$ with $deg(D)=0$, we can see the degree of the positive part $D_1$ of $D$ is equal to the degree of the negative part $D_2$ of $D$. And hence, we can easily construct a rational function $f=g/h$ such that $div(f)=D$, where $Div(g) = D_1, Div(h) = D_2$, and that $\deg g=\deg h$. This implies $f\in k(\mathbb{P}^1)^\times$. And hence, the degree map $\deg: Div(\mathbb{P}^1)\to \mathbb{Z}$ has the kernel $Prin(\mathbb{P}^1)$. Hence, by definition, $Pic(\mathbb{P}^1)\cong \mathbb{Z}$.
7. Let us compute the $h^0$ related to the divisor $D=n\infty\in Div(\mathbb{P}^1)$, where $\infty=(1:0)$. First, it can be seen that $H^0(X,n\infty)=\{f\in k(\mathbb{P}^1)^\times|div(f)+n\infty\ge 0\}$. This implies that such $f$ has only pole at $\infty$ and nowhere else, with multiplicity $i\le n$. However, the uniformizer at $\infty$ is $\frac{1}{x} (\equiv y)$. And hence, $f=\frac{g(x,y)}{x^i}$, where $\gcd(g,x)=1,g$ is homogeneous of degree $d$, and $g$ is only $cy^i$. This implies the $k$-basis for $H^0(\mathbb{P^1}, n\infty)=\{\frac{y^i}{x^i}|0\le i\le n\}$, and the dimension in this case is $n+1$. By Riemann-Roch's theorem, and 6, $h^0(X,n\infty)-h^1(X,n\infty)=1+n-g=1+n$. Hence, $h^1(X,n\infty)=0$.
If $D'\in Div(\mathbb{P}^1)$ such that $\deg(D')=\deg(D)=n$, there exists $D_1$, where $\deg(D_1)=0$, and $D'=D+D_1$. By 6, we can see that $D_1$ is principal, and hence, by 1, $h^0(X,D)=h^0(X,D')=n+1$, and $h^1(X,D)=h^1(X,D')=0$.
That means, we have computed $h^0, h^1$ for all divisor $D\in Div(\mathbb{P}^1)$.
We will visit these applications again, and further, with the more powerful theorem, the Serre's Duality.
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