One of my dreams when I finished the first course in Algebraic Geometry is that I can understand about morphisms between projective varieties. And this note is a part of my dream. We always fix $X$ as a smooth projective curve in $\mathbb{P}^n$.
1. Warm-up with divisors. Let $D:=\sum_{P\in X}n_PP\in Div(X)$, we call $D$ an effective divisor if $\forall P\in X, n_P\ge 0$. For simplicity, we often write $D\ge 0$ instead of writing $D$ is effective. For any divisor $D$ in $Div(X)$, we define
$$L(D):=\{f\in k(X)^*|div(f)+D\ge 0\}\cup\{0\}$$
It can be seen that $L(D)$ has the $k$-vector space structure, and one may ask if it has finite dimension or not. We will answer this for now. The dimension of $L(D)$ is often denoted by $l(D)$, or $H^0(D)$. But we will use the notion $l(D)$, until approaching the cohomology theory for sheaves.
One may realize if $\deg(D)<0$, then from Proposition 2.2 of the first part, for any $f\in K(X)$, $\deg(div(f)+D)=0+\deg(D)<0$, i.e. $l(D)=0$ in this case. That means, one should focus on divisor having non-negative degree. Also, from the definition of principal divisors that any divisor has no poles nor zeros must be constant functions. That means, $L(0)=k$, and as a consequence, $l(0)=1$.
Proposition 1.1. Let $D$ be any divisor in $Div(X)$, and $p\in X$ is any point, then $l(D)\ge l(D-p)\ge l(D)-1$.
Proof. It can be seen that $D-p\le D$, then for any $f\in L(D-p)$, we have $div(f)+D-p\ge 0$, this yields $div(f)+D\ge 0$, and $f\in L(D)$. Hence, $L(D-p)\subset L(D)$, and $l(D-p)\le l(D)$.
The remaining is important, let $-n$ be the coefficient of $P$ in $D$, then for any $f\in L(D)$, we can see $ord_p(f)\ge n$. And for each $f\in L(D)$, we can represent locally at $P$
$$f=\pi^{n}g_f$$
where $\pi$ is the local coordinate at $p$, and $n\le ord_f(p)$. From this, one can see $ord_p(g_f)\le 0$. We then define the following map $\phi: L(D)\to K$ sending $f$ to $g_f(P)$. This is the $k$-linear map. We will describe its kernel. If $g_f(P)=0$, then $ord_p(f)>n$, i.e. $ord_p(f)\ge n+1$, and hence, $f\in L(D-p)$. Conversely, if $f\in L(D-p)$, then $ord_p(f)\ge n+1$, and $g_f(p)=0$ for sure, i.e. $f\in L(D-p)$. That means $L(D-p)=\ker\phi$. And it follows from the basic linear algebra that $l(D)= \dim\ker\phi+\dim Im(\phi)=l(D-p)+\dim Im(\phi)\le l(D-p)+1$. (Q.E.D)
As an application of this proposition, we get
Corollary 1.2. Let $D\in Div(X)$ be a divisor of degree 0, then $l(D)=0$ or $1$. And $l(D)=1$ iff $D$ is principal.
Proof. For any point $p\in X$, we have $\deg(D-p)=-1<0$, hence $l(D-p)=0$. By Proposition 1.1, $l(D)=0$ or $1$. If $l(D)=1$, then there exists $f\in L(D),f\ne 0$ such that $div(f)+D\ge 0$, and because $\deg(div(f))=\deg(D)=0$, we can see $div(f)+D=0$, i.e. $div(f)=-D$, and $D=div(1/f)$, and $D$ is a principal divisor. Conversely, if $D=div(f)$ is a principal divisor, then one can see $div(1/f)\in L(D)$, i.e. $L(D)\ge 1$. And because $\deg(D)=0$, by our previous argument, $l(D)\le 1$. And hence, $l(D)=1$. (Q.E.D)
From Corollary 1.2, if we begin with any divisor $D$ of degree 0, then $l(D)\le 1$. And for all $p$, we have $l(D+p)\le 2$, by Proposition 1.1. That means, for all divisor $D$, by induction, one gets $l(D)\le 1+\deg(D)$. And hence, $L(D)$ is a finite dimensional $k$-vector space.
2. Maps to projective spaces via base point free divisors. We will further analyze the proof of Proposition 1.1, when $\phi$ maps $l(D)$ to $0$, then for sure, $l(D-p)=l(D)$, and in this case, we call $p$ is a base point of $D$. Otherwise, if it maps surjectively onto $K$, $l(D-p)=l(D)-1$, and in this case, $p$ is not a base point of $D$. A divisor $D$ is called base point free if it has no base point, i.e. for all $p\in X$, the map $\phi: L(D)\to K$ is surjective.
Let $D\in Div(X)$ be a base point free divisor, by our result from Section 1, $n:=l(D)$ is finite. Let $f_0,...,f_{n-1}$ be a basis for $L(D)$ as k-vector space. The assumption $D$ is base point free will yield, for any point $p\in X$, the map $\phi: L(D)\to k$ is surjective, i.e. there always exists some $f_i$ such that $\phi(f_i)=g_{f_i}(P)\ne 0$. And for all $j\ne i$, $\phi(f_j)=0$ (this is equivalent to say $(f_j)_{j\ne i}$ is the basis for $L(D-p)$). That means, locally at any point $p$, the map $\phi_D: X\to \mathbb{P}^{n-1}$ sending $p$ to $[f_0(p):f_1(p):...:f_n(p)]$ is defined. It can be seen that from a fixed base $(f_0,...,f_{n-1})$ of $L(D)$, we can obtained any other bases by acting an element from $PGL_{n-1}(k)$. And it is an automorphism of $\mathbb{P}^{n-1}$, i.e. the map $\phi_D$ is well-defined. And moreover, $\phi_D$ is a morphism from $X$ to $\mathbb{P}^{n-1}$. We will discuss in details this point further in our later note, including how to do "analytic continuation" in algebraic geometry. In short, we have the following
Proposition 2.1. Let $D\in Div(X)$ is a base-point free divisor, and $f_0,...,f_{n-1}$ is a $k$-basis for $L(D)$. Then the map $\phi_D: X\to \mathbb{P}^{n-1}$ sending $x$ to $[f_0(x):...:f_{n-1}(x)]$ is well-defined and is a morphism from $X$ to $\mathbb{P}^{n-1}$. And for all $p\in X$, there exists a basis $g_0,...,g_{n-1}$ of $L(D)$ such that $g_0(p)\ne 0$ and $g_i(p)=0$ for all $i>0$, i.e. $\phi_D$ will map $p$ to $[1:0:...:0]$ and $(g_i)_{i>0}$ is the basis for $L(D-p)$. And such a basis $(g_i)$ is called the basis with respect to $p$.
We will now detect when $\phi_D$ is an injective map. Assume that $\phi_D(p)=\phi_D(q)$, for any point $p\ne q$, and let $g_0,...,g_{n-1}$ is the basis of $L(D)$ with respect to $p$, and then $g_i(q)=0$ for all $i>0$, i.e. $(g_i)_{i>0}\in L(D-q)$, and they are linearly independent. This implies $(g_i)_{i>0}$ is a basis for $L(D-q)$. And hence, $L(D-p)=L(D-q)$. Because $p\ne q$, one can see that $L(D-p-q)=L(D-p)=L(D-q)$, and this yields $l(D-p-q)=l(D)-1$. Conversely, if $l(D-p-q)=l(D)-1$ for all $p\ne q\in X$, then it can be seen $l(D-p-q)=l(D-p)=l(D)-q$ since $D$ is base point free divisor. And this will yield, in turn $\phi_D(p)=\phi_D(q)$. Hence, we can conclude that $\phi_D$ is not an injective map iff $l(D-p-q)=l(D)-1$ for all points $p\ne q\in X$. And this leads us to the following important
Corollary 2.2. $\phi_D$ is an injective map iff $l(D-p-q)=L(D)-2$, for all point $p\ne q$ in $X$.
But it is not sufficient for our purpose, because we want give a criterion when $X$ can be embedded into $\mathbb{P}^n$, i.e. it is isomorphic to its image via the embedding. Note that not all injective morphism is an embedded. For example, let $X$ be a curve defined by $y^2z = x^3$. By simple computation, $X$ has singular point at $(0:0:1)$. Consider the map $\mathbb{P}^1\to\mathbb{P}^2$ given by the extension of the map $z \mapsto (z^2:z^3)$, i.e. $\infty\mapsto (0:1:0)$. This map is 1-1 from $\mathbb{P}^1$ to $X$. But it cannot be an isomorphism, since $X$ is singular, and $\mathbb{P}^1$ is smooth.
By extending Corollary 2.2, we have $\phi_D$ is an embedding iff $l(D-p-q)=l(D)-2$ for all points $p,q\in X$. We will discuss about this later.
In this case, because $X$ is a projective variety, any morphism from $X$ to $\mathbb{P}^n$ is closed, i.e. $\phi_D(X)$ is a closed subset of $\mathbb{P}^n$, and hence, in the case $\phi_D$ is an embedding, $\phi_D(X)\cong X$. And in this case, such a divisor is called very ample divisor. To detect very ample divisor, and how curves can map isomorphic to their images, we need the Riemann-Roch's theorem.
3. Riemann-Roch's theorem and some of its corollaries. We will state the Riemann-Roch and Riemann-Hurtwitz's formula in this section, together with some of their corollaries. We always fix $X,Y$ as two irreducible, smooth, projective curves.
Theorem 3.1. Let $X\subset\mathbb{P}^n$ be a smooth projective curve, then there exists an integer $g\ne 0$, which is called the genus of $X$ and $K\in Div(X)$ which is called canonical divisor such that for all divisor $D\in Div(X)$, we have
$$l(D)-l(K-D)=1+\deg(D)-g$$
By using this, and what we have discussed in Section 1, one can easily deduce
Corollary 3.2. If $K$ is the canonical divisor, then $l(K)=g$, and $\deg(K)=2g-2$.
One may look back my previous note about Riemann-Roch's theorem, and see how to prove the genus of $\mathbb{P}^1$ is zero. In this case, we need to describe $L(n\infty)$ for $n\ge 0$, where $\infty$ is the point $[1:0]\in \mathbb{P}^1$. And $L(D)=\{1,\frac{x_0}{x_1},...,\frac{x_0^n}{x_1^n}\}$, i.e. $l(D)=n+1=1+\deg(D)$. When $n$ is sufficient large, one can see $l(K-D)=0$, and hence $l(D)=1+\deg(D)-g$, i.e. $g=0$. In this case $\deg(K)=-2$, and $l(K)=0$. And hence, for $D=n\infty\in Div(X)$, we always have $l(D)=1+\deg(D)$, and for sure, $D$ is a very ample divisor if $n\ge 2$.
By Section 2, $\phi_{n\infty}: \mathbb{P}^1\to \mathbb{P}^n$ is an embedding sending $[x_0:x_1]$ to $[1:\frac{x_0}{x_1}:...:\frac{x_0^n}{x_1^n}]\sim [x_1^n:x_0x_1^{n-1}:...:x_0^{n-1}x_1:x_0^n]$. And $\phi_{n\infty}(\mathbb{P}^1)$ is called rational normal curve in $\mathbb{P}^n$. If $n=2$, then $\phi_{2\infty}(\mathbb{P}^1)=\{[x_0^2:x_0x_1:x_1^2]|[x_0:x_1]\in\mathbb{P}^1\}$, which is the curve $xz=y^2$ in $\mathbb{P}^2$. When $n=3$, the image of the map is exactly the twisted cubic curve, which was mentioned in the first section of our first part. At this point, we can give the following formula, but the proof will be postponed until we discuss carefully about the pull-back of hyperplane divisors.
Proposition 3.3. Let $D\in Div(X)$ be a very ample divisor, then $\deg(\phi_D(X))=\deg(D)$. In particular, the degree of the twisted cubic curve is $3$.
In the next section, we will prove that any smooth projective curve of genus 0 is isomorphic to $\mathbb{P}^1$, and give the statement of Riemann-Hurtwitz's formula.
4. Riemann-Hurtwitz's formula. Recall that in the proof of Proposition 2.2 of the first part, we mentioned about the ramification index. It is defined as follows. Let $\phi:X\to Y$ is a surjective morphism, where $X,Y$ are smooth irreducible projective curves, then for all $q\in Y$, and $p\in \phi^{-1}(q)$. Let $\pi_q, \pi_p$ be the local coordinates at $q,p$, respectively. Then the ramification index $e_p$ is defined as the highest multiplicity $\pi_q\circ\phi=\pi_p^{e_p}g$ (which is the pull-back $\phi^*(\pi_q))$, for $g(p)\ne 0$ in an open neighborhood of $p$. And very similar to what we discussed on algebraic number theory, for all point $q\in Y$, $\sum_{\phi(p)=q}e_p$ is a constant, which is called the degree of the morphism $\phi$, denoted by $\deg(\phi)$. The sum $\sum_{\phi(p)=q}e_pp\in Div(X)$ is called the pull-back of $q\in Div(Y)$, and denoted by $\phi^*(q)$. It defines a group homomorphism from $Div(Y)$ to $Div(X)$. One can easily deduce
Proposition 4.1. Let $X, Y,\phi$ be defined as above. Then $\phi$ is an isomorphism iff $\deg(\phi)=1$.
Proof. Assume that $\phi$ is an isomorphism, then for any point $q\in Y$, there exists only one $p$ such that $\phi(p)=q$, i.e. $\deg(\phi)=1$. Conversely, note that because $\phi$ is surjective, fixed $q\in Y$, we always have $e_p\ge 1$ for all $p\in X, f(p)=q$. Hence, there exists only one such $p$, i.e. $\phi$ is injective. This yields $\phi$ is bijective. Because $\phi_D(X)$ is also closed in $\mathbb{P}^n$, it is also a projective variety. And $\phi_D$ will induce an isomorphism between function field $\phi_D^*$ between $k(\phi_D(X))$ and $k(X)$. And hence, $\phi_D$ is an isomorphism. (Q.E.D)
As an application, we will prove that any smooth projective curve $X$ of genus zero is isomorphic to $\mathbb{P}^1$. Taking any $p\in X$, then by Riemann-Roch's theorem, we easily get $l(p) = 2$, and $p$ is for sure very ample, since $l(p-q-r)=0$, for all $q,r\in X$. Hence, there exists $f\in L(p)$, which is not constant, and having pole at $p$ with multiplicity 1, i.e. $div(f) = q-p$, for some $q\in X, q\ne p$, and $f$ has zero only at $q$. And one can see $L(P)$ is exactly the $k$-vector space generated by $1$ and $f$. Assume that $f=\frac{f_1}{f_2}$, then we have the map $\phi_p:X\to \mathbb{P}^1$ sending $x$ to $[1:f_1(x)/f_2(x)]\sim[f_2(x):f_2(x)]$. And hence, the pull back of $\infty\in \mathbb{P^1}$ is exactly one point $p$. And hence, the morphism is of degree 1, and it must be an isomorphism, by Proposition 4.1. Finally, we give the statement of the Riemann-Hurtwitz's formula.
Theorem 4.2. Let $X, Y$ be two smooth irreducible projective curves, and $\phi: X\to Y$ a surjective morphism. Then $2g_X-2=\deg(\phi)(2g_Y-2)+\sum_{p\in X}(e_p-1)$, where $g_X, g_Y$ are genus of $X$ and $Y$, respectively. More precisely, $K_X=\phi^*K_Y+\sum_{p\in X}(e_p-1)p$, where $K_X, K_Y$ are canonical divisors of $X$ and $Y$ respectively.
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