(Riemann-Roch's Theorem and Applications III) Hyperplane divisor associated to a morphism

Recall that in Part 1, we explain how the notion of degree of smooth projective curves can be extended to arbitrary dimension, but we have not computed the degree of such an easy example like the twisted cubic curve. In Part 2, we mention this can be done via the degree of the very ample divisor. This short part will prove this.

Let $X$ be a smooth irreducible projective curve, and $\phi: X\to \mathbb{P}^n$ is a morphism. We then know that at an open neighborhood $U$ of $q\in X$, we can represent $\phi$ as $[f_0:...:f_n]$, where $f_i\in k(X)$. Let $H$ be a hyperplane in $\mathbb{P}^n$ such that $\phi(X)\not\subset H$, and $L:=\sum_{i=0}^na_ix_i$ is the equation for $H$. At $\phi(q)$, we assume that the coordinate $x_0$ of $\phi(q)$ is non-zero. And we define the divisor $\phi^*H$, which is called the hyperplane divisor associated with the morphism $\phi$ as
$$\phi^*H:=\sum_{p\in X}ord_p(\frac{L}{x_0}\circ\phi)p$$
It can be seen easily that $\phi^*H$ is independent on the choice of the equation $L$. Let us look at more closely on the coefficient of $\phi^*H$ at a point $p\in X$. We have
$$ord_p(\frac{L}{x_0}\circ\phi)=ord_p(\frac{\sum_{i=0}^na_ix_i}{x_0}\circ[f_0:...:f_n])=ord_p(\sum_{i=0}^na_i\frac{f_i}{f_0})=ord_p(\sum_{i=0}^na_if_i)-ord_p(f_0)$$
Now, if we assume that $ord_p(f_i)=a_i$, i.e. on an open neighborhood of $p$, we can write $f_i=\pi_p^{a_i}g_i$, where $\pi_p$ is the local coordinate at $p$, and $g(p)\ne 0$. Let $D(p):=-\min a_i$. Then one can see in an open neighborhood of $p$, $[f_0:...:f_n]\sim[\pi_p^{D(p)}f_0:...:\pi_p^{D(p)}f_n]$, and hence, then there exists one coordinate $x_i$ of $\phi(p)$ is non-zero, which is exactly where $a_i$ is the minimum. In this case, one can see the zero coordinate of $\phi(p)$ is non-zero, hence, $-ord_p(f_0)=D(p)$.

From this observation, if we define a divisor $D\in Div(X)$ such that $D(p):=-\min ord_p(f_i)$, then by our previous computation, we have $\phi^*H(p) = ord_p(\sum_{i=0}^na_if_i)+D(p)$. Or equivalently, we finally obtain
$$\phi^*(H)=div(\sum_{i=0}^na_if_i)+D$$
In particular, we have $\deg(\phi^*(H))=\deg(D)$ (0.0). Before coming to the main theorem for this section, we need a following

Lemma 1. Let $\phi:X\to Y$ be a morphism between irreducible smooth projective curves. And $f$ is a regular function in $Y$. Then for all $p\in X$, we have $ord_p(f\circ\phi)=e_pord_{\phi(p)}(f)$, where $e_p$ is the ramification index at $p$.

Proof. Locally at $q:=\phi(p)$, we can represent $f=\pi_q^{n}g$, where $\pi_q$ is the local coordinate of $q$, and $g(q)\ne 0$, and $n=ord_{\phi(q)}(f)$. Due to the definition of the ramification index, locally at $p$, we have
$$f\circ\phi=\phi^*(f)=\phi^*(\pi_q^{n}h)=\phi^*(\pi_q^{e_p})\phi^*(h)=\phi_p^{ne_p}\phi^*(h)$$
where $\phi^*(h)(p)=h\circ(\phi)(p)=h(q)\ne 0$, and hence, $ord_p(f\circ\phi)=ne_p$ (Q.E.D)

We are now ready for the main theorem

Theorem 2. Let $\phi: X\to \mathbb{P}^n$ be a morphism, where $Y:=\phi(X)$ is a smooth projective curve in $\mathbb{P}^n$. Let $H$ be any hyperplane in $\mathbb{P}^n$ such that $\phi(X)\not\subset H$, then $\deg(\phi^*H)=\deg(\phi)\deg(Y)$.

Proof. It can be seen that $\phi^*H(p)=ord_p(\frac{L}{x_0}\circ \phi)=e_pord_{\phi(p)}(\frac{L}{x_0})=e_pord_{\phi(p)}(L)$. Hence
$$\deg(\phi^*H)=\sum_{p\in X}\phi^*H(p)=\sum_{p\in X}e_pord_{\phi(p)}(L)=\sum_{q\in Y}ord_q(L)\sum_{\phi(p)=q}e_p=$$
$$=\sum_{q\in Y}ord_q(L)\deg(\phi)=\deg(Y)\deg(\phi)$$
(Q.E.D)

Now, we can combine all things together to deduce

Corollary 3. Let $D\in Div(X)$ be a very ample divisor, and $l(D)=n+1$, and $\phi_D: X\to \mathbb{P}^n$ the associated morphism. Then $\deg(\phi(X))=\deg(D)$.

Proof. One can see that $\phi_D(X)\cong X$, and hence $\deg(\phi)=1$. Furthermore, one has from (0.0) that $\deg(D)=\deg(\phi^*(H))$. And hence, it follows by Theorem 2 that $\deg(\phi(X))=\deg(D)$. (Q.E.D)

Now, we come back with our example about twisted cubic curve, which is induced from the divisor $D:=3\infty$ in $\mathbb{P}^1$, and its associated map to $\mathbb{P}^3$. By our previous corollary, the twisted cubic curve must have degree 3. Also, any rational normal curve in $\mathbb{P}^n$ has degree $n$.