Intuition behind ramification index of a map between smooth curves.

I'm studying from The Arithmetic of Elliptic Curves (Silverman) and I'm having a hard time understanding the intuition behind the $$\textit{ramification index}$$ concept.

In the book, we let $$\phi:C_1 \longrightarrow C_2$$ a non constant map of smooth curves and we let $$P\in C_1$$. Then we define the ramification index of $$\phi$$ at $$P$$, denoted by $$e_\phi(P)$$ as $$e_\phi(P)=\text{ord}_P(\phi^{*}t_{\phi(P)}) \text{ ,}$$ where $$t_{\phi(P)}\in K(C_2)$$ is a uniformizer at $$\phi(P)$$.

I'm trying to break it down into parts. First of all, $$\phi^{*}t_{\phi(P)}$$ = $$t_{\phi(P)} \circ \phi$$

Now, $$\text{ord}_P(\phi^{*}t_{\phi(P)})$$ is the max $$d$$ for which $$\phi^{*}t_{\phi(P)} \in M_p^d$$, meaning the max $$d$$ for which we can express $$(t_{\phi(P)} \circ \phi)$$ as a product of maps $$f_1...f_d$$ such that $$f_i(P)=0$$ for all $$i$$, meaning that $$P$$ would be a zero of multiplicity $$d$$ for $$(t_{\phi(P)} \circ \phi)$$. It's good to see that $$(t_{\phi(P)} \circ \phi)(P)=0$$ and then, necessarily $$(t_{\phi(P)} \circ \phi)\in M_p^d$$ for $$d \geq 1$$.

The book defines $$\phi:C_1 \longrightarrow C_2$$ to be unramified at $$P$$ if $$e_\phi(P)=1$$, which means that $$(t_{\phi(P)} \circ \phi) \in M_p^1$$, in other words, we can't express $$(t_{\phi(P)} \circ \phi)$$ as a product of maps with $$P$$ being a zero for said maps. All in all, what I understand from this is that $$P$$ is a zero of multiplicity $$1$$ for $$(t_{\phi(P)} \circ \phi)$$.

If $$e_\phi (P)>1$$ we said that $$\phi$$ ramifies at $$P$$, meaning we can split $$(t_{\phi(P)} \circ \phi)$$ as the product of maps, each of them having $$P$$ as a zero. My question is, what does all of it means? Is there any geometric intuition for this definition? I "understand" the technicalities behind the definition, but I don't understand why do we define this concept and why do we define it this way.

The book goes with an example, considering the map $$\phi :\mathbb{P}^1 \longrightarrow \mathbb{P}^1$$, $$\phi([X,Y])=[X^3(X-Y)^2,Y^5]$$, it says that $$\phi$$ is ramified at the points $$[0,1]$$ and $$[1,1]$$, but I am not sure what that really means.

If anybody could explain what is the intuition behind this definition, I would be really thankful.

• $\phi$ is a morphism, $\phi^* f(Y) = f(\phi(X))$ sends the functions $f(Y)$ regular at $Q$ to the functions regular at the $P$ such that $Q = \phi(P)$. The ramification index of $\phi$ at $P$ is the order of the zero at $P$ of $f(\phi(X))$ when $f$ has a simple zero at $\phi(P)$. For a morphism of smooth projective curves we have $\sum_{P \in \phi^{-1}(Q)} e_\phi(P) = \deg(\phi)$ and $[Q] \mapsto \sum_{P \in \phi^{-1}(Q)} e_\phi(P) [P]$ is the map on divisors induced by $\phi^*$. Aug 7, 2019 at 18:48

The function field of $$\Bbb{P}^1$$ is $$k(t)$$, that is we can think to $$\Bbb{P}^1$$ as $$\overline{k} \cup \infty$$, sending $$[a:1]$$ to $$a$$.

With $$\phi : \Bbb{P^1 \to P^1},([X,Y])\to [X^3(X-Y)^2,Y^5]$$ since $$[X:Y] = [r X:rY]$$ you can set $$Y = 1, X = t$$ and obtain $$\phi([t: 1]) = [t^3(t-1)^2:1]$$ which means you are considering the morphism $$\phi : \overline{k} \cup \infty \to \overline{k} \cup \infty, t \mapsto t^3(t-1)^2$$ inducing the morphism $$\phi^*(f(t)) = f(t^3(t-1)^2)$$, $$k(t) \to k(t)$$ on the function fields.

If $$f \in k(t)$$ has a simple zero at $$a \ne \infty$$, for example $$f(t) = t-a$$, then $$\phi^*(f) = f(t^3(t-1)^2) = t^3(t-1)^2-a$$ has simple zeros at all the $$\phi^{-1}(a)$$ iff the polynomial $$P_a(t) = t^3(t-1)^2-a$$ has no double root iff $$\gcd(P_a(t),P_a'(t)) = 1$$.

You can check this fails at $$a = 1$$ and $$a=0$$.

The local ring at $$a$$ is $$k[t]_{(t-a)} = \{ \frac{f(t)}{g(t)},f,g \in k[t], g \not \in (t-a)k[t]\}$$, any function with a simple zero at $$a$$ generates its unique maximal ideal, $$\phi^*$$ sends the local ring to $$k[t^3(t-1)^2]_{(t^3(t-1)^2-a)} \subset \bigcap_{b \in \phi^{-1}(a)} k[t]_{(t-b)}$$.

For the case $$a = \infty$$ it is immediate that $$\phi$$ is totally ramified ($$e_\phi(\infty) = \deg(\phi)=5$$) because $$\phi^{-1}(\infty) = \{\infty\}$$.

For $$a \ne 0, 1,\infty$$ then $$\phi^{-1}(a)$$ contains $$5$$ distinct points where $$e_\phi(b) = 1$$.