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.

  • $\begingroup$ $\phi$ 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 $\phi^*$. $\endgroup$
    – reuns
    Aug 7, 2019 at 18:48

1 Answer 1


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 $e_\phi(b) = 1$.


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