I am trying to understand example 2.2.9 of Silverman's "Arithmetic of Elliptic Curves". In this example, Silverman considers a map
$$
\phi:\mathbb{P}^1\to \mathbb{P}^1; [X,Y]\mapsto [X^3(X-Y)^2, Y^5]
$$
and he claims that the map $\phi$ ramifies at the points $[0,1]$ and $[1,1]$ and that the ramification indices are
$$
e_\phi([0,1])=3
$$
$$
e_\phi([1,1]) = 2
$$
I am struggling to actually compute these ramification indices by hand. Here is my attempt so far…
We define the ramification index by
$$
e_\phi([0,1]) = ord_{[0,1]}(\phi^*t_{[1,1]})
$$
where $t_{[1,1]}$ is a uniformizer for $\mathbb{P}^1$ at $[1,1] = \phi([0,1])$. I think of the function field $K(\mathbb{P}^1)$ as the subfield of $K(X, Y)$ generated by the rational functions whose numerator and denominator have the same degree. Under this identification, I can regard the local ring $K[\mathbb{P}^1]_{[1,1]}$ as the subring of the function field consisting of the rational functions whose denominator does not vanish at $[1,1]$. So I think that a uniformizer at $[1,1]$ is given by $(X-Y)/Y$. I similarly think that a uniformizer at $[0,1]$ is $X/Y$, but I am not certain these are correct. I am also uncertain of how to compute
$$
\phi^*((X-Y)/Y)
$$
and to compute the order of this. Any help is greatly appreciated! 🙂
Best Answer
Mohan essentially gives the answer in his comment above. But I figured I would write it up as an answer to my question.
Note that the fibre of $\phi$ at the point $[1,0]$ is precisely the point $[1,0]$. Thus restricting $\phi$ to the affine chart given by $Y=1$ gives a map $$ \phi: \mathbb{A}^1\to \mathbb{A}^1 $$
and in this case $\phi$ is precisely the polynomial $x^3(x-1)^2$. Now in the field $K(x)$, a uniformizer for $0\in \mathbb{A}^1$ is $x$ and a uniformizer for $1\in \mathbb{A}^1$ is $x-1$. Thus, $$ ord_{[0,1]}(\phi) = ord_0(x^3(x-1)^2) = 3 $$ and similarly for the order of $\phi$ at $[1,1]$.