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! :)

  • $\begingroup$ To me it looks like $\phi$ is totally ramified at $[1,0]$ too. Are you just trying to compute the ramification indices at the two points you mention? $\endgroup$
    – Mohan
    Aug 30, 2015 at 19:09
  • $\begingroup$ @Mohan By totally ramified do you mean that each point in the fibre of the point ramifies? At the moment, I am just trying to compute the ramification indices. I would like to get my hands dirty so to speak :) $\endgroup$
    – CWcx
    Aug 30, 2015 at 19:13
  • 5
    $\begingroup$ There is only one point in the fiber of $[1,0]$, namely $[1,0]$. If you realize that, to compute ramifications at other points, you can put $Y=1$, and just calculate these for the map $\mathbb{A}^1\to\mathbb{A}^1$ given by $x^3(x-1)^2$ and then calculations are much easier to do. You can see the 3 and 2 appearing. $\endgroup$
    – Mohan
    Aug 30, 2015 at 20:17
  • $\begingroup$ Ahh, I see now! Jeez I was being stupid. Thanks for the very informative comment :) $\endgroup$
    – CWcx
    Sep 3, 2015 at 20:53

2 Answers 2


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 $[1,1]$.

  • 1
    $\begingroup$ I'm trying to understand this example now. Why is it important that the fibre of $\phi$ at $[1,0]$ is precisely the point $[1,0]$? What would happen if it were another point, or if that point had more than a single point in the fibre? $\endgroup$
    – a-lawliet
    Aug 14, 2019 at 15:30
  • $\begingroup$ @a-lawliet, see the below answer, $\phi[0,1]=[0,1]$ gives answer to your first question $\endgroup$
    – MAS
    Feb 12, 2021 at 14:24

We have $\phi[0,1]=\phi[1,1]=[0,1]$, so by definition

$$e_{\phi}[0,1] = ord_{[0:1]}(\phi^{*}t_{\phi[0,1]}) = ord_{[0:1]}(\phi^{*}t_{[0,1]}) = ord_{[0:1]}(t_{[0,1]}\circ \phi)$$

Analogously $$e_{\phi}[1,1] = ord_{[1:1]}(t_{[0,1]}\circ \phi)$$

A uniformizer in [0,1] is $t_{[0,1]}:= (x/y)$ because $ord_{[0,1]}(x/y)=1$, indeed, we will see $ord_{[0,1]}(x)=1$ and $ord_{[0,1]}(y)=0$:

First $y\in\mathcal{O}_{\mathbb{P}^1,[0,1]}$, so $ord_{[0,1]}(y)\geq 0$. Moreover $y[0,1]=1\neq 0 \Rightarrow ord_{[0,1]}(y)\leq 0$, thus $ord_{[0,1]}(y)=0$.

Second, $x[0,1]=0 \Rightarrow x\in\mathfrak{m}_{[0,1]} \Rightarrow (x)\subseteq \mathfrak{m}_{[0,1]}$ We have to check $(x)=\mathfrak{m}_{[0,1]}$. We will suppose the opposite and find a contradiction:

If $(x)\subsetneq \mathfrak{m}_{[0,1]}$ then we have a chain of prime ideals of length equal to 2 of the ring $\mathcal{O}_{\mathbb{P}^1,[0,1]}$, this is a contradiction because $dim(\mathcal{O}_{\mathbb{P}^1,[0,1]})=1$.

So we have

$$ord_{[0,1]}(x/y) = ord_{[0,1]}(x) - ord_{[0,1]}(y)=1-0=1 \Rightarrow t_{[0,1]}=x/y$$

Making a translation we will have $$t_{[1,1]} = \frac{x}{y}-1$$

Now we can compute the ramification index: $$e_{\phi}[0,1] = ord_{[0:1]}(t_{[0,1]}\circ \phi) = ord_{[0:1]}(\frac{x}{y}\circ [x^3(x-y)^2,y^5])= ord_{[0:1]}(\frac{x^3(x-y)^2}{y^5}) = ord_{[0:1]}((\frac{x}{y})^3\frac{(x-y)^2}{y^2}) = ord_{[0:1]}((\frac{x}{y})^3)+ord_{[0:1]}((\frac{x}{y}-1)^2)) = 3+0=3$$ Analogously, remembering that $t_{[1,1]} = x/y -1$:

$$e_{\phi}[0,1]) = ord_{[1:1]}(t_{[0,1]}\circ \phi) = ord_{[1:1]}((\frac{x}{y})^3)+ord_{[1:1]}((\frac{x}{y}-1)^2) = 0+2=2$$


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