Again a question about algebraic varieties ! Actually, I followed to book of Silverman "The Arithmetic of elliptic curve", and I have several questions about ramification index. For $\phi : C_1 \to C_2$ a non constant map of smooth curves, and $P \in C_1$, he's defining the ramification index of $\phi$ at $P$ as : $e_{\phi}(P) = \operatorname{ord}_{P}(\phi^*(t_{\phi(P)}))$ where $t_{\phi(P)}$ is a uniformizer at $\phi(P)$. Then, we have a proposition among which we have the formula : $\forall Q \in C_2 \; \operatorname{deg}(\phi) = \sum_{P \in \phi^{-1}(Q)} e_{\phi}(P)$.

My problem is : how to explicitly calculate $\operatorname{deg}(\phi)$ using this formula. I mean, the author gives then an example : $\phi : \mathbb{P}^1 \to \mathbb{P}^1 \; [X:Y] \mapsto [X^3(X-Y)^2:Y^5]$, and he says that $\phi$ is unramified everywhere except on $[0:1]$ and $[1:1]$ where we find : $e_{\phi}([0:1]) = 3, e_{\phi}([1:1]) = 2$. So, I tried to understand this example, but I'm stuck.

Actually, I didn't first saw why it is unramified everywhere except on $[0:1]$ and $[1:1]$, so I tried to understand what happens in $[0:1]$ and $[1:1]$ for example.

  • If $Q=[a:b], \; b \neq 0$, we have : $\mathcal{O}_{\mathbb{P}^1, Q}= k[\frac{X}{Y}]_{(\frac{X}{Y}-\frac{a}{b})}$ with maximal ideal : $\mathcal{m}_q = (\frac{X}{Y}-\frac{a}{b})k[\frac{X}{Y}]_{(\frac{X}{Y}-\frac{a}{b})}$. So, for $Q=[0:1]$ for instance, we have : $\mathcal{O}_{\mathbb{P}^1, Q}= k[\frac{X}{Y}]_{(\frac{X}{Y})}$ with maximal ideal : $(\frac{X}{Y})k[\frac{X}{Y}]_{(\frac{X}{Y})}$, and a uniformizer is then given by $\frac{X}{Y}$. But we have : $$ e_{\phi}([0:1]) = \operatorname{ord}_{[0:1]}(\frac{X}{Y} \circ [X^3(X-Y)^2 : Y^5]) = \operatorname{ord}_{[0:1]}(\frac{X^3}{Y^3}(X-Y)^2.\frac{1}{Y^2}) $$ which is not in the maximal ideal, but : $$ (\frac{X^3}{Y^5}(X-Y)^2)^2 = \frac{X^6}{Y^6}.\frac{(X-Y)^4}{Y^4} $$ seems to be on the ideal cause it's a quotient of polynomial of same degree of the form : $\frac{X}{Y} \times (*)$. So, we find : $e_{\phi}([0:1]) = 2$ ? So, obviously I'm wrong somewhere, or there is something I didn't understood.

And it's the same for the other one. So my first question is : where I'm wrong, and how to explictly determine the ramification index ?

My second question is : if we know that $\phi : C_1 \to C_2$ is given by $[x:y] \mapsto [f_1(x) : 1]$ for example, and we know in some point $P$ $f_1$ as a pole of order $n$ fixed, and otherwise $f_1$ as neither a pole nor a zero. Can we conclude that : $deg(\phi)= - \operatorname{ord}_P(f_1)) = n$ and the same if we replace the pole by a zero ? Put an other way : is there a link between the pole and zeros of the rational functions defining the map and the degree of the map ?

Sorry for the long post, and thank you by advance for enlighten me !

  • $\begingroup$ You are confusing $[0:1]$ in two different spaces. There are two point above $[0:1]$, namely $[0,1]$ and $[1,1]$. $\endgroup$
    – Mohan
    Commented Feb 28, 2020 at 16:10
  • $\begingroup$ @Mohan Yes there is two different point above [0:1], those you have given. By the formula on the 1st post, this gives $deg(\phi) = e_{\phi}([0:1])+e_{\phi}([1:1])$, and I was trying to calculate $e_{\phi}([0:1])$ in the first post (and understanding why the others points are unramified). So, what do you mean when you say I'm confusing $[0:1]$ ? I don't think I had understand where I made the confusion. $\endgroup$
    – Mathuser
    Commented Feb 28, 2020 at 17:29
  • 1
    $\begingroup$ Locally your curve is affine why do you bother with projective coordinates. In the affine coordinate your map is given by a rational function $f(t)$ in $t=X/Y$ and at $t=a$ the ramification index is the order of the zero of $f(t)-a$ at $t=a$. $\endgroup$
    – reuns
    Commented Feb 29, 2020 at 3:11
  • $\begingroup$ @reuns Yes I get it now, thank you ! $\endgroup$
    – Mathuser
    Commented Mar 2, 2020 at 22:06

2 Answers 2


Let's take $Q=[0:1]$, and always stick to the preimages of this point. Then the preimages only consist of two points $$[0: a], [a:a]\text{ where }a\not=0.$$

Near $Q\in C_2$, we can take the local neighborhood and take the uniformizer to be $t = x/y$. By the description of preimage points (non of the $y$-coordicate is 0), we could use the (same-expression) local parameter $u = x/y$ for any $P\in \phi^{-1}(Q)$. Then the pullback of $t$ is: $$\phi^*(t) = \dfrac{X^3(X-Y)^2}{Y^5} = u^3(u-1)^2.$$ This vanishes when $u=0$ or $u=1$. Thus when $u=0$, corresponding to the preimage point $[0:a]=[0:1]$, the ramification index is 3; and when $u=1$, corresponding to the preimage point $[a:a]=[1:1]$, the ramification index is 2.

if we know that $\phi:C_1\to C_2$ is given by $[x:y]\to [f_1(x):1]$ for example...

you need to give homogeneous polynomials...

  • 1
    $\begingroup$ Thank you for this detailed answer ! $\endgroup$
    – Mathuser
    Commented Mar 2, 2020 at 22:06

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 traslation we wil 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 $t_{[1,1]} = x/y -1$:

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


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