Take the 2-minute tour ×

Let $\phi:C_1\to C_2$ be a nonconstant map of two smooth curves over some algebraically closed field $K$ and let $P\in C_1$. $\phi$ gives us an induced map of fields $\phi^*:K(C_2)\to K(C_1)$, $\phi^*:f \mapsto f\circ \phi$. Silverman defines in his book on elliptic curves that the ramification degree of this map is defined as


where the order is just the normalized valuation of the DVR $K[C_1]_P$ and $t_{\phi(P)}$ denotes a uniformizer of $K[C_2]_Q$, where $Q=\phi(P)$. I'm trying to explicitly write down the connection between this and the corresponding ramification degree in the Dedekind domains, since Silverman seems to hint that they agree.

Let $\mathfrak{m}_Q$ denote the maximal ideal of $A=\phi^*K[C_2]_Q$ and let $B$ be the integral closure of $A$ in $K(C_1)$. Since, $A$ is a DVR, we know that $B$ is Dedekind and we have a factorization

$$\mathfrak{m}_QB = \mathfrak{P}_1^{e_1}\cdots \mathfrak{P}_n^{e_n}.$$

To get a nice description of the ramification degree in the setting of these Dedekind rings, I would need to show that:

  1. There's a bijection between the primes $\mathfrak{P}_i$ and the points $P\in \phi^{-1}(Q)$.

  2. If $P$ corresponds to some $\mathfrak{P}_i$, then $e_\phi(P)=e_i$.

Does anyone know how this is done? My guess is that both should be simple applications of the Nullstellensatz. I'm getting stuck here with the issue I had in a question I asked yesterday:

Trying to parse a definition in Silverman's EC book

The problem being that I don't know how to concretely compute the map $\phi^*$ and I also have trouble figuring out what $\mathfrak{m}_Q$ should look like. By the Nullstellensatz, we know that it's of the form:


I take it that in projective space $Q=[a_1,\ldots,1,\ldots,a_n]$, where the $1$ is in position $i$ corresponding to the embedding of $\mathbb{A}^n$ into $\mathbb{P}^n$? But then one would have to express $\phi^*\mathfrak{m}_Q$ explicitly and here one is again left with the problem of concretely writing down what $\phi^*$ is in order to use the Nullstellensatz in $B$.

share|improve this question

1 Answer 1

I should know this stuff more thoroughly but if my memory serves me right...in the realms of algebraic geometry we can view spec$(\mathfrak{O}_K)$ as a kind of curve. This is just the collection of prime ideals of $\mathfrak{O}_K$.

Now given an extension of number fields $L/K$ we have the map from spec$(\mathfrak{O}_L)$ to spec$(\mathfrak{O}_K)$ that sends a prime ideal $\mathfrak{P}$ to prime ideal $\mathfrak{P}\cap\mathfrak{O}_L = \mathfrak{p}$.

The pullback of a uniformizer is then the same notion as pulling back $\mathfrak{p}$ to an ideal $\mathfrak{p}\mathfrak{O}_L$ of $\mathfrak{O}_L$.

Then the notion of order at $P$ translates to the order at $\mathfrak{P}$, i.e. the ramification index.

So here $\text{ord}_P (\phi^{\star}t_{\phi(P)})$ is the same thing as $\text{ord} _{\mathfrak{P}}(\mathfrak{p}\mathfrak{O}_L)$.

share|improve this answer
That's right. In fact $\text{spec} \mathcal{O}_K$ isn't a curve but rather like a $0$-dimensional algebraic variety over $\mathbb{Q}$ (as opposed to a one-dimensional algebraic variety). –  Bruno Joyal Mar 27 '12 at 18:58
Yes, that is better! –  fretty Mar 27 '12 at 20:28

Your Answer


