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In J. Silverman's book "Arithmetic of Elliptic Curves" Chapter 2 Proposition 2.6 (a) it is considered a non constant morphism $\Phi:C_{1}→C_{2}$ between two smooth curves defined over a perfect field $K$.

I am having trouble understanding why there is no inertia degree appearing in the degree formula $\deg(\Phi)=\sum_{P\in \Phi^{-1}(Q)}e_{\Phi}(P)$ for any $Q\in C_{2}$. I don't see why in general the inertia degree should be 1 and not appear in the well known degree formula as it does in standard Number Theory.

This is what I mean: as the curves are smooth the regular functions rings are Dedekind domains and we have the unique factorization $M_{\Phi(P)}=\prod_{P′\in \Phi^{-1}(\Phi(P))}M_{P'}^{e_{\Phi}(P′)}$ being $M_{\Phi(P)}$ and $M_{P}$ the maximal ideals associated to $\Phi(P)$ and $P$ respectively (as $\Phi$ is non constant $\Phi^{*}$ is injective and embeds $K[C_{2}]$ in $K[C_{1}]$). In the degree formula the inertia degree should appear by standard Number Theory, which is not the case.

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    $\begingroup$ Why does "$K$ become algebraically closed" in characteristic $0$? $\endgroup$ – abx Aug 8 '16 at 8:39
  • $\begingroup$ It doesn't sorry for the confusion, I have edited the question. I was thinking to something else $\endgroup$ – Hair80 Aug 8 '16 at 21:45
  • $\begingroup$ Just a reminder to accept the answer if it answered your question (if for nothing else than to keep the Unanswered tab unclogged) $\endgroup$ – Will Chen Aug 19 '16 at 18:29
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The reason is that there are two ways of thinking about "points".

Let $A$ be a ring. Then, define:

  1. A scheme-theoretic/topological point of Spec $A$ is a prime ideal of $A$.
  2. A geometric/functorial point of Spec $A$ is an equivalence class of morphisms Spec $K\rightarrow$ Spec $A$, where $K$ is a field, and two morphisms: $$p_1 : \text{Spec }K_1\rightarrow \text{Spec }A,\qquad p_2 : \text{Spec }K_2\rightarrow\text{Spec }A$$ are equivalent if there exists a field $K$ containing both $K_1,K_2$ as subfields, and a morphism Spec $K\rightarrow$ Spec $A$ making the obvious diagrams commute.

Note that if $A$ is a $k$-algebra, then geometric points of Spec $A$ are just morphisms Spec $\overline{k}\rightarrow\text{Spec }A$. Further, if $A$ is a finite type $k$-algebra (for example $k[x,y]/(f)$) then a geometric point of $\text{Spec }k[x,y]/(f)$ is nothing but a pair $(a,b)\in\overline{k}^2$ satisfying the equation $f$, which is precisely how Silverman defines a "point".

Here's an example that will explain everything:

Let $k = \mathbb{Q}$, and let $A = \mathbb{Q}[x]$, $B = \mathbb{Q}[y]$, and suppose $X = \text{Spec }A$ and $Y = \text{Spec }B$. Consider the map $f : Y\rightarrow X$ given by $x\mapsto y^2$. This map is degree 2, and sends $y = a$ in $Y$ to $x = a^2$ in $X$.

First, lets consider the scheme-theoretic/topological point $q\in X$ corresponding to $x = -1$. Thus, $q$ is the prime ideal $(x+1)\subset\mathbb{Q}[x]$. It's not hard to check that the only prime of $B$ lying over $q$ is $p := (x^2+1)$, whose residue field is $\mathbb{Q}(i)$, so $p$ has "inertia degree" 2 over $q$.

Now, lets consider the geometric point $Q : $Spec $\overline{\mathbb{Q}}\rightarrow$ Spec $A$ given by sending $x\mapsto -1$. It's not hard to see that sending $y\mapsto i$ and sending $y\mapsto -i$ define two distinct morphisms $P_1,P_2 : \overline{\mathbb{Q}}\rightarrow$ Spec $B$ (ie, two distinct geometric points) which lie above $Q$.

Thus, while there is only one topological point above $x = -1$, there are two geometric points, each of which makes an appearance as a summand of the equation in Prop 2.6. This makes up for the fact that the inertia degree doesn't appear. In fact, one may define the inertia degree as the number of geometric points lying over $x = -1$ who have the same image $P\in Y$.

Ie, in the situation of our morphism $f : Y\rightarrow X$, the two equivalent ways of writing the formula would be:

  1. $deg(f) = 2\cdot e_f(p) = 2\cdot 1$ (2 is the inertia degree), or
  2. $deg(f) = e_f(P_1) + e_f(P_2) = 1 + 1$
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