I'm doing a course on elliptic curves. An isogeny $\phi:E_1 \rightarrow E_2$ induces a map $$\begin{array}{llll}\phi_*: & \mbox{Div}^0(E_1) & \rightarrow & \mbox{Div}^0(E_2) \\ \\ & \sum_{P\in E_1}n_PP & \mapsto & \sum_{P \in E_1}n_P \phi(P). \end{array}$$

Then we're told that if $f \in K(E_1)^\times$, then $$(*) \hspace{5mm} \mbox{div}(N_{K(E_1)/K(E_2)}f)=\phi_*(\mbox{div}(f)).$$

He says the proof of this goes along the lines of the proof that the norm of a principal ideal is the norm of its generator, but I can't see how this will work out. Can anyone give me a reference for a proof of this?

Also, if $f \in K(E_2)\hookrightarrow K(E_1)$, then $N_{K(E_1)/K(E_2)}f=f^n$, where $n=[K(E_1):\phi^*(K(E_2)]=\mbox{deg}(\phi)$. Am I right in thinking that in this case $(*)$ is saying that the degree of an isogeny is the size of its fibres, i.e. $\#\{P:\phi(P)=Q\}$, for any $Q \in E_2$? Why is this true when $\phi$ is not seperable?


1 Answer 1


The only reference I know for the proof of $(\ast)$ is Fulton's Book Intersection Theory (Chapter 1, Proposition 1.4(b)); this proof relies on some results in Commutative Algebra that are stated and proven in the Appendix. But the statement, there, is more general, indeed it involves a proper morphism of varieties, non necessarily curves. Perhaps there is an easier proof for smooth curves.

Yes, you are (more or less) right when you say that the degree is the size of the fibers. To be more precise, suppose we have a finite morphism $\phi:X\to Y$ between nonsingular curves over a field $k$ ($\phi$ not necessarily separable). The degree of $\phi$ can be defined to be the size of a general fiber of $\phi$. Indeed, one can prove that for general $y\in Y$ the size of $X_y$ is equal to $[K(X):\phi^\ast K(Y)]$ (the usual definition of $\deg\phi$). However, for a finite number of points in $Y$ this is not true anymore: we have $|X_y|<\deg\phi$; but the points in these "special" fibers come with multiplicity. So, what is true for every point $y\in Y$ is that $$ \deg\phi=\sum_{x\in X_y}e_{x/y}[k(x):k(y)], $$ where $e_{x/y}$ is the ramification index of $\mathcal O_{y,Y}\to\mathcal O_{x,X}$.

Saying $|X_y|=d$ means that $\phi$ is not ramified over $y$. It means that $e_{x/y}=1=[k(x):k(y)]$ for every $x\in X_y$. Instead, saying $e_{x/y}>1$ means exactly that $k(x)/k(y)$ is not separable! (I always keep in mind this correspondence: "separability" $\longleftrightarrow$ "$e=1$" $\longleftrightarrow$ "unramified".)

As an example, consider the projection of the parabola $P:x=y^2$ onto the $x$-axis: the fiber over $0$ consists of one point counted twice (and this "twice", algebraically, is reflected by the fact that $2=\dim_kk[y]/y^2$, where $k[y]/y^2$ is the coordinate ring of $P\cap \{x=0\}$).

  • $\begingroup$ Ah! I had thought my lecturer had said that if $\phi$ is separable then we only have a finite number of points whose ramification index is not 1 - I didn't realise this was true in general. Thanks for clarifying! Also I'll check out that book, thanks very much. $\endgroup$
    – porkramen
    Commented Feb 3, 2013 at 19:09

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