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Given an elliptic curve $E$ there is the multiplication by $n$ map $[n]: E \rightarrow E$.

If $K(E)$ is the fraction field then this map makes $K(E)$ a degree $n^2$ extension of itself.

My question is what are some techniques for dealing with the field theory of this extension? Can one write down generators for this extension in terms of the coefficients of E?

If not, what are popular techniques for dealing with the Galois theory of this extension?

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The Galois group is clearly $\mathbb{Z}/n \times \mathbb{Z}/n$. Given the curve $E$ in Weierstrass form, one can express the coordinates of $nP$ by using the so-called division polynomials, see en.wikipedia.org/wiki/Division_polynomials. Moreover, the roots of the $(2n + 1)$th division polynomial $\psi_{2n + 1}$ are exactly the $x$ coordinates of the points of $E[2n+1] \setminus O$. –  Francesco Polizzi Nov 10 '11 at 16:30
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The Galois group is clearly not Z/n x Z/n, but rather Sp_2(Z/n); e.g. it is S_3 for 2 torsion points. –  David Lehavi Nov 10 '11 at 19:12
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@David : Anonymous is looking at the extension between the function fields, not the one generated by torsion points. –  François Brunault Nov 10 '11 at 20:15
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1 Answer

up vote 7 down vote accepted

If $K$ is an algebraically closed field whose characteristic doesn't divide $n$, then the Galois theory of this extension is not complicated. Indeed $K(E)/[n]^* K(E)$ is then abelian with Galois group canonically isomorphic to $E[n]$. A point $P \in E[n]$ acts by translation on $K(E)$, namely $\sigma_P(f) = t_P^* f$ where $t_P : E \to E$ is the translation map, and the $\sigma_P$ are exactly the elements of the Galois group. A function $f \in K(E)$ will generate the extension if and only if its translates by $E[n]$ are pairwise distinct. It is not hard to show that, taking for example a Weierstrass model for $E$, the function $f=x$ works.

In the case $K$ is not algebraically closed one has to be careful that the extension isn't necessarily Galois anymore (it is if and only if the $n$-torsion is rational over $K$). However, regarding the question about generators, the above criterion generalizes, namely $f \in K(E)$ generates the extension if and only if its translates (which lie in $K_n(E)$, where $K_n$ is the field obtained from $K$ by adjoining the coordinates of the $n$-torsion points) are pairwise distinct. So, for example, $f=x$ is still a generator of the extension.

EDIT. In view of the OP's comments, here is an alternate way of constructing generators of the extension $K(E)/[n]^* K(E)$. These generators will have minimal polynomials of the form $X^n-[n]^* g$, for reasonable functions $g \in K(E)$.

Assume $E[n] \subset E(K)$, so that $K(E)/[n]^* K(E)$ is Galois and $\operatorname{Gal}(K(E)/[n]^* K(E)) \cong E[n]$, as explained above. By Kummer theory, this extension is generated by two $n$-th roots of suitable elements of $[n]^* K(E)$. We can find these elements as follows. Let $f \in K(E)$ such that $f^n = [n]^* g$ for some $g \in K(E)$. Taking the divisors, we see that $\operatorname{div}(f)$ is invariant by translation by $E[n]$, so it has the form

\begin{equation*} \operatorname{div}(f) = \sum_i \sum_{R \in E[n]} \lambda_i [P_i+R] \end{equation*} with $\lambda_i \in \mathbf{Z}$ and $P_i \in E$. Conversely, such a divisor is principal if and only if $\sum \lambda_i = 0$ and $n^2 \sum \lambda_i P_i = 0$ (because the sum of all $n$-torsion points is zero). The divisor of $g$ is then given by

\begin{equation*} \operatorname{div}(g) = n \sum_i \lambda_i [n P_i] \end{equation*} Note that if $n \sum \lambda_i P_i = 0$ then $f$ is already in $[n]^* K(E)$. So, letting $Q_i=nP_i$, the conditions on the divisor of $g$ are $\sum \lambda_i = 0$ and $\sum \lambda_i Q_i \in E[n] \backslash \{0\}$.

For example, for each $Q \in E[n]$, one can take $g_Q$ such that $\operatorname{div}(g_Q)=n[Q]-n[0]$. Identifying the base field of the extension with $K(E)$ and denoting the extension by $L/K(E)$, we thus get $L=K(E)(g_{Q_1}^{1/n},g_{Q_2}^{1/n})$ if and only if $(Q_1,Q_2)$ is a basis of $E[n]$. Note : the Galois action on these functions $g_{Q}^{1/n}$ is related to the Weil pairing (see for example Silverman, The arithmetic of elliptic curves). More generally one can define $g_D$ for any $D \in I$, where $I$ is the augmentation ideal of $\mathbf{Z}[E[n]]$. Then $L=K(E)(g_{D_1}^{1/n},g_{D_2}^{1/n})$ if and only if the classes of $D_1$ and $D_2$ generate $I/I^2 \cong E[n]$. Finally, there is the further flexibility in that one can choose functions $g$ whose divisors are not necessarily supported on $E[n]$.

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Thanks for the comment. I think I am looking for something more explicit in that this answer translates the difficulty of the extension $K(E)/[n]*K(E)$ into realizing $[n]*K(E)$ as some subfield which is quite complicated in terms of the Weierstrass coordinates. I am looking to describe this extension in when [n]^*K(E) is given Weierstrass coordinates. Alternatively, is there a canonical $f$ which generates $K(E)/[n]^*K(E)$ for which we can describe its minimal polynomial over $[n]^*K(E)$? Choosing $f =x$ would lead us back to the very complicated group multiplication formulas. –  Anonymous Nov 10 '11 at 16:55
    
@Anonymous : What do you mean by "when [n]^*K(E) is given Weierstrass coordinates" ? –  François Brunault Nov 10 '11 at 17:47
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@David: The OP is looking at the extension of function fields, which has degree $n^2$ (the same degree of the isogeny). See also Francois' comment after the question. –  Francesco Polizzi Nov 10 '11 at 20:26
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@Anonymous : If $K$ contains the coordinates of the $n$-torsion points, then (by Kummer theory) the extension $K(E)/[n]^* K(E)$ is abstractly isomorphic to the extension obtained from $K(E)$ by adjoining the $n$-th roots of two reasonably explicit functions (each of which having degree $n$). This point of view is related to the Weil pairing on $E[n]$. Is this what you're looking for ? It would be easier for us to help you if there is something specific you want to compute about this extension. –  François Brunault Nov 10 '11 at 21:17
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@Francois Yes, your last comment sounds like exactly what I am looking for. Could you tell me more? What I mean is by Weierstrass coordinates is that $K(E)$ has a canonical transcendental basis by Weierstrass coordinates. The exact problem I am trying to solve is that I have some polynomial with coefficients in $K(E)$, which is irreducible over $K(E)$, but split in one of the extensions determined via multiplication by $[n]$. Trying to decipher what happens by applying $[n]^*$ to the coefficients is quite complicated and by the time $n=5$ the polynomial fills up two entire computer screens. –  Anonymous Nov 11 '11 at 0:58
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