Let $E/K$ be an elliptic curve over a field $K$, with discriminant $\Delta$. Then the polynomial $x^3-\Delta$ has a root (and hence all roots since Galois) in $K(E[3])$; this can be shown laboriously through solving the 3-division polynomial (a quartic).

Is there a nicer/more intuitive way of seeing this and can you please provide a proper reference for either the above method or whatever you suggest?

  • $\begingroup$ This is a very good question, $x^3$ has a real root exactly on the real number line "if & only if" it is a cube of any constructible number $\endgroup$ – Bassam Karzeddin Dec 30 '15 at 11:54
  • $\begingroup$ "All roots since Galois" holds only if $K$ contains the cube roots of unity. (Presumably you don't allow characteristic $3$, in which case $E[3]$ behaves very differently.) $\endgroup$ – Noam D. Elkies Jan 10 '16 at 4:45

One way to think about this is via the modular curves parametrizing elliptic curves $E$ with either $E[3]$ or $\Delta^{1/3}$ rational. Note that $\Delta^{1/3}$ is rational iff $j^{1/3}$ is rational, because $j = E_4^3 / \Delta$.

Assume for simplicity that $K$ contains the cube roots of unity (because $K(E[3])$ contains them in any case thanks to the Weil pairing). The $E[3]$ curve is the modular curve $X(3)$, with a map $X(3) \to X(1)$ that forgets the $3$-torsion structure, and is Galois with group ${\rm PSL}_2({\bf Z}/3{\bf Z}) \cong A_4$.

Now $A_4$ has a normal subgroup, the "Klein $4$-group" $V_4$ consisting of the identity and the three double transpositions. (The $V$ stands for German "Vierergruppe".) So $X(3) / V_4$ is a Galois cover, call it $X'(3)$, of $X(1)$ with group $A_4 / V_4 \cong {\bf Z} / 3 {\bf Z}$, i.e. a 3:1 cyclic cover. Once $K$ contains cube roots of unity, Kummer theory says that any 3:1 cyclic cover is obtained by adjoining a cube root to the function field; here this function field is $K(j)$, so the function field $K(X'(3))$ is $K(j,F^{1/3})$ where $F$ is some rational function of $j$ that's not already a cube.

The punchline is that we can take $F=j$. Since the function field of $X'(3)$ is contained in the function field of $X(3)$, this explains why $\Delta^{1/3}$ is a rational function of the coordinates of $E[3]$.

The fact that $K(X'(3)) = K(j^{1/3})$ can almost be recovered from the ramification structure of the map $X(3) \to X(1)$. It is ramified only above $j=\infty$, $j=0$, and $j=1728$, with cycle structures $(3,1)$, $(3,1)$, $(2,2)$ respectively. Hence the cover $X'(3) \to X(1)$ is ramified only above $j=0$ and $j=\infty$, so $K(X'(3))$ must be $K((cj)^{1/3})$ for some constant $c \in K^*$. The fact that we can use $c=1$ takes a bit more work, but once we know that the cover has this form it's enough to just try some convenient $E$ with $j(E)\neq 0$ to complete the proof.


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