Cube root of discriminant of elliptic curve 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?
 A: 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.
