= Solution
The <three-isogeny descent> connecting map, identified through the <Weil pairing> with $H^1(\mathbb Q,\mu_3)=\mathbb Q^\times/(\mathbb Q^\times)^3$, is
$$
\alpha:E'(\mathbb Q)\longrightarrow
\mathbb Q^\times/(\mathbb Q^\times)^3.
$$
The long exact sequence attached to $0\to E[\phi]\to E\xrightarrow{\phi}E'\to0$ makes it a group homomorphism with
$$
\ker\alpha=\phi E(\mathbb Q).
$$
The function $f=y$ from part (b) gives the explicit formula
$$
\alpha(O)=1,
\qquad
\alpha(T)=d^{-1}=d^2\pmod{(\mathbb Q^\times)^3},
\qquad
\alpha(x,y)=y\pmod{(\mathbb Q^\times)^3}quad(P\ne O,T).
$$
At $T$, the value $d^{-1}$ is the leading coefficient of $y$ relative to the local parameter $x$, since $y(y+d)=x^3$ gives $y/x^3\to d^{-1}$. At $-T=(0,-d)$ the ordinary formula gives $-d$, whose class is the inverse of $d^{-1}$ because $-1$ is a cube.
Let $S$ be the primes dividing $d$. For a prime $\ell\notin S$, use
$$
y(y+d)=x^3.
$$
If $v_\ell(y)>0$, then $y+d$ is an $\ell$-adic unit, so $v_\ell(y)=3v_\ell(x)$. If $v_\ell(y)<0$, then $v_\ell(y+d)=v_\ell(y)$, so $2v_\ell(y)=3v_\ell(x)$. In either case $v_\ell(y)$ is divisible by three; the special values at $O$ and $T$ have the same property. Therefore
$$
\boxed{\operatorname{im}\alpha\subseteq\mathbb Q(S,3).}
$$
When $d=1$, this power-class group is trivial: a rational number whose valuation at every prime is divisible by three is a cube up to sign, and $-1=(-1)^3$. Thus $\alpha$ is trivial, its kernel is all of $E'(\mathbb Q)$, and
$$
\boxed{\phi:E(\mathbb Q)\twoheadrightarrow E'(\mathbb Q).}
$$
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