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\begin{proof}
  Let $Q \in \phi^{-1} P$. Then $\delta(P) \in
  \Hone(K, \mu_n)$ is represented by $\sigma
  \mapsto \sigma Q - Q \in E[\phi] \cong \mu_n$.
  \begin{align*}
    \ephi(\sigma Q - Q, T)
    &= \frac{g(\sigma Q - Q + X)}{g(X)}
    &&\substack{\text{for any $X \in E$} \\
    \text{avoiding zeroes and poles of $g$}} \\
    &= \frac{g(\sigma Q)}{g(Q)}
    &&\text{pick $X = Q$} \\
    &= \frac{\sigma(g(Q))}{g(Q)} \\
    &= \frac{\sigma \sqrt[n]{f(P)}}{\sqrt[n]f(P)}
  \end{align*}
  ($\phi^* f = g^n$, $f(P) = g(Q)^n$).

  But
  \begin{align*}
    \Hone(K, \mu_n)
    &\cong K^* / (K^*)^n \\
    \left(\sigma \mapsto
    \frac{\sqrt[n]{x}}{\sqrt[n]{x}}\right)
    &\mapsfrom x
  \end{align*}
  Hence $\alpha(P) = f(P) \pmod{(K^*)^n}$.
\end{proof}

\subsection{Descent by 2-isogeny}

\begin{align*}
  E:~
  & y^2 = x(x^2 + ax + b) \\
  E':~
  & y^2 = x(x^2 + a'x + b')
\end{align*}
$b(a^2 - 4b) \neq 0$, $a' = -2a$, $b' = a^2 - 4b$.
\begin{align*}
  \phi : E
  &\to E' \\
  (x, y)
  &\mapsto \left( \left( \frac{y}{x} \right)^2,
  \frac{y(x^2 - b)}{x^2} \right) \\
  \hat{\phi} : E'
  &\to E \\
  (x, y)
  &\mapsto \left( \frac{1}{4} \left( \frac{y}{x}
  \right)^2, \frac{y(x^2 - b')}{8x^2} \right)
\end{align*}
$E[\phi] = \{0, T\}$, $T = (0, 0) \in E(K)$.

$E'[\hat{\phi}] = \{0, T'\}$, $T' = (0, 0) \in
E'(K)$.

\begin{fcprop}[]
  \label{prop:16.2}
  There is a group homomorphism
  \begin{align*}
    \mathcloze{E'(K)}
    &\mathcloze[1]{\to K^* / (K^*)^2} \\
    \mathcloze{(x, y)}
    &\mathcloze[2]{\mapsto \begin{cases}
      x \bmod (K^*)^2
      & \text{if $x \neq 0$} \\
      b' \bmod (K^*)^2
      & \text{if $x = 0$}
    \end{cases}}
  \end{align*}
  with \cloze{kernel $\phi E(K)$}.
\end{fcprop}

\begin{proof}
  Either: Apply \cref{thm:16.1} with $f = x \in
  K(E')$, $g = \frac{y}{x} \in K(E)$.

  Or: direct calculation -- see \es{4}.
\end{proof}

\begin{align*}
  \alpha_E : \frac{E(K)}{\hat{\phi} E'(K)}
  &\injto K^* / (K^*)^2 \\
  \alpha_{E'} : \frac{E'(K)}{\phi E(K)}
  &\injto K^* / (K^*)^2
\end{align*}

\begin{fclemma}[]
  \label{lemma:16.3}
  $\mathcloze{2^{\rank E(K)}} = \mathcloze{\frac{|\Im \alpha_E| \cdot
  |\Im \alpha_{E'}|}{4}}$.
\end{fclemma}

\begin{proof}
  If $A \stackrel{f}{\to} B \stackrel{g}{\to} C$
  are homomorphisms of abelian groups, then there
  is an exact sequence
  \begin{picmath}
    \begin{tikzcd}
      0
      \ar[r]
      & \ker(f)
      \ar[r]
      & \ker(gf)
      \ar[r, "f"]
      \ar[draw=none]{d}[name=A, anchor=center]{}
      & \ker(g)
      \ar[rounded corners,
              to path={ -- ([xshift=2ex]\tikztostart.east)
                        |- (A.center) \tikztonodes
                        -| ([xshift=-2ex]\tikztotarget.west)
                        -- (\tikztotarget)}]{dll}
      \\
      & \coker(f)
      \ar[r, "g"]
      & \coker(gf)
      \ar[r]
      & \coker(g)
      \ar[r]
      & 0
    \end{tikzcd}
  \end{picmath}
  Since $\hat{\phi} \phi = [2]_E$, we get an exact
  sequence
  \[
    0
    \to \ub{E(K)[\phi]}_{\cong \Zbb / 2\Zbb}
    \to E(K)\tors[2]
    \stackrel{\phi}{\to} \ub{E'(K)[\phi']}_{\cong
    \Zbb / 2\Zbb}
    \to \ub{\frac{E'(K)}{\phi E(K)}}_{\cong
    \Im(\alpha_{E'})}
    \stackrel{\hat{\phi}}{\to} \frac{E(K)}{2E(K)}
    \to \ub{\frac{E(K)}{\hat{\phi} E'(K)}}_{\cong
    \Im(\alpha_E)}
    \to 0
  .\]
  Therefore
  \[
    \frac{|E(L) / 2E(K)|}{|E(K)\tors[2]|}
    = \frac{|\Im \alpha_E| \cdot |\Im
    \alpha_{E'}|}{4}
    \tag{1}
    \label{lec23eq1}
  .\]
  \gls{mwt}: $E(K) \cong \Delta \times \Zbb^r$,
  where $\Delta$ is a finite group and $r =
  \rank E(K)$.

  $\frac{E(K)}{2E(K)} \cong \frac{\Delta}{2\Delta}
  \times \left( \frac{\Zbb}{2\Zbb} \right)^r$.

  $E(K)\tors[2] \cong \Delta\tors[2]$.

  Since $\Delta$ is finite, we have that
  $\frac{\Delta}{2\Delta}$ and $\Delta\tors[2]$
  have the same order, and therefore
  \[
    \frac{|E(K) / 2E(K)|}{|E(K)\tors[2]|}
    = 2^r
    \tag{2}
    \label{lec23eq2}
  .\]
  So we are done, by using \eqref{lec23eq1} and
  \eqref{lec23eq2}.
\end{proof}

\begin{fclemma}[]
  \label{lemma:16.4}
  Assuming:
  - $K$ is a number field
  - $a, b \in \mathcal{O}_K$
  Then:
  $\Im(\alpha_E) \subset K(S, 2)$, where $S =
  \{\mathfrak{p} \mid b\}$.
\end{fclemma}

\begin{proof}
  We must show that $x, y \in K$, $y^2 = x(x^2 +
  ax + b)$ and $v_{\mathfrak{p}}(b) = 0$ then
  $v_{\mathfrak{p}}(x) \equiv 0 \pmod{2}$.

  \textbf{Case $v_{\mathfrak{p}}(x) < 0$:} \cref{lemma:9.1}
  gives that for some $r \ge 1$,
  $v_{\mathfrak{p}}(x) = -2r$ and
  $v_{\mathfrak{p}}(y) = -3r$, so done.

  \textbf{Case $v_{\mathfrak{p}}(x) > 0$:} Then
  $v_{\mathfrak{p}}(x^2 + ax + b) = 0$. So
  $v_{\mathfrak{p}}(x) = v_{\mathfrak{p}}(y^2) =
  2v_{\mathfrak{p}}(y)$.
\end{proof}

\begin{fclemma}[]
  \label{lemma:16.5}
  Assuming:
  - $b_1 b_2 = b$
  Then:
  \begin{iffc}
    \lhs
    $b_1(K^*) \in \Im(\alpha_E)$
    \rhs
    \[
      w^2 = b_1 u^4 + a u^2 v^2 + b_2 v^4
      \tag{$*$}
      \label{lec23eq3}
    \]
    is soluble for $u, v, w \in K$ not all zero.
  \end{iffc}
\end{fclemma}

\begin{proof}
  If $b_1 \in (K^*)^2$ or $b_2 \in (K^*)^2$ then
  both conditions are satisfied. So we may assume
  $b_1, b_2 \notin (K^*)^2$.

  Now note $b_1(K^*)^2 \in \Im(\alpha_E)$ if and
  only if there exists $(x, y) \in E(K)$ such that
  $x = b_1 t^2$ for some $t \in K^*$. This implies
  \[
    y^2
    = (b_1 t^2)(b_1^2 t^2 + a b_1 t^2 + b)
  \]
  hence
  \[
    \left( \frac{y}{b_1 t} \right)^2
    = b_1 t^4 + at^2 + \ub{\frac{b}{b_1}}_{= b_2}
  .\]
  So \eqref{lec23eq3} has solution $(u, v, w) =
  \left( t, 1, \frac{y}{b_1 t} \right)$.

  Conversely, if $(u,v, w)$ is a solution to
  \eqref{lec23eq3} then $uv \neq 0$ and $\left(
  b_1 \left( \frac{u}{v} \right)^2, b_1
  \frac{uw}{v^3} \right) \in E(K)$.
\end{proof}

Now take $K = \Qbb$.

\begin{example*}
  $E$: $y^2 = x^3 - x$ ($a = 0$, $b = -1$).

  $\Im(\alpha_E) = \langle -1 \rangle \subset
  \Qbb^* / (\Qbb^*)^2$.

  $E'$: $y^2 = x^3 + 4x$.

  $\Im(\alpha_{E'}) \subset \langle -1, 2 \rangle
  \subset \Qbb^* / (\Qbb^*)^2$.

  \begin{align*}
    b_1
    &= -1
    & w^2
    &= - u^4 - 4v^4 \\
    b_1
    &= 2
    & w^2
    &= 2u^4 + 2v^4 \\
    b_1
    &= -2
    & w^2
    &= -2u^4 - 2v^4
  \end{align*}
  The first and last lines are insoluble over
  $\Rbb$ (squares are non-negative). The middle
  line does have a solution: $(u, v, w) = (1, 1,
  2)$.

  Therefore $\Im(\alpha_{E'}) = \langle 2 \rangle
  \subset \Qbb^* / (\Qbb^*)^2$.

  Hence $2^{\rank E(\Qbb)} = \frac{2 \cdot 2}{4} =
  1$, so $\rank E(\Qbb) = 0$.

  So $1$ is \emph{not} a congruent number.
\end{example*}

\begin{example*}
  $E$: $y^2 = x^3 + px$, $p$ a prime which is 5
  modulo 8.
  \begin{align*}
    b_1
    &= -1
    & w^2
    &= -u^4 - pv^4
  \end{align*}
  This is insoluble over $\Rbb$, hence
  $\Im(\alpha_E) = \langle p \rangle \subset
  \Qbb^* / (\Qbb^*)^2$.

  $E'$: $y^2 = x^3 - 4px$.

  $\Im(\alpha_{E'}) \subset \langle -1, 2, p
  \rangle \subset \Qbb^* / (\Qbb^*)^2$.

  Note: $\alpha_{E'}(T') = (-4p)(\Qbb^*)^2 =
  (-p)(\Qbb^*)^2$.
  \begin{align*}
    b_1
    &= 2
    & w^2
    &= 2u^4 - 2pv^4 \\
    b_1
    &= 2
    & w^2
    &= -2u^4 + 2pv^4 \\
    b_1
    &= p
    & w^2
    &= pu^4 - 4v^4
  \end{align*}
  Suppose the first line is soluble. Then without
  loss of generality $u, v, w \in \Zbb$ with
  $\gcd(u, v) = 1$. If $p \mid u$, then $p \mid w$
  and then $p \mid v$, contradiction. Therefore
  $w^2 \equiv 2u^4 \not\equiv 0 \pmod{p}$. So
  $\left( \frac{2}{p} \right) = + 1$, which
  contradicts $p \equiv 5 \pmod{8}$.

  Likewise the second line has no solution since
  $\left( \frac{-2}{p} \right) = -1$.