I am stuck in the last step of computing the Weak Mordell-Weil group $E(\mathbb{Q})/2E(\mathbb{Q})$ and $E'(\mathbb{Q})/2E'(\mathbb{Q})$ from the two exact sequences $$ 0 \longrightarrow \frac{E'(\mathbb{Q})[\hat{\phi}]}{\phi(E(\mathbb{Q})[2])} \longrightarrow \frac{E'(\mathbb{Q})}{\phi(E(\mathbb{Q}))} {\overset{\hat{\phi}}\longrightarrow} \frac{E(\mathbb{Q})}{2E(\mathbb{Q})} \longrightarrow \frac{E(\mathbb{Q})}{\hat{\phi}(E'(\mathbb{Q}))} \longrightarrow 0. $$ and $$ 0 \longrightarrow \frac{E(\mathbb{Q})[\phi]}{\hat{\phi}(E'(\mathbb{Q})[2])} \longrightarrow \frac{E(\mathbb{Q})}{\hat{\phi}(E'(\mathbb{Q}))} {\overset{\phi}\longrightarrow} \frac{E'(\mathbb{Q})}{2E'(\mathbb{Q})} \longrightarrow \frac{E'(\mathbb{Q})}{\phi(E(\mathbb{Q}))} \longrightarrow 0 $$ when I already have the structures of $$\frac{E'(\mathbb{Q})}{\phi(E(\mathbb{Q}))} \simeq (\mathbb{Z}/2\mathbb{Z})^2$$ and $$\frac{E(\mathbb{Q})}{\hat{\phi}(E'(\mathbb{Q}))} \simeq \mathbb{Z}/2\mathbb{Z}.$$
I think it is not a really difficult thing to do but I still need some confirmation of whether my (half of the)reasoning is correct or not.
In the first exact sequence, the exactness at $\frac{E(\mathbb{Q})}{2(E(\mathbb{Q}))}$ combined with the last map's surjective-ness imply that $$ E(\mathbb{Q}/2E(\mathbb{Q}) / (\mathbb{Z}/2\mathbb{Z})^2 \simeq \mathbb{Z}/2\mathbb{Z} $$ Now what should I do?
Also how do I go on to find generators of $\frac{E(\mathbb{Q})}{2(E(\mathbb{Q}))}$, say in the case when $E: y^2 = x^3-5x^2+5x$ isogenous to $E': Y^2 = X^3 +10X^2 + 5X$ via the 2-isogeny $$ \phi: E \longrightarrow E', \hspace{1.0cm} (x,y) \mapsto \Big(\frac{y^2}{x^2}, \frac{y(5-x^2)}{x^2}\Big)$$ and $$\widehat{\phi}: E' \longrightarrow E \hspace{1.0cm} (X,Y) \mapsto \Big(\frac{Y^2}{4X^2}, \frac{Y(5-X^2)}{8X^2} \Big)$$ and when $\frac{E(\mathbb{Q})}{\hat{\phi}(E'(\mathbb{Q}))}$ is generated by $\{(0,0)\}$ and $\frac{E'(\mathbb{Q})}{{\phi}(E(\mathbb{Q}))}$ is generated by $\{(0,0), (-5,10)\}$.
I would really appreciate any kind of helpful comments and answers.
Thank you.
I will work parallely to the computations in Example 4.10, page 338 (or 303), chapter X [Computing the Mordell-Weil group], section X.4 [The Sel and Sha Groups] from Silverman, The Arithmetic of Elliptic Curves. It uses Proposition X.4.9, page 337 (or 302) in loc. cit., which gives the following over the rationals.
Let us recall the notations, and the setting of the Proposition X.4.9.
We start with an elliptic curve $E$ over $\Bbb Q$, having a $2$-torsion $\Bbb Q$-rational point, so after passing to an isomorphic elliptic curve, $E$ has the two-torsion point $(0,0)$ and the affine equation: $$ \begin{aligned} E\ &:\ y^2=x^3+ax^2+bx\ . \\ &\qquad \text{In our example: }a=-5\ ,\ b=5\ : \\ E\ &:\ y^2=x^3-5x^2+5x\ . \end{aligned} $$ Consider the two-isogenous curve $$ \begin{aligned} E'\ &:\ Y^2=X^3+AX^2+BX\ , \qquad A=-2a\ ,\ B=a^2-4b\ . \\ &\qquad \text{In our example: }A=10\ ,\ B=5\ : \\ E'\ &:\ Y^2=X^3+10X^2+5X\ . \end{aligned} $$ The isogeny $f:E\to E'$ is given by mapping the point $(x,y)=[x:y:1]$ of $E$ to the point $$ \left(\ \frac{y^2}{x^2}, \ \frac{y(b-x^2)}{x^2}\ \right) \ =\ \left[ y\ :\ (b-x^2)\ :\ \frac{x^2}y\ \right] $$ of $E'$, so $T=(0,0)=[0:0:1]$ goes to $[0:b:0]=\infty'=O'$. ($f$ kills the two torsion group generated by $T$.)
The discriminant of $E$ is $4$ times $(e_1-e_2)^2(e_1-e_3)^2(e_2-e_3)^2$, $e_1,e_2,e_3$ being the roots of $x^3+ax^2+b$, so with $e_3=0$ we get $\Delta =4(e_1e_2)^2(e_1-e_2)^2=4b^2(a^2-4b)=2^2\cdot 5^3$. Computer check:
Moreover:
The "singular" places are thus $$ S = \{\infty,2,5\}\ . $$ We have an exact sequence which asks the elements in the group in the middle $\Bbb Q(S,2)=\{\pm 1,\pm 2,\pm 5,\pm 10\}$ (taken modulo squares) (with the multiplication as operation) whether they are "part of domain or cokernel of $\delta$", $$ 0 \longrightarrow \frac{E'(\Bbb Q)}{fE(\Bbb Q)} \overset\delta\longrightarrow \{\pm 1,\pm 2,\pm 5,\pm 10\} \overset {wc}\longrightarrow WC(E/\Bbb Q) \longrightarrow 0\ . $$ Here, $\delta$ maps a generic point $P'=(X,Y)$ in the quotient of $E'(\Bbb Q)$ by Im$(f)$ to its first component $X$, taken modulo squares, and we also set the neutral element $O'$ to go to $\delta(O')=1$, and for $T'=(0,0)$, we let its image be $\delta(T')=A=(a^2-4b)$ modulo squares.
The map $wc$ maps $d$ (modulo squares) to the class of the curve $C_d$ in the Weil-Chatelet group, where $C_d$ is the following twist: $$ \begin{aligned} C_d\ :\ dw^2 = d^2-2adz^2+(a^2-4b)z^4\ .\text{ In our case:} \\ dw^2 = d^2 +10dz^2+5z^4\ . \end{aligned} $$ The $f$-Selmer group of $E$ over $\Bbb Q$, sitting in the middle of the exact sequence $$ 0 \to E'(\Bbb Q)/fE(\Bbb Q)\to \operatorname{Sel}^{(f)}(E:\Bbb Q) \to Ш(E:\Bbb Q)[f] \to 0 $$ is then isomorphic to $$ \{\ d\in\Bbb Q(S,2)\ :\ C_d\text{ has $\Bbb Q_v$-rational points at each $v\in S$ } \}\ . $$ There is furthermore a map $$ \psi:C_d\to E'\ ,\ (z,w)\to\left( \frac d{z^2},\ \frac{dw}{z^3}\right)\ . $$ It has the property of mapping a rational point $P$ of $C_d$ to a point $\psi(P)\in E'(\Bbb Q)$, which goes via $\delta$ to $d$ modulo squares.
We compute now the given example.
$\boxed{d=5}$: The point $T'=(0,0)\in E'(\Bbb Q)$ is mapped to $a^2-4b=B=5$, so $5$ is in the $f$-Selmer group.
$\boxed{d=2}$: $C_2$ is the curve $2w^2=4+20z^2+5z^4$. Suppose there is a point $(z,w)$ of this curve in $\Bbb Q_2$. The $2$-order of $2w^2$ is odd, so it can not be negative, because on the RHS we have only coefficients of even $1$-order, and even $z$-powers. So the point is defined over $\Bbb Z_2$. Since $2w^2$ is divisible by two, we have $z=2Z$. This implies $w=2W$, and we get $2W^2=1+20Z^2+20Z^4$, contradiction.
$\boxed{d=-1}$: $C_{-1}$ is the curve $-w^2=1-10z^2+5z^4$. By inspection, and by chance for $z=1$ we obtain $w=2$, so the rational point $(1,2)$ is on $C_{-1}$. It is mapped by $\psi$ to $(-1, -2)$, which is a point of $E'(\Bbb Q)$.
This implies that $\operatorname{Sel^{(f)}}(E:\Bbb Q)$, seen inside of $\Bbb Q(S,2)=\langle -1,2,5\rangle$ (modulo squares) is isomorphic to $\langle -1,5\rangle$.
This implies that $E'(\Bbb Q)\to \operatorname{Sel^{(f)}}(E:\Bbb Q)$ is surjective, and that $Ш(E:\Bbb Q)[f]=0$.
The same investigation for $E$ replaced by $E'$, and the homogenous spaces $$ C'_d\ :\ dw^2=d^2-20dz^2+80z^4\ . $$
$\boxed{d=5}$: The point $(0,0)$ is mapped to $A^2-4B=10^2-4\cdot 5=80\equiv 5$ modulo squares, so $5$ is in the Selmer group for the dual.
$\boxed{d<0}$: There are no points locally over $\Bbb R$, since the RHS in the definition of $C'd$ is positive.
$\boxed{d=2}$: $C_2$ is the curve $2w^2=4-40z^2+80z^4$. The $2$-order of $2w^2$ is odd, so it can not be negative, else it should be the order of the RHS, which comes from $80z^4$, and is even. Thus a potential $\Bbb Q_2$-rational point is defined over $\Bbb Z_2$. We have then $w=2W$, but then the RHS is not divizible by $8$ in $\Bbb Z_2$. So $2$ is not in the Selmer group.
We have obtained for $g=\hat f$ the identification $\operatorname{Sel}^{(g)}(E:\Bbb Q)\cong\langle 5\rangle$, and $Ш(E':\Bbb Q)[g]=0$.
So far we have $$ \begin{aligned} \frac{E'(\Bbb Q)}{fE(\Bbb Q)} & \cong \langle 5,-1\rangle\cong(\Bbb Z/2)^2=\Bbb F_2^2 \\ &\cong \Bbb F_2(0,0)\oplus \Bbb F_2(-1,-2) \ ,\ (0,0)\to 5\ ,\ (-1,-2)\to-1\ , \\ \frac{E(\Bbb Q)}{gE'(\Bbb Q)} & \cong \langle 5\rangle\cong(\Bbb Z/2)=\Bbb F_2 \\ &\cong \Bbb F_2(0,0) \ ,\ (0,0)\to 5\ . \end{aligned} $$ We have recovered the same information from the OP, but with the additional data on generators, and rational points on the curves, the non-torsion point $(-1,-2)$ being found by chance.
The sequence $$ 0 \longrightarrow \frac{E'(\mathbb{Q})[g]}{f(E(\mathbb{Q})[2])} \longrightarrow \frac{E'(\mathbb{Q})}{fE(\mathbb{Q})} \overset g\longrightarrow \frac{E(\mathbb{Q})}{2E(\mathbb{Q})} \longrightarrow \frac{E(\mathbb{Q})}{gE'(\mathbb{Q})} \longrightarrow 0 $$ becomes $$ 0 \longrightarrow (\Bbb F_2(0,0)\oplus \Bbb F_2(-1,-2))[g] \longrightarrow \Bbb F_2(0,0)\oplus \Bbb F_2(-1,-2) \overset g\longrightarrow \frac{E(\mathbb{Q})}{2E(\mathbb{Q})} \longrightarrow \Bbb F_2(0,0) \longrightarrow 0 $$ We compute $g=\hat f$ on the torsion point $T'=(0,0)$, (well, $g$ was designed to kill the two-torsion coming from $T'$,) and on the point $Q=(-1,-2)$, getting
So $T'\to 0$, and $Q\to P=(1,1)$. The above sequence is than $$ 0 \longrightarrow \Bbb F_2(0,0) \longrightarrow \Bbb F_2(0,0)\oplus \Bbb F_2(-1,-2) \overset g\longrightarrow \frac{E(\mathbb{Q})}{2E(\mathbb{Q})} \longrightarrow \Bbb F_2(0,0) \longrightarrow 0 $$ and we cancel the $\Bbb F_2(0,0)$ part "on the left side", getting $$ 0 \longrightarrow \Bbb F_2(-1,-2) \overset g\longrightarrow \frac{E(\mathbb{Q})}{2E(\mathbb{Q})} \longrightarrow \Bbb F_2(0,0) \longrightarrow 0 $$ This gives the middle term $\cong \Bbb F_2^2$, generated by $P=g(Q)=(1,1)$ and $T=(0,0)$. The two-torsion of $E(\Bbb Q)$ is $\Bbb Z/2$, generated by $T$. We have so far $ E(\Bbb Q) \cong \Bbb Z\oplus \Bbb Z/2 $. It turns out that ($P$ is not a non trivial odd multiple of a generator, so) $$ E(\Bbb Q)=\Bbb Z P\oplus \Bbb Z T \cong \Bbb Z\oplus \Bbb Z/2\ . $$
In a similar manner we can analyze the "other side".
The sequence $$ 0 \longrightarrow \frac{E(\mathbb{Q})[f]}{g(E'(\mathbb{Q})[2])} \longrightarrow \frac{E(\mathbb{Q})}{gE'(\mathbb{Q})} \overset f\longrightarrow \frac{E'(\mathbb{Q})}{2E'(\mathbb{Q})} \longrightarrow \frac{E'(\mathbb{Q})}{fE(\mathbb{Q})} \longrightarrow 0 $$ becomes $$ 0 \longrightarrow \Bbb F_2(0,0)[f] \longrightarrow \Bbb F_2(0,0) \overset f\longrightarrow \frac{E'(\mathbb{Q})}{2E'(\mathbb{Q})} \overset\cong\longrightarrow \Bbb F_2(0,0)\oplus \Bbb F_2(-1,-2) \longrightarrow 0\ . $$ The isogeny $f$ kills the torsion point $T=(0,0)$, so $\Bbb F_2(0,0)[f] =\Bbb F_2(0,0)$ cancel each other giving immediately the isomorphism $\cong$. We have so far $ E'(\Bbb Q) \cong \Bbb Z\oplus \Bbb Z/2 $. It turns out that $Q=(-1,-2)$ is a generator of the $E'(\Bbb Q)/\text{(Torsion)}\cong\Bbb Z$ part, $$ E'(\Bbb Q)=\Bbb Z Q\oplus \Bbb Z T' \cong \Bbb Z\oplus \Bbb Z/2\ . $$