Let $E,E'$ be a pair of elliptic curves defined over $\mathbb{Z}$. Let $T\_p[E], T\_p[E']$ be their associated ($p$-adic) Tate modules. These are Galois representations for the absolute Galois group of $\mathbb{Q}$, which we denote by $G$, denote them by $\rho,\rho'$, respectively.


Let $\overline{\rho},\overline{\rho'}$ be their associated residual representations and assume that $\overline{\rho}\cong\overline{\rho'}$. Is it true that $H^1(G,T\_p[E]) \cong H^1(G,T\_p[E'])$?


Thanks in advance!



Galois cohomology of Tate modules
Let $S$ be a finite set containing all places where an elliptic curve $E$ has bad reduction as well as $p$ and $\infty$. Write $T$ for $T\_pE$ and $G\_S$ for the Galois group of the maximal extension of $\mathbb{Q}$ unramified outside $S$. Then, for any $k>0$ the exact sequence $0\to T \to^{[p^k]} T \to E[p^k]\to 0$ is exact. Since $T^{G\_S}=0$, we get an isomorphism $H^1(G\_S,T)[p^k] \cong E(\mathbb{Q})[p^k]$ and a short exact sequence
$$ 0\to H^1(G\_S,T)/p \to H^1(G\_S, E[p]) \to H^2(G\_S,T)[p] \to 0.$$
If $p>3$, then $E(\mathbb{Q})[p^{\infty}]\cong H^1(G\_S,T)\_{\mathrm{tors}} = H^1(\mathbb{Q}, T)\_{\mathrm{tors}}$ only depends on the $G$-fixed part of $E[p]$. Therefore the torsion subgroup are isomorphic groups for $E$ and $E'$ in this case. For $p=2$ or $p=3$ or for larger number fields that fails.


The short exact sequence shows that the $\mathbb{Z}\_p$-rank of $H^1(G\_S,T)$ may also depend on more than just the Galois module $E[p]$, since there is no reason that the dimension of $H^2(G\_S,T)[p]$ as an $\mathbb{F}\_p$-vector space does not depend on this alone.


Here is an explicit counter example. Take $p=3$ and the curves $E$ <https://www.lmfdb.org/EllipticCurve/Q/20449g1/> and $E'$ <https://www.lmfdb.org/EllipticCurve/Q/20449d2/> . They have isomorphic $3$-torsion as they are the twist by $D=-143$ of a well known example of such a pair.
Now $E$ has rank $2$ and, if you believe BSD or if you are willing to calculate more than on that page, the $3$-primary part of the Tate-Shafarevich group has order $1$. Instead $E'$ has rank $0$ but its Tate Shafarevich group has order $9$.
Consider the Cassels sequence
$$ 0\to \operatorname{Sel}(T) \to H^1(G\_S,T) \to \bigl(E(\mathbb{Q}\_3)\otimes \mathbb{Q}\_3/\mathbb{Z}\_3\bigr)^{\vee} \to \operatorname{Sel}(E[p^{\infty}])^{\vee}, $$
where ${}^{\vee}$ stands for the Pontryagin dual, the Selmer groups are the projective and the direct limit of the usual $3^k$-Selmer groups.
For the curve $E$, we have $\operatorname{Sel}(T) \cong \mathbb{Z}\_3^2$, $\operatorname{Sel}(E[3^{\infty}])^{\vee}\cong \mathbb{Z}\_3^2$, while the local term is free of rank $1$ with the map to the right being injective since the points of infinite order in $E(\mathbb{Q})$ will not reduce to torsion point locally. We conclude that $H^1(G\_S,T)$ is free of rank $2$ for $E$.


For $E'$ instead, the group $\operatorname{Sel}(T')$ is trivial, while $\operatorname{Sel}(E'[3^{\infty}])^{\vee}$ is isomorphic to $\bigl(\mathbb{Z}/3\mathbb{Z}\bigr)^2$. The local term is still free of rank $1$. Hence, no matter whether the right hand map is injective or not, we have that $H^1(G\_S, T')$ is free of rank $1$ for $E'$.


Unsurprisingly, the $3$-Selmer groups $\operatorname{Sel}(E[3])$ are isomorphic of dimension $2$ over $\mathbb{F}\_p$, just for one of the curves this comes from global points while for the other is it the non-trivial Tate-Shafarevich group.


