Hi, this is only a partial answer to (3), that was too long to be a comment.

In a recent post of mine, Felipe Voloch pointed out a very useful tameness criterion proved by Gross ("A tameness criterion for galois representations..." Duke J. 61 (1990) on page 514).

In your case, $E$ is good ordinary at $p$ and the criterion for tameness applies. If your sequence $(\ast)$ splits, then $E(\mathbb{Q}_p)$ has a point of order $p$, and therefore $\operatorname{Gal}(\mathbb{Q}_p(E[p])/\mathbb{Q}_p)$ is diagonalizable (and, in fact, of the form $[\ast,0;0,1]$). It follows from Gross' criterion that $j(E)\equiv j_0 \bmod p^2$, where $j_0$ is the $j$-invariant of the canonical lift of the reduction of $E$.

Unfortunately, this is only a necessary condition, as $\operatorname{Gal}(\mathbb{Q}_p(E[p])/\mathbb{Q}_p)$ may be diagonalizable but without the lower right corner being trivial.

Hi, this is only a partial answer to (3), that was too long to be a comment.

In a recent post of mine, Felipe Voloch pointed out a very useful tameness criterion proved by Gross ("A tameness criterion for galois representations..." Duke J. 61 (1990) on page 514).

In your case, $E$ is good ordinary at $p$ and the criterion for tameness applies. If your sequence $(\ast)$ splits, then $E(\mathbb{Q}_p)$ has a point of order $p$, and therefore $\operatorname{Gal}(\mathbb{Q}_p(E[p])/\mathbb{Q}_p)$ is diagonalizable (and, in fact, of the form $[\ast,0;0,1]$). It follows from Gross' criterion that $j(E)\equiv j_0 \bmod p^2$, where $j_0$ is the $j$-invariant of the canonical lift of the reduction of $E$.

Unfortunately, this is only a necessary condition, as $\operatorname{Gal}(\mathbb{Q}_p(E[p])/\mathbb{Q}_p)$ may be diagonalizable but without the lower right corner being trivial.

Hi,

In a recent post of mine, Felipe Voloch pointed out a very useful tameness criterion proved by Gross ("A tameness criterion for galois representations..." Duke J. 61 (1990) on page 514).

In your case, $E$ is good ordinary at $p$ and the criterion for tameness applies. If your sequence $(\ast)$ splits, then $E(\mathbb{Q}_p)$ has a point of order $p$, and therefore $\operatorname{Gal}(\mathbb{Q}_p(E[p])/\mathbb{Q}_p)$ is diagonalizable (and, in fact, of the form $[\ast,0;0,1]$). It follows from Gross' criterion that $j(E)\equiv j_0 \bmod p^2$, where $j_0$ is the $j$-invariant of the canonical lift of the reduction of $E$.

Unfortunately, this is only a necessary condition, as $\operatorname{Gal}(\mathbb{Q}_p(E[p])/\mathbb{Q}_p)$ may be diagonalizable but without the lower right corner being trivial.

Hi, this is only a partial answer to (3), that was too long to be a comment.

In a recent post of mine, Felipe Voloch pointed out a very useful tameness criterion proved by Gross ("A tameness criterion for galois representations..." Duke J. 61 (1990) on page 514).

In your case, $E$ is good ordinary at $p$ and the criterion for tameness applies. If your sequence $(\ast)$ splits, then $E(\mathbb{Q}_p)$ has a point of order $p$, and therefore $\operatorname{Gal}(\mathbb{Q}_p(E[p])/\mathbb{Q}_p)$ is diagonalizable (and, in fact, of the form $[\ast,0;0,1]$). It follows from Gross' criterion that $j(E)\equiv j_0 \bmod p^2$, where $j_0$ is the $j$-invariant of the canonical lift of the reduction of $E$.

Unfortunately, this is only a necessary condition, as $\operatorname{Gal}(\mathbb{Q}_p(E[p])/\mathbb{Q}_p)$ may be diagonalizable but without the lower right corner being trivial.