What is LSB and MSB?

By 2, the image lies in a maximal subgroup other than the Borels. Unless you are in the exceptional cases, which cannot occur for all $p$, the image is in the normaliser of a split or a non-split Cartan group. Therefore you are looking for a quadratic extension of $K$ and an abelian extension above it such that the other conditions are verified.

There are only three quadratic extensions in this case and one is unramified.

For $m=2$. the composition of two distinct quadratic extensions of $\mathbb{Q}_p$ with odd $p$ is never totally ramified. So if $p\equiv 3 \pmod{4}$ then we have $e=f=2$.

Yes. Proposition VII.5.5 in Silverman's Arithmetic of elliptic curves.

That curve has complex multiplication by $\mathbb{Z}[i]$ so, $a_p=0$ for all primes $p\equiv 3 \pmod{4}$. That means that the number of affine points is $p$ and the group order is $p+1$.

I say "partial answer" because I hope someone knows the historical details of what was really done.

The truly analogue question in char $p$ is when the coefficient field is finite. But all finite extensions of such a field are again isomorphic to a Laurent power series ring over a (possibly larger) finite field.

1. This can be deduced from the 9-term Cassels-Poitou-Tate exact sequence for $n$-Selmer groups then take limits. Probably done in other places, but Coates-Sujatha's "Galois cohomology of elliptic curves" does it. 2. When the analytic rank is zero... $E(\mathbb{Q})\to E(\mathbb{Q}_p)$ is injective for any $p$ and the profinite completion doesn't do anything to it. The exactness at the local term is the hard part of the global duality statement and I don't think the additional assumption will simplify the proof. Cohomology of number fields, or the original work by Cassels are best for that

Also there is no reason that the map from the profinite completion of $E(\mathbb{Q})$ to the profinite completion of $E(\mathbb{Q}_p)$ is injective. For instance if $E$ has rank $2$, then the profinite completion of $E(\mathbb{Q})$ contains two copies of $\mathbb{Z}_p$, while the local version is a sum os a finite group with a single copy of $\mathbb{Z}_p$.

Without assuming finiteness of the Tate-Shafarevich group, the last non-zero term has to be replaced by the dual of the projective limit of $n$-Selmer groups as $n$ runs over all natural numbers. I don't think you could hope for the exactness at the term before if you have the wrong group on its right.

Also your example doesn't have trace equal to $a_p=-2$.

... and it has a very good answer there.

In " then it is birationally equivalent to an elliptic curve", what is "it" ? The above equation for the fixed $u$ seen as a curve in the variables $D$ and $v$ ?

Obvious remark: The discriminant of $f$ is a square times the discriminant of $K$, so all primes dividing the discriminant of $f$ to an odd power are ramified. Otherwise I recommend to study Chapter 6 in Cohen's A Course in Computational Number Theory.

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