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Suppose $p\geq 5$ is a prime, and $C$ a genus-one curve, defined over $\mathbf{Q}$. Is there always an extension $K/\mathbf{Q}_{p}$ whose degree divides a power of $6$, so that $C(K)$ is not empty? (I posted that question before, but it was badly formulated. I hope there is no mistake this time.)

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    $\begingroup$ I don't think so -- the Weil-Chatelet group of an elliptic curve over Q_p has lots of p-power torsion, I think so choosing C to represent a p-torsion class I don't think you can kill this class by restriction to a prime-to-p extension. But maybe I'm confused! $\endgroup$
    – JSE
    Commented Mar 31, 2016 at 21:03
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    $\begingroup$ Let $r>2$ be a prime, and let $p$ be any prime that is congruent to $1$ modulo $r$. Then $\mathbb{F}_p$ contains a primitive $r^\text{th}$ root. Thus, there are symbol algebras of dimension $r^2$ and period=index $r$ over $\mathbb{Q}_p$. For the associated Severi-Brauer variety $P$ of relative dimension $r-1$, via Hensel's lemma, you can find a degree $r$, genus $1$ curve $C$ in $P$ that "deforms" a cyclic polygon of $r$ lines. If $K/\mathbb{Q}_p$ is a field extension such that $C(K)$ is not empty, then $K$ splits the Brauer class. Hence $K$ has degree divisible by $r$. $\endgroup$ Commented Apr 1, 2016 at 9:37

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I am just posting my comment above as an answer. As JSE points out, surely there are more methodical approaches to the cohomology groups $H^1_{ \acute{e}t }(\text{Spec}(\mathbb{Q}_p),E)$. However, already the Brauer group of a field gives many nontrivial elements of these cohomology groups. Let $r$ be an integer. Let $Q$ be a field that contains a primitive $r^\text{th}$ root of unity $\zeta$. Let $a,b\in Q^\times\setminus (Q^\times)^r$ be units. There is a symbol algebra that is the quotient of the free associative $Q$-algebra generated by symbols $u$ and $v$ by the left-right ideal generated by the following relations, $$ u^r = a, \ v^r = b, \ vu = \zeta uv.$$ When $r$ equals $2$, this is the usual quaternion algebra. When $Q$ does not contain a primitive $r^\text{th}$ root of unity, there is a generalization of this notion called a cyclic algebra that serves a similar role (and this should extend the examples below to primes $p$ that are not congruent to $1$ modulo $r$). The symbol algebra $A$ is a $Q$-vector space of dimension $r^2$ with monomial basis $u^mv^n$ for $0\leq m,n < r$.

For the symbol algebra $A$, there is a $Q$-scheme $P$ whose Yoneda functor on the category of affine $Q$-schemes $\text{Spec}(R)$ is naturally isomorphic to the functor of left ideals in $A\otimes_Q R$ that are locally direct summands of $A\otimes_Q R$ of constant rank $r$. For every field extension $K/Q$ such that $A\otimes_Q R \cong \text{End}_R(V)$ for a locally free $R$-module $V$ of rank $r$, then $P\times_{\text{Spec}(Q)}\text{Spec}(R)$ is isomorphic to the projective space $\mathbb{P}_R(V) = \text{Proj} \text{Sym}^\bullet_R V^\vee$ via the natural transformation that associates to every locally direct summand $L\subset V$ of rank $1$ the annihilator of $L$ as a left ideal of $\text{End}_R(V)$. In particular, the Picard group of $P\otimes_Q Q^{\text{sep}}$ is $\mathbb{Z}$ with trivial Galois action. The obstruction to finding an element in $\text{Pic}(P)$ whose base change is a generator of $\mathbb{Z}$ is an element in the Brauer group $\text{Br}(Q)$ that equals (up to a sign) the Brauer class of the algebra $A$.

In particular, assuming $\overline{a}$ has order $r$ in $K^\times/(K^\times)^r$, then $K=Q[\alpha]/\langle \alpha^r - a\rangle$ is a field extension of $Q$ of order $r$ such that $A\otimes_Q K$ is isomorphic to a matrix algebra. Thus, there exists a $Q$-morphism $u:\text{Spec}(K) \to P$ whose base change to $P\otimes_{\text{Spec}(Q)}\text{Spec}(K)\cong \mathbb{P}^{r-1}_K$ consists of $r$ distinct, linearly independent $K$-points that are cyclically permuted by the action of the Galois group $\text{Aut}(K/Q) \cong \mu_r(Q)$. Having chosen a generator $\zeta$ of the group of $r^\text{th}$ roots of unity, there is a unique curve in $P\otimes_{\text{Spec}(Q)}\text{Spec}(K)$ whose image in $\mathbb{P}^{r-1}_K$ is the union of the $r$ lines obtained as the spans of consecutive pairs from among the $r$ cyclically ordered points. This is a projective, geometrically connected, nodal curve of arithmetic genus $1$. Since it is unique, it satisfies the descent condition to equal the base change $C'_Q\otimes_{\text{Spec}(Q)}\text{Spec}(K)$ of a curve $C'_Q \subset P$. Finally, the Hilbert scheme over $K$ parameterizing such curves in $P$ is smooth at this $Q$-point, and the unique irreducible component of the Hilbert scheme parameterizing this $Q$-point has a dense open subset $U$ that parameterizes smooth, projective, geometrically connected curves in $P$ of genus $1$.

In the special case that $Q$ equals $\mathbb{Q}_p$, then $Q$ is what is called a "big" or "large" or "ample" field by various authors such as Florian Pop, János Kollár, etc. The point is, because of Hensel's lemma applied to an integral model of the Hilbert scheme over $\text{Spec}(\mathbb{Z}_p)$, the open set $U$ has $\mathbb{Q}_p$-points. Each such $\mathbb{Q}_p$-point gives a smooth, projective, connected curve $C_Q\subset P$ that is of arithmetic genus $1$. For every field extension $L/\mathbb{Q}_p$ such that $C_Q(L)$ is nonempty, in particular $P(L)$ is nonempty. The field extensions that "split" the symbol algebra $A$ all have degree divisible by the order $r$ of $[A]$ in the Brauer group.

Finally, by local class field theory (or Grothendieck's exposes on the Brauer group), $\text{Br}(\mathbb{Q}_p)[r]$ is a cyclic group of order $r$ generated by the class of a symbol algebra. Thus, there exists a smooth, projective, geometrically connected, genus $1$ curve over $\text{Spec}(\mathbb{Q}_p)$ such that for every finite field extension $K/\mathbb{Q}_p$ with $C(K)$ nonempty, $r$ divides the degree of $K/\mathbb{Q}_p$.

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