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Jason Starr
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There are already two great answers, but I want to post an answer that works over all local fields, such as $\mathbb{Q}_p$, based on an alternative philosophy. Instead of starting with an elliptic curve $(E,0)$ and studying torsors $X$ for that curve that have large index, first we start with a "simpler" ambient scheme $Y$ that manifestly has large index, and then we try to find a genus $1$ curve $X$ in that scheme.

Open Problem. For every Severi-Brauer variety $Y$ over a field $K$, i.e., for every $K$-scheme $Y$ such that $Y\times_{\text{Spec}\ K}\text{Spec}\ \overline{K}$ is $\overline{K}$-isomorphic to $\mathbb{P}^{n-1}_{\overline{K}},$ does there exist a genus $1$ $K$-curve $X$ and a $K$-morphism $X\to Y?$

This problem was explicitly stated as part of an open problem session for the conference, "Ramifications in Algebra and Geometry", cf. Problems 2 and 3 of the following: http://www.mathcs.emory.edu/RAGE/RAGE-open-problems.pdf

There are positive results for general Severi-Brauer varieties and general fields for small values of the integer $n$. One such result was reported by David Saltman at a seminar in Fall 2016 at Stony Brook University:

https://www.math.stonybrook.edu/deptcalendar/event.php?ID=3964&Date=2016-11-02

For some reason, the link to the Stony Brook University calendar is demanding a password(!), so here is a link to a seminar announcement for a similar seminar by David Saltman at NYU.

https://math.nyu.edu/dynamic/calendars/seminars/algebraic-geometry-seminar/846/

Further positive results are in an article of A. J. de Jong and Wei Ho.

MR3091612
de Jong, Aise Johan; Ho, Wei
Genus one curves and Brauer-Severi varieties.
Math. Res. Lett. 19 (2012), no. 6, 1357–1359.
https://arxiv.org/abs/1207.4810

The following proposition is an answer to Problem 2, and is also an answer to Problem 3 for cyclic algebras over a "big" field.

Proposition. For every Severi-Brauer $K$-variety $Y$ arising from a cyclic $K$-algebra of rank $n^2$, there exists a unique $\text{Aut}(Y)$-orbit of the Hilbert $K$-scheme of $Y$ parameterizing nodal, elliptic normal curves whose (geometric) irreducible components are lines, and this component has a $K$-point parameterizing such a curve $X_0.$ If $K$ is a "big" field in the sense of Florian Pop, e.g., the fraction field of a Henselian DVR, then there are also smooth elliptic normal curves $X$ in $Y$ obtained as deformations of $X_0.$ If the period of the cyclic algebra equals $n$, then every curve in $Y$ has index divisible by $n$.

Corollary. For every local field $K$, e.g., $\mathbb{Q}_p$ or $\mathbb{F}_p((t)),$ for every order-$n$ element in the Brauer group $\text{Br}(K)\cong \mathbb{Q}/\mathbb{Z},$ there exists a cyclic $K$-algebra $A$ of rank $n^2$ representing this class, and there exists a smooth, geometrically connected, genus-$1$ $K$-curve $X$ in $Y$ whose index is divisible by $n$.

Before giving the proof of the proposition, recall the definition of cyclic algebras, as in Roquette's beautiful history of class field theory.

MR2222818 (2006m:11160)
Roquette, Peter
The Brauer-Hasse-Noether theorem in historical perspective.
Schriften der Mathematisch-Naturwissenschaftlichen Klasse der Heidelberger Akademie der Wissenschaften, 15.
Springer-Verlag, Berlin, 2005. vi+92 pp. ISBN: 3-540-23005-X
https://www.mathi.uni-heidelberg.de/~roquette/brhano.pdf

Let $K$ be a field, and let $n$ be an integer $\geq 2.$ Let $L/K$ be a finite field extension of degree $n$ that is Galois with cyclic Galois group $\langle \sigma \rangle.$ Let $a\in K^\times$ be an element. The cyclic algebra over $K$ associated to $L$ and $a$ is the $L$-vector space of dimension $n$, $$A(L/K,\sigma,a) := L\cdot 1 \oplus L\cdot u \oplus \dots \oplus L\cdot u^{n-1}, $$ with the unique $K$-central algebra structure determined by the relations, $$1\cdot x = x = x\cdot 1, \ \ u^n = a\cdot 1, \ \ u\cdot b = \sigma(b) \cdot u,$$ for every $x\in A(L/K,\sigma,a)$ and for every $b\in L.$ This is a central simple $K$-algebra.

The Severi-Brauer variety of $A(L/K,\sigma,a)$ is the smooth, projective $K$-scheme $Y$ that represents the functor on $K$-schemes $T$ associating to $T$ the set of all left ideals in $A(L/K,\sigma,a)\otimes_K \mathcal{O}_T$ that are locally direct summands of (free) rank $n.$ There is an evident inclusion of $L$ in $A(L/K,\sigma,1)$ as $L\cdot 1.$ That inclusion induces a left ideal in $A(L/K,\sigma,a)\otimes_K L$ giving an $L$-point of $Y.$ The Galois orbit of this $L$-point, $\Gamma \subset Y$, is a smooth closed $K$-subscheme of dimension $0$ and length $n.$

In particular, the index of $Y$, i.e., the least positive length of a $0$-dimensional closed $K$-subscheme of $Y$, equals $n.$ So for cyclic algebras of period $n,$ i.e., the order of the corresponding Brauer class in the Brauer group of the field, the index equals the period, irrespective of the field $K.$ The Merkurjev-Suslin Theorem, and the refinements by Merkurjev, imply that for every prime integer $n$, the $n$-torsion subgroup of the Brauer group is generated (as a group) by the classes of cyclic algebras of rank $n^2$ as above.

Even when every Brauer element of order $n$ is a linear combination of classes of cyclic algebras of rank $n^2$, a typical order-$n$ element of this group is not represented by a cyclic algebra. There are many examples of central simple algebras whose index is much larger than the period. (The period divides the index, and they have the same list of prime factors. The exponent is the smallest integer $e$ such that the index divides the $e^{\text{th}}$ power of the period.) It is an open problem to relate the symbol length of a Brauer class (i.e., the fewest number of cyclic $n$-algebras necessary to generate a specified $n$-torsion Brauer class) and the index of the class. Here is one example.

Open Problem. For a function field $K$ of a surface over an algebraically closed field, for which the period always equals the index by de Jong's Period-Index Theorem, is every Brauer class represented by a cyclic algebra?

Proof of the Proposition. For the Severi-Brauer variety $Y$ of a cyclic algebra, there is a unique $\text{Aut}(Y)$-orbit of the parameter space of nodal, elliptic normal curves whose geometric irreducible components are lines: the $n$-gon of lines is uniquely determined by the unordered $n$-tuple of nodes $\Gamma,$ and any two linearly nondegenerate ordered $n$-tuples of points in $\mathbb{P}^{n-1}$ are projectively equivalent (the set of such projective equivalences is naturally a torsor for a torus of rank $n-1$). The claim is that this orbit has a $K$-point parameterizing such a curve $X_0.$

Here is the construction. For the Galois orbit $\Gamma\subset Y$ constructed above, the automorphism $\sigma$ restricts to an automorphism of $\Gamma.$ There is a unique minimal closed subscheme $X_0\subset Y$ that contains $\Gamma,$ whose geometric irreducible components are lines, and such that for every geometric point $p:\text{Spec}\kappa \to \Gamma$, there is a $\kappa$-irreducible component of $X_0$ that contains both $p$ and $\sigma(p).$ Concretely, after base change to $L$, the union of the $n$ lines $\Lambda_r=\text{span}(\sigma^r(p),\sigma^{r-1}(p))$, $r=0,\dots,n-1,$ is Galois-invariant, hence equals the base change of a $K$-curve $X_0\subset Y$.

The $K$-curve $X_0$ is geometrically connected and geometrically reduced. The curve $X_0$ is nodal: the point $\sigma^r(p)$ is contained in two irreducible components $\Lambda_r$ and $\Lambda_{r+1}.$ The arithmetic genus of $X_0$ equals $1$. Geometrically, $X_0$ is an elliptic normal curve in $\mathbb{P}^{n-1},$ i.e., it is linearly nondegenerate and linearly normal (necessarily of degree $n$). In fact, any two such curves in $\mathbb{P}^{n-1}$ are conjugate under the group $\text{Aut}(\mathbb{P}^{n-1})$ of projective linear transformations. Thus, there is a unique $\text{Aut}(X)$-orbit of such curves $X_0$ in $Y.$

Finally, an obstruction group for infinitesimal deformations of a curve $X$ in $Y$ with ideal sheaf $\mathcal{I}$ is $$O_{X,Y}=\text{Ext}^1_{\mathcal{O}_X}(\mathcal{I}/\mathcal{I}^2,\mathcal{O}_X).$$ This is compatible with base change from $K$ to $L$, where $Y_0$ equals a union of $n$ lines. Since the normal bundle of $\Lambda_r$ equals $\mathcal{O}(1)^{\oplus (n-2)}$, and since even after twisting down by $\sigma^r(p)$ and $\sigma^{r-1}(p)$, the twisted sheaf $\mathcal{O}(-1)$ on the line has vanishing $h^1$, it follows that the obstruction group is the zero group, and infinitesimal deformations smooth all nodes. Thus, the Hilbert $K$-scheme parameterizing elliptic normal curves in $Y$ is smooth at the point parameterizing $X_0,$ and the unique irreducible component of the Hilbert scheme containing this point has a dense open subscheme $U$ parameterizing smooth elliptic, normal curves.

If the field $K$ is "ample" or "big" in the sense of Florian Pop, then there are $K$-points of $U$. QED

Proof of the Corollary By Hensel's Lemma, the fraction field of every Henselian DVR is "big". For a local field $K$ such as $\mathbb{Q}_p$ or $\mathbb{F}_p((t)),$ every period-$n$ element in the Brauer group $\text{Br}(K)\cong \mathbb{Q}/\mathbb{Z}$ is represented by a cyclic $K$-algebra of rank $n^2$ by the Brauer-Hasse-Noether-(Albert) Theorem and Hasse's Structure Theorem. By the proposition, there exist $K$-points of $U$ parameterizing smooth elliptic normal curves $X$ in $Y$. QED

Jason Starr
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