What is the simplest example (or perhaps best reference) for the fact that there are genus $1$ curves (over a field of your choice  or if you wish, over $\mathbb{Q}$, to make it more exciting) with no points of degree less than $n$? Brian Conrad gave a slick answer here: https://math216.wordpress.com/2011/04/22/fourteenthpost/ but I wonder if there is something that can be explained more simply. My expectation is that this isn't possible, because describing genus one curves extrinsically is hard, but there may be a clever trick I am missing.

2$\begingroup$ Just take a genus $1$ curve mapping to a SeveriBrauer scheme of dimension $n1.$ Every zerocycle in the SeveriBrauer scheme has degree divisible by $n.$ $\endgroup$– Jason StarrNov 18, 2017 at 17:30

$\begingroup$ @JasonStarr: What construction are you using to make such a genus1 curve? $\endgroup$– nfdc23Nov 18, 2017 at 17:57

3$\begingroup$ @nfdc23. I am not claiming that every SeveriBrauer scheme has a genus $1$ curve mapping to it (I think that de Jong  Ho and also Saltman have studied this question). However, I can construct curves in this way. For instance, there are constructions of nodal curves of genus $1$ in every SeveriBrauer variety arising from a "cyclic" algebra (of arbitrary period) over every field. If the field is a $p$adic field (or any "ample" / "large" field in the sense of Florian Pop), then those nodal curves deform over that field, in that SeveriBrauer variety to a smooth genus $1$ curve. $\endgroup$– Jason StarrNov 18, 2017 at 18:19

$\begingroup$ Maybe someone can make this work based on the $n=3$example that $X^3+5\,Y^3+25\,Z^3=0$ is a genus $1$ curve without points over $\mathbb{Q}$. Let $p$ be a prime. Try to find a polynomial $f\in\mathbb{Z}_p[X,Y,Z]$ such that $C:X^n+p\,Y^n+p^2\,Z^n+p^3\,f(X,Y,Z)=0$ has genus $1$ by imposing $n(n3)/2$ double points. Maybe this works when $n$ is prime. $\endgroup$– Chris WuthrichNov 19, 2017 at 13:09
3 Answers
How about a universal example?
Let $E$ be an elliptic curve over any field $k_0$ and let $L$ be a degree $n$ line bundle on $E$. Then the actions of $n$torsion points of $E$ by translation preserves $L$, and hence these automrophisms act by projective linear automorphisms of $H^0(E,L)$, giving a map $E[n] \to PGL_n$. Let $k= k_0( PGL_n / E[n])$ and let $E'$ be the generic fiber of $(E \times PGL_n) / E[n]$ over $PGL_n/ E[n]$, a genus one curve.
I claim $E'$ has no $k$points of degree $d<n$. Suppose it has one, then there is a degree $d$ divisor on $E$ defined over $k_0(PGL_n)$ such that acting by an automorphism in $E[n] \subseteq PGL_n$ and translating by the same point in $E[n]$ returns the original point. However because $PGL_n$ is rational, $k_0(PGL_n)$ is a rational function field, so every $k_0$point of $E$ is a $k$point of $E$. So in fact there is a degree $d$ divisor invariant by translation by $n$torsion points, which is impossible unless $d$ is a multiple of $n$.

$\begingroup$ Is there an abuse of notation in referring to "$(E \times {\rm{PGL}}_n)/E[n]$" as an elliptic curve? The group ${\rm{PGL}}_n$ is rather noncommutative... $\endgroup$– nfdc23Nov 18, 2017 at 17:54

$\begingroup$ @nfdc23. Presumably you see this, but probably Will Sawin meant the generic fiber of the morphism $(E\times PGL_n)/E[n] \to (PGL_n/E[n]).$ $\endgroup$ Nov 18, 2017 at 17:57

2$\begingroup$ @nfdc23 Yes, I was abusing notation (and also mistyped something). Jason Starr is of course completely correct as to what I meant. I've now edited it to be less abusive. $\endgroup$ Nov 18, 2017 at 19:25
Let $K = \mathbf{C}((t))$. For each integer $n \geq 1$, let $K_n = K(t^{1/n})$ be the unique degree $n$ extension of $K$. We shall show that there exist genus $1$ curves $C_n$ over $K$ with no rational point over $K_m$ for $m < n$.
(The argument is inspired by Conrad's in the link above for number fields, except we replace all deep number theory input by relatively easy Galois theory over $K$. It still uses the identification of genus 1 curves as a torsors and some Galois cohomology, so it might not count as "simple", depending on your definition.)
Say $E$ is an elliptic curve over $\mathbf{C}$, and let $A = E_K$ be the base change. We shall show that for each $n \geq 1$, there exist $A$torsors $C_n$ over $K$ that are split (i.e., have a rational point) over $K_n$ but not over $K_m$ for $m < n$.
Recall that $A$torsors over $K$ are classified by $H^1(K,A)$ (Galois cohomology). We shall show the following: for each $n \geq 1$, there exist $A$torsors $x_n \in H^1(K,A)$ with order exactly $n$. To see why this formulation implies what we want, one simply notes that the restriction map $res:H^1(K,A) \to H^1(K_m,A)$ and the trace map $tr:H^1(K_m,A) \to H^1(K,A)$ satisfy $tr \circ res = m$. Thus, if an $A$torsor $x \in H^1(K,A)$ is split by restriction to $K_m$, then $m \cdot x = 0$, so $x$ has order dividing $m$. In particular, if $x$ has order exactly $n$, then one must have $n \mid m$.
Now we must construct these elements $x_n \in H^1(K,A)$. For this, we make a simple observation first: for each $n \geq 1$, the inclusion $A[n] \to A$ induces an injective map on $H^1(K,)$; this observation is proven at the end. Granting this, it is enough to find elements of order exactly $n$ in $H^1(K,A[n])$. But now $A[n] \simeq E[n] \simeq (\mathbf{Z}/n)^2$ as Galois modules because all torsion points were already defined over $K$. So the group $H^1(K,A[n])$ identifies with $H^1(K, \mathbf{Z}/n)^2$. As the absolute Galois group of $K$ is just a $\widehat{\mathbf{Z}}$ (with the $\mathbf{Z}/m$quotient corresponding to $K_m$), we can conclude that $H^1(K,A[n]) \simeq \mathrm{Hom}(\widehat{\mathbf{Z}}, (\mathbf{Z}/n)^2) \simeq (\mathbf{Z}/n)^2$, which certainly has elements of order exactly $n$.
It remains to check that $H^1(K,A[n]) \to H^1(K,A)$ is injective. This is equivalent to checking that $A(K)$ is divisible. Set $V = \mathbf{C}[[t]]$ and $\mathcal{A} = E_V$ for the base change. By the valuative criterion, we have $A(K) = \mathcal{A}(V)$, so it is enough to prove divisibility of $\mathcal{A}(V)$. But $\mathcal{A}(V) = \lim_m \mathcal{A}(V/t^m)$, so we can approach this in stages. Now $\mathcal{A}(V/t) \simeq E(\mathbf{C})$ is certainly divisible. The transition map $\mathcal{A}(V/t^m) \to \mathcal{A}(V/t^{m1})$ is surjective (by smoothness), and the kernel is a $\mathbf{C}$vector space (as it is a copy of the tangent space to $E$ at some point) and thus uniquely divisible. One can then easily inductively lift the divisibility of $\mathcal{A}(V/t)$ to $\lim_m \mathcal{A}(V/t^m)$, as wanted.
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 SeveriBrauer 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 $\mathbb{P}^{n1}_{\overline{K}},$ 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/RAGEopenproblems.pdf
There are positive results for general SeveriBrauer 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=20161102
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/algebraicgeometryseminar/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 BrauerSeveri 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 "large" field.
Proposition. For every SeveriBrauer $K$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 "large" 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 BrauerHasseNoether theorem in historical perspective.
Schriften der MathematischNaturwissenschaftlichen Klasse der Heidelberger Akademie der Wissenschaften, 15.
SpringerVerlag, Berlin, 2005. vi+92 pp. ISBN: 354023005X
https://www.mathi.uniheidelberg.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 $n$, $$A(L/K,\sigma,a) := L\cdot 1 \oplus L\cdot u \oplus \dots \oplus L\cdot u^{n1}, $$ 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 SeveriBrauer 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 MerkurjevSuslin 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 PeriodIndex Theorem, is every Brauer class represented by a cyclic algebra?
Proof of the Proposition. For the SeveriBrauer 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 $\mathbb{P}^{n1}$ are projectively equivalent (the set of such projective equivalences is naturally a torsor for a torus of rank $n1$). 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 $\Lambda_r=\text{span}(\sigma^r(p),\sigma^{r1}(p))$, $r=0,\dots,n1,$ is Galoisinvariant, 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 $\mathbb{P}^{n1},$ i.e., it is linearly nondegenerate and linearly normal (necessarily of degree $n$). In fact, any two such curves in $\mathbb{P}^{n1}$ are conjugate under the group $\text{Aut}(\mathbb{P}^{n1})$ 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 $\mathcal{O}(1)^{\oplus (n2)}$, and since even after twisting down by $\sigma^r(p)$ and $\sigma^{r1}(p)$, 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 "large" 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 "large". 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 BrauerHasseNoether(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