# degree five genus one curves without rational points?

Let $$X$$ be a smooth genus one curve over $$k$$. I don't call it elliptic curve because it will have no rational points.

By index of $$X$$ we mean the smallest degree of a closed point on $$X$$; equivalently by Riemann-Roch that's the same as the smallest positive degree of a divisor, or the greatest common divisor of all degrees of closed points.

Of course, the index of a curve equals one if the field is algebraically closed, so we consider non-closed fields here. For example, a smooth cubic curve in $$\mathbb{P^2}$$ can have index one or three.

Here is the question: can one characterize fields $$k$$ which admit genus one curves of index $$d$$?

I am especially interested in the $$d = 5$$ case; then the model of such curve over algebraic closure is a linear section of the Grassmannian $$\mathrm{Gr}(2,5)$$. For which fields do we have genus one curves of index $$5$$? Is there a way to figure this out without writing explicit equations?

UPDATE: I actually don't fully understand the case of number fields. Are all number fields allowed?

• Are you asking for a list of such fields? Every function field of transcendence degree at least one over an uncountable algebraically closed field (every $d>1$ works). Nov 21, 2020 at 18:44
• This article is relevant to understanding the case $d=5$ : Tom Fisher, Minimisation and reduction of 5-coverings of elliptic curves Algebra & Number Theory (2013) dx.doi.org/10.2140/ant.2013.7.1179 Nov 23, 2020 at 11:21
• I think all number fields are allowed, see this article by William Stein: wstein.org/papers/index Nov 23, 2020 at 12:05
• .. and finite fields will never have that property as all genus one curves over a finite field have a point. Nov 23, 2020 at 14:10
• Well if there is curve of index $5$ then certainly there is a degree $5$ extension of $k$, given by the field over which a rational point is obtained. But the converse need not hold: over a finite field there is an extension of degree $5$, but every curve of genus $1$ has a rational point. Nov 24, 2020 at 18:02

I'll address the case $$d = 5$$ over any number field, without recourse to Gross-Zagier formulas and Tate-Shafarevich groups. If $$X$$ has index 5, then it has order 5 in $$H^1(k,E)$$, hence comes from $$H^1(k,E)$$, where $$E$$ is the Jacobian. I assume the converse is not true (presumably some 5-torsion classes come from torsors of index 25), but I'll give examples of cohomology classes which are visibly of index 5.
Over any field there exists at least one elliptic curve with a rational point of order 5, or by duality, a subgroup (scheme) isomorphic to $$\mu_5$$. Over $$\mathbb{Q}$$ there is even an elliptic curve (of conductor 50 I believe) that contains both. In the first case, consider the induced map $$\mathrm{Hom}(G_k,\mathbb{Z}/5\mathbb{Z}) = H^1(k,\mathbb{Z}/5\mathbb{Z}) \stackrel{i}{\to} H^1(k, E)$$ In the dual case, consider $$H^1(k, \mu_5) \simeq k^{\times}/k^{\times 5}$$. The kernel of $$i$$ comes from the $$k$$-rational points in the kernel of the dual isogeny, so it has $$\mathbb{F}_5$$-dimension at most 1, in either case. The image of $$i$$ gives you torsors which have index 1 or 5 (since they are twists of a 5-isogeny). The index is 1 precisely if the cohomology class "comes from rational points", i.e if its in the image of $$E(k)/5E(k)$$ under the coboundary map. If $$k$$ is a number field, then the latter is a finite group, whereas $$H^1(k, \mathbb{Z}/5\mathbb{Z})$$ or $$H^1(k, \mu_5)$$ are both infinite (see the explicit descriptions above). So there are infinitely many index 5 PHS $$X$$.