6
$\begingroup$

Any pointed nodal (ie, proper semistable with a specified rational point lying in the smooth locus) curve of arithmetic genus 1 over a field $k$ must be irreducible and has precisely 1 node, which must be rational over the base field, and its normalization must be $\mathbb{P}^1_k$.

Over an algebraically closed field, all such curves are isomorphic to the compactification of the plane curve $y^2 = x^3 + x^2$, say with the marked point at $P = (0,0)$.

Over a general field $k$, is there a classification of pointed nodal curves of arithmetic genus 1 (such curves would essentially be twists of $y^2 = x^3 + x^2$)?

$\endgroup$
  • 2
    $\begingroup$ It is not true that the normalization must be $\mathbb{P}^1_k$. It is true that the normalization is a smooth, genus $0$ curve whose base change to the separable closure of $k$ is isomorphic to the base change of $\mathbb{P}^1_k$. Thus, the set of nodal curves of arithmetic genus $1$ over $k$ maps to the set of "conics" over $k$, i.e., it maps to the $2$-torsion subgroup of the Brauer group of $k$. $\endgroup$ – Jason Starr Jul 18 '17 at 21:47
  • 2
    $\begingroup$ @JasonStarr doesn't the marked point force the normalization to have a rational point, and hence be $\mathbb{P}^1_k$? (Perhaps I should have said the marked point must lie in the smooth locus) $\endgroup$ – stupid_question_bot Jul 18 '17 at 22:24
  • $\begingroup$ You are correct: I did not notice the marked point. If you throw in the marked point, then the remaining invariant is the separable, degree $2$ field extension of $k$. So instead of mapping to the $2$-torsion subgroup of the Brauer group of $k$, you are mapping to $k^\times /(k^\times)^2$ (assuming that the characteristic is not $2$). $\endgroup$ – Jason Starr Jul 18 '17 at 23:22
  • $\begingroup$ When you say nodal' and semistable', do you require that every P^1 has at least 2 special points? The terminology is not uniformly applied; I think semistable' is very ambiguous (cf Liu/de Jong vs the enumerative geometry literature), and nodal' usually puts no restriction on the automorphism group, in which case your curve need not be irreducible. $\endgroup$ – David Holmes Jul 19 '17 at 14:21
3
$\begingroup$

I am just posting my comment as an answer. The closed substack $\Delta$ of $\overline{\mathcal{M}}_{1,1}$ parameterizing pointed, nodal, stable curves of arithmetic genus $1$ is naturally equivalent to the classifying stack $BC_2$, where $C_2$ is the cyclic group of order $2$. Thus, for every scheme $S$ (or algebraic space), the set of equivalence classes of $1$-morphisms from $S$ to $\Delta$ is bijective to the set of isomorphism classes of $C_2$-torsors over $S$, i.e., finite, degree $2$, étale covers of $S$ (possibly disconnected). For a field $k$ of characteristic $\neq 2$, the set of isomorphism classes is bijective to $k^\times/(k^\times)^2$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.