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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 $y^2 = x^3 + x^2$)?

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    $\begingroup$ 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, 2017 at 21:47
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    $\begingroup$ @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, 2017 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, 2017 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, 2017 at 14:21

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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 $k^\times/(k^\times)^2$.

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