Is there an explicit construction of the moduli space $M_{1,0}/\mathbb{Q}$ of genus $1$ curves whose set of $R$-points, for a $\mathbb{Q}$-algebra $R$, is the set of isomorphism classes of genus $1$-curves over $R$?
$\begingroup$
$\endgroup$
2
-
6$\begingroup$ This moduli space is not a variety or a scheme, which you can see for instance since the map from $K$-points to $L$-points is not an injection for $L/K$ a field extension. What type of answer do you hope for? $\endgroup$– WojowuCommented Sep 14 at 14:26
-
2$\begingroup$ I am curious if it forms an algebraic stack and may be realized as a quotient of a scheme by a group, and whether apart from finitely many points perhaps with non-trivial automorphisms (where the action isn't free) it is representable by a scheme. $\endgroup$– kindasortaCommented Sep 14 at 20:20
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
2
The functor of isomorphism classes of genus 1 curves over $R$ is not representable by a scheme or algebraic space since it's not an fppf sheaf. Descent fails for two reasons: there exist nontrivial twists, and there exist families of genus 1 curves $X \to S$ over a scheme $S$ such that $X$ is not a scheme but is a scheme étale locally on $S$.
Instead, we can consider the pseudofunctor $\mathcal{M}_1$ which associates to $R$ the groupoid of algebraic spaces smooth and proper over $R$ with fibers curves of genus 1. Then $\mathcal{M}_1$ is an algebraic stack which has a relatively explicit construction.
Let $\mathcal{E} \to \mathcal{M}_{1,1}$ be the universal elliptic curve. Then $\mathcal{M}_1$ is the quotient stack $$ \mathcal{M}_1 \cong [\mathcal{M}_{1,1}/\mathcal{E}] $$ where $\mathcal{E}$ acts trivially. The map $\mathcal{M}_1 \to \mathcal{M}_{1,1}$ sends a genus 1 curve to its Jacobian and the isomorphism is expressing the fact that every genus 1 curve is a torsor for its Jacobian.
$\mathcal{M}_{1,1}$ can also be expressed as the quotient of a scheme by a group using the Weierstrass equation so combining the two, we can write $\mathcal{M}_1$ as the quotient of a scheme by a group. Edit: this last claim is false, see Will Sawin's answer.
answered Sep 14 at 20:32
-
$\begingroup$ I am skeptical of the last sentence. First you seem to be working with the quotient of a scheme over $S$ by a group scheme over $S$, where $S$ is itself a stack $\mathcal M_{1,1}$, which I don't think is what is usually meant by the quotient of schemes by groups. Second it is not true that $(X/G)/H$ can always be expressed as the quotient of $X$ by some extension of $G$ by $H$. It may be in this case but one needs to check. $\endgroup$ Commented Sep 14 at 20:52
-
$\begingroup$ @WillSawin Yeah I think you are right to be skeptical. I originally thought the extension should just be the product in this case but that doesn't seem right. $\endgroup$ Commented Sep 14 at 20:56
$\begingroup$
$\endgroup$
With regards to your questions in the comments
I am curious if it forms an algebraic stack and may be realized as a quotient of a scheme by a group, and whether apart from finitely many points perhaps with non-trivial automorphisms (where the action isn't free) it is representable by a scheme.
It is an algebraic stack.
One doesn't usually talk about the set of $R$-points of a stack - they have a groupoid of $R$-points.
It cannot be a quotient of a scheme by a group, because the stabilizer of each point in the quotient $X/G$ is a subgroup of $G$. The stabilizer of an elliptic curve $E$ in the moduli stack of elliptic curves is the automorphism group of $E$, which includes $E$, so $E$ would have to be a subgroup of $G$ for all $E$. This is impossible since $E$ would have to lie in the kernel of all maps from $G$ to affine group schemes, which is necessarily commutative, and only finitely many $E$ can lie in that.
Every point has nontrivial automorphisms so no open set is representable by a scheme.
