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Sorry about this question which is not on a research level. But I am very confused about this "first" example of coarse moduli space of genus 1 curves.

The moduli space I talk about here is the coarse moduli space, using the language of schemes and the moduli functor. By definition, this moduli space should be a scheme. For genus 1 curves (smooth and proper), I have the "intuition" that this moduli $\mathcal{M}_{1}$ is the "Affine line $\mathbb{A}^1$".

More precisely, as I read in the definition, the moduli functor is a function $F: Schemes \rightarrow Sets$. For a given scheme $S$, $F(S)$ is the set of all (smooth and proper) $C/S$ that has smooth (geometrical) fibers of genus 1. And the coarse moduli space is a scheme that is "closed" to the (non-exist) fine moduli space.

But, myabe it is a stupid question, what is this scheme $\mathbb{A}^1$ at all? Many notes or books don't explan this, and others even write this moduli space as $\mathbb{A}^1_k$, for a field $k$, which I don't know where this field $k$ comes from. $\mathcal{M}_{1}$ should not be an algebraic variety over a fixed field $k$, if I understand correctly.

I hope that some of you are kind to explain to me this "elementary" question. Otherwise I have a big difficulty to study the moduli space.

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    $\begingroup$ Could you please be more precise? Over what type of base schemes are you considering your families of genus $1$ curves. The "universal" version of the affine line is $\text{Spec}(\mathbb{Z}[x])$. But if you work with varieties over a specified field $k$, you could instead work with $\text{Spec}(k[x])$. $\endgroup$
    – Jason Starr
    Commented Apr 6, 2015 at 12:10
  • $\begingroup$ @JasonStarr : See the 3rd paragraph. In this setting, you mean the scheme we have is $\mathrm{Spec}(\mathbb{Z}[x])$? Could you specify the setting for a specified field $k$? What is the moduli functor $F$ this time? Like $F : Schemes/k \rightarrow Sets$, for a $k$-scheme $X$, $F(X)$ is the set of all smooth and proper $C/X$ such that each fiber is of genus 1? $\endgroup$
    – user565739
    Commented Apr 6, 2015 at 12:35

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So here's the issue: $\mathcal{M}_1$ is not an algebraic variety. It's only a stack. But you're looking at the coarse moduli space, which is roughly the algebraic space that most closely approximates the stack, and in this case, it's actually a smooth affine variety.

Here's some context and details: let $k$ be a field not of characteristic 2 or 3 (there's a more complex proof in those cases, but I'm going for simple.) Then, you can write every genus 1 curve as $y^2=x(x-1)(x-\lambda)$ for some $\lambda\neq 0,1$, which gives you what's often called the $\lambda$-line $\mathbb{A}^1_k\setminus\{0,1\}$ which parameterizes a family containing every genus 1 curve up to isomorphism.

So then you want to check which curves are isomorphic, and this problem is solved by the $j$-invariant, which gives a map $\mathbb{A}^1_k\setminus \{0,1\}\to \mathbb{A}^1_k$. This latter line, the $j$-line, is the coarse moduli space, and it is not too hard to show that, from a definition of coarse moduli space of a functor (your functor is flat families of smooth genus 1 curves).

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  • $\begingroup$ I have a feeling that this don't answer my question. Maybe it is because I didn't ask my question precisely. I have made some details, if you want to look it again. $\endgroup$
    – user565739
    Commented Apr 6, 2015 at 12:24

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