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Let $Y(3)$ be the fine moduli space (say, over $\mathbb{C}$) representing elliptic curves equipped with a full level 3 structure. Abstractly, there are 24 such structures for any elliptic curve, but thanks to every elliptic curve having $[-1]$ as an automorphism, there are generically only 12 equivalence classes. Thus, the natural map $Y(3)\rightarrow\text{Spec }\mathbb{C}[j]$ to the $j$-line is generically 12-to-1.

On the other hand, the "forget structure" map $Y(3)\rightarrow\mathcal{M}_{1,1}$ is finite etale and has geometric fibers of size 24, so I would think of it as a 24-to-1 map.

And yet, I can only imagine the composition $Y(3)\rightarrow \mathcal{M}_{1,1}\rightarrow \text{Spec }\mathbb{C}[j]$ to be the same as the 12-to-1 map described above.

Please, so that my christmas isn't ruined trying to figure out if I've been operating on a huge misunderstanding for the last year, can someone tell me that this apparent contradiction is because the morphism from $\mathcal{M}_{1,1}$ to the $j$-line isn't representable and hence it doesn't make much sense to talk about degree? (I've unravelled the definition and I think I've convinced myself that the map isn't representable, but I would still like to confirm that this is the cause of the apparent contradiction)

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    $\begingroup$ The map $j:M_{1,1} \rightarrow \mathbf{A}^1$ is indeed proper and etale but not representable, just like the map to the coarse moduli space from any separated DM stack (of finite type over a noetherian ring) when the stack isn't an algebraic space. (The map $Y(n) \rightarrow M_{1,1}$ is the stack quotient by the non-free action of ${\rm{GL}}_2(\mathbf{Z}/n\mathbf{Z})$ whereas $Y(n) \rightarrow \mathbf{A}^1$ is the scheme quotient by that same non-free action.) Ho-ho-ho, merry Christmas. $\endgroup$
    – user74230
    Commented Dec 25, 2014 at 6:11

2 Answers 2

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If the map from $M_{1, 1}$ to the $j$-line can be said to have a degree, that degree should be $\frac{1}{2}$, which makes everything work out. The reason is that its fibers are generically not a finite set but a finite groupoid, namely $\text{pt}/\mathbb{Z}_2$ (corresponding to the $-1$ automorphism), which has groupoid cardinality $\frac{1}{2}$.

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  • $\begingroup$ Presumably one can make this precise by defining a stacky fundamental class living in rational cohomology, but I don't know the details. $\endgroup$ Commented Jan 3, 2015 at 7:49
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This is to record User74230's answer to your question.

The map $j:M_{1,1}→\mathbb A^1$ is indeed proper and etale but not representable, just like the map to the coarse moduli space from any separated DM stack (of finite type over a noetherian ring) when the stack isn't an algebraic space. (The map $Y(n) \to M_{1,1}$ is the stack quotient by the non-free action of $GL_2(\mathbb Z/n\mathbb Z)$ whereas $Y(n)\to\mathbb A^1$ is the scheme quotient by that same non-free action.) Ho-ho-ho, merry Christmas.

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  • $\begingroup$ I don't think that $j$ is etale. $\endgroup$
    – Tom Graber
    Commented Dec 28, 2014 at 9:45
  • $\begingroup$ @TomGraber Thank you for your comment. I think User74230 meant to write "quasi-finite" instead of "etale". This makes me wonder: aren't all etale morphisms representable? $\endgroup$ Commented Dec 29, 2014 at 9:52

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