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Let $\mathcal{M}_{1, 1}$ be the stack of elliptic curves. Its coarse moduli space is $\mathbb{A}^1_{\mathbb{Z}}$ with the map $\mathcal{M}_{1, 1} \rightarrow \mathbb{A}^1_{\mathbb{Z}}$ given by the $j$-invariant. I have heard that this map is ramified (i.e. not etale) at points of $\mathbb{A}^1_{\mathbb{Z}}$ at which $j = 0$ or $j = 1728$. How does one prove this?

I know that these are precisely the points where the automorphism group jumps, but I don't see how to use this. I am familiar with the argument that proves that $\mathcal{M}_{1, 1} \rightarrow \mathbb{A}^1_{\mathbb{Z}}$ is etale away from $j = 0$ and $j = 1728$, the essential point being that the automorphism functor is the etale group $\mathbb{Z}/2\mathbb{Z}$ on this locus.

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    $\begingroup$ One way is to notice that at these points, the automorphisms act non-trivially on the tangent space to the stack. This implies that the differential of j is zero at these points, and therefore j is not etale. $\endgroup$ – t3suji Sep 22 '15 at 6:29
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    $\begingroup$ The map $\mathcal{M}_{1,1}\to\mathbb{A}^1$ is not representable. What's (your :) ) definition of "étale" and of "ramified"? mathoverflow.net/questions/224124/… $\endgroup$ – Qfwfq Nov 21 '15 at 17:28
  • $\begingroup$ @Qfwfq: The stack $\mathcal{M}_{1, 1}$ is DM, so etaleness of the map at a point $s \in \mathbb{A}^1$ is well-defined by requiring etaleness of the composed map $X \rightarrow \mathbb{A}^1$ at some (equivalently, any) preimage $x \in X$ of $s$ where $X \rightarrow \mathcal{M}_{1, 1}$ is an etale cover by a scheme. Then, a map is ramified over $s$ if it is not etale over $s$. $\endgroup$ – O-Ren Ishii Dec 24 '15 at 22:58
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I would guess that the following argument should work.

By definition, the map $\mathbb{H} \to \mathcal{M}_{1,1} = [SL_2\mathbb{Z} \setminus \mathbb{H}]$ is unramified. However, the map $\mathbb{H} \to M_{1,1} = SL_2\mathbb{Z} \setminus \mathbb{H}$ is ramified precisely at those two points. Consequently, the ramification has to come from the map $\mathcal{M}_{1,1} \to M_{1,1}$.

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    $\begingroup$ This argument would work if the question were phrased in the language of stacks over complex analytic spaces, instead of schemes over the integers. If you replace the upper half-plane with the $SL_2(\mathbb{Z}/3\mathbb{Z})$-cover that comes from full level 3 structure, then I think the argument works more generally (but I certainly wouldn't say "by definition"). $\endgroup$ – S. Carnahan Sep 22 '15 at 6:29
  • $\begingroup$ Oh, silly me, I didn't notice the $\mathbb{Z}$ in the question... $\endgroup$ – Simon Rose Sep 22 '15 at 7:07
  • $\begingroup$ @S.Carnahan forgive my ignorance, but what if 3 is not invertible? $\endgroup$ – David Roberts Sep 22 '15 at 7:54
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    $\begingroup$ @DavidRoberts If 3 is not invertible, then the cover is not étale. In that case, you might as well calculate the $j$-invariant of a first-order deformation of the Artin-Schreier curve $y^2 = x^3 - x$. $\endgroup$ – S. Carnahan Sep 23 '15 at 9:00

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