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I've heard that the étale fundamental group of the moduli stack of elliptic curves (over $\mathbb{Z}$) is trivial. Is there an easy proof of that? (Note that there are plenty of étale covers once one inverts a prime $p$, given by taking elliptic curves with some form of a level $p^n$ structure.)

More generally, I'd be interested in how it works out for the stack of cubic curves which are allowed either to be smooth or to have a nodal singularity.

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I believe the paper `Horizontal divisors on arithmetic surfaces associated with Belyĭ uniformizations' by Ihara proves that the fundamental group of $\mathbb{P}^1\setminus \{0,1,\infty\}_{\mathbb{Z}}$ is trivial. I'm not sure because I rarely think about fundamental groups of stacks, but the result you want might follow from this. –  Minhyong Kim Aug 19 '12 at 17:36
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Upon further thought, maybe not. –  Minhyong Kim Aug 19 '12 at 17:53
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I dunno, that implication sounds reasonable to me! –  JSE Aug 19 '12 at 20:38
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up vote 22 down vote accepted

Yes, there is a proof which is long but one might consider to be "easy" after digesting it.

Let's first show that the question (of triviality of connected finite etale covers) for the moduli stack $M_1$ is equivalent to its counterpart for the "Deligne-Rapoport" compactification $\overline{M}_1$ (a regular proper DM stack), since the method of the harder direction will be used in our argument for the case of $M_1$. The easier direction is that if the case of ${M}_1$ is known then we can settle the case of $\overline{M}_1$. It suffices to show that if a normal noetherian DM (or Artin) stack $X$ has a dense open substack $U$ with no nontrivial connected finite etale cover then the same holds for $X$. It suffices to show more generally that if $U$ is a dense open substack of $X$ and $X' \rightarrow X$ is a finite etale cover and $s:U \rightarrow X'|_U$ is a section over $U$ then $s$ uniquely extends over $X$. The uniqueness allows us to work over a smooth scheme chart, so we're reduced to the well-known case when $X$ is a scheme (though one can also adapt to the case of stacks the proof in the scheme case via chasing connected components).

The more interesting direction is the converse: the case of $\overline{M}_1$ implies the case of $M_1$. For this we just have to show that for any connected finite etale cover $T$ of $M_1$ the connected finite normalization $\overline{T} \rightarrow \overline{M}_1$ (which is flat since $\overline{M}_1$ is a regular DM stack of dimension 2) is etale around the closed complement $\infty$ of $M_1$ in $\overline{M}_1$. Since $\overline{M}_1$ is regular and $\infty$ is a relative Cartier divisor that is regular with generic characteristic 0, it follows from the relative Abyhankar Lemma (Exp. XIII, SGA1 for schemes, easily adapts to DM stacks by usual etale-localization stuff) that $\overline{T}$ is "relatively tamely ramified" along $\infty$. Thus, the etale-local structure near $\infty$ is given by an $e$th-root extraction of a local generator of the ideal of the connected substack $\infty$ with $e$ a unit along $\infty$. But (as in Saito's argument mentioned by Minhyong) ${\rm{Spec}} \mathbf{Z}$ supports points of every possible prime residue characteristic, so $e$ has no prime factors and hence $e = 1$, so $\overline{T}$ is etale over $\overline{M}_1$ as desired.

Now we directly attack the case of $M_1$ (though one can also solve the case of $\overline{M}_1$ directly in a more illuminating "topological" manner over $\mathbf{C}$ after some preliminaries with the triviality of $\pi_1({\rm{Spec}}(\mathbf{Z}))$ to handle geometric connectivity of connected components, but that gets caught up in "foundational" issues related to analytification of stacks). We shall initially work with the open substack $M$ of elliptic curves with $j \ne 0, 1728$ on fibers; i.e., the stack of elliptic curves whose automorphism scheme is the constant group $\langle \pm 1 \rangle$ (exercise: equivalent to impose this condition on automorphism groups of geometric fibers, since that constant group has no nontrivial automorphisms); one might say this is the moduli stack of elliptic curves with "no extra automorphisms". I claim that $M$ has exactly one nontrivial connected finite etale cover, a scheme cover of degree 2, and we'll use this to bootstrap to get a handle on the entire moduli stack.

Let $Y$ be the open subscheme of $\mathbf{A}^1_{\mathbf{Z}} = {\rm{Spec}}(\mathbf{Z}[j])$ defined by $j(j-1728)$ being a unit. Early in Silverman's first book on elliptic curves you'll find an elliptic curve $E$ over $Y$ with $j$-invariant $j$, and I claim that the corresponding morphism $f:Y \rightarrow M$ is a finite etale $\mathbf{Z}/(2)$-torsor. The $j$-invariant of the universal elliptic curve over $M$ defines a morphism $j:M \rightarrow Y$, so it makes sense to form the elliptic curve $j^{\ast}(E)$ over $M$. One sees that $f$ is precisely the Isom-stack between $j^{\ast}(E)$ and the universal elliptic curve over $M$, and this Isom-stack is a torsor for the automorphism functor of the universal elliptic curve over $M$, which is to say for the constant group $\mathbf{Z}/(2)$ over $M$, because any two elliptic curves with no extra automorphisms are isomorphic etale-locally on the base if they have the same $j$-invariant (exercise in deformation theory, etc.).

Let's grant that $\pi_1(Y) = 1$, and see how to conclude. Then at the end we will prove $\pi_1(Y) = 1$. Consider a connected finite etale cover $q:M' \rightarrow M$ of degree $> 1$. I claim it is isomorphic to $f$. Consider the pullback $Y' \rightarrow Y$ of $q$ along $f:Y \rightarrow M$. Since $Y$ has trivial fundamental group, this pullback splits as a disjoint union of copies of $Y$. Choosing such a component of the pullback defines a morphism $s:Y \rightarrow M'$ over $M$. But $f$ and $q$ are finite etale maps, so $s$ is also a finite etale map. But $M'$ is connected, so the open and closed image of $s$ is full; i.e., $s$ identifies $Y$ as a finite etale cover of $M'$, so we conclude that $M'$ is sandwiched inside the degree-2 finite etale cover $f$. But $q$ has degree $> 1$, so it follows that $q = f$ as desired.

OK, now we can solve the original problem for $M_1$ (conditional on the triviality of $\pi_1(Y)$). The moduli stack $M_1$ is regular and connected with $M$ a dense open substack that we have just seen has exactly one nontrivial connected finite etale cover. Let $M'_1 \rightarrow M_1$ be a connected finite etale cover with degree $> 1$. We seek a contradiction. The restriction over $M$ is a finite etale cover $M' \rightarrow M$ of degree $> 1$, and since $M'$ is open in the connected regular stack $M'_1$ it must also be connected. Thus, $M'$ is $M$-isomorphic to $Y$ (over $M$ via $f$). Hence, the elliptic curve $E$ over $Y$ extends to an elliptic curve $E'_1$ over $M'_1$ (namely, the pullback of the universal elliptic curve over $M_1$!).

The integral structure has done its job, and now to get the contradiction we consider a connected etale scheme neighborhood $(S,s)$ of a point $\xi$ with $j=0$ (or $j=1728$) on the DM stack $M'_1$ considered over $\mathbf{Q}$. We extend $S$ to a smooth connected complete curve $\overline{S}$ (with constant field that might be larger than ${\mathbf{Q}}$, but that won't matter for the ramification considerations we are about to undertake). Clearly $\overline{S}$ is a finite flat cover of the projective $j$-line over $\mathbf{Q}$ and at the point $s$ over $j=0$ (or $j=1728$) it has ramification degree 4 or 6 (I can't remember which is which) because of etaleness over $M_1$ and the fact that $M_1$ over the $j$-line has ramification over $j=0$ (or $j=1728$) equal to 4 or 6 (due to deformation theory considerations). By design, there is an elliptic curve over the open curve $S$ (namely, the pullback of the elliptic curve $E'_1$ over $M'_1$ that extends the elliptic curve $E$ over $Y$) whose discriminant in the function field of $\overline{S}$ is $(j(j-1728))^{-1}$ (well-defined up to 12th powers of nonzero elements, of course). But the "good reduction" at $s \in S$ forces the discriminant of any model over the function field of $S$ to have valuation at $s$ that is a multiple of 12, whereas for $(j(j-1728))^{-1}$ this valuation is $-4$ or $-6$ (since $S$ at $s$ has ramification over the $j$-line equal to 4 or 6). This is a contradiction, so $M_1$ has no nontrivial connected finite etale cover, assuming $\pi_1(Y) = 1$.

Finally, we prove $\pi_1(Y) = 1$. Note that $Y$ is the open complement in $\mathbf{P}^1_{\mathbf{Z}}$ of the union of the sections $\infty$, $j=0$, and $j=n$ with $n = 1728$. We will now work with any nonzero integer $n$. Since $\infty$ is disjoint from the others, by Saito's argument with the relative Abhyankar's Lemma as explained above, we see that any finite etale cover of $Y$ has normalization over that projective $j$-line over $\mathbf{Z}$ that is etale over $\infty$. Hence, it suffices to show that the open complement of $j(j-n)=0$ in $\mathbf{P}^1_{\mathbf{Z}}$ has trivial $\pi_1$. Making the change of coordinates $t = 1/j$ (which moves $j = 0$ out to $\infty$), this open complement is identified with the open complement $U_n$ in the affine $t$-line $\mathbf{A}^1_{\mathbf{Z}}$ of the locus $nt=1$. So it is enough to prove that $\pi_1(U_n) = 1$. Equivalently, we claim that $U_n$ has no nontrivial Galois connected finite etale covers. This will rest on three special facts about $\mathbf{Z}$: the triviality of $\pi_1({\rm{Spec}}(\mathbf{Z}))$, the triviality of ${\rm{Pic}}(\mathbf{Z})$, and the smallness of the group of roots of unity in $\mathbf{Z}$. Beware that $U_n(\mathbf{Z})$ is empty when $n \not\in \mathbf{Z}^{\times}$.

Let $h:V \rightarrow U_n$ be a Galois connected finite etale cover with degree $> 1$, so over $\mathbf{Q}$ we get a nontrivial connected finite etale cover $V'$ of the $\mathbf{Q}$-fiber $U'_n$ of $U_n$. We first claim that $V'$ must be geometrically connected over $\mathbf{Q}$. Since we're in characteristic 0, this amounts to the condition that $V'$ has constant field $\mathbf{Q}$. If we let the number field $K$ be its constant field then by normality of $V$ it follows that $h$ factors through $(U_n)_{O_K}$, with $V \rightarrow (U_n)_{O_K}$ necessarily surjective. This forces $(U_n)_{O_K}$ to be etale over $U_n$ (since $h$ is a finite etale cover), so since $U_n$ is fpqc over ${\rm{Spec}}(\mathbf{Z})$ it follows that ${\rm{Spec}}(O_K)$ is etale over ${\rm{Spec}}(\mathbf{Z})$. This forces $K = \mathbf{Q}$, as desired.

By the coordinate change $x = t/n$ we identify $U'_n$ with ${\rm{GL}}_1$, so $V'$ is a geometrically connected cover of ${\rm{GL}}_1$ over $\mathbf{Q}$. Thus, for $d = {\rm{deg}}(h) > 1$ we see that over an algebraically closed extension $k$ of $\mathbf{Q}$ the map $h_k$ is identified with the endomorphism $x^d$ of ${\rm{GL}}_1$. That is, the map $h'$ induced by $h$ between $\mathbf{Q}$-fibers is a "$\mathbf{Q}$-form" of the $\mu_d$-torsor ${\rm{GL}}_1$ over ${\rm{GL}}_1 = U'_n$. The set of isomorphism classes of such forms is given by $${\rm{H}}^1({\rm{GL}}_1,\mu_d) = (\mathbf{Q}^{\times}/({\mathbf{Q}}^{\times})^d) \times x^{\mathbf{Z}/d\mathbf{Z}}$$ (since ${\rm{GL}}_1$ has trivial Pic and has unit group $\mathbf{Q}^{\times} x^{\mathbf{Z}}$). Explicitly, for $q \in \mathbf{Q}^{\times}$ and $j \in \mathbf{Z}$ the finite etale cover of ${\rm{GL}}_1 = U'_n$ associated to the class of $(q,j \bmod d)$ is given by the covering equation $y^d = q x^j$. As a covering of ${\rm{GL}}_1$ with coordinate $x$, this has geometric covering group $\mu_d(\overline{\mathbf{Q}})$ via scaling on $y$, so by inspection this geometric covering group action is not defined over $\mathbf{Q}$ (i.e., the automorphism group scheme for the covering is not a constant group over $\mathbf{Q}$, or in other words not all of these geometric automorphisms are defined over $\mathbf{Q}$) except when $d = 2$. Ah, but recall that we arranged for $h$ to be a Galois covering, so in our setting with the covering $h'$ the geometric covering group must be defined entirely over $\mathbf{Q}$ (as a constant group). In particular, $h$ must have degree $d = 2$. Also, the geometric connectedness of the covering forces ${\rm{gcd}}(j,d) = 1$.

To summarize, we have proved that $V \rightarrow U_n$ viewed over $\mathbf{Q}$ is given by $y^2 = q(t/n)$ for some $q \in \mathbf{Q}^{\times}$. By changing $y$ by a $\mathbf{Q}^{\times}$-scaling (as we may certainly do), we can change $q$ by any square multiple we wish, so we can arrange that $q/n$ is equal to a squarefree integer $r$. Then $V$ is identified with the normalization of $\mathbf{Z}[t][1/(nt-1)]$ in the $(nt-1)$-localization of $\mathbf{Z}[y,t]/(y^2 - rt)$. Using that $r$ is a squarefree integer, we claim that $\mathbf{Z}[y,t]/(y^2 - rt)$ is normal (in contrast with the situation for $\mathbf{Z}[y]/(y^2 - r)$ when $r$ is odd!). This is clear after inverting 2, so by Serre's homological criterion the only issue is to check the normality at the generic points in characteristic 2, which is to say that the maximal ideal of the local ring at these points is principal. If $r$ is even (so it is twice an odd integer) then $y$ lies in such primes, so $(2,y)$ is the only such prime and $t$ isn't in this prime. Hence, in such cases $y$ is a local generator (as the equation $rt = y^2$ with $t$ a local unit and $r$ twice an odd integer makes $2$ a local unit multiple of $y^2$). If $r$ is odd then $(2)$ is itself prime because the reduction of $y^2 - rt$ modulo 2 is the element $y^2 - t \in \mathbf{F}_2[t,y]$ that is irreducible.

We conclude that
$$V = {\rm{Spec}}(\mathbf{Z}[y,t]/(y^2 - rt))_{nt-1}$$ over $U_n = {\rm{Spec}}(\mathbf{Z}[t])_{nt-1}$. Ah, but this is not etale over $U_n$, since passing to characteristic 2 turns this into a dense open piece of a purely inseparable quadratic cover in characteristic 2. Contradiction, so $\pi_1(U_n) = 1$. QED

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I hadn't gone through things carefully as you did, but this is the kind of issue I was worried about. I seem to recall that the disjointness of the divisors was rather important in Takeshi Saito's argument cited in Ihara's paper. But I can't remember how it goes anyways. –  Minhyong Kim Aug 20 '12 at 2:46
    
Thanks! This is an awesome answer. Even modulo the $j/1728$ thing it was very helpful, and I'm accepting it. –  Akhil Mathew Aug 20 '12 at 14:35
    
I have fixed the earlier incomplete analysis of $\pi_1(Y)$, so now its triviality is completely proved. (The comments above refer to the earlier version with a gap that is now gone.) –  user22479 Aug 21 '12 at 13:45
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This is an awesome answer. –  DamienC Aug 21 '12 at 16:15
    
OK -- wow! Thanks again. There's a lot here, and I haven't digested most of it yet. I hope I'll be able to make an intelligent comment about this after I've grokked Abyankhar's lemma. –  Akhil Mathew Aug 22 '12 at 0:47
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