There's a natural map $f:\overline{\mathcal{M}}_{1,1}\to \overline{M}_{1,1}\cong \mathbb{P}^1$ from the stack of elliptic curves to the coarse space. Both spaces have $Pic=\mathbb{Z}$ hence $f^*:\mathbb{Z}\to\mathbb{Z}$ is an homomorphism. What homomorphism? My guess is: $x \mapsto 24 x$ since the generator of the stack is the Hodge class, which has degree 1/24. Do you agree?

1$\begingroup$ That seems reasonable. But how do you know that $Pic \mathcal M_{1,1} = \mathbb Z$? You have a proof or reference? $\endgroup$– JoëlCommented Nov 1, 2013 at 17:06

2$\begingroup$ If I remember correctly, Mumford only does it over a field not of characteristic 2 or 3. The general reference is Fulton and Olsson: arxiv.org/abs/0704.2214 $\endgroup$– Lennart MeierCommented Nov 1, 2013 at 23:14

2$\begingroup$ It is easy to see that the 12'th tensor power of the Hodge bundleand no smaller tensor powerdescends to the coarse moduli space: consider the possible automorphism groups of elliptic curves and their actions on the tangent space at the identity. $\endgroup$– nafCommented Nov 2, 2013 at 11:09

2$\begingroup$ The general statement is, I think, the following: Let $f:\mathcal{X} \to X$ be the map from a DMstack to its coarse moduli stack, $L$ a line bundle on $\mathcal{X}$ and $n$ the lowest common multiple of the orders of all automorphism groups. Then $f_*L^{\otimes n}$ is a line bundle and $L^{\otimes n} \to f^*f_*L^{\otimes n}$ an isomorphism. At least, I know a proof in the case that $\mathcal{X}$ is separated and of finite type over a noetherian base scheme. $\endgroup$– Lennart MeierCommented Nov 4, 2013 at 15:17

1$\begingroup$ I really would love to see a proof. Do you have a reference? $\endgroup$– IMeasyCommented Nov 4, 2013 at 20:34
2 Answers
I believe the number is 12.
I will assume the characteristic of the base field is not 2 or 3 so that I can use $\overline{\mathcal{M}}_{1,1} \simeq \mathbf{P}(4,6)$. Recall that $\mathbf{P}(4,6)$ is constructed by dividing $V = \mathbf{A}^2 \smallsetminus \{ (0,0) \}$ by the weight $(4,6)$action of $\mathbf{G}_m$.
The line bundle $\mathcal{O}(1)$ (which coincides with the Hodge bundle and generates the Picard group) can be constructed as the equivariant line bundle $V \times \mathbf{A}^1$ on $V$ with $\mathbf{G}_m$ acting by weights $(4,6,1)$. Then $\mathcal{O}(4)$ has a canonical section $g_4$ and $\mathcal{O}(6)$ has the section $g_6$. In $\mathcal{O}(12)$ we have the two sections
$1728 g_4^3$
$\Delta = g_4^3  27 g_6^2$.
This linear series defines $j : \mathbf{P}(4,6) \rightarrow \mathbf{P}^1$.
Note that $j$ has degree $1/2$ because of the generic automorphism on $\mathbf{P}(4,6)$. That is, $j_\ast j^\ast$ is multiplication by $1/2$. Thus the Hodge class $\lambda$ satisfies
$\int_{\overline{\mathcal{M}}_{1,1}} \lambda = \int_{\mathbf{P}(4,6)} c_1(\mathcal{O}(1)) = \frac{1}{12} \int_{\mathbf{P}(4,6)} c_1(j^\ast \mathcal{O}(1)) = \frac{1}{12} \int_{\mathbf{P}^1} j_\ast j^\ast c_1(\mathcal{O}(1)) = \frac{1}{24}$.
Another thing that may be confusing here is that $j$ is generically unramified, so that a local equation for a point in $\mathbf{P}^1$ pulls back under $j$ to a local equation in $\overline{\mathcal{M}}_{1,1}$. Thus $j^\ast \Delta = \delta$ (if $\Delta$ denotes the boundary in the coarse moduli space and $\delta$ the boundary in the stack).

$\begingroup$ One can use specialisation to deduce the result over any field: both the Hodge bundle and $O_{\mathbb{P}^1}(1)$ are defined over $Spec(\mathbb{Z})$ and generate the relevant Picard groups $\endgroup$– nafCommented Nov 8, 2013 at 6:30
Let us choose $\Delta_{irr}$ the point representing the class of a nodal curve as generator of $Pic(\overline{M}_{1,1})\cong\mathbb{Z}$. Let $\delta_{irr}$ be the corresponding boundary divisor in $\overline{\mathcal{M}}_{1,1}$, and let $f:\overline{\mathcal{M}}_{1,1}\rightarrow\overline{M}_{1,1}$ be the canonical morphism between the stack and its coarse moduli space.
Since a nodal curve $[C]\in\Delta_{irr}$ of arithmetic genus $1$ has two automorphisms (identity and elliptic involution) by Proposition $3.92$ of "HarrisMorrison Moduli of curves" we have: $$\delta_{irr}=\frac{1}{Aut(C)}f^{*}\Delta_{irr} = \frac{1}{2}f^{*}\Delta_{irr}.$$ Now, by Theorem 6.9 of Hain's notes:
http://arxiv.org/pdf/0812.1803v2.pdf
the Hodge class $\lambda$ generates $Pic(\overline{\mathcal{M}}_{1,1})$ and furthermore we have $$\mathcal{O}_{\overline{\mathcal{M}}_{1,1}}(\delta_{irr}) = 12\lambda\in Pic(\overline{\mathcal{M}}_{1,1}).$$ Finally $$f^{*}\Delta_{irr} = 2\delta_{irr} = 2\cdot12\lambda = 24\lambda$$ and the homomorphism $$f^{*}:Pic(\overline{M}_{1,1}) = \left\langle\Delta_{irr}\right\rangle\rightarrow Pic(\overline{\mathcal{M}}_{1,1}) = \left\langle\lambda\right\rangle$$ is given by $n\mapsto 24n$ as you predicted.
I guess that the underlying fact is that $\overline{\mathcal{M}}_{1,1}\cong\mathbb{P}(4,6)$. In order to pass to the coarse moduli space $\overline{M}_{1,1}\cong\mathbb{P}^{1}$ you have to take into account the two points of $\mathbb{P}(4,6)$ with stabilizers $\mathbb{Z}_{4}$ and $\mathbb{Z}_{6}$ ($lcm(4,6) = 12$) but also the fact that $\mathbb{P}(4,6)$ is not wellformed meaning that $4,6$ are both divided by $2$. So the general point of $\mathbb{P}(4,6)$ has stabilizer $\mathbb{Z}_{2}$ which corresponds to the elliptic involution of the general elliptic curve.

$\begingroup$ As pointed out by Jonathan Wise, your appeal to Proposition 3.92 of HarrisMorrison is not correct. $\endgroup$– nafCommented Nov 7, 2013 at 11:31