Let $\mathcal X$ be a smooth proper finite type Deligne-Mumford stack over $\mathbb C$ that is generically a scheme. Let $X$ be its coarse moduli space.

If $\mathcal X$ can be defined over $\overline{\mathbb Q}$, then $X$ can be defined over $\overline{\mathbb Q}$. This is because "the base-change of the coarse moduli space is the coarse moduli space of the base-change".

Does the converse hold?

What if we drop the properness condition?

My motivation is mere curiosity.


2 Answers 2


No. Even if $X$ is "defined over $\overline{\mathbb{Q}}$" in the sense that $X$ is isomorphic to $X_0\otimes_{\overline{\mathbb{Q}}} \mathbb{C}$ for some variety $X_0$ over $\overline{\mathbb{Q}}$, nonetheless the stack $\mathcal{X}$ may not be defined over $\overline{\mathbb{Q}}$.

For instance, let $X_0$ be $\mathbb{P}^1_{\overline{\mathbb{Q}}}$ so that $X$ is $\mathbb{P}^1_{\mathbb{C}}$. Now let $t_0,t_1,t_2\in \mathbb{P}^1_{\mathbb{C}}(\mathbb{C})\setminus\{0,1,\infty\}$ be three distinct points such that at least one is transcendental, e.g., $2$, $3$, $\pi$. Denote by $$f:Y\to X,$$ the genus $2$, hyperelliptic curve branched over $$\{0,1,\infty,t_0,t_1,t_2\}.$$ There is an action of $\mathbb{Z}/2\mathbb{Z}$ on $Y$ via the hyperelliptic involution. Let $\mathcal{X}$ be the quotient Deligne-Mumford stack, $$\mathcal{X} = [Y /(\mathbb{Z}/2\mathbb{Z})].$$ The coarse moduli space is $X$, via the unique morphism $\pi: \mathcal{X}\to X$ that factors $f$. Of course $X$ is defined over $\overline{\mathbb{Q}}$. However, the branch divisor of the morphism $\pi$ is not defined over $\overline{\mathbb{Q}}$ as a subscheme of $X$. Thus, also $\mathcal{X}$ is not defined over $\overline{\mathbb{Q}}$.

Added by OP:

It might be useful to (slightly) generalize the above construction.

Let $X=\mathbb P^1_{\mathbb C}$ and let $f:Y\to X$ be a hyperelliptic curve with branch locus $D$ (and $Y$ of positive genus). Write $U=X\backslash D$. Let $\mathcal X$ be the stack $[Y/G]$. Then the unique morphism $\pi:\mathcal X \to X$ that factors $f:Y\to X$ is the coarse moduli space of $\mathcal X$. Suppose that $U$ can't be defined over Qbar. Then $Y\to X$ (and $Y$) can't be defined over $\overline{\mathbb Q}$ (as the hyperelliptic involution is unique). Therefore, $\mathcal X\to X$ and $\mathcal X$ can't be defined over Qbar.

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    $\begingroup$ It seems to me that hyperelliptic is overkill and you could do exactly the same thing in the elliptic case, ramified at $0,1,\infty,\pi$. $\endgroup$ May 12, 2014 at 0:07
  • $\begingroup$ @Ari: What you added contains a mistake. "Bonus: we can find $\mathcal{X}$ such that $Y$ and $X$ are defined over $\mathbb{Q}$, but $\mathcal{X}$ isn't. (Just take $Y = X = \mathbb{P}^1$)". That is impossible. This is the reason that I chose the example I chose, where the morphism $\pi$ is "characteristic". $\endgroup$ May 12, 2014 at 12:58
  • $\begingroup$ I suppose the characteristic aspect of the hyperelliptic involution is the answer to my question: while it marginally complicates the construction, it simplifies the proof. $\endgroup$ May 12, 2014 at 21:51
  • $\begingroup$ A cheaper example would be to take (a non-trivial) root stack of an effective divisor on $\mathbb{P}^1$ corresponding to a transcendental point. This stack would not be defined over $\bar{\mathbb{Q}}$ since its non-stacky locus isn't. $\endgroup$ Aug 14, 2014 at 8:23
  • $\begingroup$ @DanielBergh. "A cheaper example ..." Since $\text{Aut}(\mathbb{P}^1)$ acts triply transitively, you would need to use at least 4 points. $\endgroup$ Aug 14, 2014 at 9:48

Let $\mathcal{X}$ be an algebraic stack over $k$. A coarse moduli space for $\mathcal{X}$ over $k$ is a morphism $\pi:\mathcal{X}\rightarrow X$, where $X$ is a scheme such that:

  • the morphism $\pi$ is universal for morphisms to schemes;
  • $\pi$ induces a bijection between $|\mathcal{X}|$ and the closed points of $X$, where $|\mathcal{X}|$ denotes the set of isomorphism classes in $\mathcal{X}$.

If $\mathcal{X}$ admits a coarse moduli space $\pi:\mathcal{X}\rightarrow X$ then this is unique up to unique isomorphism. A separated algebraic stack has a coarse moduli space which is a separated algebraic space.

Let $\mathcal{X}$ be a separated stack admitting a scheme $X$ as coarse moduli space $\pi:\mathcal{X}\rightarrow X$. The map $\pi$ is universal for morphisms in schemes, that is for any morphism $f:\mathcal{X}\rightarrow Y$, with $Y$ scheme, there exists a unique morphisms of schemes $g:X\rightarrow Y$ such that $f = g\circ\pi$.

Now, if $\mathcal{X}$ is defined over $k$ we have a morphism $f:\mathcal{X}\rightarrow Spec(k)$. By universality there is a morphism of schemes $g:X\rightarrow Spec(k)$ such that $g\circ\pi = f$. In particular $X$ is defined over $k$.

Conversely, if $X$ is defined over $k$ then there is a morphism $g:X\rightarrow Spec(k)$. By composition we get a morphism $f:=g\circ\pi:\mathcal{X}\rightarrow X$. Therefore $\chi$ is defined over $k$.

For instance both the stack $\overline{\mathcal{M}}_{g,n}$ and its coarse moduli space $\overline{M}_{g,n}$ parametrizing Deligne-Mumford stable curves are defined over $\mathbb{Z}$. In particular over any commutative ring and hence over any field.

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    $\begingroup$ I think I'm missing the argument. Assume that the coarse moduli space $X$ has a model $X_0$ over $\overline{\mathbb {Q}}$. Where exactly do you show that $\mathcal X$ has a model over $\overline{\mathbb {Q}}$? $\endgroup$ May 10, 2014 at 23:08
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    $\begingroup$ @CamSar your definition of "$X$ is defined over $\overline{\mathbb Q}$" is not correct. $\endgroup$ May 10, 2014 at 23:48
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    $\begingroup$ Indeed, every $X$ over $\mathbb C$ has a morphism to $\operatorname{Spec}\overline{\mathbb Q}$ since there is a morphism $\operatorname{Spec}\mathbb C\to\operatorname{Spec}\overline{\mathbb Q}$. $\endgroup$ May 10, 2014 at 23:49
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    $\begingroup$ If $X$ is a scheme over $\mathbb C$, then "$X$ is defined over $\overline{\mathbb Q}$" means there exists a scheme $Y$ over $\overline{\mathbb Q}$ and an isomorphism $X\cong Y\times_{\overline{\mathbb Q}}\mathbb C$ (and similarly for stacks, etc.). $\endgroup$ May 10, 2014 at 23:51

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