# Do Deligne-Mumford curves also have rational functions

If $X$ is a curve over a field of characteristic zero, then $X$ has a rational function, i.e., a finite morphism to the projective line.

Question. Suppose that $X$ is a Deligne-Mumford (or just algebraic) stack of dimension one over a field of characterisic zero. Does there exist a finite morphism to some $\mathbf P^1(n,m)$?

What if $X$ is assumed to be a quotient stack?

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I think you had better assume that $X$ is smooth and that there is a dense open substack of $X$ that is a scheme. A singular one-dimensional DM stack can have non-Abelian stabilizer groups. Also, you could always take a non-Abelian gerbe over a curve. – Jason Starr Sep 10 '13 at 15:13

## 1 Answer

I suspect the following works. Let $\mathcal{X}$ be a proper, smooth, finite type Deligne-Mumford stack over $k$ that is one-dimensional and that has a dense open substack $U$ that is a scheme. Let $u:\mathcal{X}\to X$ be the coarse moduli space. Let $\{p_1,\dots,p_r\}\subset X$ be the complement of $U$. For each point $p_i$ of $D$, let the order of the corresponding stabilizer be $n_i$. Let $n$ be the least common multiple over every integer $n_i$. Define $m_i = n/n_i$.

Consider the divisor $D_0$ on $X$ that is $$m_1\underline{p}_1 + \dots m_r\underline{p}_r + n\underline{q}_1+ \dots + n\underline{q}_s,$$ where $q_1,\dots,q_s$ are points different from $p_1,\dots,p_r$. For $s$ sufficiently large, $D_0$ is linearly equivalent to an effective Cartier divisor $D_\infty$ that is disjoint from $D_0$. These two divisors together define a morphism $f:X\to \mathbb{P}^1$ such that the preimage of $\{0\}$ is $D_0$ and the preimage of $\{\infty\}$ is $D_\infty$.

Now let $v:\mathbb{P}(1,n)\to \mathbb{P}^1$ be the coarse moduli space, where the unique stacky point of $\mathbb{P}(1,n)$ maps to $0$. Since $u^{-1}f^{-1}(\{0\})$ is divisible by $n$ as an effective divisor on $\mathcal{X}$, it seems to me that $f\circ u$ factors through $v$, i.e., there is a unique $1$-morphism $F:\mathcal{X}\to \mathbb{P}(1,n)$ such that $v\circ F$ is $2$-equivalent to $f\circ u$.

Since $f$ is finite, the only thing to check is that $F$ is representable. Of course $F$ is representable away from the stacky points $p_i$. Since $n_i$ divides $n$, also $F$ should be representable at $p_i$.

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Hi, is it 'easy' to write down an example of a morphism $\mathcal X \to X \to \mathbf P^1$ which does not factor through a 1-morphism $F:\mathcal X \to \mathbf P^1(n,m)$ for any $n,m$? – Keesjan Sep 11 '13 at 8:26
@Keesjan: Usually it is best to ask a follow-up question as a separate question on MO. I am not sure I understand your follow-up question: if I take $n=m=1$, then the morphism $v:\mathbb{P}(n,m)\to \mathbb{P}^1$ is an isomorphism. – Jason Starr Sep 11 '13 at 12:20
My apologies for the unusual etiquette. My follow-up question is actually related to your answer. In fact, you construct a finite morphism $\mathcal X\to X \mathbf P^1$ which factors through some $\mathbf P^1(n,m)$. So this is probably very stupid, but why bother finding a factorization? Doesn't the morphism $\mathcal X\to X \mathbf P^1$ already answer the question? Is it not a "rational function"? Hope I'm making sense... – Keesjan Sep 11 '13 at 13:00
@Keesjan: You know what you were asking about better than I do. However, my interpretation of a "finite morphism" of stacks is a morphism that is representable by finite morphisms. If $\mathcal{X}$ has stacky points, then no morphism $\mathcal{X}\to \mathbb{P}^1$ is representable. The construction above produces a representable morphism $\mathcal{X}\to \mathbb{P}(1,n)$. – Jason Starr Sep 11 '13 at 15:37
Thnx for this. My silly question was aiming at the statement that "if $\mathcal X$ has stacky points, then no morphism $\mathcal X\to \mathbf P^1$ is representable". Thank you very much for your answer. :) – Keesjan Sep 11 '13 at 15:53