# Moduli of curves in characteristic zero

Let $K$ be a field of characteristic zero, and let $\overline{K}$ be its algebraic closure. Let $\overline{M}_{g,n}(K)$ and $\overline{M}_{g,n}(\overline{K})$ be the coarse moduli spaces parametrizing Deligne-Mumford stable curves over $K$ and $\overline{K}$ respectively.

Is it true that $\overline{M}_{g,n}(K)\times_{Spec(K)}Spec(\overline{K})\cong \overline{M}_{g,n}(\overline{K})$ ?

My question is related to "What is $M_g$ over a finite field, really?" regarding a similar issue in positive characteristic.

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You are using non-standard notations. The notation $\overline{M}_{g,n}(K)$ usually denotes the set of $K$-valued points of the scheme $\overline{M}_{g,n}$, not a (finite type) $K$-scheme. You could ask, is the natural morphism of $\overline{K}$-schemes, $$\overline{M}_{g,n,\overline{K}} \to \overline{M}_{g,n,K}\times_{\text{Spec}(K)} \text{Spec}(\overline{K})$$ an isomorphism? The answer to this is yes, because the formation of the coarse moduli space is compatible with flat base change. I recommend reading Keel-Mori for a detailed discussion of this property. – Jason Starr Jul 11 '14 at 15:06

Let $S$ be a scheme and let $\mathcal{X}$ be an Artin stack that is locally of finite presentation over $S$ and has ﬁnite inertia stack $I_S(\mathcal{X})$. There exists a coarse moduli space $\pi:\mathcal{X}\rightarrow X$, such that:
• The structure map $X\rightarrow S$ is separated if $\mathcal{X}\rightarrow S$ is separated, and it is locally of finite type if $S$ is locally noetherian.
• The map $\pi$ is proper and quasi-ﬁnite.
Moreover, if $X^{'}\rightarrow X$ is a ﬂat map of algebraic spaces then $\pi:\mathcal{X}^{'}:=\mathcal{X}\times_{X}X^{'}\rightarrow X^{'}$ is a coarse moduli space.
In you case the map $Spec(\overline{K})\rightarrow Spec(K)$ is flat. Therefore $\overline{M}_{g,n}^{K}\times_{Spec(K)}Spec(\overline{K})$ is a coarse moduli space for the stack $\overline{\mathcal{M}}_{g,n}^{\overline{K}}$. Since the coarse moduli space of $\overline{\mathcal{M}}_{g,n}^{\overline{K}}$ is unique up to a unique isomorphism you get $\overline{M}_{g,n}^{K}\times_{Spec(K)}Spec(\overline{K})\cong \overline{M}_{g,n}^{\overline{K}}$.