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I've seen, on several occasions, papers whose purpose it is to construct a moduli space over $\mathbb{Z}$ for a moduli problem for which a moduli space over $\mathbb{C}$ was already constructed. Let's give things names:

Let $X$ be a (coarse should be enough) moduli space over $\mathbb{Z}$ for some moduli problem. Let $Y_{\mathbb{C}}$ be the moduli space over $\mathbb{C}$ for the same moduli problem. In these papers it seems that they often say that $X\times Spec(\mathbb{C})\cong Y_{\mathbb{C}}$ (that the moduli space they constructed, when viewed over $\mathbb{C}$, is just the previously constructed moduli space over $\mathbb{C}$).

$X(\mathbb{C})$ should be in correspondence with the $\mathbb{C}$-objects of our moduli problem. $Y_{\mathbb{C}}(\mathbb{C})$ should also be in correspondence with the $\mathbb{C}$-objects of our moduli problem. However, $Y_{\mathbb{C}}(\mathbb{C})$ seems to mean sections of the structure morphism $Y_{\mathbb{C}}\rightarrow Spec(\mathbb{C})$ rather than any morphism $Spec(\mathbb{C})\rightarrow Y_{\mathbb{C}}$. This seems incongruous with the fact that it should be in correspondence with $X(\mathbb{C})$, since those can be quite weird indeed! The image of the morphism $Spec(\mathbb{C})\rightarrow X$ can be the generic point of any dimensional subscheme, whereas the image of any map in $Y_{\mathbb{C}}(\mathbb{C})$ must correspond to a maximal point of $Y_{\mathbb{C}}$.

Hopefully, you braved through the confusing notation. I feel like I'm missing something. Can any of you explain this weird phenomenon?

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Related: mathoverflow.net/questions/18747/… Andrea Ferretti and others gave very interesting answers. –  Qfwfq Jul 19 '11 at 21:19
    
Your basic confusion seems to be between Zariski closure over $\mathbb{C}$ and Zariski closure over a sub-field. A point in $Y_{\mathbb{C}}(\mathbb{C})$, if not defined over $\mathbb{Q}$, will have a strictly bigger $\mathbb{Q}$-rational Zariski closure when viewed as a point of $X$. –  Keerthi Madapusi Pera Jul 19 '11 at 23:54

1 Answer 1

up vote 4 down vote accepted

In general, maps over $W$ from $W$ to $X \times W$ are the same as maps from $W$ to $X$, by the universal property of products. Here, $W = \operatorname{Spec} \mathbb{C}$. I think one possible reason for the apparent incongruity is that lots of complex points of the base-changed space end up more generic after dropping to the original space, because they aren't defined over $\mathbb{Q}$.

Let's consider an explicit example of what happens when we compose a map from $W$ to $W \times X$ along the projection to $X$, when $X = \mathbb{A}^1$. In the base change, we have ideals in $\mathbb{C}[x]$ like $(x-\pi)$ that aren't defined over the integers. These ideals lie in the preimage of $(0) \subset \mathbb{Z}[x]$ under the base change map, since the intersection with the subring is zero.

In higher dimensions, we can have $\mathbb{C}$-points with transcendental coordinates that still satisfy algebraic equations, hence lie on subvarieties, and these will map to the respective generic points under projection.

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Right you are... Boy, math knocks me out sometimes... –  Makhalan Duff Jul 19 '11 at 20:10
    
So in fact, for every $Spec(\mathbb{C})\rightarrow Y_{\mathbb{C}}$ (even not sections of the structure morphism, even ones whose image is the generic point of some non-trivial complex subvariety!) induces a section of the structure morphism $Y_{\mathbb{C}}\rightarrow Spec(\mathbb{C})$ (because it is also a map $Spec(\mathbb{C})\rightarrow X$) and in particular a maximal point of $Y_{\mathbb{C}}$. How mysterious! –  Makhalan Duff Jul 19 '11 at 20:20

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