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Given two complex varieties over a common base, I can take their fiber product in the category of varieties, or I can take their fiber product in the category of schemes and then take the reduced subscheme. I have heard, that these two operations yield the same. Has someone a reference?

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up vote 6 down vote accepted

Here is a proof of this fact:

Let $\mathcal C$ be the category of (not necessarily irreducible) complex varieties. Then $\mathcal C$ can be identified with the category of reduced finite type $\mathbb C$-schemes.

Let $\mathcal D$ be the category of all finite type $\mathbb C$-schemes.

Then obviously $\mathcal C$ is a full subcategory of $\mathcal D$, and the inclusion $\mathcal C \subset \mathcal D$ has a right adjoint, namely passage to the underlying reduced subscheme. General nonsense (i.e. an easy categorical argument) then shows that if $X\to S$ and $Y \to S$ in $\mathcal C$ are two morphisms, the fibre product in $\mathcal C$ can be computed by first computing the fibre product in the bigger category $\mathcal D$, and then applying the right adjoint to the inclusion. That is, the fibre product in the category of varieties over $\mathcal C$ is equal to the reduced subscheme of the fibre product in $\mathcal D$ (which coincides with the fibre product in the category of all schemes, just because the fibre product of morphisms of finite type $\mathbb C$-schemes is again finite type over $\mathbb C$).

I'm not sure of a reference. Because the proof is easy when you have the right categorical framework, it is the kind of thing that is well-known to experts but whose proof is not necessarily written down explicitly.

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Okay, thank you very much :) –  Jan Nov 8 '10 at 21:28
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