Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 7 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.

share|cite|improve this answer
Okay, thank you very much :) – Jan Nov 8 '10 at 21:28

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.