1
$\begingroup$

Let $X,Y,S$ are schemes, and $f:X\to S, g:Y \to S$ are morphisms, does fiber product $X\times_S Y$ always exist in the usual sense(for example as defined in Hartshorne )?

Here is an interesting thing. If $f(X) \cap g(Y)=\emptyset$, there is no way to have their product. Say letting $X=S=Spec(k[x]_x), Y=Spec(k)$, and $g$ corresponding to the ring map $f(x)\in k[x] \to f(0)\in k$, geometrically, $g$ is the embedding of a point to $0 \in \mathbb{A}^1=S$, and $f$ is the open immersion $\mathbb{A}^1-0 \to \mathbb{A}^1$, they satisfy above property.

Do we need to add some restriction to make fiber product always exists?

  • 0
    @Soarer Thank you for your answer, I feel more comfortable with products now.2011-09-13

0 Answers 0