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I am trying to gain a better understanding of the notion of fibre products of schemes. Two major applications that I've began to study are:

1) Making an $S$-scheme $X$ into an $S'$-scheme via a morphism $S' \rightarrow S$, and

2) Obtaining the notion of a fibre: For a point $p$ of $S$, the canonical morphism $\operatorname{Spec}k(p) \rightarrow S$ induces the scheme $\operatorname{Spec}k(p) \times_S X$ over $\operatorname{Spec}k(p)$ whose underlying topological space coincides with the fibre over $p$.

What is the effect of base change on the fibres? Say, I have $f: X \rightarrow S$, $g: S' \rightarrow S$ and $p \in S, p' \in g^{-1}(p)$. Denote $X' = X \times_S S'$ and let $f': X' \rightarrow S'$ be the base change of $f$. Then there are two fibres

$\begin{align} f^{-1}(p) &= \operatorname{Spec}k(p) \times_S X\\ f'^{-1}(p') &= \operatorname{Spec}k(p') \times_{S'} X' \end{align}$

and there is an induced morphism $\operatorname{Spec}k(p') \times_{S'} X' \rightarrow \operatorname{Spec}k(p) \times_S X$ by the universal property of the fibre product. Will this be an isomorphism in general, or at least in certain cases? I am mostly curious about what happens if $g$ is a nonconstant map of smooth curves over $k$.

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Let $A\to B\to C, D\to C$ be morphisms in any category, then there is a general natural equivalence: $$A\times_B(B\times_C D) \stackrel {\simeq}{\longrightarrow } A\times_C D,$$ which can be easily checked by the universal property of fiber product.

So in the above case we have: $$f'^{-1}(p') = \mathrm{Spec}\,k(p')\times_{S'} (S' \times_S X) \simeq \mathrm{Spec}\,k(p') \times_S X \\\simeq \mathrm{Spec}\,k(p') \times_{\mathrm{Spec}k(p)}(\mathrm{Spec}\,k(p)\times_S X)= \mathrm{Spec}\,k(p')\times_{\mathrm{Spec}k(p)}f^{-1}(p).$$

The fiber $f^{-1}(p)$ can be any given scheme over $k(p)$, (for every $X$ over $k(p)$, the fiber of the morphism $X\to \mathrm{Spec}\,k(p)\to S$ over $p$ is equal to $X$) so your problem reduces to consideration of base changes over the field extension: $k(p')/k(p)$. And this base extension is always isomorphism iff $k(p) \to k(p')$ is isomorphism i.e. $k(p) = k(p')$. (You can see it by assuming $f^{-1}(p) = \mathrm{Spec}\,k(p)$.)

In the case that $S'\to S$ is morphism of smooth (irreducible) curves, $p$ and $p'$ are both either a close point ($k(p), k(p')$ finite over $k$) or both generic ($k(p), k(p')$ f.g with tr.deg =1 over $k$). So you should only check that $[k(p'):k(p)] =1 $ for the isomorphism of fibers in the general case. In the second case this is equivalent to the bi-rationality of $g$ and in the first case this condition is obviously true if $k$ is assumed to be algebraically closed.

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  • $\begingroup$ Thank you so much! That was very in-depth and clear. $\endgroup$ – Paul Apr 14 '15 at 17:53

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