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Let $X,Y_1, Y_2$ be noetherian schemes over $\mathbb{C}$ and $Y_1,Y_2$ be integral schemes. Let $f: X \to Y_1 \times_{\mathbb{C}}Y_2$ be a morphism and $X_0$ be its generic fibre (i.e. fibre over the generic point of $Y_1 \times Y_2$).

On the other hand, we have morphism $g : X \to Y_1 \times Y_2 \to Y_1$, and let $X'$ be the generic fibre of $g$. Finally, we have morphism $h: X' \to X \to Y_1 \times Y_2 \to Y_2$ and the generic fibre $X''$ of $h$. Is it true that $X_0 \cong X''$?

On the level of rings, this is to show the isomorphism between $(A\otimes_B K(B)) \otimes_C K(C)$ and $A \otimes_{B\otimes C} K(B \otimes C)$, where $K(-)$ is the quotient field of rings.

I feel the above statement is true, but unable to give a proof.

Li Yutong
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    Consider the example $A=B\otimes_{\mathbb C} C$, and $B=C=\mathbb C[t]$. – Cantlog Mar 26 '14 at 17:14
  • It's not even true that a tensor product of domains is a domain! – Leopoldo Mar 26 '14 at 17:15
  • We are in a very special case here, though. The tensor product of two integral domains over an algebraically closed field is an integral domain: see here. – Zhen Lin Mar 26 '14 at 17:28
  • @Cantlog Than you so much! I see what you mean -- so the statement is false. But I was curious how did you get this counter-example. I can image what is the fibre over closed points but I don't have good intuition about fibre over generic point. – Li Yutong Mar 27 '14 at 01:07
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    @LiYutong: the generic point of $Y_1\times Y_2$ is unique and its "coordinates" are the respective generic points of $Y_1$, $Y_2$. But there are a lot of points in $Y_1\times Y_2$ having these coordinates. They form a space of positive dimension if $Y_1, Y_2$ are of positive dimension. In other words, $X_0$ is the generic fiber, but $X''$ is the fiber over all points of $Y_1\times Y_2$ whose coordinates are generic points of $Y_1$ and $Y_2$. – Cantlog Mar 27 '14 at 15:59

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