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Suppose k is algebraically closed. X and Y are two normal varieties over k. Is it necessarily still normal?

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up vote 4 down vote accepted

The answer is yes.

In general one can define a normal morphism of schemes f:X -> Y to be a flat morphism such that for every y \in Y the fibre over y is geometrically normal.

Then we have the following theorem on normality and base change (see EGA Ch 2 IV 6.14.1)

Let g: Y' -> Y be a normal morphism of locally noetherian schemes. Then for every normal Y-scheme X the fibre product X x_Y Y' is normal.

Over an algebraically closed field flatness and geometric normality reduce to just being normal so the result follows.

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Thank you! You argument is definitely true. By the way, the definition of normal morphism seems should be flat and every fiber geometrically normal. But it may also be true that if a variety is geometrically normal, the under any base change it remains to be normal, not necessarily by a field extension's base change. –  Taisong Jing Oct 24 '09 at 5:30
    
You are right... I fixed the answer up - in particular no need to mention affines here and the answer is yes. –  Greg Stevenson Oct 24 '09 at 7:42
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Are you asking about products or fiber products? If you're asking about fiber products, the answer is no. For example, you can have two smooth surfaces in A³ whose intersection is a nodal cubic (see the picture on this page of Hartshorne). This intersection is the fiber product of the two surfaces over A³.

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