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Suppose $k$ is an algebraically closed field, and $X$, $Y$ are two normal varieties over $k$. Is the product $X \times Y$ necessarily still normal?

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2 Answers 2

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The answer is yes.

In general one can define a normal morphism of schemes $f:X \rightarrow Y$ 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 $g: Y' \rightarrow Y$ be a normal morphism of locally noetherian schemes. Then for every normal $Y$-scheme $X$ the fibre product $X \times_Y Y'$ is normal.

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

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  • $\begingroup$ 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. $\endgroup$
    – TJCM
    Oct 24, 2009 at 5:30
  • $\begingroup$ You are right... I fixed the answer up - in particular no need to mention affines here and the answer is yes. $\endgroup$ Oct 24, 2009 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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    $\begingroup$ There's another easy and explicit example: take the squaring map $\mathbb{A^1} \rightarrow \mathbb{A}^1$ ($k[x^2] \rightarrow k[x]$ on rings -- which is flat!) and take its fiber product with itself to get $k[x, y]/x^2 - y^2$ which is not normal. In reference to Greg's answer, the squaring map is not geometrically normal at 0. $\endgroup$ Feb 12, 2017 at 22:51

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