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Let $X,Y$ be schemes over a field $k$. I know that $X \times_k Y$ is not equal to the catesian product $X \times Y$. For example $\mathbb{A}_{k}^2 \neq \mathbb{A}_{k} \times \mathbb{A}_{k}$.

So, I don't understand example II.4.0.1 in Hartshorne's book.

This example explain that if $X$ is the affine line with double origin points, then $X\times_k X$ is the affine plane with double axes and four origin points... in this case, why $X\times_k X= X\times X$? and my additional question is that in general case, is there condition that $X \times_S Y = X\times Y$?, where $S$ is an arbitrary scheme?

Arnaud D.
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Sang Cheol Lee
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  • See also this similar question: http://math.stackexchange.com/questions/322188/how-to-compute-the-topological-space-of-fibered-product-of-schemes/322256#322256 –  Jul 30 '13 at 06:21

1 Answers1

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In this example, it is still not true that $X\times_k X = X\times X$ as topological spaces, since for example on the complement of the origin(s), the two are not equal. This follows for the same reason as your previous example: the Zariski topology on $\mathbb A^2$ is not equal to the product topology on $\mathbb A^1\times\mathbb A^1$.

What is true in both cases is that there is a natural correspondence between closed points of the fibre product and products of closed points. This follows because we know that maximal ideals of $K[x,y]$ have the form $(x-a,y-b)$ for $a,b\in K.$

Andrew
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  • thanks Andrew. then, hartshorne book is wrong? – Sang Cheol Lee Jul 30 '13 at 04:18
  • Dear @SangCheolLee, no, Hartshorne is not wrong. You might prefer to look at Shafarevich Basic Algebraic Geometry 2, p. 44-5, which goes into a bit more detail on this example. Cheers – Andrew Jul 30 '13 at 12:15