Let $k$ be an algebraically closed field. Consider the topological space $B= \operatorname{Spec}k[X]$ with usual Zariski topology. Then is it true that with Zariski topology, $B \times B$ is homeomorphic with $ \operatorname{Spec} k[X,Y]$ ?
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2$\begingroup$ What do you mean by $B \times B$? Is this the space with the product topology or the product of varieties? In the former case what you say is false in general, in the latter it's true since $k[X] \otimes_k k[Y] \cong k[X,Y]$ $\endgroup$– Exit pathApr 16, 2018 at 19:11
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$\begingroup$ @leibnewtz: I mean product of varieties. Why is it true ? Could you please explain (as an answer) or give some references ? Thanks .. $\endgroup$– userMay 16, 2018 at 17:34
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$\begingroup$ @leibnewtz: In the latter its false in general; rather than a product of varieties, you have in mind the fiber product over $\mathrm{Spec}(k)$ (or equivalently, a product of "varieties over $k$") $\endgroup$– user14972May 16, 2018 at 18:54
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$\begingroup$ @Hurkyl: could you please give some explanations ? $\endgroup$– userMay 16, 2018 at 19:32
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$\begingroup$ @Hurkyl Yes, you're right. I was thinking of the product in the category of varieties over $k$. $\endgroup$– Exit pathMay 16, 2018 at 22:02
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1 Answer
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This is a bit too long for a comment, so I'll add it here. The fiber product of two schemes $X$ and $Y$ over $k$ satisfies a universal property: It is the pullback of the diagram $X \to Spec(k) \leftarrow Y$. In other words, if we have a $k$ scheme $Z$ with maps of $k$ schemes $Z \to X$ and $Z \to Y$ making the obvious diagram commute, we get a unique map $Z \to X \times_k Y$.
It follows that to show $Spec(A \otimes_k B) \cong Spec(A) \times_k Spec(B)$ we just need to show that $Spec(A \otimes_k B)$ satisfies the universal property above. Since the $Spec$ functor induces a contravariant equivalence of categories between $k$-algebras and affine schemes over $k$, we need to show that $A \otimes_k B$ satisfies the dual universal property in the category of $k$-algebras. Try to show that a map $A \otimes_k B \to R$ of $k$ algebras is the same as a pair of maps of $k$-algebras $A \to R$ and $B \to R$. Use this to conclude the desired isomorphism.
