# Under Zariski topology , $\operatorname{Spec} k[X,Y] \cong \operatorname{Spec} k[X] \times \operatorname{Spec}k[Y]$?

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 $\operatorname{Spec} k[X,Y]$ ?

• 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 $k[X] \otimes_k k[Y] \cong k[X,Y]$ Apr 16, 2018 at 19:11
• @leibnewtz: I mean product of varieties. Why is it true ? Could you please explain (as an answer) or give some references ? Thanks ..
– user
May 16, 2018 at 17:34
• @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 $k$")
– user14972
May 16, 2018 at 18:54
• @Hurkyl: could you please give some explanations ?
– user
May 16, 2018 at 19:32
• @Hurkyl Yes, you're right. I was thinking of the product in the category of varieties over $k$. May 16, 2018 at 22:02

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.