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

Question

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]$ ?

Answer 1

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.