# Product of affine varieties is the product of topological spaces

Let $k$ be an algebraically closed field, and $A, B$ affine $k$-algebras. We can define a functor $\mathfrak F$ from the category of affine $k$-algebras to that of affine algebraic varieties, by making $Max(A)$ into a topological space (for $g \in A$, set $D(g)$ to be the set of $\mathfrak m \in Max(A)$ for which $g \not\in \mathfrak m$; this is a basis for a topology of which every open set is a finite union). We then define a sheaf $\mathcal O_A$ on $Max(A)$ in a natural way, and obtain the affine $k$-algebra $(Max(A), \mathcal O_A) = \mathfrak F(A)$.

I know how to prove that the functor $\mathfrak F$ induces an antiequivalence of categories. I know that the tensor product $A \otimes_k B$ is a coproduct in the category of affine $k$-algebras, so it follows that $\mathfrak F(A \otimes_k B)$ is a product in the category of affine varieties. But I have also read that the cartesian product $Max(A) \times Max(B)$, along with the natural projections onto $Max(A), Max(B)$, can serve as the underlying set (usually not in the product topology) of the product of affine varieties $\mathfrak F(A)$ and $\mathfrak F(B)$. Is there an easy way to see this?

I'm trying to do things as "categorically" as possible, in particular avoiding noncanonical descriptions of affine $k$-algebras (as quotients of polynomial rings by radical ideals) where I can. Any help would be greatly appreciated.

• For schemes it's false, at least. – Matt Samuel Mar 28 '15 at 22:28
• You should write «can serve as the underlying SET», not «can serve as the underlying TOPOLOGICAL SPACE». – Mariano Suárez-Álvarez Mar 28 '15 at 22:34
• The functor sending an affine variety (in the classical sense) to its set of points is representable, so it preserves all limits. In particular, it preserves products. – Zhen Lin Mar 28 '15 at 22:56

## 1 Answer

For affine varieties over algebraically closed fields, the Nullstellensatz guarantees that the maximum spectrum functor can be identified with the functor $\text{Hom}(\text{Spec } k, -)$, which obviously preserves finite products; in fact it preserves all limits.

Over a non-algebraically closed field this fails, and instead of the maximum spectrum functor it's much nicer to look at the functor $\text{Hom}(\text{Spec } \bar{k}, -)$ together with the action of the absolute Galois group $\text{Gal}(\bar{k} / k)$ on it.