# Product of quasi-projective varieties is quasi-projective

This has been asked here and here before, but the first one does not have a complete answer, and the second somehow is aiming at showing it is irreducible (even though I see no reason why this should be the case). It is exercise 5.3.2 of Smith's Algebraic Geometry.

The product is the Segre product, and the definition of quasi-projective varieties we are working with is a locally closed subset of projective space (intersection of open and closed Zariski subsets).

What I have so far, as per the theorem proven in the chapter, is that the Segre map is an isomorphism, and since morphisms of varieties are continuous, the map is a homeomorphism (1). So this question becomes a topology problem.

Now, if $$X$$ and $$Y$$ are two quasi-projective varieties, they can be written as $$U \cap S$$ and $$V \cap T$$ respectively, where $$U, V$$ are open and $$S, T$$ closed. Their product $$X \times Y = (U \cap S) \times (V \cap T) = (U \times V) \cap (S \times T)$$, which under the product topology (2) is the intersection of an open and closed set. The Segre map then preserves this topology, and so we should be done.

My problems with this argument are numbered above.

1. Can we just say this? I feel a bit uncomfortable with just saying this outright, but I can't think of a counterexample. Maybe I just lack confidence with all the material so far though.
2. I read somewhere that we can't just randomly put the product topology in most situations and expect everything to work nicely, so I'm not sure if it works so well here.

Any pointers on if I'm wrong would be great, or if there's a better way to attack this problem :)

Your proof is not correct since it assumes that the topology on $$X\times Y$$ is the product topology of the Zariski topologies on $$X$$ and $$Y$$. This is definitely not the case (in the book you are reading there should be an exercise illustrating this).

A good way is to use the Segre embedding as other answers suggested: take $$X=U_X\cap Z_X\subseteq \mathbf{P}^n$$ and $$Y=U_Y\cap Z_Y \subseteq \mathbf{P}^m$$ for appropriate open sets $$U_X,U_Y$$ and closed sets $$Z_X,Z_Y$$. Then there is a morphism

$$\sigma_{m,n} : \mathbf{P}^n\times \mathbf{P}^m \to \mathbf{P}^{mn + n +m}$$

which is a closed embedding. Hence is it enough to show that $$S=\sigma_{m,n}(X\times Y)$$ is locally closed and irreducible.

Let $$\pi_1,\pi_2$$ be the two projection maps onto the two factors of $$\mathbf{P}^n\times \mathbf{P}^m$$. These maps are morphisms and note that

$$W = \sigma_{m,n}(\pi_1^{-1}(X)\cap \pi_2^{-1}(Y))$$

Each of the two sets on the right side are locally closed in $$\mathbf{P}^n\times \mathbf{P}^m$$ since $$X,Y$$ are quasi-projective and the $$\pi_i$$ are continuous. Indeed

$$W= \sigma_{m,n}(\pi_1^{-1}(U_X\cap Z_X)\cap \pi_2^{-1}(U_Y\cap Z_Y))=\\ = \sigma_{m,n}(\pi_1^{-1}(Z_X)\cap \pi_2^{-1}(Z_Y))\cap \sigma_{m,n}(\pi_1^{-1}(U_X)\cap \pi_2^{-1}(U_Y))$$

is intersection of a closed set and an open set (recall that the Segre map is a homeomorphism). Irreducibility can be shown by contradiction.

• Why do we want it to be irreducible? Commented Dec 9, 2019 at 15:44