# Product of affine varieties vs. product of (quasi-)projective varieties

Suppose that $F_1,\ldots,F_r\in k[T_1,\ldots, T_n]$ and $G_1,\ldots,G_s\in k[S_1,\ldots,S_m]$ where $k$ is an algebraically closed field. Clearly $X=V(F_1,\ldots,F_r)\subseteq\mathbb A^n_k$ and $Y=V(G_1,\ldots,G_s)\subseteq\mathbb A^m_k$ are two affine varieties and $X\times Y\subseteq\mathbb A^{n+m}_k$ has a natural structure of affine variety as follows:

$$X\times Y=V(F_1,\ldots,F_r,G_1,\ldots,G_s)$$

For the product of (quasi-)projective variety we need the Segre embedding to define a structure of variety, and I don't understand the motivation. Why a straightforward argument as the above doesn't work for (quasi-)projective varieties?

Note that you write $X \times Y \subset \mathbb A^{m+n}$, and so you seem to be using almost without thinking about the isomorphism $\mathbb A^m\times \mathbb A^n \cong \mathbb A^{m+n}$.

Forget about the equations, and just consider $\mathbb P^m \times \mathbb P^n$. What straightforward way might you suggest to think of this as a variety?
• Probably this is a stupid comment, but what about $\mathbb P^{m+n}$? – Dubious Oct 20 '13 at 18:08
• @Galoisfan: Dear Galoisfan, Of course if this were correct, then the whole question of products of (quasi)-projective varieties would be easy, and we wouldn't need the theory of the Segre embedding. But is it correct? Why don't you think about the topological spaces $\mathbb CP^1$ and $\mathbb C P^2$? Is the latter homeomorphic to the product of two copies of the former? Regards, – Matt E Oct 20 '13 at 18:10
• @Galoisfan: Alternatively, can you find a non-constant morphism $\mathbb P^2 \to \mathbb P^1$? (This is just a computation, based directly on the definitions.) If not, then clearly $\mathbb P^2$ is not isomorphic to $\mathbb P^1 \times \mathbb P^1$ (since whatever the latter eventually is defined to be, it will have projections onto its two factors). Regards, – Matt E Oct 20 '13 at 18:11