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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?

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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 $\mathbb A^m\times \mathbb A^n \cong \mathbb A^{m+n}$.

Okay, now for your question:

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?

Once you can do this case, you can do any quasi-projective case. So you should think about the role of the Segre embedding in this case.

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  • $\begingroup$ Probably this is a stupid comment, but what about $\mathbb P^{m+n}$? $\endgroup$
    – Dubious
    Commented Oct 20, 2013 at 18:08
  • $\begingroup$ @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, $\endgroup$
    – Matt E
    Commented Oct 20, 2013 at 18:10
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    $\begingroup$ @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, $\endgroup$
    – Matt E
    Commented Oct 20, 2013 at 18:11
  • $\begingroup$ @Galoisfan: P.S. I'm not meaning to be obtuse by being so oblique. I just got the impression from the way you wrote the question that you hadn't thought about what the product of projective spaces looks like, and I'm encouraging you to think about a bit for yourself. $\endgroup$
    – Matt E
    Commented Oct 20, 2013 at 18:13
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    $\begingroup$ @Galoisfan: Dear Galoisfan, So now you have a good answer to your question. (And don't worry, you're certainly not the first AG student to wonder why there is all this fuss about the Segre embedding!) Cheers, $\endgroup$
    – Matt E
    Commented Oct 21, 2013 at 0:16

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