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In general, the product topology on two (quasiprojective) varieties is not the same as the topology of the product variety given by the Segre embedding. This is something I've often seen asserted is true, but I'm having difficulties writing down a complete proof of this fact. What I want to prove is

Let $V \subseteq \mathbb{P}^m$, $W \subseteq \mathbb{P}^n$ be varieties with the Zariski topology. Then the product topology on $V \times W$ is not the same as the topology of the Segre product $\sigma(V \times W)$, unless one of the varieties is simply a finite set of points.

This is an early exercise in Harris' Algebraic geometry - a first course (Exer. 2.22) and in Smith's An Invitation to Algebraic Geometry (Exer. 5.3.4). This is likely a standard fact, but regardless I indicate the kind of partial progress I've made below (possibly red herrings) - feel free to skip the walls of text below.


Firstly, this fact is easily "proved" when explicit varieties $V, W$ are given. For example, if $V = W = \mathbb{P}^1$, one can simply use the diagonal in $\mathbb{P}^1 \times \mathbb{P}^1$ as an example of a Zariski-closed set in $\sigma(\mathbb{P}^1 \times \mathbb{P}^1)$ that is not closed in the product topology. From this and other examples, I believe that in the cases that the topologies differ, the topology on $\sigma(V \times W)$ is finer than the product topology.

Next, if neither $V$ nor $W$ is a finite set of points, then $\dim V, \dim W > 0$, so one may assume by restricting to a subvariety that $V$ and $W$ are actually one-dimensional irreducible varieties. In this case, the closed sets of $V \times W$ are well-known: the closed sets in $V$ and $W$ are the empty set, the whole set, or a finite number of points, and so the closed sets in $V \times W$ are the arbitrary intersections of finite unions of the products of such sets.

At this point, I see several different avenues that may lead to a proof. The first is more of an geometric approach. We want to find a set that is closed in $\sigma(X \times Y)$ but not $X \times Y$; might not some "tilted" hyperplane section do the trick? Or perhaps we can mimick the example with $V = W = \mathbb{P}^1$. The diagonal may not make sense in general, but if we have a nonconstant morphism $V \to W$ or $W \to V$, then we can use the graph of this morphism in place of the diagonal to conclude. We also know that the closed sets in $\sigma(V \times W)$ are the zero locus of bihomogeneous polynomials.

Another approach is to work in the (homogeneous) coordinate rings. That is, we wish to find a proper (prime) ideal containing $I(\sigma(V \times W))$ where we do not simply add in "products" of polynomials in $k[V]$ and $k[W]$. The fact that we are working with one-dimensional irreducible varieties means that the transcendence degree of $k[\sigma(V \times W)]$ is two, so one simply needs to quotient out by one transcendental element that does not correspond to a closed set already in the product topology $V \times W$. Finally, one may exploit the fact that the Segre product is the categorical product in the category of varieties and the topological product is the categorical product in the category of spaces. If the Segre product is the categorical product in the category of spaces too and the projection maps are the same in this category, then the "unique" morphism from $X \times Y$ to $\sigma(X \times Y)$ is $\sigma$ on the level of sets, so one simply needs to show that $\sigma$ is not continuous.

Unfortunately, I am unable to turn any of these vague ideas into a complete proof. Since this result is so basic, I fear that I am overlooking something simple, and I would greatly appreciate it if someone could tell me what it is I'm missing.

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  • $\begingroup$ I think you've missed the point, though: it's not important that the product of varieties very rarely has the product topology; rather, it's more important to know that it doesn't happen in all cases! $\endgroup$
    – Zhen Lin
    Commented May 18, 2013 at 17:24
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    $\begingroup$ @ZhenLin Yes, I agree. I have counterexamples to show that the topologies do not coincide (indeed, given explicit varieties one can easily show the topologies are different), but the exercises I referred to seem to suggest that the product of varieties never has the product topology unless one of them is a finite set of points, and I would like to see a proof of this. $\endgroup$
    – JHF
    Commented May 18, 2013 at 17:33
  • $\begingroup$ I'm not sure of how to prove this but I just wanted to point out that as you suspected, the topology on the Segre product would indeed be finer than the product topology since the projects are regular and therefore continuous. $\endgroup$
    – Seth
    Commented Mar 2, 2014 at 22:15

1 Answer 1

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Suppose $V$ and $W$ are both infinite. As you note, we may assume $V$ and $W$ are both irreducible curves, and we may similarly assume that $V$ and $W$ are both affine. Now just take any nonconstant regular function $f:V\to k$ and any nonconstant regular function $g:W\to k$, and consider the regular function $h(v,w)=f(v)-g(w)$ on $V\times W$. Since $V$ and $W$ are irreducible curves, the fibers of $f$ and $g$ are finite, and also all but finitely many of the fibers must be nonempty. This means that if $C\subset V\times W$ is the vanishing set of $h$, then $C$ is infinite, but the two projections $C\to V$ and $C\to W$ both have finite fibers. No such set can be closed in the product topology (any infinite closed set in the product topology must contain a set of the form $\{v\}\times W$ or a set of the form $V\times\{w\}$).

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