1
$\begingroup$

I have a few questions about products. I think I understand it but would like to hear some additional insights.

i) If $X\subseteq \mathbb{A}^n$ and $Y\subseteq \mathbb{A}^m$ are affine varieties we define $X\times Y$ in the usual sense, as a subset of $\mathbb{A}^{n+m}$. It turns out that $X\times Y$ is a variety as well. But it is not necessarily the case that $X\times Y$ is homeomorphic to the product topology of $X$ and $Y$.

ii) If $X\subseteq \mathbb{P}^n$ and $Y\subseteq \mathbb{P}^m$ are quasi-varieties, in the projective sense, we define their product via the Segre embedding. First we start with the map $\sigma: X\times Y \to \mathbb{P}^{nm+n+m}$ given by $\sigma(x_i,y_j) = (x_iy_j)$. Second it turns out that the image of $\sigma$ is quasi-projective and so we rather work with $\sigma(X\times Y)$. I ask this because I see people say, "consider $X\times Y$ ..." , are they really saying $\sigma(X\times Y)$?. It is confusing because the categorical product is not really the usual product of sets. Maybe some notation, like $X\times'Y$ would be better?

$\endgroup$
2
$\begingroup$

If $X,Y$ are classical varieties of any type (affine, projective,...) over an algebraically closed field , their product is denoted $X\times Y$.
It is the categorical product and the pleasant surprise is that its underlying set is the product of that of the factors: $|X\times Y|=|X|\times |Y|$.
Unfortunately the product topology is not suitable for $X\times Y$ and the correct topology is the Zariski topology.
It indeed corresponds in the projective case to the topology inherited under the Segre embedding $\sigma: X\times Y \to \mathbb{P}^{nm+n+m}$ from the target projective space $\mathbb{P}^{nm+n+m}$.

As you may be aware, a huge foundational revolution occurred about 55 years ago under the leadership of the late Grothendieck and the basic object of study of algebraic geometry, the variety, was replaced by the more general scheme.
The categorical product of two schemes still exists but its construction is rather more complicated than it was for varieties.
And the properties of the product can be rather counterintuitive.
In particular the underlying set of the product of two schemes has only a distant relation to that of the factors: it can be much smaller (even empty!) or much larger (infinite for two one-point schemes !) than the product of the underlying sets.

Edit: more technical results
At Fredrik's request I'll give two examples for users already knowledgeable of basic scheme theory.

a) If $X\to Spec(k),X'\to Spec(k')$ are schemes defined over fields $k,k'$of different characteristics their absolute product $X\times X'=X\times_{Spec(\mathbb Z)} X'=\emptyset$ is empty.
The simplest example is given by Fredrik in his comment: $$X=Spec(\mathbb F_2)\to Spec(k)=Spec(\mathbb F_2),\quad X'=Spec(\mathbb F_3)\to Spec(k')=Spec(\mathbb F_3)$$
b) The scheme $X=Spec(\mathbb C)$ has only one point but the product $X\times X$ is infinite and even has infinite Krull dimension by Grothendieck's best hidden result!

$\endgroup$
4
  • $\begingroup$ Dear Georges, excellent answer as usual, but I'm interested in those pathologies (?) you mention at the end. What are examples of schemes with empty product (ah, maybe $Spec \mathbb Z/2 \times Spec \mathbb Z/3$?)? And what are examples of spaces with infinite product? (ah, maybe $Proj k[T] \times Proj k[T]$?). $\endgroup$ Dec 13 '14 at 10:15
  • 1
    $\begingroup$ Dear @Fredrik: thank you for the kind words. I have ll written an edit to address your (excellent) questions. $\endgroup$ Dec 13 '14 at 11:22
  • 1
    $\begingroup$ Probably worth explaining what's going on in example (b): In this context, it makes more sense to think of the category of schemes over a particular base than just the category of schemes. If we don't specify a base, then that basically means we're implicitly working with schemes over $\mathbb{Z}$. So the "product" $\operatorname{Spec}(\mathbb{C}) \times \operatorname{Spec}(\mathbb{C})$ is what would more likely be called the fibered product of those schemes over $\mathbb{Z}$. Of course $\operatorname{Spec}(\mathbb{C})$ would more often be considered a scheme over $\mathbb{C}$. $\endgroup$ Dec 13 '14 at 12:18
  • $\begingroup$ The point is that obviously $\mathbb{C} \otimes_{\mathbb{C}} \mathbb{C} \cong \mathbb{C}$, whereas $\mathbb{C} \otimes_{\mathbb{Z}} \mathbb{C}$ is something horrific. Anyway, hopefully that saves some poor student from having to chase back a long sequence of definitions to figure out what's going on there. $\endgroup$ Dec 13 '14 at 12:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.