0
$\begingroup$

I was reading a bit of Hartshorne, and I know that the product of quasi-projective varieties is again a quasi-projective variety.

I should hope that taking products is associative, but I am unsure. What I mean by this is that if $X, Y, Z$ are all quasi-projective varieties over some algebraically closed $k$, is it always true that $(X\times Y)\times Z\cong X\times (Y\times Z)$ so that we can not bother with the parentheses?

I searched around, but everything led me back to 3.16 in Hartshorne.

$\endgroup$
  • 6
    $\begingroup$ Yes. This is true of all categorical products. In fact there is a distinguished such isomorphism. $\endgroup$ – Qiaochu Yuan Jan 6 '15 at 2:58
3
$\begingroup$

Let me expand upon Qiaochu's answer.

It is a fact that $X \times Y$ is a categorical product, meaning that it has a universal property, namely this: it comes equipped with two maps, called the projections, $\pi_1:X \times Y \to X$ and $\pi_2:X \times Y \to Y$ satisfying the following property. For any other object $Z$ and maps $\alpha:Z \to X$ and $\beta:Z \to Y$, there is a unique map $\gamma:Z \to X \times Y$ such that $\pi_1 \circ \gamma = \alpha$ and $\pi_2 \circ \gamma = \beta$.

This ensures that products are unique up to unique isomorphism. This is done, in sketch, as follows: any two products both satisfy the universal property of products, hence there are unique maps to both of them. Composing these unique maps we get a unique map from one of the proposed products to itself. But the identity map also satisfies this property, and by uniqueness, the composition must be the identity. Hence they are isomorphic.

This is enough to show that products are associative. One does this by showing that two orderings of the parentheses satisfy the universal property, hence they must be isomorphic by the above argument.

$\endgroup$

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.