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Recently, I learn some algebraic geometry and notice that fibre product appears everywhere. I feel like that fibre product plays the same role as Cartesian product of variety in classical algebraic geometry. But I can't figure out that why one wants to use fibre product instead of Cartesian product.

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    $\begingroup$ I would suggest that you try writing down the Cartesian product of say, $\mathrm{Spec} \mathbb{Q}(\sqrt{2})_{\mathrm{Spec} \mathbb{Q}} \mathrm{Spec} \mathbb{Q}(\sqrt{2}) $ and the corresponding fiber product to see why they are different. $\endgroup$ – Mohan Jul 12 '17 at 3:02
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    $\begingroup$ The product of two bundles (in the category of bundles over some space) is basically the same thing as pullback (in the category of spaces) of the corresponding diagram of spaces. $\endgroup$ – Hurkyl Jul 12 '17 at 4:57
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    $\begingroup$ It's just that the fibered product over $S$ is the Cartesian product in the category of schemes over $S$. $\endgroup$ – Kevin Carlson Jul 12 '17 at 5:26
  • $\begingroup$ Fibre products appear very naturally in very concrete situations: If $N\subseteq M$ and $f:X \to M$, then the pre-image of $N$ is $f^{-1}(N) \cong N\times_M X$ (with respect to the inclusion map $\iota: N \to M$). $\endgroup$ – AnonymousCoward Jul 12 '17 at 18:34
  • $\begingroup$ Also, when studying principal bundles (which underly much of modern geometry), one can prove that every principal bundle over a space $X$ is the pullback (fibered product) of a universal bundle $EG\to BG$ by a map $f:X \to BG$. $\endgroup$ – AnonymousCoward Jul 12 '17 at 18:37
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I don't have any intuitive reason for why the fibered product is the "right" choice for schemes, but here is some algebraic formalism that shows why it works for affine schemes, at least.

Recall that we have an anti-equivalence between the categories of affine schemes and algebras. Let $R$ be a ring and $A$ and $B$ be commutative $R$-algebras. The tensor product $\otimes_R$ is the coproduct in the category of $R$-algebras, so we get a commutative diagram

$\hspace{4.25cm}$ enter image description here

that is universal in the following sense: if we have the same diagram with $A \otimes_R B$ replaced by another $R$-algebra $C$, then there is a unique $R$-algebra homomorphism $A \otimes_R B \to C$ such that the diagram commutes.

By contravariance, for the fibered product $\DeclareMathOperator{\Spec}{Spec} \Spec(A) \times_{\Spec(R)} \Spec(B) := \Spec(A \otimes_R B)$ we get the same diagram with all the arrows reversed:

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and the "dual" universal property yields a morphism $\Spec(C) \to \Spec(A) \times_{\Spec(R)} \Spec(B)$. If you tilt your head 45 degrees and ignore the $\Spec(R)$ portion of the diagram, you'll recognize this diagram as the one used to define the universal property of the product in a category. (The arrows to $\Spec(R)$ come from the fact that we are working with $\Spec(R)$-schemes that each come equipped with a morphism to $\Spec(R)$.)

This shows that the fibered product $\times_{\Spec(R)}$ is the product in the category of affine schemes. To even construct the fibered product for general schemes requires some technical glueing arguments---the proof of their existence given at the Stacks Project begins with "Please skip this proof"---but I hope this gives you some idea why the fibered product works in this role.

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Varieties are defined over a predetermined field, which sticks around when you pass to their product. Schemes are not intrinsically defined over anything in particular (well okay, they are over $\mathbb{Z}$); if you are working with a scheme over a field, you need the fiber product to "remember" that you are working over that field.

For varieties $X$ and $Y$ over an algebraically closed field $k$, you're lucky that you can place a canonical variety structure on the cartesian product $X \times Y$ which also happens to be the product of $X$ and $Y$ in the category of varieties.

If you view varieties as schemes (e.g. if $X = \mathfrak m\textrm{-Spec}(A)$ is an affine variety for a finitely generated algebra $A$ over your field $k$, you instead think of $X$ as all of $\textrm{Spec}(A)$) and you want the analogous product of $X$ and $Y$ (viewed as a scheme), the literal cartesian product is now useless. You will now need the fiber product $X \times_{\textrm{Spec}(k)} Y$.

If $X$ and $Y$ are affine varieties, say $X$ is as above and $Y = \mathfrak m\textrm{-Spec}(A)$, then to view the product variety $X \times Y$ as a scheme, you will need to think of it as $\textrm{Spec}(A \otimes_k B)$, with which there is no suitable identification with the cartesian product $\textrm{Spec}(A) \times \textrm{Spec}(B)$.

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