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Let $X$ and $Y$ be two schemes over a field $k$, and let $X \times_{k} Y$ be their fiber product over $k$.

If $X \times_{k} Y$ has a finite number of connected components, then the same is true for $X$ and $Y$. This is easy in the affine case, as $A \hookrightarrow A \otimes_{k} B$, and in general by AG, as stated in Goertz & Wedhorn, Corollary 5.45.

I wonder about the converse: if $X$ and $Y$ have a finite number of connected components, is the same true for $X \times_{k} Y$?

Since $X$ is the sum of its connected components, I think we can reduce the question to the case where both $X$ and $Y$ are connected.

Could anyone answer this or provide a counterexample, preferably with affine schemes?

Edit: $X$ and $Y$ are non empty.

Martin Brandenburg
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Kaneki1013
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    Your statement in the second paragraph is wrong (consider empty $X$). – Martin Brandenburg Dec 08 '24 at 23:09
  • Thanks for the remark. I'm only interested by non empty schemes. – Kaneki1013 Dec 08 '24 at 23:20
  • I'm wondering about something like $A = \mathbb{Q}(\sqrt{2}, \sqrt{3}, \sqrt{5}, \ldots), B= \mathbb{R}, k = \mathbb{Q}$. In this case, I feel like $A \otimes_{k} B$ should split into an infinite direct sum of copies of $R$, since you can write $A$ as $\mathbb{Q}[X_{1}, X_{2}, \ldots]/\langle X_{1}^{2}-2, X_{2}^{2}-3, \ldots \rangle$, and so I think (?) we should have $\mathbb{R}[X_{1}, X_{2}, \ldots]/\langle X_{1}^{2}-2, X_{2}^{2}-3, \ldots \rangle$. But maybe I'm making some mistake in the details... – Alex Wertheim Dec 08 '24 at 23:47
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    @AlexWertheim that won't work - the fiber product of two affine schemes over an affine scheme is again an affine scheme and hence quasi-compact; in the specific example of $\operatorname{Spec} k^{\oplus\Bbb N}$ one gets the Stone-Cech compactification of $\Bbb N$, see here. – KReiser Dec 08 '24 at 23:52
  • @KReiser yeah, you're right, thanks. Now that I think about it, $\bigoplus_{\mathbb{N}} \mathbb{R}$ isn't even a ring anyway. I'm not totally sure what $A \otimes_{k} B$ is isomorphic to in my example (product, maybe?), but in any event, it's not a counterexample here. – Alex Wertheim Dec 08 '24 at 23:57
  • Probably one example is $A = B=\mathbb{C},k = \mathbb{Q}$. by This question, $A \otimes_{k} B$ should be isomorphic to an uncountable product of copies of $\mathbb{C}$. but i dont know how to find the isomorphism. – Kaneki1013 Dec 09 '24 at 00:23
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    Again, that won't work - the fiber product of two affine schemes over an affine scheme is an affine scheme, and any affine scheme is quasicompact, so it has finitely many connected components. If you want to find an example, you must pick at least one of $X,Y$ to be non-quasicompact (hence non-affine). – KReiser Dec 09 '24 at 01:30
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    I wouldn’t be so sure of that. I think taking the square of $\operatorname{Spec} \overline{\mathbb{Q}}$ should give an interesting counterexample: the underlying space is homeomorphic to the absolute Galois group, which is infinite, totally disconnected, but nonetheless compact Hausdorff. – Zhen Lin Dec 09 '24 at 08:26
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    @KReiser it's not true that compact spaces have finitely many connected components, there are affine schemes with infinitely many connected components. Any non-Noetherian von Neumann regular ring, e.g. an infinite product of fields, provides such an example. – Lukas Heger Dec 09 '24 at 11:48
  • @LukasHeger ah, perhaps it is my turn to learn - I would have thought that the open cover consisting of all the connected components would have shown that a quasi-compact space can only have finitely many connected components - how does this argument fail? – KReiser Dec 09 '24 at 14:33
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    @KReiser there's no reason why the connected components should be open, in general they are not. If the space is locally connected (e.g. a topological manifold), then connected components are open, but schemes are not that nice topologically in general. – Lukas Heger Dec 09 '24 at 14:40
  • Here's a more familiar example of a space where connected components are not open (although it's not quasi-compact): $\Bbb Q$ with the subspace topology – Lukas Heger Dec 09 '24 at 14:41
  • @LukasHeger oh wow, what an oversight on my part. Apologies for the bad information in the comments! – KReiser Dec 09 '24 at 14:42
  • @ZhenLin I read about this homeomorphism before,but i have so little knowledge about infinite Galois theory. To use this as an counterexample it should be proved that the profinite topology in the absolute Galois group is not discrete? – Kaneki1013 Dec 09 '24 at 15:37

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To sum up: when $X$ and $Y$ are affine over a common base $k$, then $X \times_k Y$ is also affine over $k$, hence quasicompact; however, even if $X$ and $Y$ have finitely many connected components, $X \times_k Y$ may not.

For example, consider the case $X = Y = \operatorname{Spec} \overline{\mathbb{Q}}$, $k = \mathbb{Q}$. Then $X$ and $Y$ are just points, and $X \times_k Y = \operatorname{Spec} {\overline{\mathbb{Q}} \otimes_\mathbb{Q} \overline{\mathbb{Q}}}$, of course, but $\overline{\mathbb{Q}} \otimes_\mathbb{Q} \overline{\mathbb{Q}}$ is an extremely complicated ring: ring homomorphisms $\overline{\mathbb{Q}} \otimes_\mathbb{Q} \overline{\mathbb{Q}} \to \overline{\mathbb{Q}}$ correspond to pairs of automorphisms of $\overline{\mathbb{Q}}$, so (skipping lots of steps) we can identify prime ideals of $\overline{\mathbb{Q}} \otimes_\mathbb{Q} \overline{\mathbb{Q}}$ with elements of the absolute Galois group, and this identification extends to a homeomorphism of topological spaces. But (the underlying space of) the absolute Galois group has infinitely many connected components.

It is tempting to say that a quasicompact scheme has finitely many connected components, but the (counter)example above is a compact Hausdorff space yet has (uncountably) infinitely many connected components. (It is a profinite set.) The reason for this failure of intuition is that connected components need not be open in a general topological space.

To avoid pathologies (or, to be more open minded, surprises) like this, you should assume $X$ and $Y$ are of finite type over $k$. Then $X \times_k Y$ is also of finite type over $k$, and schemes of finite type over a field always have finitely many connected components.

Zhen Lin
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