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Let $k$ be an algebraically closed field and $X,Y$ varieties (i.e. integral, separated schemes of finite type over $k$). Is the fibre product $X \times_k Y$ necessary irreducible or integral?

I have another more or less related question:

If $A,B$ are finitely generated $k$-algebras which are also integral domains, is $A \otimes_k B$ an integral domain?

Li Zhan
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  • In addition to the ways below, another way to prove irreducibility is suggested in Ravi Vakil's FOAG:

    9.5.L. EASY EXERCISE.: Show that a scheme $X$ is irreducible if and only if there exists an open cover $X = \cup U_i$ with $U_i$ irreducible for all i, and $U_i \cap U_j \neq ∅$ for all i, j.

     – Tim kinsella May 05 '21 at 18:05

2 Answers2

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a) Yes, the product $X\times_k Y$ of two varieties over an algebraically closed field $k$ is a variety. To prove it, reduce to affine varieties and for those use:

b) If the field $k$ is algebraically closed and if $A$ and $B$ are $k$-algebras without zero divisors, then their tensor product $A\otimes_kB$ also has no zero divisors.
(Whether $A$ or $B$ is finitely generated is irrelevant.)
This is proved in Iitaka's Algebraic Geometry, page 97 (Lemma 1.54)

Edit: And if $k$ is not algebraically closed...
... all is not lost!

Let $k$ be a field and $R$ be a $k$-algebra which is a domain and with fraction field $K$.
If the extension $k\to K$ is separable and if $k$ is algebrically closed in $K$, then for all $k$-algebras $S$ which are domains, the $k$-algebra $R\otimes_k S$ will be a domain.
(Beware that separable means universally reduced. If $k\to K$ is algebraic, separability coincides with the usual notion.)
With luck, one of $A$ or $B$ may play the role of $R$ and $A\otimes_k B$ will be a domain.

As an illustration, any intermediate ring $k\subset R\subset k(T_1,\cdots,T_n)$ (where the $T_i$'s are indeterminates) satisfies the conditions above and will remain a domain when tensorized with a domain.

Bibliography
For the product of algebraic varieties I recommend Chapter 4 of Milne's online notes.
For the tensor product of fields, you might look at Bourbaki's Algebra, Chapter V, §17.

New Edit
At Li's request I'll show that $X\times Y$ is irreducible.
Let $$X\times Y=F_1\cup F_2$$ with $F_i$ closed and consider the sets $X_i=\lbrace x\in X\mid \lbrace x \rbrace\times Y\subset F_i\rbrace$.
Since $$\lbrace x \rbrace\times Y=[(\lbrace x \rbrace\times Y)\cap F_1]\cup [(\lbrace x \rbrace\times Y)\cap F_2]$$ we see, by irreducibility of $\lbrace x \rbrace\times Y$ (isomorphic to $Y$), that each vertical fiber $\lbrace x \rbrace\times Y$ is completely included in (at least) one of the $F_i$'s. In other words, $$X=X_1 \cup X_2.$$ It suffices now to show that the $X_i$'s are closed, because by irreducibility of $X$ we will then have $X_1=X$ (say) and thus $X\times Y=F_1$: this will prove that indeed $X\times Y$ is irreducible.
Closedness of $X_i$ is proved as follows:
Choose a point $y_0\in Y$. The intersection $(X\times \lbrace y_0 \rbrace)\cap F_i$ is closed in $X\times \lbrace y_0 \rbrace$ and is sent to $X_i$ by the isomorphism $X\times \lbrace y_0 \rbrace \stackrel {\cong}{\to}X$. Hence $X_i\subset X$ is closed. qed.

Georges Elencwajg
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  • Could you please say something more about showing $X \times Y$ is irreducible by reducing to the affine case? I could not figur out how to do that. – Li Zhan May 31 '12 at 23:52
  • Dear Li, that part is purely topological: I have added a new edit giving a detailed proof. – Georges Elencwajg Jun 01 '12 at 08:52
  • Dear @Georges Elencwajg, a typo, ${x}\times F_i$ should be $F_i$. – wxu Jun 01 '12 at 10:08
  • Dear @wxu, I have corrected the typo. Thank you very much for your attentive reading. – Georges Elencwajg Jun 01 '12 at 10:22
  • Dear @Georges Elencwajg, I got it! Thank you so much, I cannot appreciate more! – Li Zhan Jun 01 '12 at 10:24
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    Dear Li: you are right, my proof was not complete and I have modified it. Thanks for your attentive reading. Notice however that I had not written, have never written and will never write that the projection of a product onto a factor is closed (unless the other factor is complete) : I am too scared that the hyperbola $xy=1$ (which is an old acquaintance of mine ) would come and bite me! – Georges Elencwajg Jun 01 '12 at 16:45
  • I fully understand it now. I appreciate your constant help very much! – Li Zhan Jun 02 '12 at 12:03
  • Dear Li, thank you very much: nothing is more pleasant than being read by a user mercilessly checking details (and pointing to insuficient arguments!) and yet showing appreciation . – Georges Elencwajg Jun 02 '12 at 12:17
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    It is not clear to me why $(X\times {y_0})\cap F_i$ gets sent to $X_i$. Could you elaborate on this step? (It seems to me that we only get that the image contains $X_i$) – E.O. Feb 11 '18 at 20:18
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    @E.O. you can take the intersect of all the images as you vary $y_0$ to obtain $X_i$. – suncup224 Apr 20 '19 at 04:48
  • @GeorgesElencwajg , by $\lbrace x \rbrace\times Y$ do you mean the scheme theoretic fiber of $X\times Y$ over x? If so are you sure it's homeomorphic to Y? E.g. if X and Y are spectra of fields whose tensor product is not a field then it is false for reasons of cardinality. But maybe you make use of the algebraically closed condition? – Tim kinsella Oct 23 '20 at 04:31
  • Also I just want to remark that another way to prove irreducibility (following the local calculation you've explained) is suggested in Ravi Vakil's FOAG:

    9.5.L. EASY EXERCISE.: Show that a scheme $X$ is irreducible if and only if there exists an open cover $X = \cup U_i$ with $U_i$ irreducible for all i, and $U_i \cap U_j \neq ∅$ for all i, j.

     – Tim kinsella Oct 23 '20 at 04:39
  • @GeorgesElencwajg Where do you use the "affine case" in your topological proof? – Jérémy Blanc Oct 27 '20 at 10:18
  • @Jérémy Blanc The topological proof doesn't use the affine case but it supposes that you already know that $X \times_k Y$ is a variety, which is the non-trivial part that requires the affine case. I have also used implicitly the fact that the underlying set of the product is the product of the underlying sets. This is certainly false in a more general context: think of $ \operatorname {Spec} \mathbb C \times_\mathbb R \operatorname {Spec} \mathbb C $ which has two points although the factors are singleton sets. – Georges Elencwajg Oct 27 '20 at 17:11
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As mentioned by Georges Elencwajg, the fiber product is also integral if $k$ is algebraically closed. In general, some author require a variety over any field $k$ should be geometrically integral. In this way the fiber product of varieties is also variety. As for the affine case, I just want to mention an example that $\mathbb{C} \otimes_{\mathbb{R}} \mathbb{C}$ is not a domain. In fact it is isomorphic to $\mathbb{C} \times \mathbb{C}$.

user26857
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Joy-Joy
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