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?


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.

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.

  • $\begingroup$ 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. $\endgroup$ – Li Zhan May 31 '12 at 23:52
  • $\begingroup$ Dear Li, that part is purely topological: I have added a new edit giving a detailed proof. $\endgroup$ – Georges Elencwajg Jun 1 '12 at 8:52
  • $\begingroup$ Dear @Georges Elencwajg, a typo, $\{x\}\times F_i$ should be $F_i$. $\endgroup$ – wxu Jun 1 '12 at 10:08
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    $\begingroup$ 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! $\endgroup$ – Georges Elencwajg Jun 1 '12 at 16:45
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    $\begingroup$ 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 $X_i$) $\endgroup$ – E.O. Feb 11 '18 at 20:18

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 $\mathbb{C} \times \mathbb{C}$.

  • $\begingroup$ Thanks a lot! That example helps me clarify initially uncorrect image of fibre product. $\endgroup$ – Li Zhan Jun 1 '12 at 0:10

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