We use the definitions of this question. Let $\mathrm{Aff}(k)$ be the category of affine $k$-varieties. Let $X, Y$ be objects of $\mathrm{Aff}(k)$. Does the product $X\times Y$ exist in $\mathrm{Aff}(k)$?

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    $\begingroup$ I'm not sure what you're asking. The product of $X\subset\mathbf A^m$ and $Y\subset\mathbf A^n$ can be defined as a subvariety $X\times Y\subset\mathbf A^{m+n}$. Alternatively, if you define an affine variety as the spectrum of an affine algebra, you can take $\mathrm{spec}(A)\times\mathrm{spec}(B) = \mathrm{spec}(A\otimes B)$. Is that it? $\endgroup$ – jathd Nov 29 '12 at 9:01
  • $\begingroup$ @jathd Did you see the definition of an affine $k$-variety in the linked question? $\endgroup$ – Makoto Kato Nov 29 '12 at 9:14
  • $\begingroup$ I didn't look closely enough at the other post. But it seems to me it defines affine varieties in a way essentially equivalent to saying that they're spectra of affine $k$-algebras (with their structure sheaf). $\endgroup$ – jathd Nov 29 '12 at 9:50
  • $\begingroup$ @jathd So the category $Aff(k)$ is anti-equivalent to the category of reduced $k$-algebras of finite type. Is a tensor product of reduced $k$-algebras necessarily reduced? $\endgroup$ – Makoto Kato Nov 29 '12 at 10:02

There is antiequivalence of categories between affine schemes and rings, sending a scheme $X=Spec(A)$ to its ring of global functions $A=\Gamma(X, \mathcal O_X)$.
In this correspondence the product $Spec(A)\times Spec(B)$ of two schemes correspond to the tensor product $A\otimes B$ of their rings.
You can "restrict" this antiequivalence to $k$- schemes and thus obtain an antiequivalence between affine $k$-schemes and $k$-algebras sending the product (over $Spec(k)$ ) $X\times_k Y$ of two such $k$-schemes $X,Y$ to the tensor product algebra $\Gamma(X, \mathcal O_X)\otimes _k \Gamma(X, \mathcal O_Y)$.

And now the fun begins!

Given that $X$ and $Y$ have some property, their product may or may not have that property.
For example the product of two integral $k$-schemes is an integral scheme if $k$ is algebraically closed.
For more general $k$ this result fails: for example over $\mathbb R$ we have $Spec(\mathbb C)\times_\mathbb R Spec(\mathbb C)=Spec(\mathbb C )\sqcup Spec(\mathbb C)$, which is reducible, hence not integral although its factors are integral . (See also the last section of this post).
And if $k$ is an imperfect field of characteristic $p$ with a purely inseparable extension $K=k(a)$ satisfying $a\notin k$ but $a^p \in k$, the scheme $Spec(K)\times_k Spec(K)$ is not reduced and hence not integral although its factors are integral .
[Non reducedness follows from:
$\sqrt [p]{a}\otimes 1-1\otimes \sqrt [p]{a} \neq 0$ but $(\sqrt [p]{a}\otimes 1-1\otimes \sqrt [p]{a} )^p= a\otimes 1-1\otimes a =0$ ]

A detailed study of these questions can be found in EGA Chapter IV (Seconde Partie) §4, in which field theory plays a crucial role.

An ideological position.
Your linked post refers to the language of Weil's Foundations.
Like most contemporary algebraic geometers I consider that Grothendieck's scheme theory has relegated Weil's theory to an interesting episode in the history of algebraic geometry but has no place in contemporary mathematics, so that I don't want to have anything to do with that language.

The answer to Makoto's ACTUAL question!
Surprisingly, the full subcategory of schemes with objects reduced schemes has finite products: the product of two reduced schemes $X,Y$ is the reduction $(X\times Y)_{red} $ of their product $X\times Y $ in the category of schemes .
Essentially, this is because the morphism $X\to X\times Y$ factors through $(X\times Y)_{red} $ if $X$ is reduced and similarly for $Y$ .
The same result holds for reduced finitely generated $k$-schemes.

An intriguing example
Consider the purely inseparable extension $K=k(a)$ above ($a\notin k, a^p\in k$).
The coproduct $K\sqcup K$ in the category of $k$-algebras is $K\otimes_k K \cong K[S]/(S-a)^p$, a non-reduced $k$-algebra.
However the coproduct in the category of reduced $k$-algebras also exists and is $K\sqcup'K=K$ (with both morphisms $K\to K\sqcup'K=K$ being the identity).
The reason of this enigmatic result is that given a reduced $k$-algebra $L$, there is at most one morphism of $k$- algebras $K\to L$, and this morphism must send $a\in K$ to the unique $p$-th root $l\in L$ of $a^p\in k$ .
By taking $Spec$'s, you will get an intriguing example of product in the category of reduced affine $Spec (k)$-schemes.

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    $\begingroup$ "so that I don't want to have anything to do with that language." 1+ $\endgroup$ – Martin Brandenburg Nov 29 '12 at 9:54
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    $\begingroup$ @MakotoKato: You can study classical varieties using scheme-theory, so what is your point? Or do you know a problem about classical varieties, which can be solved in Weil's language, while it cannot be solved within the framework of schemes? That would be a different story. $\endgroup$ – Nils Matthes Nov 29 '12 at 11:42
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    $\begingroup$ @NilsMatthes My point is that classical varieties are more natural and/or more geometric than schemes. Think about $\mathbb{Q}$-varieties(in this case we can take $\mathbb{C}$ as $\Omega$). When the base field is algebraically closed, it's easy to see that irreducible varieties and integral separated scheme of finite type over an algebraically closed field are essentially the same. (to be continued) $\endgroup$ – Makoto Kato Nov 29 '12 at 13:08
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    $\begingroup$ However, when the base field is not algebraically closed, it's not so easy to see geometrically reduced separated schemes of finite over a field are essentially the same as classical varieties defined over the same field. Here comes the notion of $k$-varieties which is a generalization of Weil's varieties. I think this point of view can help students understand better what schemes are. $\endgroup$ – Makoto Kato Nov 29 '12 at 13:08
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    $\begingroup$ @NilsMatthes "You can study classical varieties using scheme-theory" Because you know how to translate the former language to the latter one(when the base field is algebraically closed, it is not difficult). In other words, you need to know the both languages in the first place. $\endgroup$ – Makoto Kato Nov 29 '12 at 13:40

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