# Infinite products exist *only* for affine schemes?

Infinite products don't exist in the category of schemes (see Jonathan Wise's answer here). However, all limits of affine schemes exist in the category of schemes (and are affine). I would like to know if affine schemes are the only ones which have this property, thus even sharpening the result. For example, we may ask the following:

Let $X$ be a scheme such that all powers $X^I$ exist in the category of schemes. Can we deduce that $X$ is affine?

I hope that the answer will be "yes". You may work in the category of $k$-schemes for a field $k$ and assume that $X$ is integral. You may also assume that all limits constructed from $X$ exist, for example all equalizers of morphisms $X^I \to X^J$, etc. Feel free to add other assumptions as well.

Jonathan has already made the following comment (which I cannot fill with details):

I believe the proof can be modified to show that if a product of a collection of quasi-compact schemes is a scheme then the product of some collection of all but finitely many of them is affine. Assuming they are flat over Z or something, I suspect this will be impossible unless all of those schemes are affine. Certainly an infinite product of projective lines is not affine.

Some time ago, I've already proven the following result (see here): Assume that $(X_i)$ is a family of $S$-schemes such that their fiber product $P$ exists in the category of schemes. If $Q$ denotes the fiber product of the $X_i$ in the category of locally ringed spaces (which exists and can be described explicitly), then the canonical morphism $P \to Q$ is bijective and the stalk maps are isomorphisms. But I don't know how to get the topology of $P$. This would be very helpful to show that $P$ does not exist.

But as a first step one would have to show that $Q$ is not a scheme, which is already hard in general. Here is an idea: Let $X \neq \emptyset$ be a $k$-scheme and assume that the LRS fiber product $Q=X^I$ ($I$ infinite set) is a scheme. Then the explicit construction implies that there are open subsets $U_i \subseteq X$, such that $U_i = X$ for almost all $i$, and that $\prod_i U_i$ is an affine scheme. Let $U$ be the (finite) product of the $U_i \neq X$. Then $U$ is a scheme, and $U \times X^I$ is an affine scheme. Under suitable finiteness conditions (?), a combination of Serre's criterion and the Künneth formula (see here) would imply that $X^I$ is an affine scheme. But then $X^I \cong X \times X^I$ shows with the same argument that $X$ is affine. But this all works only if $X$ and $X^I$ are quasicompact and quasiseparated.

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@Scott: Although you've deleted your answer, I think this is interesting: I've also tried to prove $\prod_{i \in I} \coprod_{j \in J_i} X_{i,j} = \coprod_{t \in \prod_{i \in I} J_i} \prod_{i \in I} X_{i,t(i)}$, so that in particular we can construct products of infinite sums of affine schemes, but it did not work out: There is just a map from the right to the left, but it is not an isomorphism. Thus, in particular I don't know if your example ($X = \coprod_{i \geq 0} \text{Spec}(k)$) works. –  Martin Brandenburg May 20 '11 at 8:08
Note that in the category of topological spaces, there is a canonical continuous bijection $\coprod_{t \in \prod_{i \in I} J_i} \prod_{i \in I} X_{i,t(i)} \to \prod_{i \in I} \coprod_{j \in J_i} X_{i,j}$, but it is only a homeomorphism in trivial cases. –  Martin Brandenburg May 20 '11 at 10:02
Martin, if the product of the sums of affine schemes existed then clearly it would have a canonical map from that sum of products. Are you asserting that there are cases where the product exists but the map is not an iso? Or only that there are cases where if the product did exist then the map would not be an iso? –  Tom Goodwillie May 20 '11 at 12:58
@Tom: In the comment, I'm just saying that the functors of points are not identical, without adressing existence issues. –  Martin Brandenburg May 20 '11 at 13:02

I will try to argue that if an product of quasi-compact schemes is a representable by a scheme, then all but finitely many of them must be affine.

Edit: I rewrote the argument a little to address the comments. I hope the argument might finally be satisfying...

Lemma. If $X_i$, $i \in I$ are quasi-compact schemes over a field $k$, then $\prod X_i$ has a closed point (i.e., there is a map from the spectrum of some field $K$ to $X$ that is representable by closed immersions) whose image in each of the $X_i$ is also closed.

Proof. For each $i$, let $x_i$ be a closed point of $X_i$ (which exists because the $X_i$ are quasi-compact). Let $K$ be a compositum of the residue fields of the $x_i$ (i.e., the residue field at a closed point of $Z = \prod_i x_i$, which exists because $Z$ is affine). Let $X$ be the product of the $X_i$. Then we get a map $\mathrm{Spec} \: K \rightarrow X$, which we have to show is a closed embedding.

It's enough to argue that the map $Z \rightarrow X$ is a closed embedding. But if $Y$ is a scheme over $X$ then $Y_Z$ can be obtained as the intersection of closed subschemes $Y_{x_i} = Y \times_{X_i} x_i$, which is a closed subscheme. $\Box$

Suppose that the product of schemes $X_i$, indexed by $i \in I$, is representable by a non-empty scheme $X$. Assume that each $X_i$ is quasi-compact, so that for each $i$, we can find a Zariski cover $Y_i \rightarrow X_i$ of $X_i$ by an affine scheme $Y_i$. Then we get a map $Y = \prod Y_i \rightarrow X$ and the source is an affine scheme. Let $x$ be a closed point of $X$. Let $U$ be an affine neighborhood of $x$ in $X$. Let $y$ be a $K$-point of $Y$ above $x$, and let $V$ be a distinguished affine open neighborhood of $y$ in $Y$, contained in the pre-image of $U$, and defined by the invertibility of some $f \in \Gamma(Y, {\cal O}_Y)$. Since $f$ can be expressed using functions defined on finitely many of the $Y_i$, the set $V$ must be the pre-image of a distinguished open affine $V' \subset \prod_{i \in I'} Y_i$ where $I'$ is a finite subset of $I$. Let $I''$ be the complement of $I'$ in $I$ and let $X' = \prod_{i \in I'} X_i$ and define $X''$, $Y'$, $Y''$ analogously. Then $V' \times Y''$ is an affine open neighborhood of $y$ in $Y$, contained in $V$. If $U'$ denotes the image of $V'$ in $X'$ then $U' \times X''$ is an open neighborhood of $x$ in $X$, contained in $U$. Let $x'$ be the projection of $x$ to $X'$, which we can safely assume is a closed point of $X'$, hence a closed point of $U'$, and let $K$ be the residue field of $x'$. Therefore $\{ x' \} \times X''$ is a closed subscheme of $U$ and is therefore affine. But $\{ x' \} \times X'' = K \otimes_k X''$ so by fpqc descent for affine schemes over $k$ (SGA1.VIII.2) it follows that $X''$ is affine.

Replacing $I$ with $I''$ we can now assume that $X$ is affine. Before proving that each $X_i$ is affine, I claim that each $X_i$ is separated. We must show that for any scheme $Y$ and any two maps $Y \rightrightarrows X_i$, the locus in $Y$ where the maps agree is closed. Let $x$ be a closed $K$-point of $\prod_{j \not= i} X_j$ for some extension $K$ of $k$. Then we can extend the maps $Y \rightrightarrows X_i$ to maps $Y \otimes_k K \rightrightarrows X$. Since $X$ is separated, the locus in $Y \otimes_k K$ where these maps agree is closed. But this is the pullback via the map $Y \otimes_k K \rightarrow Y$ of the locus where the maps $Y \rightrightarrows X_i$ agree, so this latter subset of $Y$ must be closed by fpqc descent (see SGA1.VIII.4).

If $X$ is affine and non-empty then it is separated, so each $X_i$ is separated. Therefore $Z := \prod_{j \not= i} X_j$ is separated for each $i$. (Note that we don't yet know that $Z$ is representable by a scheme, but it still makes sense to claim that $Z$ is separated as a functor: for any scheme $T$ and any two elements $a$ and $b$ of $Z(T)$, the locus in $T$ where $a$ and $b$ agree is closed.) Let $z$ be a closed $K$-point of $Z$. Then $X_i \otimes_k K = X \times_Z \{z\}$ is closed in $X$, hence is affine because we assumed $X$ to be affine. This implies $X_i$ is affine by fpqc descent (see SGA1.VIII.2).

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For the last question, you can argue as follows. Assume all $X_i$ are nonempty $k$-schemes, and $X$ is a scheme. Let us first show that all $X_i$ are separated if $X$ is. Fix $i\in I$. The projection $X\to \prod_{j\neq i} X_j$ is a separated morphism, hence has separated fibers. Since the target is nonempty it has a $k'$-valued point for some extension $k'$ of $k$. Hence $X_i\otimes_k k'$ is separated, and so is $X_i$ by descent. Assume now that $X$ is affine: since all $X_i$ are separated, we can repeat the argument with "separated" replaced by "affine". –  Laurent Moret-Bailly May 25 '11 at 7:48
Thanks. You both assume that also the fiber product with respect to a cofinite subset of $I$ exist, right? Of course this is automatic if we only consider powers of a fixed scheme. @Jonathan Wise: I assume that $x$ means a $k$-point of $W'$ (which is then mapped to $\prod_{i \in I'} X_i$), right? But why does it have to exist? In general, there are only $k'$-points for some $k'/k$, and $k$ alg. closed does not help to reduce to $k'=k$. –  Martin Brandenburg May 25 '11 at 10:04
@Laurent Moret-Bailly: In the case that the fiber product is affine, you seem to use that the projection (away from one factor) is affine, why is this so? Perhaps we first have to prove that $\prod_{j \neq i} X_i$ is separated? But I don't know how it works. –  Martin Brandenburg May 25 '11 at 12:59
@Martin: I had overlooked the ''cofinite subproduct'' bit. If $I'$ is finite and $I''$ is its complement, we can pick a point $x$ in $\prod_{i\in I'}X_i$ and conclude that $\left(\prod_{i\in I''}X_i\right)\otimes_k \kappa(x)$ is a scheme if $X$ is. This at least answers the objection if each $X_i$ has a rational point. –  Laurent Moret-Bailly May 25 '11 at 13:37
@Martin: I did treat the separated case first. –  Laurent Moret-Bailly May 25 '11 at 13:43

I believe (Edit: don't believe) that the following is true:

If for each $i\in I$ the scheme $X_i$ is a filtered union of affine open subschemes (i.e. if the union of any two affine opens is contained in some affine open) then the product $\prod_{i\in I}X_i$ exists.

Proof: Take the direct limit, over all ways $V:i\mapsto V_i$ of choosing an affine open subscheme in each $X_i$, of the affine scheme $\prod_{i\in I}V_i$. The point is that this gives the right $Hom(Y, -)$ when $Y$ is affine (hence quasicompact) because any map $Y\to X_i$ must factor through some affine open $V\subset X_i$.

Edit: This direct limit of schemes might not exist, as pointed out in the comments. In any case where it fails to exist, it appears to me that the product does not exist. Reason: There are compatible maps $\prod_{i\in I}V_i\to\prod_{i\in I}X_i$ inducing a natural map of sets from $colim_V(Hom(Y,\prod_{i\in I}V_i))$ to $Hom(Y,\prod_{i\in I}X_i)=\prod_{i\in I}Hom(Y,X_i)$ for a scheme $Y$. When $Y$ is affine, this is an isomorphism because $Hom(Y,X_i)$ is then the colimit of $Hom(Y,V)$ over affine $V\subset X_i$. I'm saying that if the product of these direct limits of affines exists, then the direct limit of products of affines must also exist and must coincide with it.

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Why is this direct limit a scheme? If, for each $i$, you have an inclusion $V_i\hookrightarrow V'_i$ of affine open subschemes, we do get a monomorphism $\prod_{i\in I} V_i\to\prod_{i\in I}V'_i$, but this won't be an open immersion in general. –  Laurent Moret-Bailly May 20 '11 at 17:25
Good point. So when this is not a scheme, does that show that in this case the product of schemes does not exist? –  Tom Goodwillie May 20 '11 at 17:57
If I understand your edit correctly, you want to deflect the problem to the non-existence of a certain categorical filtered colimit. But in your proof you use some property which a categorical filtered colimit might not have, namely that every map from a quasicompact scheme into it has to factor over one of the involved schemes. –  Martin Brandenburg May 21 '11 at 5:28
I see what you mean. Here's what I meant to say. If $\prod_{i\in I}X_i$ exists then it turns to be also a colimit (over all functions $i\mapsto V_i$) of $\prod_{i\in I}V_i$. Reason: whenever you have compatible maps $\prod_{i\in I}V_i\to Z$ for all such functions then you get a map $\prod_{i\in I}X_i\to Z$, because it's enough to define that map on open affines of $\prod_{i\in I}X_i$ and each of these actually is in some $\prod_{i\in I}V_i$, because its projection to $X_i$ is inside some affine open, which you can choose to be $V_i$. Does that work? –  Tom Goodwillie May 21 '11 at 12:25
No because you again use arguments which are - a priori - only valid if we already knew that the products exist and behave in the usual way. –  Martin Brandenburg May 23 '11 at 14:32