Thinking of arbitrary tensor products of rings, $A=\otimes\_i A\_i$ ($i\in I$, an arbitrary index set), I have recently realized that $Spec(A)$ should be the product of the schemes $Spec(A\_i)$, a priori in the category of affine schemes, but actually in the category of schemes, thanks to the string of equalities (where $X$ is a not necessarily affine scheme)


$$ Hom\_{Schemes} (X, Spec(A))= Hom\_{Rings}(A,\Gamma(X,\mathcal O))=\prod\_ {i\in I}Hom\_{Rings}(A\_i,\Gamma(X,\mathcal O)) $$


$$ =\prod\_ {i\in I}Hom\_{Schemes}(X,Spec(A\_i))$$


Since this looks a little too easy, I was not quite convinced it was correct but a very reliable colleague of mine reassured me by explaining that the correct categorical interpretation of the more down to earth formula above is that the the category of affine schemes is a reflexive subcategory of the category of schemes.
(Naturally the incredibly category-savvy readers here know that perfectly well, but I didn't at all.)


And now I am stumped: I had always assumed that infinite products of schemes don't exist and I realize I have no idea why I thought so!


Since I am neither a psychologist nor a sociologist, arguments like "it would be mentioned in EGA if they always existed " don't particularly appeal to me and I would be very grateful if some reader could explain to me what is known about these infinite products.



Arbitrary products of schemes don't exist, do they?
Let me rephrase the question (and Ilya's answer). Given an arbitrary collection $X\_i$ of schemes, is the functor (on affine schemes, say)


$Y \mapsto \prod\_i Hom(Y, X\_i)$


representable by a scheme? If the $X\_i$ are all affine, the answer is yes, as explained in the statement of the question. More generally, any filtered inverse system of schemes with essentially affine transition maps has an inverse limit in the category of schemes (this is in EGA IV.8). The topology in that case is the inverse limit topology, by the way.


It is easy to come up with examples of infinite products of non-separated schemes that are not representable by schemes. This is because any scheme has a locally closed diagonal. In other words, if $Y \rightrightarrows Z$ is a pair of maps of schemes then the locus in $Y$ where the two maps coincide is locally closed in $Y$.


Suppose $Z$ is the affine line with a doubled origin. Every distinguished open subset of an affine scheme $Y$ occurs as the locus where two maps $Y \rightrightarrows Z$ agree. Let $X = \prod\_{i = 1}^\infty Z$. Every countable intersection of distinguished open subsets of $Y$ occurs as the locus where two maps $Y \rightarrow X$ agree. Not every countable intersection of open subsets is locally closed, however, so $X$ cannot be a scheme.


Since the diagonal of an infinite product of separated schemes is closed, a more interesting question is whether an infinite product of separated schemes can be representable by a scheme. Ilya's example demonstrates that the answer is no.


Let $Z = \mathbf{A}^2 - 0$. This represents the functor that sends $Spec A$ to the set of pairs $(x,y) \in A^2$ generating the unit ideal. The infinite product $X = \prod\_{i = 1}^\infty Z$ represents the functor sending $A$ to the set of infinite collections of pairs $(x\_i, y\_i)$ generating the unit ideal. Let $B$ be the ring $\mathbf{Z}[x\_i, y\_i, a\_i, b\_i]\_{i = 1}^\infty / (a\_i x\_i + b\_i y\_i = 1)$. There is an obvious map $Spec B \rightarrow X$. Any (nonempty) open subfunctor $U$ of $X$ determines an open subfunctor of $Spec B$, and this must contain a distinguished open subset defined by the invertibility of some $f \in B$. Since $f$ can involve at most finitely many of the variables, the open subset determined by $f$ must contain the pre-image of some open subset $U'$ in $\prod\_{i \in I} Z$ for some finite set $I$. Let $I'$ be the complement of $I$. If we choose a closed point $t$ of $U'$ then $U$ contains the pre-image of $t$ as a closed subfunctor. Since the pre-image of $t$ is $\prod\_{i \in I'} Z \cong X$ this shows that any open subfunctor of $X$ contains $X$ as a closed subfunctor.


In particular, if $X$ is a scheme, any non-empty open affine contains a scheme isomorphic to $X$ as a closed subscheme. A closed subscheme of an affine scheme is affine, so if $X$ is a scheme it is affine.


Now we just have to show $X$ is not an affine scheme. It is a subfunctor of $W = \prod\_{i = 1}^\infty \mathbf{A}^2$, so if $X$ is an affine scheme, it is locally closed in $W$. Since $X$ is not contained in any closed subset of $W$ except $W$ itself, this means that $X$ is open in $W$. But then $X$ can be defined in $W$ using only finitely many of the variables, which is impossible.


**Edit:** Laurent Moret-Bailly pointed out in the comments below that my argument above for this last point doesn't make sense. Here is a revision: Suppose to the contrary that $X$ is an affine scheme. Then the morphism $p : X \rightarrow X$ that projects off a single factor is an affine morphism. If we restrict this map to a closed fiber then we recover the projection from $Z$ to a point, which is certainly not affine. Therefore $X$ could not have been affine in the first place.


