Infinite products of schemes usually do not exist. For example in Examples, Section 110.56 it is shown that an infinite product of copies of \mathbf{P}^1 is not even an algebraic space.
On the other hand, infinite products of affine schemes do exist and are affine. Using Schemes, Lemma 26.6.4 this corresponds to the fact that in the category of rings we have infinite coproducts: if I is a set and R_ i is a ring for each i, then we can consider the ring
Given another ring A a map R \to A is the same thing as a collection of ring maps R_ i \to A for all i \in I as follows from the corresponding property of finite tensor products.
Lemma 32.3.1.slogan Let S be a scheme. Let I be a set and for each i \in I let f_ i : T_ i \to S be an affine morphism. Then the product T = \prod T_ i exists in the category of schemes over S. In fact, we have
and the projection morphisms T \to T_{i_1} \times _ S \ldots \times _ S T_{i_ n} are affine.
Proof. Omitted. Hint: Argue as in the discussion preceding the lemma and use Lemma 32.2.2 for existence of the limit. \square
Lemma 32.3.2. Let S be a scheme. Let I be a set and for each i \in I let f_ i : T_ i \to S be a surjective affine morphism. Then the product T = \prod T_ i in the category of schemes over S (Lemma 32.3.1) maps surjectively to S.
Proof. Let s \in S. Choose t_ i \in T_ i mapping to s. Choose a huge field extension K/\kappa (s) such that \kappa (s_ i) embeds into K for each i. Then we get morphisms \mathop{\mathrm{Spec}}(K) \to T_ i with image s_ i agreeing as morphisms to S. Whence a morphism \mathop{\mathrm{Spec}}(K) \to T which proves there is a point of T mapping to s. \square
Lemma 32.3.3. Let S be a scheme. Let I be a set and for each i \in I let f_ i : T_ i \to S be an integral morphism. Then the product T = \prod T_ i in the category of schemes over S (Lemma 32.3.1) is integral over S.
Proof. Omitted. Hint: On affine pieces this reduces to the following algebra fact: if A \to B_ i is integral for all i, then A \to \otimes _ A B_ i is integral. \square
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