Infinite products of schemes usually do not exist. For example in Examples, Section 110.55 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. 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.
\[ T = \mathop{\mathrm{lim}}\nolimits _{\{ i_1, \ldots , i_ n\} \subset I} T_{i_1} \times _ S \ldots \times _ S T_{i_ n} \]
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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