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Here is another example with a rigorous proof (which is a collaboration with "owk").

Example. Let $R$ be a discrete valuation ring, $I$ an infinite set. Glue two copies of $\text{Spec}(R)$ along the generic point to get a $R$-scheme $X$. Then in the category of $R$-schemes the power $X^I$ does not exist.

Proof: Write $\text{Spec}(R) = {\eta,\mathfrak{m}}$, \{\eta,\mathfrak{m}\}$, where$\eta$is the generic point and$\mathfrak{m}$is the special point. Let$K$be the quotient field and$k$the residue field of$R$. Assume$P = X^I$exists in the category of$R$-schemes. For an$R$-scheme$T$, a$T$-valued point of$X$corresponds to an open covering$T = T_1 \cup T_2$such that$T_1 \cap T_2 = T_{\eta}$; this also gives a description of the$T$-valued points of$P$. . If we apply this to$K$-schemes or$k$-schemes, we see$X \times_R K = \text{Spec}(K)$and$X \times_R k = \text{Spec}(k) \coprod \text{Spec}(k) = \text{Spec} k[x]/(x^2-x)$. Now the reduction$X(R) \to X(k)$is bijective: It maps$(\text{Spec}(R),{\eta}), (\text{Spec}(R),\{\eta\}), ({\eta},\text{Spec}(R))$\{\eta\},\text{Spec}(R))$ to $(\text{Spec}(k),\emptyset), (\emptyset,\text{Spec}(k))$. From $P(T)=X(T)^I$ we deduce that also $P(R) \to P(k)$ is bijective.

Since fibers may be described by fiber products and fiber products commute with fiber products by general nonsense, we get as $K$-schemes

$P_{\eta} = (X \times_R K)^I = \text{Spec}(K)^I = \text{Spec}(K)$.

Let us denote the unique point in $P_{\eta}$ also by $\eta$. As $k$-schemes, we get

$P_{\mathfrak{m}} = (X \times_R k)^I = \text{Spec}(k[(x_i)_{i \in I})/(x_i^2-x_i)_{i I}]/(x_i^2-x_i)_{i \in I})$.

We see that $P_{\mathfrak{m}}$ is homeomorphic to ${0,1}^I$, \{0,1\}^I$, in particular it is not discrete. Remark that$P_{\mathfrak{m}}$is not open in$P$since otherwise we would get the contradiction$P(R)=\emptyset$. Also remark that$P_{\mathfrak{m}}$may be identified with$P(k)$, on which$\text{Aut}(P)$acts transitively. Next we want to show that$\eta$is a generic point of$P$. If not, let$U$be a nonempty open subset of$U$with$\eta \notin U$. Then$U \subseteq P_{\mathfrak{m}}$and it follows that$P_{\mathfrak{m}}$is the union of the$\sigma(U)$,$\sigma \in \text{Aut}(P)$, and therefore open, contradiction. Since$P_{\mathfrak{m}}$is not discrete, there is some nonempty open subset$\text{Spec}(A) \subseteq P$which contains two points$p_1,p_2 \in P_{\mathfrak{m}}$. They induce$p_1,p_2 \in P(k) \cong P(R)$. Since$R$is local,$p_1,p_2$are induced by$p_1,p_2 \in \text{Spec}(A)(R)$. But now$\text{Spec}(A)(R) \subseteq \text{Spec}(A)(K) = P(K)= {\eta}$, \{\eta\}$, thus $p_1=p_2$, contradiction. -qed

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Here is another example with a rigorous proof (which is a collaboration with "owk").

Example. Let $R$ be a discrete valuation ring, $I$ an infinite set. Glue two copies of $\text{Spec}(R)$ along the generic point to get a $R$-scheme $X$. Then in the category of $R$-schemes the power $X^I$ does not exist.

Proof: Write $\text{Spec}(R) = {\eta,\mathfrak{m}}$, where $\eta$ is the generic point and $\mathfrak{m}$ is the special point. Let $K$ be the quotient field and $k$ the residue field of $R$. Assume $P = X^I$ exists in the category of $R$-schemes.

For an $R$-scheme $T$, a $T$-valued point of $X$ corresponds to an open covering $T = T_1 \cup T_2$ such that $T_1 \cap T_2 = T_{\eta}$; this also gives a description of the $T$-valued points of $P$. If we apply this to $K$-schemes or $k$-schemes, we see $X \times_R K = \text{Spec}(K)$ and $X \times_R k = \text{Spec}(k) \coprod \text{Spec}(k) = \text{Spec} k[x]/(x^2-x)$. Now the reduction $X(R) \to X(k)$ is bijective: It maps $(\text{Spec}(R),{\eta}), ({\eta},\text{Spec}(R))$ to $(\text{Spec}(k),\emptyset), (\emptyset,\text{Spec}(k))$. From $P(T)=X(T)^I$ we deduce that also $P(R) \to P(k)$ is bijective.

Since fibers may be described by fiber products and fiber products commute with fiber products by general nonsense, we get as $K$-schemes

$P_{\eta} = (X \times_R K)^I = \text{Spec}(K)^I = \text{Spec}(K)$.

Let us denote the unique point in $P_{\eta}$ also by $\eta$. As $k$-schemes, we get

$P_{\mathfrak{m}} = (X \times_R k)^I = \text{Spec}(k[(x_i)_{i \in I})/(x_i^2-x_i)_{i \in I})$.

We see that $P_{\mathfrak{m}}$ is homeomorphic to ${0,1}^I$, in particular it is not discrete. Remark that $P_{\mathfrak{m}}$ is not open in $P$ since otherwise we would get the contradiction $P(R)=\emptyset$. Also remark that $P_{\mathfrak{m}}$ may be identified with $P(k)$, on which $\text{Aut}(P)$ acts transitively.

Next we want to show that $\eta$ is a generic point of $P$. If not, let $U$ be a nonempty open subset of $U$ with $\eta \notin U$. Then $U \subseteq P_{\mathfrak{m}}$ and it follows that $P_{\mathfrak{m}}$ is the union of the $\sigma(U)$, $\sigma \in \text{Aut}(P)$, and therefore open, contradiction.

Since $P_{\mathfrak{m}}$ is not discrete, there is some nonempty open subset $\text{Spec}(A) \subseteq P$ which contains two points $p_1,p_2 \in P_{\mathfrak{m}}$. They induce $p_1,p_2 \in P(k) \cong P(R)$. Since $R$ is local, $p_1,p_2$ are induced by $p_1,p_2 \in \text{Spec}(A)(R)$. But now $\text{Spec}(A)(R) \subseteq \text{Spec}(A)(K) = P(K)= {\eta}$, thus $p_1=p_2$, contradiction.