Suppose instead that a finite set $F$ belongs to $\mathcal U$. If none of its singleton subsets belonged to $\mathcal U$, all their complements would belong to $\mathcal U$, and intersecting those complements with $F$ would put the empty set in $\mathcal U$. Hence $\{n\}\in\mathcal U$ for some $n\in F$.

Upward closure then puts every subset containing $n$ in $\mathcal U$, while no subset omitting $n$ can belong to it. Therefore
$$
\mathcal U=\{A\subseteq\mathbb N:n\in A\},
$$
the <principal ultrafilter> at $n$. Together with part i, this proves the dichotomy.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 144 / 2 / b / ii / Solution

Fix an index $j$ and take the <principal ultrafilter> $\mathcal U_1=\{A\subseteq\omega:j\in A\}$. Evaluation at the $j$th coordinate gives
$$
\prod_i\mathbb F_i/\mathcal U_1\cong\mathbb F_j,
$$
so the ultraproduct has characteristic $p_j$. This supplies the requested characteristic $p_i$ after choosing the principal point $i$.

Solved by gpt-5.6-sol high.

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 144 / 2 / a / Solution

{wiki=Ultrafilter#Principal_ultrafilters}

The principal ultrafilter at $x\in X$ is $\{A\subseteq X:x\in A\}$. An ultraproduct by a principal ultrafilter is isomorphic to the factor indexed by $x$.


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 Incoming links: Principal ultrafilter