A $\kappa$-complete prime ideal can't be second-order definable

Question

In "A note on the Combinatorial Principles $\diamondsuit(E)$" Devlin proves that ($\diamondsuit(E)\implies\exists$ disjoint $F,G\subseteq E$ s.t. $\diamondsuit(F)$ and $\diamondsuit(G)$ both hold) for subsets of $\omega_1$. In the proof he uses that $\omega_1$ is not measurable. To get this result for any regular uncountable $\kappa$ he states, that there can't be a $\kappa$-complete prime ideal, which is second-order definable. Is this a simple fact? Where could i find a proof for this?

Answer 1

You can find this embedded in the proof that a measurable cardinal is $\Sigma^2_0$-indescribable (in fact $\Pi^2_1$-indescribable!), which means that any second-order property (no matter how complex!) is reflected on a club below the measurable cardinal.

In particular, the existence of a $\kappa$-complete ultrafilter would reflect down, even if $\kappa$ is the least measurable cardinal.

The simplest proof I can think of for this fact uses ultrapowers, or some other elementary embedding. This should probably be in Jech "Set Theory", and definitely in Kanamori's book as Proposition 6.5 (originally due to Hanf and Scott).