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"$0^\#$ exists" is an informally stated large cardinal axiom that is to be understood as "there is an uncountable set of Silver indiscernibles", "every uncountable cardinal is a Silver indiscernible", "there is a nontrivial elementary embedding $j:L\to L$ consistent with the axiom of choice and with $j$ definable in $V$" and many other equivalent formulations.

$0^\#$ is a (real number coding of) an infinite subset of $\mathbb N$ and thus ZFC can trivial prove its existence as it can construct $\mathcal P(\mathbb N)$ and thus indirectly trivially prove the existence of $0^\#$. This is why I gave clarification for the axiom "$0^\#$ exists" above. Similarly, ZFC can prove the existence of $\omega_1$ and thus indirectly prove the existence of all countable ordinals, which includes some Silver indiscernibles. Talking about "the first Silver indiscernible" in $L_{\omega_1}$ is also tricky, as to my understanding Silver indiscernibles are those ordinals, whose properties cannot be used to define uniquely identify them in $L$ (any $L$-definable property of a Silver indiscernible is true for all the rest of them in $L$).

KP cannot do either of those as it does not have the powerset axiom, nor can it even talk about uncountable sets. Thus adding large cardinal axioms is not necessary in order to strengthen KP consistency-wise, and large unrecursive countable ordinal axioms suffice.

My question is about where KP + "Silver indiscernibles exist", without any mention of uncountable sets, would fall relative to other theories or extensions of KP, consistency-wise, and if that extension even makes sense.

How about ZFC-(Powerset)+"Silver indiscernibles exist"?

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    $\begingroup$ I'm a bit confused. How are you phrasing "Silver indiscernibles exist"? (Also, you're mixing theories and structures in a way that confuses me; one thing I can say with confidence is that (assuming $0^\sharp$ exists) no individual Silver indiscernible is definable in $L_{\omega_1}$ ($\not=L_{\omega_1^L}$), and consequently the set of indiscernibles isn't definable in $L_{\omega_1}$ either. $\endgroup$
    – Noah Schweber
    Commented Apr 5, 2022 at 21:50
  • $\begingroup$ I vaguely recall that the existence of $0^\sharp$ is equivalent to the existence of a countable ordinal $\alpha$ such that $L_\alpha \models \mathsf{ZFC}$ and $L_\alpha$ contains an infinite indiscernible sequence. This statement makes perfect sense in $\mathsf{KP}$. How meaningful this would be in a weaker context than $\mathsf{ZFC}$ is unclear to me, though. $\endgroup$
    – James E Hanson
    Commented Apr 5, 2022 at 21:57
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    $\begingroup$ @JamesHanson That statement is actually weaker than the existence of $0^\#.$ By $\Sigma^1_2$ absoluteness, if $0^\#$ exists (and one really needs much less, e.g. an $\omega$-Erdos cardinal), your statement holds in $L$. You might be thinking of the fact that $0^\#$ exists iff there is an $L_\alpha$ with an uncountable set of order indiscernibles. $\endgroup$
    – Gabe Goldberg
    Commented Apr 5, 2022 at 22:43
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    $\begingroup$ The existence of $0^\#$ can be formulated in the language of second-order arithmetic (i.e., referring only to hereditarily countable objects) as the $\Sigma^1_3$ statement that there is a remarkable wellfounded EM set. $\endgroup$
    – Gabe Goldberg
    Commented Apr 5, 2022 at 22:50
  • $\begingroup$ I remember a talk by Ralf Schindler from quite a few years ago about formulation $0^\#$ in third(?) order arithmetic. $\endgroup$
    – Asaf Karagila ♦
    Commented Apr 5, 2022 at 22:53

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