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Suppose 0# exists.

It is clear that every order preserving map from the indiscernibles to the indiscernibles gives an elementary embedding from $L$ to $L$. Furthermore, following lemmas 18.7 and 18.8 of Jech, if $\alpha$ is an infinite infinite limit ordinal, an increasing map from alpha to beta gives an elementary embedding from $L_{i_\alpha}$ to $L_{i_\beta}$, where $i_\alpha$ is the $\alpha$-th indiscernible. This is because $L_{i_\alpha}$ equals the Skolem hull in itself of the first $\alpha$ indiscernibles. However, I am not clear on the following points.

1) Is it the case that for a finite successor ordinal, n, $L_{i_n}$ is necessarily equal to the Skolem hull in $L_{i_n}$ of the first n indiscernibles? Jech only proves this result for infinite ordinals.

2) Is it possible that there could be an elementary embedding from $L$ to $L$, or from $L_{i_\alpha}$ to $L_{i_\beta}$ ($\alpha, \beta$ may be finite or infinite), that does not always map indiscernibles to indiscernibles? This sounds weird, but I'm not convinced it's impossible. As far as I know, there's no formula in $L$ that defines "$\alpha$ is a Silver indiscernible." (In fact there is no such formula -- see Andreas Blass's comment below.)

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  • $\begingroup$ Just a quick comment about the last sentence in the question: There is certainly no formula defining in $L$ the notion of Silver indiscernible. Proof: If we had such a formula, we could express in $L$ the property "$\alpha$ is the first Silver indiscernible", thereby discerning the first from all the other Silver indiscernibles. $\endgroup$
    – Andreas Blass
    Commented Aug 20, 2011 at 1:55
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    $\begingroup$ The second question, at least for the case of L to L, also has a negative answer, i.e., the indiscernibles have to map to themselves. I will post an answer later today or tomorrow (unless someone else does that before me). $\endgroup$
    – Ali Enayat
    Commented Aug 20, 2011 at 21:40
  • $\begingroup$ Norman: I deleted my answer since, as pointed out by Andreas Blass my formulation contains a fatal flaw, and since I am travelling, I don't have the wherewithal to fix it in the near future; I will check back upon my return. $\endgroup$
    – Ali Enayat
    Commented Aug 23, 2011 at 0:10

2 Answers 2

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The answer to question 1 is no. Let $n$ be a finite ordinal, and consider the structure $M$ with universe $L_{i_n}$, with constant symbols for the smaller Silver indiscernibles $i_0,\dots,i_{n-1}$ as well as symbols for the membership relation $\in$ and the usual, $L$-definable Skolem functions. This structure $M$ is constructible. (This is where it's essential that $n$ is finite.) So the Skolem hull, the smallest elementary substructure $N$ of $M$, is constructibly countable. But $M$ itself is very large in the sense of $L$, since Silver indiscernibles like $i_n$ are constructibly inaccessible (and much more). Therefore $N$ is not all of $M$.

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  • $\begingroup$ The same idea shows that, if we take $0^{\#}$, a theory with a countable infinity of constant symbols for indiscernibles, and restrict to just a finite number $n$ of indiscernibles, then this fragment of $0^{\#}$ is in $L$. $\endgroup$
    – Andreas Blass
    Commented Aug 20, 2011 at 19:44
  • $\begingroup$ The same argument shows that the hull of the first $\beta=\alpha+1$ many indiscernibles, even when $\alpha$ is infinite, does not exhaust $L_{i_\beta}$, since it will be contained in the hull of $L_{i_\alpha+1}$, which is in $L$ and much smaller than $i_\beta$ in $L$. So probably Norman meant to say that Jech proves (1) only for limit ordinals. $\endgroup$
    – Joel David Hamkins
    Commented Aug 21, 2011 at 0:20
  • $\begingroup$ Joel, I disagree, if you have any infinite set of indiscernibles (and enough combinatorics to talk about satisfaction for set models) then you can define $0^\#$. $\endgroup$
    – François G. Dorais
    Commented Aug 21, 2011 at 2:29
  • $\begingroup$ @Francois: I think Joel's argument is OK. Although the first line of his comment mentions the hull of an infinite set of indiscernibles, which $L$ can't see, he ultimately uses the fact that this hull is included in the hull of $L_{i_\alpha+1}$, and $L$ does see this. Its constructible cardinality is "only" $i_\alpha$, so it's smaller than $L_{i_{\alpha+1}}$. $\endgroup$
    – Andreas Blass
    Commented Aug 21, 2011 at 2:53
  • $\begingroup$ Yes, I meant it as Andreas explains. The point is that the hull of a set of indiscernibles does not include that set, but is determined by the union of the hull of all its finite subsets. So it is contained in the hull of any larger set. $\endgroup$
    – Joel David Hamkins
    Commented Aug 21, 2011 at 2:58
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The answer to Q2 is 'No'. Suppose $j:L\rightarrow L$ is a non-trivial elementary embedding. We use the following fact:

$\bullet$ $cp(j)$ (the first ordinal moved by $j$) is always a Silver indiscernible.

Now let $I$ be the class of Silver indiscernibles, and $\delta \in I$ but $j(\delta)\notin I$ for a contradiction. Let $H$ be the Skolem hull in $L$ of $j(\delta)\cup j$''$I\backslash (\delta +1)$. $H$ is isomorphic to $L$. If $j(\delta)\notin H$ but $\pi:H \rightarrow L$ is the transitive collapse, then $\pi^{-1}:L\rightarrow L$ is non-trivial with critical point $j(\delta)$. Hence, by the Fact, $j(\delta)$ must be in $H$. Then we see that for some $\vec \xi <j(\delta)$ some $\overrightarrow{j(\zeta)} > j(\delta)$ with $\vec \zeta \in I\backslash (\delta +1)$ that

$L \models $ ''$\exists \vec \xi < j(\delta)( j(\delta) = t(\vec \xi ,\overrightarrow{j(\zeta)}))$''.

for some term $t$. But then:

$L \models $ ''$\exists \vec \xi < \delta( \delta = t(\vec \xi, \overrightarrow{\zeta}))$''

is a definition of the indiscernible $\delta$ from larger indiscernibles and smaller ordinals, which is impossible. (This works for the variant of the question, taking embeddings between sets, if $\alpha, \beta$ are limit ordinals.)

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  • $\begingroup$ I don't think the argument goes through for embeddings between sets (the variant of the question). In fact, let $\alpha$ be a limit ordinal and work in $L[G]$ where $G$ collapses $\iota_\alpha$ to $\omega$, and let $j:L_{\iota_\alpha}\to L_{\iota_\alpha}$ be elementary with $j\in L[G]$ and $\mathrm{crit}(j)=\iota_0$. Let $\left<\kappa_n\right>_{n<\omega}$ be the critical sequence of $j$. Then there is $n$ such that $\kappa_n$ is not an indiscernible: otherwise, since $\left<\kappa_n\right>_{n<\omega}\in L[G]$, we get $0^\#\in L[G]$. @PhilipWelch $\endgroup$
    – Farmer S
    Commented Jun 25, 2021 at 16:33
  • $\begingroup$ @Farmer: I do agree with you on the countable set version, I was too sweeping, and yes it properly completes the original question 2. (I should add that alpha (and beta) are uncountable.) Thank you! $\endgroup$
    – Philip Welch
    Commented Jun 25, 2021 at 20:50

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