Relation between indiscernibles for $L$ and for $L[A]$

It is known that $L\models 2^\kappa=\kappa^+$, and that for a set of ordinals $A$ we know that $L[A]\models \exists\lambda\forall\kappa>\lambda(2^\kappa=\kappa^+)$.

In this sense, there is some similarity between $L$ and $L[A]$. Both models have definable well orderings, and both models have a very nice sense of minimality deep within them.

Assuming $0^\sharp$ exists we have the class of Silver indiscernibles from which can define $L$ (using the definable Skolem functions, and the Skolem hull of $I$). Assuming that $A^\sharp$ exists we have a similar class for $L[A]$ as well.

Denote by $I$ the Silver indiscernibles and $I_A$ the corresponding class of $L[A]$.

Is there any intersection between them? Is there some $\alpha$ such that $I\setminus\alpha=I_A\setminus\alpha$?

If the answer to both questions is no in the general case, can we say anything on particular cases known?

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There are several issues.

• First, of course, when $A\in L$ then $L[A]=L$ and consequently $I_A=I$.

• In any case, there will be large overlap in $I_A$ and $I_B$, since they are both class clubs, and hence intersect in a closed unbounded class of ordinals.

• If $0^\sharp\in L[A]$, then every cardinal of $L[A]$ is a cardinal in $L[0^\sharp]$, and hence is an $L$-indiscernible, but of course, not every cardinal of $L[A]$ is an $L[A]$-indiscernible. So the eventual agreement you requested does not generally hold.

• In any case, $I_A\subset I$.

• It can be that $I_A=I$ even when $A\notin L$. For example, I believe that this is the case when $A$ is an $L$-generic Cohen real added by forcing, since one can lift any $j:L\to L$ to $j:L[A]\to L[A]$.

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Thanks, Joel. I did not think about the club-ness of the class. I did have in mind the last point you had. Thanks for the quick answer! :-) – Asaf Karagila Aug 15 '11 at 0:18

It can also be the case that $I_A$ is a periodic (but club) subclass of $I$: by Jensen Coding one can define (necessarily by class forcing) reals $a\subset\omega$ with $0^\sharp \notin L[a]$ and that there are countable ordinals $\alpha,\beta$ so that for all $\tau\in On$ the $\tau$'th silver indiscernible for $a$ is the $\alpha + \beta\cdot\tau$'th indiscernible for $L$.

Also: this must be a class forcing, for any set forcing $P\in L$, and any $G$, $P$-generic over $L$, the $L[G]$-indiscernibles are exactly the $L$-indiscernibles from some point on.

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