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?



Relation between indiscernibles for $L$ and for $L[A]$
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 $j:L[A]\to L[A]$.


