TITLE: If $0^{\sharp}$ exists, then every uncountable cardinal in $V$ is as large as it can be in $L$.
QUESTION [5 upvotes]: According to Wikipedia, if $0^{\sharp}$ exists, then every uncountable cardinal in $V$ satisfies every large cardinal property in $L$ that can be realized in $L$, e.g. weak compactness, total ineffability, etc.  It's easy enough to see why every uncountable in $V$ will be inaccessible, or even Mahlo, in $L$.
How can one see that some of the slightly larger large cardinal properties (e.g. weak compactness, total ineffability, etc.) are satisfied in $L$ by the uncountable cardinals in $V$?  Is there a good reference for some of these results?

REPLY [9 votes]: If $0^\sharp$ exists, then every uncountable cardinal
$\kappa$ of $V$ is one of the Silver indiscernibles in $L$,
and this implies that $L_\kappa$ is an elementary substructure of $L$. This implies
that $\kappa$ is a limit cardinal in $L$ and therefore,
since some of the indiscernibles are regular, that $\kappa$
is inaccessible in $L$. Every order-preserving map on the
indiscernibles induces an elementary embedding $j:L\to L$,
and thus every indiscernible is the critical point of such
a $j$. From this, it follows that every such $\kappa$ has
the tree property in $L$, because if $T$ is any
$\kappa$-tree in $L$, then $j(T)$ has nodes on the
$\kappa$-th level, which gives you a $\kappa$-branch in
$T$. But being inaccessible and having the tree property is
equivalent to being weakly compact, so every such $\kappa$
is weakly compact in $L$. You can get other properties by
arguing with the embedding like this.