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  • $\begingroup$ Not to be petty about terminology, but aren't the theorems about $0^\#$ and the stationary class where $\sf GCH$ fails are more of corollaries to the first theorem? $\endgroup$ Commented Nov 9, 2013 at 13:07
  • $\begingroup$ Yes, that's right. I guess I'll edit, once I'm done with breakfast... $\endgroup$ Commented Nov 9, 2013 at 13:09
  • $\begingroup$ Bon appetit! :-) $\endgroup$ Commented Nov 9, 2013 at 13:09
  • $\begingroup$ One approach which might help to solve the case for $L[c]$ where $c$ is Cohen, is to see if $M\subseteq V$ is a model of $\sf ZF$ then $j\colon V\to L$ can be restricted to $M$ as well. In that case, I feel, we may be able to conclude that if $M\subseteq V$ is a model of $\lnot\sf AC$ then there is no such $j$. This will exclude Cohen extensions of $L$, but not Sacks or Miller extensions which are minimal. $\endgroup$ Commented Nov 9, 2013 at 13:23
  • 2
    $\begingroup$ As a corollary to your corollary - if there is such $j:V\rightarrow L$ then $GCH$ hold everywhere (except maybe at $\aleph_0$). Since $0^\#$ doesn't exist, we can use the covering lemma: let $\lambda$ be uncountable cardinal in $V$. $j^{\prime\prime} \lambda \subset B \in L$ where $|B|^V=\lambda$. Now $|B|^L = \mu < (\lambda^+)^V$ but by the proof of the first theorem, $|P^V (\lambda)| \leq |P^L(B)| = (\mu^+)^L$, so $(2^\lambda)^V = (\mu^+)^L \leq (\lambda^+)^V$. $\endgroup$ Commented Nov 10, 2013 at 6:41