Followup to last question: When is $f$ an elementary embedding in $L$? Following up on my previous question, I've also been thinking about why the existence of $0^{\#}$ doesn't imply that $L$ satisfies the existence of a Reinhardt cardinal. 
$0^{\#}$ implies the existence of a nontrivial elementary embedding $j:L\rightarrow L$. Using some properties of Godel operations, one can show that any such $j$ is included in $L$. Using the fact that $L\models\varphi[x]$ iff $L\models(L\models\varphi[x])$, one can show that any such $j$ is an elementary embedding in $L$. Finally, using the fact that $L$ is an inner model, one can show that $j$ is nontrivial in $L$. Why is this not downward absolute to $L$?
When is a function $f$ an elementary embedding in $L$?
Given a transitive inner model $M$ of $ZF$, what requirements are there that $M$ satisfies the existence of a Reinhardt cardinal?
Unlike the previous question, the piece of information I am missing here isn't that the domain of the embedding in question is not a model of ZFC (in this question it is $L$). So what information am I missing?
 A: A key missing ingredient is amenability. Suppose $M,N$ are transitive models of enough set theory and $j\!:M\to N$ is elementary. We say that $j$ is amenable to $M$ if and only if $j\upharpoonright M\cap V_\alpha\in M$ for all $\alpha.$
For instance, any elementary embedding $j\!:V\to M$ is amenable.
What Kunen's theorem establishes in terms of this notion is that, provably in $\mathsf{ZFC}$, if $M$ is a transitive model of set theory, and $j\!:M\to M$ is elementary, then $j$ is not amenable to $M$.
Now, if $j\!:L\to L$ is elementary, what we conclude is not that set theory is inconsistent (we do not get that there are Reinhardt cardinals in $L$) but rather that $j$ is not amenable to $L$. 
It is a useful exercise to make this explicit (without directly appealing to Kunen's result). For instance, if $\alpha>\operatorname{cp}(j)$ is a cardinal, then $j\upharpoonright L_\alpha\notin L$. One way to see this is to note that otherwise the $L$-ultrafilter on $\kappa=\operatorname{cp}(j)$ derived from $j$ (that is, $\{X\in \mathcal P(\kappa):\kappa\in j(X)\}$) would be an element of $L$, since it is definable from $j\upharpoonright L_\alpha$. This would imply, for example, that $0^\sharp\in L$.  

A few years ago, I gave a series of talks at the 15th Latin American Symposium in Mathematical Logic, in Bogotá, on determinacy and inner models. Amenability is one of the topics I covered, and the above is based on my notes.
