For ordinals $\alpha<\beta$, we say $\alpha<\_{el}\beta$, if there is an elementary embedding with domain $L\_\beta$ and critical point $\alpha$.


Is $<\_{el}$ transitive?



Is this relation about elementary embedding transitive?
EDIT: If the codomain $N$ is allowed to be illfounded, then the answer is yes.
(Here if $\kappa=\mathrm{crit}(j)$ then this will mean that $\kappa\subseteq N$, but it might be that $N$ is illfounded and $\kappa$ is exactly the wellfounded part of the codomain, in which case $\kappa\notin N$.)


For let $\kappa\_0<\beta\_0=\kappa\_1<\beta\_1$ be ordinals and $j\_i:L\_{\beta\_i}\to N\_i$ be elementary (for $i=0,1$) with $\mathrm{crit}(j\_i)=\kappa\_i$. Let $U\_0$ be some non-principal ultrafilter derived from $j\_0$ (let $x\in N\_0$ such that $N\_0\models$"$\alpha\in x\in j\_0(\kappa\_0)$" for each $\alpha<\kappa\_0$, and let $U\_0$ be the filter over $\kappa\_0$ derived from $x$). Then $U\_0$ is an $L\_{\beta\_0}$-ultrafilter over $\kappa\_0$ which is $L\_{\beta\_0}$-$\kappa\_0$-complete (i.e. closed under length ${<\kappa\_0}$-intersections of sequences $\left<X\_\alpha\right>\_{\alpha<\gamma}\in L\_{\beta\_0}$). And because $\kappa\_1=\mathrm{crit}(j\_1)$, $\kappa\_1$ is a (regular) cardinal in $L\_{\beta\_1}$. Therefore $\mathcal{P}(\kappa\_0)\cap L\_{\beta\_1}\subseteq L\_{\beta\_0}=L\_{\kappa\_1}$. But then $U\_0$ is also an $L\_{\beta\_1}$-ultrafilter over $\kappa\_0$ which is $L\_{\beta\_1}$-$\kappa\_0$-complete. So letting $N=\mathrm{Ult}(L\_{\beta\_1},U\_0)$ and $j:L\_{\beta\_1}\to N$ the ultrapower map, then $j$ is elementary and $\mathrm{crit}(j)=\kappa\_0$.




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EDIT 2: On the other hand, if the codomain is required to be wellfounded, and $0^\sharp$ exists, the answer is no. For let $\kappa\_0$ be the least $L$-indiscernible and let $\pi:L\_{\omega\_1}\to L\_{\omega\_1}$ be elementary with $\mathrm{crit}(\pi)=\kappa\_0$. (Here $\omega\_1$ is as computed in $V$.)


There is an elementary $\sigma:L\_{\kappa\_0}\to L\_{\kappa\_0}$. For there is an elementary $\sigma':L\_{\kappa\_\omega}\to L\_{\kappa\_\omega}$ where $\kappa\_\omega$ is the $\omega$th $L$-indiscernible, for example the one with $\sigma'(\kappa\_n)=\kappa\_{n+1}$. But the existence of some such $\sigma'$ is forced over $L$ by $\mathrm{Coll}(\omega,\kappa\_\omega)$, and so by indiscernibility, the existence of a $\sigma$ as stated is forced over $L$ by $\mathrm{Coll}(\omega,\kappa\_0)$, and therefore there is actually some such embedding $\sigma$ in $V$.


So letting $\kappa=\mathrm{crit}(\sigma)$, then $\kappa<\_{\mathrm{el}}\kappa\_0<\_{\mathrm{el}}\omega\_1$. But I claim $\kappa\not<\_{\mathrm{el}}\omega\_1$.
For since $\kappa<\kappa\_0$, $\kappa$ is not an $L$-indiscernible. But if $j:L\_{\omega\_1}\to L\_{\lambda}$ is elementary (where $\lambda$ is some ordinal) then $\mathrm{crit}(j)$ is an $L$-indiscernible. (This is a standard fact. The key is that if $U$ is the normal ultrafilter derived from $j$, then $\mathrm{Ult}(L,U)$ is wellfounded.)




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These observations leave open the version where the codomain is required to be wellfounded, but $0^\sharp$ does not exist.


