Posting "Large cardinals and constructible universe" mentions that $\omega_1^L < \omega_1$ if we assume Ramsey cardinal.
My question can we have more downslides like for example $\omega_2^L < \omega_1$? Can we have the whole of $L$ being countable? I mean like taking some suitable version of $\sf MK$, with or without choice, and axiomatize that $|L|=\aleph_0$? If those are possible, what are the respective consistency strengths?
Since $\omega_1^V\subseteq L$ (simply because $L$'s construction goes through all the ordinals), we obviously can't have $L$ be countable. On the other hand, if $0^\sharp$ exists then every uncountable cardinal in $V$ has all the large cardinal properties which are compatible with $L$; in particular, $\omega_{1}^V$ is (inaccessible, Mahlo, weakly compact, etc.)$^L$. Interestingly there is a kind of converse to this, namely the covering theorem: either $0^\sharp$ exists or $L$ is "fairly good" at analyzing uncountable sets.
