I'm interested in the existence and properties of an analogue version of $L$ for models of ZF$^-$ (ZF without the power set axiom), which for simplicity I'll call $L^-$. By "analogue" I mean the least inner model of a set theory universe $V$ (same ordinals and transitive).


1. Assuming $V$ is a model of ZF$^-$, does a least inner model exist, and is it unique? Intuitively I'd answer that as "yes", though I don't know how to prove it.
2. Assuming "yes" for question 1, and assuming $V$ is a model of ZF (note here that $V$ **does** satisfy the power set axiom), is $L^- = L$ (obviously $L$ itself is a model of ZF$^-$, so $L^- \subseteq L$)?


Assuming existence and uniqueness, I strongly **suspect equality**, though it isn't obvious to me. If we look at the power set axiom as a large cardinal axiom (postulating the existence of some large cardinal with a certain property), then this just means that $L$ "inherits" these "large cardinals" from the universe. On the other hand, if $L^-$ is actually a proper subclass of $L$, then a host of new questions about this $L^-$ spring to mind.


3. In a universe where $V$ is a model of ZF$^-$ only, does $L^-$ satisfy AC? This question seems to be a little bit trickier than for $L$, because of our non-constructive definition of $L^-$.



Least inner model of ZF without power set axiom
Your intuition is correct, for we have $L^-=L$.


Inside any model $V$ of $\text{ZF}^-$, we can still build $L$. You don't need the power set axiom to construct $L$. And furthermore, the resulting inner model $L$ will satisfy $\text{ZF}^-$. 


From this, it follows that $L^-=L$, since if $W$ is a transitive model of
$\text{ZF}^-$ containing all the ordinals, then $W$ will construct the same $L$ as we did in $V$, and so $L\subset W$, verifying leastness as desired.


