Space complexity and cryptography
Maybe there is an encryption scheme out there that can be 'broken' in polynomial time, but only with super-polynomial space.


That possibility can be excluded; if an algorithm uses $F(x)$ time (for any $F$), then that algorithm can be implemented using $O(F(x))$ space.  The reason is fairly obvious; in $F(x)$ steps, the algorithm can access at most $\ell F(x)$ memory locations (assuming a 'step' accesses at most $\ell$ memory locations, where $\ell$ is some constant).  So, by tracking which memory locations the algorithm actually touches, and just tracking those (which another constant multiplier in terms of overhead), and so this modified algorithm uses (at most) a constant factor more memory than the total number of steps.