According to wikipedia, a spectral space is defined to be homeomorphic to the spectrum of some commutative ring. They form a category $Spec$ where we take morphisms to be those whose preimage of open and quasi-compact sets are also as such. An alternative (ring-independent) characterisation is:
It is compact ordered topological space satisfying the priestley separation condition
Since we have a contravariant functor $F:CRing \rightarrow Spec$; an algebra, which is a morphism of commutative rings $A \rightarrow B$, under application of the functor, goes to $FB \rightarrow FA$, a morphism of Spectral spaces.
Now, what is the topological condition on $FB \rightarrow FA$ when $A \rightarrow B$ (understood as an algebra) is
Free?
Separable?