Ostrowski's theorem give the answer for valuations, but is there a complete classification of (at least separated) topologies on Q (compatible with the field operations, obviously)?



Topologies on the field of rationals 
According to [this link](http://www.encyclopediaofmath.org/index.php/Topological_field), there are as many field topologies on $\mathbb{Q}$ as there are subsets of $\mathbb{R}$, so I doubt there is a classification. A reference seems to be Wieslaw's book "topological fields" (it doesn't seem to be on google books, unfortunately).


PS: note that there is only one non-Hausdorff field (or ring) topology on a field (the two open sets one) since the closure of 0 is an ideal, and the others are completely regular, as any Hausdorff group topology. Also there is only continuum many metrizable topologies on $\mathbb{Q}$, so most of the topologies referred to in the above link are quite pathological, non first countable for instance.


PPS: searching "number of field topologies" in MathSciNet returns the following references


Podewski, Klaus-Peter
The number of field topologies on countable fields.
Proc. Amer. Math. Soc. 39 (1973), 33–38. 


Kiltinen, John O.
On the number of field topologies on an infinite field.
Proc. Amer. Math. Soc. 40 (1973), 30–36. 


and they are both freely accessible (thanks AMS!) [here](http://www.ams.org/journals/proc/1973-039-01/S0002-9939-1973-0311633-9/home.html) and [here](http://www.ams.org/journals/proc/1973-040-01/S0002-9939-1973-0318118-4/home.html).


Concerning the proof for a countable field $K$, Podewski manages to define a continuum $\mathcal{G}$ of (metrizable) field topologies on $K$ such that the suprema of any two distinct susbsets of $\mathcal{G}$ are distinct field topologies.


The details are somewhat complicated, but the idea is quite natural. Suffice it to say
that the metrizable field topologies on $K$ are parameterized --- via fundamental sequences of neighbourhoods of $0$ --- by chains $G$ in a partially ordered set $P$ of "conditions" (is it forcing in disguise?), which specify for a finite number of elements of $K$ wether they belong or not to the $n$-th neighborhood in the sequence.


Strangely, for uncountable fields $K$ the proof is easier, using valuations instead of chains of conditions, and the fact that the transcendence degree of $K$ over the prime subfield is the cardinality of $K$.


