TITLE: What does ramification have to do with separability?
QUESTION [10 upvotes]: Does ramification have anything to do with inseparability? It feels like an extension of Q in which p ramifies should somehow correspond to an extension of F_p(t). Does totally ramified <--> purely inseparable?
In fact, saying an irreducible polynomial f(x) is inseparable is the same as saying that f(x) ramifies when we extend Q[x] to L[x], where L is the splitting field of f(x).
By correspond, I generally mean taking an extension of Q defined by a root of p(x)=r, where r is a rational making the extension nontrivial, and then extending F_p(t) by a root of p(x)=t. It's interesting because then this extension of F_p(t) corresponds to a number of extensions of Q (this is the same thing when you do Galois theory by looking at fundamental groups of branched coverings of C. Then do you look at etale fundamental groups of objects associated to these function fields over finite fields?).

REPLY [3 votes]: Here's a candidate definition for "total ramification" which jibes quite well with pure inseparability: 
A finite map f:X -> Y is totally ramified at y if the scheme theoretic fibre X_y -> y is a universal homeomorphism. 
If k is a field, and A is a finite k-algebra, then A is totally ramified over k in the above sense if and only if a) A is local, and b) the last condition holds after all base changes on k. If k has characteristic 0, this is equivalent to requiring that A be local with residue field k. This means that in the case X and Y are curves over a field of characteristic 0, this gives the usual notion. On the other hand, a finite extension L/K of fields is totally ramified in the above sense iff L/K is geometrically connected iff L/K is purely inseparable.