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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?).

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I should also add that saying that an extension $L/K$ is separable is equivalent to saying that the extension $L[x]/K[x]$ of Dedekind domains is unramified - that is, irreducible polynomials in $K[x]$ do not have any repeated factors in $L[x]$. – David Corwin Aug 23 '10 at 23:50

Let (A,m) and (B,n) be local rings, with a local map f : A --> B. (The condition that f is local means that f^{-1}(n) = m.) Also, assume that f obeys a finiteness condition called "essentially of finite type"; I'll ignore this. By definition, f is unramified if (1) Bm=n and (2) B/n is a seperable extension of A/m. Condition (1) is usually the hard part to verify, but in this answer I will concentrate on condition (2) and try to provide some intuition for why this condition is included.

Let f:A --> B be a map of rings. I might need some finite generation hypothesis; I'm not sure. Then f is unramified if and only if the following is true: For every prime ideal p in A, the tensor product B \otimes_{A} \bar{Frac(A/p)} is isomorphic to a direct sum of several copies of \bar{Frac(A/p)}. Here \bar indicates algebraic closure.

Tensoring with the algebraic closure of the residue field at a prime is called "taking the geometric fiber" over that prime, in algebraic geometry. So the geometric statement is that a map is unramified if and only if all of its geometric fibers are reduced and of dimension 0. (Again, modulo any finiteness hypothesis I may have forgotten.)

The point here is that, if L/K is a separable algebraic field extension, then L \otimes_K \bar{K} is isomorphic to \bar{K}^[L:K]. For an inseparable extension, this tensor product has nilpotents. (Specifically, if t is in L but not in K, and t^p=u is in K, then (t-u^{1/p}) will become nilpotent in the tensor product.) So the geometric fiber will not be reduced for such an extension.

While the definition of unramified requires separability, in the sense explained above, there is no implication in the other direction.

I used the early parts of deJong's notes as a reference when writing this.

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The preceding error also occurs in the second paragraph. Also, I think it should be stressed that mB = n is usually the hardest part of verifying a map is unramified! – Bhargav Nov 2 '09 at 17:34
Thanks for all the comments. I have revised the answer. – David Speyer Nov 2 '09 at 18:02

In the related context of dominant maps of curves (ie, extensions of function fields), a map which is ramified at every point must be inseparable. For example, for P^1 in characteristic p, the map z \mapsto z^p is ramified everywhere, and corresponds to the inseparable extension k(z^p) \to k(z). On the other hand, a separable map must be generically smooth and thus ramified at only finitely many points. I believe the converse also holds (every inseparable map is everywhere ramified), but I'm not entirely sure.

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Dear @Eric, Indeed, an inseparable map has to be everywhere ramified (or maybe better to say nowhere unramified) because, as you say, the smooth ($=$étale$=$unramified) locus is open, so if it doesn't contain the generic point, it has to be empty. Six years later... – Keenan Kidwell Apr 4 at 16:29

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.

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I don't think this is true.

Any quadratic field can be defined by taking a root of x^2 = d. Your recipe, if I understand it right, would say to look at the polynomial x^2 - t over F_2(t). This polynomial is inseparable, so your hypothesis would imply that 2 ramifies in ANY quadratic extension of Q, which is false.

As to the more conceptual question, I don't know of any results that explicitly define some canonical extension of a function field corresponding to any extension of Q. In other words, I don't think there's any way to talk about ramification of number fields in terms of inseparability of characteristic p extensions of function fields. If one sticks to function fields themselves and asks what ramification has to do with separability, the answer is that they are certainly not the same thing.

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Hmm okay but 2 ramifies in a large class of quadratic extensions of Q. Also, ramification and separability are both related to the existence of nilpotents in algebras. – David Corwin Nov 2 '09 at 1:44

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