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Let (A,m) and (B,n) be local rings, with a local map f : A --> Btaking m to n. (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, if f is unramified then if (1) Bm=n and (2) B/n is a seperable extension of A/m. It Condition (1) is interesting usually the hard part to see verify, but in this answer I will concentrate on condition (2) and try to provide some intuition for why this condition is includedin the definition.

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 implies separablerequires 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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Let (A,m) and (B,n) be local rings, with a map f : A --> B taking m to n. By definition, if f is unramified then B/n is a seperable extension of A/m. It is interesting to see why this is included in the definition.

Let f:A --> B be a map of rings. I might need some finite generation hypothesis; I'm not sure. Then f is ramified 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 unramified implies separable, there is no implication in the other direction.

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

2 added 2 characters in body

Let (A,m) and (B,n) be local rings, with a map f : A --> B taking m to n. By definition, if f is unramified then B/n is a seperable extension of A/m. It is interesting to see why this is included in the definition.

Let f:A --> B be a map of rings. I might need some finite generation hypothesis; I'm not sure. Then f is ramified 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 ramified 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 unramified implies separable, 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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