Page 258 - DMTH403_ABSTRACT_ALGEBRA

Unit 27: Separable Extensions

Consider now the extension of scalars up to K of the trace map Tr K(a)/K : K(a)  K: Notes

Tr  id  Tr : K  K(a)  K  K K.

K k(a)/K K K
Tr is the trace map on K  K K( ) as a K -vector space.

Since tensoring with a field extension preserves injectivity and surjectivity of a linear map,

Tr K(a)/K is onto  Tr is onto; Tr K(a) =K  Tr = 0

Since K()  K[X] /((X)) as K-algebras, K( ) K[X]/( (X))   as K -algebras, and thus is isomorphic
to the direct product of the rings K [X]/ (X p m –  i ). The trace is the sum of the traces to K on each

K[X]/(X p m –  i ). Let’s look at the trace from K[X]/(X p m –  i ). to K .

m
m
In K [X], X p m –  = (X –  i ) . Then K[X]/(X p m –  i ) = K[Y]/(Y ), where Y = X   . If m = 0, then
p
p
i
i
m
p
K [Y ] / (Y ) = K , so the trace to K is the identity. If m > 0, any element of K [Y ]/ (Y ) is the
m
p
sum of a constant plus a multiple of Y , which is a constant plus a nilpotent element (since Y
r
m
p
mod Y is nilpotent). Any constant in K [Y ]/ (Y ) has trace 0 since pm = 0 in K (because
p
m
m > 0). A nilpotent element has trace 0. Thus the trace to K of any element of K [Y ] / (Y ) is 0.
p
To summarize, when  is separable over K (i.e., m = 0), the trace map from K() to K is onto since
it is onto after extending scalars to K . When a is inseparable over K (i.e., m > 0), the trace map
is identically 0 since it vanishes after extending scalars.
Theorem 1 implies Theorem 4.
Proof. Set L = K, L = K( ) = L ( ), and more generally L = K( ,.... ) = L ( ) for i  1. So we
0
i
i-1
1
i
1
i
0
1
1
have the tower of field extensions
K = L  L  L  ...  L  L = L.
2
r
r-1
0
1
By transitivity of the trace,
Tr L/K = Tr L 1 /L 0 o Tr L 2 /L 1 o o Tr r L /L r–1
Since  is separable over K and the minimal polynomial of  over L divides its minimal
i
i-1
i
polynomial over K,  is separable over L . Therefore Tr Li-1 ( )/L : L  L is onto from the
i-1
i-1
i
i-1
i
i
proof of Theorem 1, so the composite map TrL/K : L  K is onto. Therefore L/K is separable by
Theorem 1.
Corollary: Theorem 1 implies Theorem 5.
Proof: By Theorem 1 and the hypothesis of Theorem 5, both Tr L/F and Tr F/K are onto. Therefore,
their composite Tr L/K is onto, so L/K is separable by Theorem 1.
Proof: We will begin with the case of a simple extension L = K(). Let (X) be the minimal
polynomial of  over K, so L  K[X]/((X)) as K-algebras and
K  K L  K[X]/  (X) 
as K -algebras. This ring was considered in the proof of Theorem 1, where we saw its structure
is different when (X) is separable or inseparable. If (X) is separable in K[X], then K[X]/( K (X))
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