Page 264 - DMTH403_ABSTRACT_ALGEBRA

Unit 27: Separable Extensions

Theorem 9: Let A be a finite-dimensional K-algebra, L/K be a field extension, and B = L A be Notes
K
the base extension of A to an L-algebra. For a  A, Tr (1a) = Tr (a).
B/L A/K
Proof: Let e ,....., e be a K-basis of A. Write ae = n i 1 c e , so the matrix for m in this basis is (c ).
j 
ij
1
n
ij i
a

The tensors 1  e ,....., 1  en are an L-basis of B, and we have
1
n
(1  a) (1  e) = 1  ae = c (1  e ),
j 
i
ij
j
i 1

so the matrix for m 1a on B is the same as the matrix for m on A. Thus Tr A/K (a) = Tr B/L (1  a).
a
Remark: Because m 1a and m have the same matrix representation, not only are their traces the
a
same but their characteristic polynomials are the same.
Theorem 10: Let A be a finite-dimensional K-algebra. For any field extension L/K, the base
extension by K of the trace map A  K is the trace map L  A  L. That is, the function
K
idTr A/K : LK A  L which sends an elementary tensor x a to xTr A/K (a) is the trace map
Tr (LKA)/L .
Proof: We want to show Tr (LKA)/L (t) = (idTr A/K )(t) for all t  L A. The elementary tensors
K
additively span L A so it succes to check equality when t = x  a for x  K and a  A. This means
K
we need to check Tr (LKA)/K (x  a) = xTr A/K (a).
Pick a K-basis e ,..., e for A and write ae =  n i 1 c e with c  K. The elementary tensors
ij i
j
n
1
ij

1  e ,..., 1  e are an L-basis of LK A and
n
1
n n
x
j 
i 
(x  a)(1  e )   ae  c (x  e )  c x(1  e )
i
ij
j
ij
i 1 i 1


by the definition of the L-vector space structure on L A. So the matrix for multiplication by
K
x  a in the basis {1  e } is (c x), which implies
ij
i
n n
Tr (L KA)/L (x  a)   c x  x  cii  xTr A /K (a).
ii

i 1 i 1


Derivations
A derivation is an abstraction of differentiation on polynomials. We want to work with
derivations on fields, but polynomial rings will intervene, so we need to understand derivations
on rings before we focus on fields.
Let R be a commutative ring and M be an R-module (e.g., M = R). A derivation on R with values
in M is a map D : R  M such that D(a + b) = D(a) + D(b) and D(ab) = aD(b) + bD(a). Easily, by
induction D(a ) = na D(a) for any n  1. When M = R, we will speak of a derivation on R.
n
n-1
Example: For any commutative ring A, differentiation with respect to X on A[X] is a derivation
on A[X] (R = M = A[X]).
Example: Let R = A[X] and M = A as an R-module by f(X)a := f(0)a. Then D: R  M by D(f)
= f’(0) is a derivation.
 X .
i
D
Example: Let D : R  R be a derivation. For f(X)  i  a X in R[X], set f (X) = i  D a i i
i
This is the application of D coefficentwise to f(X). The operation f  f is a derivation on R[X] (to
D
check the product rule, it suffices to look at monomials).
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