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Unit 27: Separable Extensions
Theorem 12: Let L/K be a finite extension of fields, and D: K K be a derivation. Suppose Notes
a L is separable over K, with minimal polynomial (X) K[X]. That is, (X) is irreducible in
K[X], () = 0, and () 0. Then D has a unique extension from K to a derivation on the field
K(), and it is given by the rule
D
D(f()) = f ( ) f'( ) D(a)
'(a)
for any f(X) K[X].
Proof: The rule (B.1) looks bizarre at first. To make it less so, we start by assuming D has an
extension to K(), and prove by a direct computation that it must be given by the indicated
formula. For any K(), write = f(), where f(X) = r i=0 c X and c K. Then
i
i
i
r
D
D() = D(f()) = (D(c ) i + c (i i-1 D( ))) = f ( ) + f'( )D( ).
i
i
i=0
Taking f(X) = (X) to be the minimal polynomial of over K, f() = 0, so if D has an extension to
K() then (B.2) becomes
0 = () + ()D(),
D
which proves (since () 0) that D() must be given by the formula ()/(). Plugging this
D
formula for D(), shows D() must be given by the formula. Since was a general element of
K(), this proves D has at most one extension to a derivation on K().
Now, to show the formula works, we start over and define
D
D(f()) : = f (a) f'( ) D ( ) .
'( )
We need to show this formula is well-defined.
Suppose f () = f () for f ; f K[X]. Then f (X) f (X) mod (X), say
1
1
1
2
2
2
f (X) = f (X) + (X)k(X)
1
2
for some k(X) K[X]. Differentiating both sides with respect to X in the usual way,
f (X) = f (X) + (X)k(X) + (X)k(X).
1
2
Evaluating at X = ,
f () = f () + ()k().
2
1
Since () 0, we divide by () and multiply through by () to get
D
f' ( ) D (a) f' (a) D ( ) D ( )k( ).
1
' a 2 '( )
We want to add f ( ) to both sides. First, apply D to the coefficients in (B.3), which is a derivation
D
1
on K[X], to get
D
D
D
f (X) f (X) (X) (X) D (X)k(X).
2
1
k
Therefore,
D
D
f (a) f (a) D ( )k(a).
2
1
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