Special and general types
A theorem in Galois cohomology due to David Hilbert.
There are actually two versions of Hilbert’s theorem 90, one multiplicative and the other additive. We begin with the multiplicative version.
Before embarking on the proof, we recall from the article projective resolution that if is a finite cyclic group of order , then there is a projective resolution of as trivial -module:
where the map is induced from the trivial group homomorphism (hence is the map that forms the sum of all coefficients of all group elements), and where , are multiplication by special elements in , also denoted , :
The calculations in the proof that follows implicitly refer to this resolution as a means to defining (in the case ), by taking cohomology of the induced cochain complex
Let be an element of the group algebra, and denote the action of on an element by exponential notation . The action of the element is
which is precisely the norm . We are to show that if , then there exists such that .
By lemma 1 below, the homomorphisms are, when considered as elements in a vector space of -valued functions, -linearly independent. It follows in particular that
is not identically zero, and therefore there exists such that the element
is non-zero. Using the fact that , one may easily calculate that , as was to be shown.
Now we give the additive version of Hilbert’s theorem 90:
Under the same hypotheses given in 1, and regarding the additive group as a -module, we have
The trace of an element is defined by
We want to show that if , then there exists such that . By the theorem on linear independence of characters (following section), there exists such that ; notice belongs to the ground field since . Put
One may then calculate that
where in the second line we used .
Independence of characters
The next result may be thought of as establishing “independence of characters” (where “characters” are valued in the multiplicative group of a field):
A single obviously forms a linearly independent set. Now suppose we have an equation
where , and assume is as small as possible. In particular, no is equal to , and . Choose such that . Then for all we have
Dividing equation 2 by and subtracting from it equation 1, the first term cancels, and we are left with a shorter relation
which is a contradiction.
A corollary of this result is an important result in its own right, the normal basis heorem.
(Will write this out later. I am puzzled that all the proofs I’ve so far looked at involve determinants. What happened to the battle cry, “Down with determinants!”?)