## Forms in odd degree extensions and self-dual normal bases.(English)Zbl 0729.12006

Let A be a finite dimensional algebra over a field K, with K-linear involution J: $$A\to A$$. The involution can be extended to an involution $$J_ L$$ of $$A_ L=A\otimes_ KL,$$ for any field extension L/K. Let $$N(L)=\{a\in A_ L:\;aJ_ L(a)=1\}$$ and let $$K_ s$$ be a separable closure of K. Let $$H^ 1(K,N)$$ denote the Galois cohomology set $$H^ 1(Gal(K_ s/K), N(K_ s)).$$ The authors prove the following theorem, which answers a question posed by J.-P. Serre [Cohomologie galoisienne des groupes algébriques linéaires, Colloque Théor. Groupes Algébr., Bruxelles 1962, 53-68 (1962; Zbl 0145.175)]: If L is a finite extension of odd degree of K, then the canonical map $$H^ 1(K,N) \to H^ 1(L,N)$$ is injective. This result has several applications. One of them is concerned with systems of bilinear forms. If two systems of bilinear forms over K become isomorphic over an extension of odd degree, then they are already isomorphic over K.
A basis $$\{e_ 1,...,e_ n\}$$ of the K-vector space L is said to be self-dual if $$Tr_{L/K}(e_ ie_ j)=\delta_{ij}.$$ Let $$G=Gal(L/K)$$. If the set $$\{g(x): g\in G\}$$ is a basis of L over K (for a fixed x), then it is called a normal basis. The authors show that any finite Galois extension of odd degree has a self-dual normal basis. For finite abelian extensions, the converse statement is also true. These results are valid for fields of characteristic not 2. If char K$$=2$$ and L/K is a finite abelian extension, then L has a self-dual normal basis if and only if the exponent of G is not divisible by 4. [Cf. also the first author, Indag. Math. 51, 379-383 (1989; Zbl 0709.12004)).
Reviewer: M.Kula (Katowice)

### MSC:

 12F05 Algebraic field extensions 11E04 Quadratic forms over general fields 11R21 Other number fields 12F10 Separable extensions, Galois theory 11E39 Bilinear and Hermitian forms 11E72 Galois cohomology of linear algebraic groups

### Citations:

Zbl 0145.175; Zbl 0709.12004
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