Studying the proof of Hilbert's 90 theorem modern version, I went through this lemma:given a Galois finite extension $K \subset L$ and an $L$ algebra $A$,we define the $(A,K)$ forms as the $K$ algebras $B$ s.t $B \otimes L \cong A$. This forms are classified up to isomorphisms,by $H^1(Gal(L/K),Aut(A))$.

The key part of the lemma was to prove that given a vector space $V$ on $L$ and an action of the Gal group s.t $\sigma(\lambda v)=\sigma(\lambda)\sigma(v)$ ,there is an isomorphism between $V^{Gal} \otimes L \cong V$.

I would like to get a proof using Galois normal basis theorem,but I'm not managing to use that.


Your presentation of the problem is a bit restricted (see below). As I understand it, a finite Galois extension $L/K$ and an $L$-algebra $A$ are given, $G=Gal(L/K)$ acts naturally on $Aut(A)$, and you want to show that the $(A, K)$-forms (as you define them) are classified up to isomorphism by $H^1 (G, Aut(A))$. But since $Aut(A)$ is not abelian, note that your $H^1$ is not a cohomology group, it's just a set (see e.g. Serre's "Galois Cohomology", chap.I,§5).

You want a proof of Speiser's lemma using the normal basis theorem: If $V$ is an $L$-vector space equipped with a semi-linear $G$-action, i.e. such that $ g(\lambda.v)= g(\lambda).g(v)$ for all $g \in G, \lambda \in L, v \in V$, then the natural map $\lambda: V^{G} {\otimes}_K L \to V$ is an isomorphism.

Proof. Consider the natural map $\lambda_{L}: (V{\otimes}_K L)^{G} \to V {\otimes}_K L$, where $G$ acts on $L$ trivially and on $V$ as in the lemma. By definition , $(V {\otimes}_K L)^G \cong V^{G} {\otimes}_K L$, hence $\lambda_{L}$ can be identified with the map $ (V^{G} {\otimes}_K L){\otimes}_K L \to V {\otimes}_K L$ obtained by tensoring with $L$, and $\lambda$ is an isomorphism iff $\lambda_{L}$ is. Looking at the tensor product $L {\otimes}_K L$ where $G$ acts trivially on the first factor and naturally on the second, one sees that $L {\otimes}_K L \cong \oplus_g L{e_g}$, where $G$ acts on the RHS by permuting the basis elements ${e_g}$ of $L$ (check by writing $L=K[X]/(f)$ and applying the Chinese Remainder thm.). This decomposition induces a decomposition of the $L {\otimes}_K L$ -module $V {\otimes}_K L$ as a direct sum of $L$-vector spaces ${e_g}(V {\otimes}_K L)$. Identifying ${e_1}(V {\otimes}_K L)$ with an $L$-vector space $W$ with trivial $G$-action, one obtains an $L$[G] - module isomorphism $V {\otimes}_K L \cong W^n$, where $n$ is the order of $G$ and $G$ acts on the RHS by permuting the factors. Modulo this identification, the elements of $(V {\otimes}_K L)^G$ correspond to diagonal elements $(w,..., w)$ of $W^n$, and multiplication by $e_g$ in $V {\otimes}_K L$ corresponds to setting the components of a vector in $W^n$ to $0$ except the one indexed by $g$. It follows that the $L {\otimes}_K L$ -submodule of $V {\otimes}_K L$ generated by $(V {\otimes}_K L)^G$ contains all the elements corresponding to the vectors of $W^n$ of the form $(0, 0, ..., w, 0,.., 0)$. This shows the surjectivity of $\lambda_L$, and its injectivity is an obvious consequence of the injectivity of the diagonal map $W \to W^n$. QED

Actually, the preliminaries in your OP could be more clearly reformulated in terms of tensor spaces as in Serre, chap.III, §1.1, op. cit. An $L/K$ -form $(V, x)$, is a pair where $V$ is a finite dimensional $K$-vector space and $x$ a tensor on $V$ of type $(p, q)$, i.e. $x \in V^{\otimes p}\otimes {V^*}^{\otimes q}$, where $V^*$ is the dual space of $V$. The case $x=0$ (of any type) corresponds to a mere vector space $V$; the case of type $(1, 2)$ corresponds to a $K$-bilinear map $V {\otimes}_K V \to V$, i.e. to a $K$-associative algebra as in your OP. For a given type $(p, q)$, the notion of $L$-isomorphic $L/K$-forms $(V,x)$ and $(W, y)$ is defined in an obvious way, and for a fixed $L/K$-form $(V, x)$, Speiser's lemma allows to construct a bijection between the set of $L/K$- forms which are $L$- isomorphic to $(V,x)$ and the cohomology set $H^1 (G, Aut(V,x))$. In the trivial case $x=0$ and dim $V = n$, one has Speiser's theorem $H^1 (G, GL_n (L))= 0$, which is a non commutative generalization of Hilbert's 90 . For a detailed proof, see e.g. Serre's "Local Fields", chap.X, §2) ./.


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