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QUASIASSOCIATIVE ALGEBRAS 141
Proof. The proof is similar to the proof in the associative case.
Given X = [xij] ∈ Mn,φ(K), suppose that x11 is nonzero. Then it can
be used as a pivot: multiplying the matrix E (−x2,1 φ(1−1,1,1−1) ) by X we 21φ x1,1 φ(2,1−1 ,1.1−1 )
obtain a new matrix with zero (21) entry. So if we multiply X by the matrices E (−xk,1 φ(1−1,1,1−1) ), k = 2,...,n, we obtain a new matrix which has all
k1φ x1,1 φ(k,1−1 ,1.1−1 )
the entries in the first column equal to zero, except the (11) entry which is x11. Now we repeat the process with the second column of this new matrix, playing the diagonal element of this column the role of pivot. And we carry on like this. What can fail? One of the entries we want to use as pivot is zero. This is fixed either by multiplying X by appropriate matrices Pijφ in the beginning of the process (interchanging rows), or by moving to the next column (in case there are no elements in the column that can be used as pivots). In the end we obtain from X a matrix U in row echelon form. So we conclude that there are deformed permutation matrices Pi1j1φ,Pi2j2φ,...,Pikjkφ and deformed elementary ma- trices Ei1j1φ(α1),Ei2j2φ(α2),...,Eirjrφ(αr), where ip > jp, p = 1,··· ,r and j1 ≤j2 ≤···≤jr, suchthat
Eirjrφ(αr)(Eir−1jr−1φ(αr−1)(...(Ei1j1φ(α1)(Pi1j1φ(Pi2j2φ(...(PikjkφX)...))...)=U. As we know that Eijφ(−α)(Eijφ(α)Y ) = Y, we can multiply both members of
this equation by Eirjrφ(−αr),...,Ei1j1φ(−α1), and we get Pi1j1φ(Pi2j2φ(...(PikjkφX)...)) =
Ei1j1φ(−α1)(Ei2j2φ(−α2)...(Eir−1jr−1φ(−αr−1)(Eirjrφ(−αr)U))...). Now the result follows from the previous lemma.
3. Quasiassociative algebras with semisimple null part
In this section we will recall some results on quasiassociative algebras with semisimple null part, obtained in [1] and [6], which will be used in the next section to classify simple quasiassociative algebras. This classification is based on the structure of quasiassociative division algebras.
The quasiassociative algebra A = ⊕g∈GAg is said to be a quasiassociative division algebra if any nonzero homogeneous element has a right and a left inverse. In this case, it is easy to prove that Ae is a division associative algebra and Ag is an Ae-bimodule satisfying Ag = Aeug = ugAe, for any nonzero ug ∈Ag,anyg∈G.