Page 114 - Textos de Matemática Vol. 44

104 N. KRUPNIK AND Y. SPIGEL
Proposition 6.10. Let F0(A,B) be a polynomial of some degree m, and (6.9) F (A, B) = (F0(A, B) + αE)(F0(B, A) + βE),
where α, β are some constants. If
(6.10) (F0(A,B)+αE)(F0(B,A)+βE)=(F0(A,B)+βE)(F0(B,A)+αE), then F(A,B) can be represented in the form
(6.11) F (A, B) = G(A, B)G(B, A) + δE,
where δ is a constant.
Proof. If α = β then equality (6.9) gives representation (6.11). Let α ̸= β. In this case it follows from (6.10) that F0(A,B) = F0(B,A). Using this equality it is not difficult to check that
(6.12) F (A, B) = (F0(A, B) + γE)2 + δE = G(A, B)G(B, A) + δE,
where γ = (α + β)/2 and δ = −(α + β)2/4.
Theorem 6.11. Let a non-trivial polynomial F (A, B) satisfy the conditions of Theorem 2.1 with deg F1 = 2 = deg F2. Then it admits a representation
(6.13) F (A, B) = G(A, B)G(B, A) + δE.
For the proof of this theorem we need the following
Lemma 6.12. Let a polynomial F(A,B) of third degree be represented as a product F1(A, B)F2(A, B) of two homogeneous polynomials (1 ≤ deg Fj ≤ 2) . If the polynomials satisfy equality (2.2), then
(6.14) F (A, B) = constant (A + B)3
The proof is simple and we omit it. Now we prove Theorem 6.11.
Proof of Theorem 6.11. We represent the polynomials F1(A, B), F2(A, B) as the sums
(6.15) Fj (A, B) = Hj (A, B) + Kj (A, B) (j = 1, 2),
where Hj are homogeneous polynomials of second degree and Kj are linear
polynomials. Recall that
(6.16) F1(A, B)F2(A, B) = F2(B, A)F1(B, A). It follows from (6.15), (6.16) that
H1(A, B)H2(A, B) = H2(B, A)H1(B, A).