A problem on bilinear maps ( problem U 228 in Mathematical Reflections )

Problem Let LK be a separable algebraic extension of fields. Solutions of the problem We give three solutions to the above problem. First solution First solution of the problem: We first prove the following lemma: Lemma 1. Under the conditions of the problem, let x ∈ L, a ∈ V and b ∈ W be arbitrary. Let α = h (a, b) and β = h (a, xb) − xh (a, b). Then, every positive n ∈ N satisfies h (a, x n b) = x n α + nx n−1 β. Proof of Lemma 1. Let us prove that (2) holds for every positive n ∈ N. We will prove this by strong induction over n: Induction step 1 : Let N ∈ N be positive. Assume that (2) holds for every positive n ∈ N satisfying n < N. We must then prove that (2) holds for n = N. The equality (2) holds for n = 1 (since h a, x 1 =x b = h (a, xb) = x =x 1 h (a, b) =α + h (a, xb) − xh (a, b) =β=1β=1x 1−1 β (since 1x 1−1 =1x 0 =1 and thus 1=1x 1−1) = xα + 1x 1−1 β). In other words, if N = 1, then (2) holds for n = N. Hence, if N = 1, the induction step is already completed. Thus, for the rest of the induction step, we can WLOG assume that N = 1. Assume this. 1 A strong induction does not need an induction base.