“…We have: ann(N ) = v 6 , v 2 , v 3 ∈ ann 2 (N ) and e 1 ∈ ann 3 (N ). Since e 2 4 , e 2 5 ∈ v 2 , v 3 , v 6 then the possible types of N are: [1,4,1], [1,3,2] • dim (U 3 ⊕ U 1 ) 2 = 1 =⇒ α 46 = 0. N ≃ N 6,20 (α, β) : e 2 1 = e 2 + e 3 , e 2 2 = e 6 , e 2 3 = −e 6 , e 2 4 = α(e 2 + e 3 ), e 2 5 = βe 6 , e 2 6 = 0 with α, β ∈ F * .…”