1995
DOI: 10.1007/bf02572626
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Isotropy of quadratic forms over the function field of a quadric

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Cited by 78 publications
(60 citation statements)
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References 7 publications
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“…One could take η to be a complementary form since φ α | k(φα) is hyperbolic. Here we see η| k(φα) is anisotropic by the following result of Hoffmann (Theorems 1 in [2]). If q, q are anisotropic forms over k such that dim(q) > 2 n−1 > dim(q ), then q | k(q) is anisotropic.…”
Section: Remarksmentioning
confidence: 99%
See 1 more Smart Citation
“…One could take η to be a complementary form since φ α | k(φα) is hyperbolic. Here we see η| k(φα) is anisotropic by the following result of Hoffmann (Theorems 1 in [2]). If q, q are anisotropic forms over k such that dim(q) > 2 n−1 > dim(q ), then q | k(q) is anisotropic.…”
Section: Remarksmentioning
confidence: 99%
“…For a quadratic form ξ let us denote as Q ξ the quadric defined by ξ, e.g., Q α = Q φα . Theorem 6.1 (Rost [11], Hoffmann [2]) Let ξ be a subform of the Pfister form φ α of dim(ξ) = 2 n−1 + s, s > 0 (i.e., ξ is a neighbour of φ α ). Let η be the complementary form (φ α = ξ ⊕ η).…”
Section: Glmentioning
confidence: 99%
“…In collaboration with Mammone [15] we extended to characteristic 2 a theorem of Hoffmann on the isotropy of quadratic forms [5]. Here is our result.…”
Section: Definition 55 Any Quadratic Form Of Dimension ≤ 1 Is Calledmentioning
confidence: 80%
“…We will use induction on m. If we are in the situation of (i), the anisotropy of ϕ F (ψ) follows from [H1,Theorem 1]. Note that if m = 0, then we are certainly in the situation of (i).…”
Section: Lemma 2 Let ϕ 0 Be a Pfister Neighbor Of Dimensionmentioning
confidence: 99%
“…If ψ ∈ C 2 (K, P ), then dim ϕ ≤ 2 k+1 < dim ψ and the anisotropy of ϕ K(ψ) follows from [H1,Theorem 1].…”
Section: Remarks and Examplesmentioning
confidence: 99%