On the spinor norm and A_{0}(X, K_{1}) for quadrics

by Markus Rost

This is a TEXed version (Sept. 1996) of the original preprint Sept. 1988.

I. Introduction

1) We denote by K_{n}F the n-th Milnor K-group of field (for convenience). For X/F projective there is a complex

⨁_{v^{∈}X_{(1)}} K_{n+1}K(v) ⟶^{d} ⨁_{v^{∈}X_{(0)}} K_{n}K(v) ⟶^{N} K_{n}F

where d is given by the tame symbol and N is given by the norm map in Milnor K-Theory. We denote coker d by A_{0}(X, K_{n}) and by N_{X} : A_{0}(X, K_{n}) → K_{n}F the induced norm map.

2) In this note all fields have characteristic different from 2. Let φ : V → F be a quadratic module. We denote by X_{φ} ⊂ IPV the associated projective quadric hypersurface and we put N_{φ} = N_{X_{φ}}. Note that X_{cφ} = X_{φ}. Let C(φ), C_{0}(φ) be the (even) Clifford algebra of φ and consider V as a subspace of C(φ) in the usual way. The special Clifford group of φ is defined as

SΓ(φ) = {α ^{∈} C_{0}(φ) | αV α^{−1} = V in C(φ)}.

One has a commutative diagram

� � � � � �  _{sn} �  _{sn} �  _{2} � �  1 F^{∗}/(F^{∗})^{2} F^{∗} F^{∗}  1 SO(φ) SΓ(φ) F^{∗} SΓ

where F^{∗} ⊂ SΓ(φ) is central and SΓ(φ) acts on (V, φ) by α(v) = αvα^{−1}; the spinor norm sn is given by sn(α) = α^{t}α. If dim φ = 2, the SΓ(φ) = C_{0}(φ)^{∗}, C_{0}(φ) is the quadratic extension of F defined by the discriminant of φ and sn is given by the norm for C_{0}(φ)/F. If dim φ = 3, then SΓ(φ) = C_{0}(φ)^{∗}, C_{0}(φ) is a quaternion algebra and sn is induced by the reduced norm for the algebra C_{0}(φ) | F. If φ_{0} is a subform of φ (i.e. φ_{0} = φ | V_{0} for some subspace V_{0} of V), then SΓ(φ_{0}) ⊂ SΓ(φ).

3) In this note we construct a natural homomorphism ω̃_{φ} : SΓ(φ) ⟶ A_{0}(X_{φ}, K_{1}) such that N_{φ} ∘ ω̃_{φ} = sn.

ω̃_{φ} is surjective (at least if F has no odd extensions). Therefore, if one investigates the injectivity of N_{φ}, one is let to consider the kernel of the spinor norm.

4) An element α ^{∈} SΓ(φ) is called plane, if α ^{∈} SΓ(φ_{0}) for a 2-dimensional subform φ_{0} of φ. It is known that SΓ(φ) is generated by plane elements (Dieudonné). We denote by \overline{SΓ(φ)} the quotient of SΓ(φ) by its commutator subgroup and elements of the form αβ^{−1}, where α, β ^{∈} SΓ(φ) are plane such that sn(α) = sn(β). Let \overline{sn} : \overline{SΓ(φ)} → F^{∗} be the homomorphism induced by sn. An element of \overline{SΓ(φ)} is called plane, if it equals ᾱ for some plane α ^{∈} SΓ(φ). It turns out that ω̃_{φ} factors through a homomorphism ω_{φ} : SΓ(φ) → A_{0}(X_{φ}, K_{1}).

5) Since N ∘ ω_{φ} = \overline{sn} and ω_{φ} is surjective, we have

Theorem 1. If every element of \overline{SΓ(φ)} can be written as a product of two plane elements, then N_{φ} is injective in degree 1.

All forms φ, for which I can prove the injectivity of N_{φ} satisfy the the hypothesis of Theorem 1. We have

Proposition 2. φ satisfies the hypothesis of Theorem 1 in the following cases

i) dim φ ≤ 5

ii) φ = ρ ⊗ (ψ ⊕ ⟨c⟩), where ρ is a Pfister form, ψ is a Pfister neighbor and c ^{∈} F^{∗}

iii) φ = ψ ⊕ c⟨1, 1⟩, where ψ is a Pfister neighbor and c ^{∈} F^{∗}.

Recall that a Pfister neighbor is a form of type ψ_{0} ⊕ bψ_{1} where ψ_{0} is a Pfister form and ψ_{1} is a subform of ψ_{0}. Note that every Pfister neighbor (hence every Pfister form) is included in case ii).

6) The perhaps simplest type of quadratic forms not covered by i), ii) or iii) are 6dimensional forms φ such that C_{0}(φ) has (maximal) index 4 over its center. We have

Proposition 3. There exists a field F and 6-dimensional quadratic form φ over F, such that N_{φ} is not injective. φ can be chosen to have discriminant 1.

This result is based on a relation between KerN_{φ} and SK_{1}(C_{0}(φ)) which is obtained by Swan’s computation of K_{1}(X_{φ}). To give an explicit example, let A = D(a, b) ⊗ D(ā, b̄) a tensor-product of two quaternion algebras such that |SK_{1}A| > 2. Let

φ = ⟨−a, −b, ab⟩ ⊕ −⟨−ā, −b̄, āb̄⟩

be the associated form (X_{φ} is the Grassmanian for submodules of A of rank 8). By Swan we have K_{1}(X) = (K_{1}F)^{4} ⊕ K_{1}A ⊕ K_{1}A. Consider

j : A_{0}(X_{φ}, K_{1}) = H^{4}(X; K_{5}) ⟶ K_{1}(X) ⟶ K_{1}A

where the last map is given by projection to one of the factors. One may check that Nrd ∘ j = 2N_{φ}. Hence j(ker N_{φ}) ⊂ SK_{1}A. One can show that SK_{1}(A)/j(KerN_{φ}) is of order at most 2; it is generated by j(u) for any u such that N_{φ}(u) = −1.

II. The special Clifford group

Remark added in July 1996 : In the following I seem to care only on anisotropic forms. The much simpler case of isotropic forms should considered in the very beginning.

Let φ : V → F be a quadratic module. For α ^{∈} SΓ(φ) let

supp α = {v ^{∈} V; α(v) = v}^{⊥}.

Note that α ^{∈} SΓ(φ | supp (α)). Clearly

supp α = 0 ⟺ α ^{∈} F^{∗}
dim supp α ≤ 2 ⟺ α is plane .

Since the product of any two reflections in O(φ) is a plane rotation, we have the following consequence of the theorem of Cartan-Dieudonné.

Proposition 4. Any element of SΓ(φ) can be written as product of [\frac{dim φ}{2}] plane elements. □

Consequently dim(supp α) ≠ 1, 3 for α ^{∈} SΓ(φ). Two plane elements α, β are called to be linked if dim(supp α + supp β) ≤ 3. In this case αβ is again plane, because supp αβ ⊂ supp α + supp β and dim(supp αβ) = 3 is impossible. Note that α and β commute, if supp α ⊥ supp β.

Theorem 5. Let G be the free group on the set of all plane elements of SΓ(φ). Denote by g_{α} the generator corresponding to α. Then

G ⟶ SΓ(φ), g_{α} ⟶ α

is surjective and its kernel is the normal subgroup generated by elements of the form

R_{1})  g_{α}g_{β}g_{αβ}^{−1} if α and β are linked.
R_{2})  [g_{α}, g_{β}]  if supp (α) ⊥ supp (β).

For v_{1}, v_{2} ^{∈} V such that φ(v_{1}) = φ(v_{2}) we denote by ε(v_{2}, v_{1}) ^{∈} SΓ(φ) the trivial element if v_{1} = v_{2}, otherwise any plane element such that ε(v_{2}, v_{1})(v_{1}) = v_{2}. Note that a plane α ^{∈} SΓ(φ) is linked with ε(α(v), v) for any v ^{∈} V.

Remark added in Jan 1998 : The plane element ε(v_{2}, v_{1}) is assumed to have support in the subspace generated by v_{1}, v_{2}.

Let Ĝ be the quotient of G by the relations R_{1}, R_{2} and denote by ĝ the image of g_{α} in Ĝ. Theorem 5 states that Ĝ → SΓ(φ), ĝ → α is bijective. Surjectivity follows from Proposition 4. In the following we proof the injectivity (I don’t know, whether and in how far this question has been considered in the literature).

Lemma 6.

ĝ_{α}ĝ_{β}ĝ_{α}^{−1} = ĝ_{αβα^{−1}} for plane α, β ^{∈} SΓ(φ).

Proof. Let W = (supp α)^{⊥} ∩ supp β.

If dim W = 2, then supp β ⊂ (supp α)^{⊥}. Therefore αβα^{−1} = β and the lemma follows from R_{2}.

If dim W < 2, then there exists a nonzero v ^{∈} W^{⊥} ∩ supp β. Then α, β are linked with γ = ε(α(v), v), hence by R_{2}:

ĝ_{α}ĝ_{β}ĝ_{α}^{−1}ĝ_{αβα^{−1}} = ĝ_{α′}ĝ_{β′}ĝ_{α′}^{−1}ĝ_{α′β′α^{′−1}}

with α′ = αγ^{−1} , β′ = γβγ^{−1}.

Note that γ fixes W, because v ^{∈} W^{⊥} and α fixes W. This shows W ⊂ W′ = (supp α′)^{⊥} ∩ supp β′. Moreover α(v) = γ(v) ^{∈} W′\W, hence dim W′ > W and we are left with the case dim W = 2 after eventually repeating this argument. □

End of the proof of Theorem 5: Let α_{1}, …, α_{N} ^{∈} SΓ(φ) be plane such that α_{1} … α_{N} = 1. We show ĝ_{α_{1}} … ĝ_{α_{N}} = 1 by induction on dim φ. Let v ^{∈} V be any anisotropic vector, let v_{i} = α_{i} … α_{n}(v), v_{N+1} = v = v_{1}, let γ_{i} = ε(v_{i}, v_{i+1}) and β_{i} = α_{i}γ_{i}^{−1}. α_{i} and γ_{i} are linked, because α_{i}(v_{i+1}) = v, thus ĝ_{α_{i}} = ĝ_{β_{i}} ĝ_{γ_{i}}.

Put δ_{i} = γ_{1} … γ_{i}, (δ_{0} = 1); δ_{i} is plane and is one choice of ε(v, v_{i+1}). Hence γ_{i} and δ_{i−1} are linked and therefore ĝ_{δ_{i}} = ĝ_{γ_{1}} … ĝ_{γ_{i}}.

Let ρ_{i} = δ_{i−1}β_{i}δ_{i−1}^{−1}. Then g_{δ_{i−1}}g_{β_{i}}g_{δ_{i−1}}^{−1} = g_{ρ_{i}} by the lemma. Taking things together we find

g_{α_{1}} … g_{α_{N}} = g_{β_{1}}g_{γ_{1}}g_{β_{2}} … g_{β_{N}} g_{γ_{N}} = g_{ρ_{1}} … g_{ρ_{N}} g_{ρ_{N}} .

Now, we are done, because ρ_{1} … ρ_{N} and δ_{N} fix v. □

We have the following consequence of Theorem 5.

Corollary 7. Let Ḡ be the free abelian group generated by all plane elements of SΓ(φ). Denote by ḡ_{α} the generator corresponding to α. Then

Ḡ ⟶ \overline{SΓ(φ)}, ḡ_{α} ⟶ ᾱ

is surjective and its kernel is generated by elements of the form

R̄_{0})  ḡ_{α}ḡ_{β}^{−1}  if sn(α) = sn(β) R̄_{1})  ḡ_{α}ḡ_{β}ḡ_{αβ}^{−1} if α and β are linked.  □

The corollary is the basis of our construction of an epimorphism ω_{φ} : SΓ(φ) → A_{0}(X, K_{1}) described in the next section.

The rest of this section is devoted to the proof of Proposition 2. Clearly Proposition 2 i) follows from Proposition 4.

Let

D(φ) = {φ(v); v ^{∈} V anisotropic} ⊂ F^{∗}

and

N(φ) = {sn(α); α ^{∈} SΓ(φ) plane } ⊂ F^{∗}.

N(φ) consists just of all norms from quadratic extensions K/F such that φ_{K} is isotropic. Define

∑_{φ} : N(φ) ⟶ \overline{SΓ(φ)}, ∑_{φ}(sn(α)) = ᾱ

∑_{φ} is welldefined by the very definition of S\overline{Γ(φ)}. ∑_{φ} is injective, since sn is a left inverse. ∑_{φ}(N(φ)) generates \overline{SΓ(φ)} by Proposition 4.

For a subform φ_{0} of φ, we denote by i_{∗} : \overline{SΓ(φ_{0})} → \overline{SΓ(φ)} the homomorphism induced by the inclusion SΓ(φ_{0}) ⊂ SΓ(φ).

Lemma 8. Let φ be a quadratic form. Then

i) D(φ) · D(φ) = N(φ).

ii) If φ represents 1, then D(φ) ⊂ N(φ).

iii) Let v ^{∈} V and α ^{∈} SΓ(φ) plane, such that φ(v) = 1 and v ^{∈} supp α. Then sn(α) ^{∈} D(φ).

iv) Let v ^{∈} V and α ^{∈} SΓ(φ). Then α = α_{1} … α_{n} with α_{i} plane and v ^{∈} supp α_{i}.

v) If φ represents 1, then

∑_{φ}(φ(v_{1})φ(v_{2})) = ∑_{φ}(φ(v_{1})) ∑_{φ}(φ(v_{2}))

for any anisotropic v_{1}, v_{2} ^{∈} V. (The left hand side is defined by i)). vi) Suppose ψ represents 1 and let φ = ψ ⊕ ⟨b⟩. Then

\overline{SΓ(φ)} = i_{∗}(\overline{SΓ(ψ)}) · ∑_{φ}(D(φ)).

Proof. It is easy to check i) - iii) for dim φ = 2 and iv) - vi) for dim φ = 3. One may however reduce to these cases by restriction to appropriate subforms. This is obvious except for iv) and vi). For iv) one has to consider only plane elements α by Proposition 4; hence one may replace φ by φ | (supp α + vf), which is of dimension ≤ 3.

For vi) the reduction can be done as follows. Write (V, φ) = (W, ψ) ⊕ (F, ⟨b⟩) and let v_{0} = (0, 1) ^{∈} W ×F = V. For a given α ^{∈} SΓ(φ) put β = ε(α(v_{0}), v_{0}). Then β^{−1}·α fixes v_{0} and is therefore contained in SΓ(ψ). Hence it remains to show β̄^{∈} i_{∗}(SΓ(ψ)) · ∑_{φ}(D(φ)) for which one may restrict to φ | V′ with V′ = supp β + F v_{1}, where v_{1} ^{∈} V is such that φ(v_{1}) = 1. □

Lemma 9. Let ψ = ⟨1⟩ ⊕ ψ′ be a Pfister form, let ψ̄′ be a subform of ψ′ and let ψ̄ = ⟨1⟩ ⊕ ψ̄′. Moreover let ζ be an arbitrary form representing 1 and let b ^{∈} F^{∗}. Put φ = (ψ ⊗ ζ) ⊕ ⟨b⟩ and φ̂ = φ ⊕ bψ̄′ = (ψ ⊗ ζ) ⊕ bψ̄. Then

i_{∗} : \overline{SΓ(φ)} ⟶ \overline{SΓ(φ̂)}

is surjective.

Proof. Since φ̂ represents 1 we know that ∑φ̂(D(φ̂)) generates \overline{SΓ(φ̂)} by Lemma 8 i). Note that

D(φ̂) ⊂ D(φ) · D(ψ̄) ⊂ D(φ) · D(φ) = N(φ)

since ψ is multiplicative and ζ represents 1. Hence

∑φ̂(D(φ̂)) ⊂ ∑φ̂(N(φ)) ⊂ i_{∗}(\overline{SΓ(φ)})

by Lemma 8 v). □

Lemma 10. Let φ be a Pfister neighbor, i.e. φ = ψ ⊕ bψ̄ where ψ is a Pfister form, ψ̄ is a subform of ψ and b ^{∈} F^{∗}. Then \overline{sn} : \overline{SΓ(φ)} → F^{∗} is injective and has image N(φ) = D(ψ ⊗ ≪−b≫).

Proof. Let ρ = ψ ⊗ ≪−b≫. Then D(ρ) is a group as for every Pfister form and therefore D(ρ) = N(ρ) by Lemma 8 i). Since φ_{K} is isotropic if and only if ρ_{K} is isotropic, we have N(φ) = D(ρ). Therefore N(φ) is a group, hence Im sn = N(φ) = D(ρ).

Injectivity of \overline{sn}: If φ is a Pfister form (i.e. ψ̄ = ψ), then ∑_{φ} is a homomorphism in view of Lemma 8 v) and is a left inverse to \overline{sn}.

In the general case one may apply Lemma 9 with ζ = ⟨1⟩ to reduce to case ψ̄ = ⟨1⟩. In this case we find by Lemma 8 vi) and the above remarks for Pfister forms:

\overline{SΓ(φ)} = i_{∗}(\overline{SΓ(ψ)}) · ∑_{φ}(D(φ)) = ∑_{φ}(N(ψ)) · ∑_{φ}(D(φ)).

Hence every element in \overline{SΓ(φ)} can be written as product of two plane elements and we are done. □

Proof of Proposition 2 ii). By Lemma 9 we may replace φ = ρ ⊗ (ψ ⊕ ⟨c⟩) by φ̃ = (ρ ⊗ ψ) ⊕ ⟨c⟩. Since ρ ⊗ ψ is itself a Pfister neighbor, we may assume φ = ψ ⊕ ⟨c⟩. By Lemma 8 vi) and Lemma 10 we have

\overline{SΓ}(φ) = ∑_{φ}(N(ψ)) · ∑_{φ}(D(φ)).  □

Proof of Proposition 2 iii). Write (V, φ) = (W, ψ)+(F ×F, ⟨c, c⟩) and let v_{0} = (0, 0, 1), v_{1} = (0, 1, 0) ^{∈} V = W ×F ×F. For given α ^{∈} SΓ(φ) let β = ε(α(v_{0}), v_{0}) and γ = ε(β^{−1}α(v_{1}), v_{1}). Then δ = γ^{−1}β^{−1}α fixes v_{0} and v_{1}, hence δ ^{∈} SΓ(ψ) and δ̄ ^{∈} ∑_{φ}(N(ψ)) by Lemma 10. By Lemma 8 iii) we have sn(β), sn(γ) ^{∈} D(cφ) and therefore β̄ · γ̄ ^{∈} ∑_{φ}(D(cφ)) · ∑_{φ}(D(cφ)) = ∑_{φ}(N(φ)). Hence ᾱ = β̄ · γ̄ · δ̄ is in ∑_{φ}(N(ψ)) · ∑_{φ}(N(φ)). □

III. The map \overline{SΓ(φ)} ⟶ A_{0}(X_{φ}, K_{1})

We will use the following

Theorem 11.

i) A_{0}(X_{φ}, K_{0}) ↪ K_{0}F for arbitrary φ
ii) A_{0}(X_{φ}, K_{n}) ↪ K_{n}F for isotropic φ
iii) A_{0}(X_{φ}, K_{1}) ↪ K_{1}F for dim φ = 3.

i) is proved in [Merkuriev, Suslin; On the norm homomorphism in degree 3].

ii) follows from i) by the norm principle.

iii) is one of the main points in the proof of Hilbert Satz 90 for K_{2} for quadratic extensions. It stands at the heart of our construction. It shows that for a 3-dimensional form φ the groups \overline{SΓ(φ)} = SΓ(φ)/[SΓ(φ), SΓ(φ)] and A_{0}(X_{φ}, K_{1}) are naturally isomorphic, because \overline{sn} and N_{φ} are injective and have the same image in F^{∗} = K_{1}F.

Theorem 12. For quadratic forms φ over F there exists unique homomorphisms

ω_{φ} : \overline{SΓ(φ)} ⟶ A_{0}(X_{φ}, K_{1})

such that

i) If dim φ = 3, then ω_{φ} = N_{φ}^{−1} ∘ \overline{sn}.

ii) If φ_{0} is a subform of φ, then

�  ^{ω_{φ_{0}}} �_{i_{∗}} �  _{ω_{φ}} �  _{i_{∗}} A_{0}(X_{φ}, K_{1}) \overline{SΓ(φ)}  A_{0}(X_{φ_{0}}, K_{1}) \overline{SΓ(φ_{0})}

is commutative.

Proof. Let Ḡ be as in Corollary 7 and define

ω̂_{φ} : Ḡ ⟶ A_{0}(X_{φ}, K_{1})

as follows. For α ^{∈} SΓ(φ) plane choose a subform φ_{0} of φ of dimension 2 such that α ^{∈} SΓ(φ_{0}). (φ_{0} is unique if α ^{∉} F^{∗}). Then α ^{∈} SΓ(φ_{0}) = K_{1}F (X_{φ_{0}}) = A_{0}(X_{φ_{0}}, K_{1}) and we define ω̂_{φ}(ḡ_{α}) to be the image of α under A(X_{φ_{0}}, K_{1}) → A(X_{φ}, K_{1}). This definition does not depend on the choice of φ_{0} because of Theorem 11 i). In order to prove Theorem 12, it suffices to show that ω̂_{φ} vanishes on the relations R̄_{0}), R̄_{1}) in Corollary 7. This is clear for R̄_{1}) because of Theorem 11 iii) and follows for R̄_{0}) from Proposition 12. Let v_{1}, v_{2} ^{∈} X_{(0)} be two points of degree 2, let F_{i} = K(v_{i}) and let α_{i} ^{∈} F_{i}^{∗} such that N_{F_{1}|F} (α_{1}) = N_{F_{2}|F} (α_{2}). Then [{α_{1}}, v_{1}] − [{α_{2}}, v_{2}] is in the image of

⨁_{v^{∈}X_{(1)}}K_{2}K(v) ⟶^{d} ⨁_{v^{∈}X_{(0)}}K_{1}K(v).

Proof. Let F_{0} = F_{1} ⊗_{F} F_{2}. We have

α_{i} = {β} + {N_{F_{0}|F_{i}}(γ)}

for some β ^{∈} F^{∗} and γ ^{∈} F_{0}^{∗} (take β = (tr α_{1} + tr α_{2})^{−1}, γ = α_{1} + α_{2} in the generic case). Hence

[{α_{1}}, v_{1}] − [{α_{2}}, v_{2}] = {β} · (v_{1} − v_{2}) + cor_{F_{1}|F} (u_{1} − u_{2})

where u_{1}, u_{2} ^{∈} ⨁_{v^{∈}(X_{F_{1}})_{(0)}}K_{1}K(v) are given by

u_{1} = [N_{F_{0}|F_{1}}{γ}, ṽ_{1}],

with ṽ_{1} a rational point of X_{F_{1}} and

u_{2} = [γ, Ṽ_{2}]

with ṽ_{2} the point over v_{2}. Now {β} · (v_{1} − v_{2}) ^{∈} Im d, because A_{0}(X_{φ}, K_{0}) ↪ K_{1}F.

Furthermore, over F_{1} we have N_{φ}(u_{1} − u_{2}) = 0, hence u_{1} − u_{2} ^{∈} Im d, because X_{F_{1}} has a rational point. □

Proposition 13. If F has no extension of odd degree, then ω_{φ} : \overline{SΓ(φ)} → A_{0}(X_{φ}, K_{2}) is surjective.

It follows from Knebusch’s norm principle that Im (N_{φ} ∘ ω_{φ}) = Im N_{φ}. One may use the proof of Knebusch’s norm principle to show that ω_{φ} is surjective in general. Since this is a bit tedious I omit a proof here.

Proof of Theorem 1. Since A_{0}(X_{φ}, K_{1}) ↪ K_{1}F for isotropic φ, we have 2Ker N_{φ} = 0 by a transfer argument. Hence we may assume that F has no odd extension, again using transfers. But then by Proposition 13 and the very definition of \overline{SΓ(φ)}:

Ker N_{φ} = ω_{φ}(Ker \overline{sn}) = 0.  □

For a quadratic form put D_{1}(φ) = Im sn = Im N_{φ}. For the proof of Proposition 13 we need the following lemma which can be deduced also from the arguments in [Merkuriev, Suslin; On the norm homomorphism in degree 3].

Lemma 14. Let dim φ = 4 and let H/F be a quadratic extension. Then for every u ^{∈} D_{1}(φ_{H}) there exists (over F) two 3-dimensional subforms φ′, φ′′ of φ such that u ^{∈} D_{1}(φ′_{H}) · D_{1}(φ′′_{H}).

Proof. (sketch) Write φ = ⟨−a, −b, ab, c⟩. It is easy to check that

D_{1}(φ) = Nrd (D(a, b) ⊗ F (\sqrt{c})) ∩ F^{∗} ⊂ F (\sqrt{c})^{∗}.

Using this for φ_{H} one finds

u = Nrd (d) · (1 + c Nrd (d′))

for some d, d′ ^{∈} D(a, b) ⊗_{F} H with d′ + \overline{d′} = 0. Now put φ′ = ⟨−a, −b, ab⟩, then Nrd (d) ^{∈} D_{1}(φ′_{H}). It is not hard to find ā, b̄ ^{∈} F^{∗} such that D(a, b) ≃ D(ā, b̄) and 1 + cNrd (d′) ^{∈} Nrd (D(āc, b̄c) ⊗_{F} H). Since φ̄ = ⟨−āc, −b̄c, ab̄, c⟩ has the same even Clifford algebra as φ, we know that φ̄ is similar to φ by quadratic form theory. Hence ⟨−āc, −b̄c, āb̄⟩ is similar to a subform φ′′ of φ. Now we are done because 1 + c Nrd (d′) ∈ D_{1}(φ′′_{H}).

Consequence. Let H/F be a quadratic extension. Then

cor_{H/F} (Im ω_{φ_{H}}) ⊂ Im ω_{φ}.

Proof. We may assume dim φ = 4. For α ^{∈} SΓ(φ_{H}) plane there exists by Lemma 14 subforms φ′, φ′′ of φ of dimension 3 and α′ ^{∈} SΓ(φ′_{H}), α′′ ^{∈} SΓ(φ′′_{H}) such that sn(α) = sn(α′) sn(α′′). We know that \overline{sn} is injective, hence

cor_{H/F} (ω_{φ_{H}}(α)) = cor_{H/F} (ω_{φ_{H}}(\overline{α′}) + ω_{φ_{H}}(\overline{α′′})).

Now cor_{H/F} (ω_{φ_{H}}(\overline{α′})) is in the image of A_{0}(X_{φ′}, K_{1}) → A_{0}(X_{φ}, K_{1}). But we know that ω_{φ′} is surjective, hence cor_{H/F} (ω_{φ_{H}}(\overline{α′})) ^{∈} Im ω_{φ}. Similarly cor_{H/F} (ω_{φ_{H}}(\overline{α′′})) ^{∈} Im ω_{φ}. □

Proof of Proposition 13. By the consequence we have the norm principle for Im (ω_{φ}) if F has no odd extensions. Since A_{0}(X_{φ}, K_{1}) is generated by corestrictions from splitting fields K of φ and since A_{0}(X_{φ_{K}}, K_{1}) = Im ω_{φ_{K}} we conclude A_{0}(X_{φ}, K_{n}) = Im ω_{φ}. □