A Variant of a Theorem by Springer^{∗}

I. Panin, U. Rehmann

December 6, 2003

The main aim of the preprint is to prove the following result which is a variant of the Springer theorem [La] for quadratic spaces over local rings.

Theorem. Let R be a local Noetherian domain which has an infinite residue field and let S = R[T]/(F (T)) be an integral extension which is étale over R. Let (V, q) be a quadratic space over R such that the S-quadratic space (V ⊗_{R} S, q ⊗_{R} S) contains a hyperbolic plane ℍ_{S}. If the degree of the polynomial F (T) is odd, then the space (V, q) contains a hyperbolic plane already over R.

This theorem is one of the main ingredient in the proof of the following result proved in [P].

0.1 Theorem. Let R be a regular local ring, K its field of fractions and (V, φ) a quadratic space over R. Suppose R contains a field of characteristic zero. If (V, φ)⊗_{R} K is isotropic over K, then (V, φ) is isotropic over R, that is there exists a unimordular vector v ∈ V with φ(v) = 0.

It is well-known that any finite étale extension S of R has the form S = R[T]/(F (T)), where F (T) is a monic separable polynomial. If A is a semi-local ring and (W, ϕ) is a quadratic space over A, then W contains a hyperbolic plane if and only if W contains a unimodular isotropic vector w. A vector w is called unimodular if w can be taken as the first vector w_{1} of a free A-base w_{1}, …, w_{n} of the A-module W.

1 Preliminaries

In this section we formulate two results which will be used in the proof of the main theorem. These two results will be proved in Section 3. We need to fix certain notation

k is an infinite field (char(k) ≠ 2)

f(t) is a monic separable polynomial of degree n over k

l = k[t]/(f(t)) is a separable k-algebra θ = t mod f(t) is an element of the algebra l (W, ϕ) is a quadratic space over k (W_{l}, ϕ_{l}) is the quadratic space (W ⊗_{k} l, ϕ ⊗_{k} l) over l W [t]^{(m)} = W · 1 ⊕ W · t ⊕ … ⊕ W · t^{m−1} ⊂ W [t] = W ⊗_{k} k[t] k[t]^{(m)} = k · 1 ⊕ k · t ⊕ … ⊕ k · t^{m−1} ⊂ k[t] ev : W [t]^{(n)} → W_{l} is a map given by

∗Both authors were supported by INTAS-99-0817

ev(v_{0} + v_{1}t + … + v_{n−1}t^{n−1}) = v_{0} + v_{1}θ + … + v_{n−1}θ^{n−1}

v(θ) =^{def} ev(v(t)) for any v(t) ∈ W [t]^{(n)}

ϕ^{(n)} : W [t]^{(n)} → k[t]^{(2n−1)} is the map given by

ϕ^{n}(v_{0} + v_{1}t + … + v_{n−1}t^{n−1}) = ϕ(v_{0}) + 2(v_{0}, v_{1}) · t + ϕ(v_{1}) · t^{2} + … + ϕ(v_{n−1}) · t^{2n−2} 1.1 Proposition. Suppose the quadratic space (W_{l}, ϕ_{l}) contains a hyperbolic plane ℍ_{l} as a direct summand. Then there exists a unimodular isotropic vector w ∈ W_{l} and for any such a vector w there exists an element v(t) ∈ W [t]^{(n)} satisfying the following conditions (1) ϕ^{(n)}(v(t)) ∈ k[t]^{(2n−1)} is a polynomial of degree 2n − 2

(2) ϕ^{(n)}(v(t)) is a separable polynomial

(3) ⟨w, v(θ)⟩ ∈ l^{∗}, where l^{∗} ⊂ l is the group of units of l

(4) ϕ_{l}(v(θ)) = 0 in l 1.2. Let A be a semi-local ring and let Ā = A/Rad(A) and let (U, ψ) be a quadratic space over A and suppose (U, ψ) contains a hyperbolic plane ℍ_{A} as a direct summand. We will use ”bar” for the reduction modulo the radical Rad(A). For example, (Ū, ψ̄) is the quadratic space (U/Rad(A) · U, ψ̄) over Ā. 1.3 Lemma. Let u ∈ U be a unimodular isotropic vector. Then for any isotropic vector v ∈ Ū with the property ⟨v, ū⟩ ∈ A^{∗} there exists a vector v ∈ U such that

(1) v̄ = v in Ū and

(2) ψ(v) = 0. Proof. Let v_{0} be any vector in U with v̄_{0} = v. Then take

v = \frac{−⟨v_{0}, v_{0}⟩}{2⟨v_{0}, u⟩} · u + v_{0}.

Then v is the desired isotropic vector. In fact, ψ(v) = −⟨v_{0}, v_{0}⟩ + ⟨v_{0}, v_{0}⟩ = 0 and v̄ = v̄_{0}, because ⟨v̄_{0}, v̄_{0}⟩ = ⟨v, v⟩ = 0. Since v̄_{0} = v one has v̄ = v. Lemma is proved.

1.4 Lemma. Let k[t] be the polynomial ring over the field k and let (W, ϕ) be the quadratic space over k. Let w(t) = w_{0} + w_{1} · t + … + w_{n−1} · t^{n−1} be an element of W [t] with w_{i} ∈ W. Suppose the polynomial ϕ^{(n)}(w(t)) ∈ k[t] is separable and let g(t) ∈ k[t] be an irreducible polynomial dividing ϕ^{(n)}(w(t)). Then w(t) does not vanish modulo g(t).

Proof. If w(t) vanishes modulo g(t), then w(t) = g(t) · u(t) for an element u(t) ∈ W [t]. In this case one has

ϕ^{(n)}(w(t)) = ϕ^{(n)}(g(t) · u(t)) = g(t)^{2} · ϕ^{(n)}(u(t)) ∈ k[t].

This relation contradicts with the separability of ϕ^{(n)}(w(t)). Thus w(t) does not vanish modulo g(t). Lemma is proved.

1.5 Lemma. Let A be a semi-local ring and let (U, ψ) be a quadratic space over A. Let v ∈ U be a unimodular isotropic vector. Then one can split a hyperbolic plane of (U, ψ), i.e.,

(U, ψ) ≅ (U′, ψ′) ⊥ ℍ.

Proof. It is easy.

2 The proof of the main theorem

2.1 Theorem. Let R be a local Noetherian domain which has an infinite residue field of characteristic different of 2. Let S = R[T]/(F (T)) be an integral extension which is étale over R. Let (V, q) be a quadratic space over R such that the S-quadratic space (V ⊗_{R} S, q ⊗_{R} S) contains a hyperbolic plane ℍ_{S}. If deg F (T) is odd then the space (V, q) contains a hyperbolic plane ℍ_{R}.

Proof. Let 𝔪 be the maximal ideal of R and let k be the residue field R/𝔪 of R. Let l = S/𝔪S = k[t]/(f(t)), where f(t) = F [T] mod 𝔪. Since S is étale over R so is the k-algebra l. In particular the k-algebra l is separable.

We will write (V_{R}, q_{R}) for (V, q) and write (V̄ , q̄) for the reduction modulo 𝔪 of the R-quadratic space (V_{R}, q_{R}). Let (V_{S}, q_{S}) be the scalar extension of (V_{R}, q_{R}) up to S and let (V̄_{l}, q̄_{l}) be the reduction modulo 𝔪S of the S-quadratic space (V_{S}, q_{S}). Clearly, (V̄_{l}, q̄_{l}) = (V̄ , q̄) ⊗_{k} l. Set now (W, ϕ) = (V̄ , q̄). Then (W ⊗_{k} l, ϕ ⊗_{k} l) = (V̄_{l}, q̄_{l}) and we will write (W_{l}, ϕ_{l}) for (V̄_{l}, q̄_{l}).

By the hypotheses of the theorem the space (W_{l}, ϕ_{l}) contains a hyperbolic plane ℍ_{l} as a direct summand. Thus we are under the hypotheses of Proposition 1.1. So using notation of Section 1 one can find a vector w ∈ W_{l} (unimodular and isotropic) and an element v(t) = v_{0} · 1 + v_{1} · t + … + v_{n−1} · t^{n−1} ∈ W [t]^{(n)} satisfying the following conditions

(1) ϕ^{(n)}(v(t)) ∈ k[t] has degree 2n − 2

(2) ϕ^{(n)}(v(t)) is a separable polynomial over k

(3) ⟨w, v(θ)⟩ ∈ l^{∗} is an invertible element of l (4) ϕ_{l}(v(θ)) = 0 ∈ l, where θ = t mod f(t) is the element of l

Recall now that the quadratic space (W_{l}, ϕ_{l}) is the reduction modulo 𝔪S of the quadratic space (V_{S}, q_{S}). Thus by Lemma 1.3 one can lift the element v = v(θ) ∈ W_{l} up to a unimodular isotropic vector v ∈ V_{S}.

Since S = R[T]/(F (T)) we have V_{S} = V_{R}⊗_{R}R[T]/(F (T)). If we set θ = T mod F (T), then one can find elements v_{0}, v_{1}, …, v_{n−1} ∈ V_{R} such that v = v_{0}·1+v_{1}·θ+…+v_{n−1}·θ^{n−1}. Consider now the element v(T) = v_{0} · 1 + v_{1} · T + … + v_{n−1} · T^{n−1} in V_{R} ⊗_{R} R[T] and consider the diagram

[figure: W_{l}; W × … × W; ev; ϕ^{(n)}; k × … × k; disc; k; V̄_{l}; V̄ × … × V̄; ev; q̄^{(n)}; k × … × k; disc; k; V_{S}; V_{R} × … × V_{R}; Ev; q_{R}^{(n)}; R × … × R; Disc; R]

where Ev(u_{0}, …, u_{n−1}) = u_{0} · 1 + u_{1} · θ + … + u_{n−1} · θ^{n−1} ∈ V_{S} and q_{R}^{(n)}(u_{0}, …, u_{n−1}) = q_{R}(u_{0}) + 2 · (u_{0}, u_{1}) · T + q_{R}(u_{1}) · T^{2} + … + q_{R}(u_{n}) · T^{2n−2} and Disc(a_{0}, a_{1}, …, a_{2n−2}) is the discriminant of the polynomial a_{0} + a_{1}T + … + a_{2n−2}T^{2n−2}. The vertical arrows are the canonical maps.

Clearly, this diagram commutes and maps Ev, ev are isomorphisms. In particular, v_{i} = v_{i} mod 𝔪 for the components v_{0}, …, v_{n−1} (resp. v_{0}, …, v_{n−1}) of the element v(t) ∈ W [t] (resp. v(T) ∈ V_{R}[T]). Thus the reduction modulo 𝔪 of the polynomial q_{R}^{(n)}(v(T)) coincides with the polynomial ϕ^{(n)}(v(t)). The last polynomial is separable and has degree 2n − 2 by the choice of v(t).

Since v(θ) = v by the very choice of the element v(T) and since v is q_{R}-isotropic one has the relation q_{R}(v(T)) = 0. Thus q_{R}(v(T)) vanishes modulo F (T) in the ring R[T], i.e., there exists a polynomial H(T) in R[T] such that

q_{R}^{(n)}(v(T)) = F (T) · H(T).  (∗)

Taking the reduction modulo 𝔪 we get the relation

ϕ^{(n)}(v(t)) = f(t) · h(t),  (∗∗)

where h(t) is the reduction of H(t) modulo 𝔪. Since deg ϕ^{(n)}(v(t)) = 2n−2 and deg f(t) = n one gets that deg h(t) = n − 2. Since deg H(T) ≤ n − 2 we see that deg H(T) = n − 2 and H(T) is monic (the highest coefficient is invertible). Further ϕ(v(t)) is separable. Thus h(t) is separable as well. This shows that the R-algebra S′ = R[T]/(H(T)) is an étale R-algebra.

Denote by A the class of T modulo H(T) in the ring S′. Then the vector v(A) = v_{0} · 1 + v_{1} · A + … + v_{n−1} · A^{n−1} of the quadratic S′-space (V_{S′}, q_{S′}) is isotropic. In fact, q_{S′}(v(A)) = F (A) · H(A) = 0 on S′. If the vector v(A) is unimodular then one can split a hyperbolic plane ℍ_{S′} of the quadratic space (V_{S′}, q_{S′}) over S′ (see Lemma 1.5). So in this case we constructed a finite étale extension S′ = R[T]/(H(T)) such that

deg H(T) = deg F (T) − 2

and such that the space (V_{S′}, q_{S′}) contains a hyperbolic plane ℍ_{S′} as a direct summand. Repeating this procedure several times we finally get a direct hyperbolic summand of the quadratic space (V_{R}, q_{R}) itself. Thus to complete the proof of the theorem it remains to check that the vector v(A) ∈ V_{S′} is unimodular.

For this denote by k′ the ring S′/𝔪S′ and observe that k′ = k[t]/(h(t)), where as above h(t) is the reduction modulo 𝔪 of the polynomial H(t). Further denote by V̄_{k′} the k′-module V̄ ⊗_{k} k′ and consider the commutative diagram (with the same elements v(T) and v(t) as above in this proof)

[figure: v(α); ∈; V̄_{k′}; V̄ [t]^{(n)}; \overline{Ev}; v(t); ∋; v(A); ∈; V_{S′}; V_{R}[T]^{(n)}; Ev; v(T); ∋] .

Here Ev(u_{0} + u_{1}T + … + u_{n−1}T^{n−1}) = u_{0} + u_{1} · A + … + u_{n−1} · A^{n−1} and \overline{Ev}(w_{0} + w_{1} · t + … + w_{n−1} · t^{n−1}) = w_{0} + w_{1} · α + … + w_{n−1} · α^{n−1} and α = t mod h(t) ∈ k′. To check that v(A) is unimodular it suffices to verify that v(α) is unimodular. Observe that V̄_{k′} = V̄ ⊗_{k} k[t]/(h(t)). Let h(t) = h_{1}(t) · . . . · h_{r}(t) be the decomposition of h(t) in a product of irreducible polynomials. Since h(t) is separable one has h_{i}(t) ≠ h_{j}(t) for i ≠ j. Then V̄_{k′} = ∏_{i=1}^{r} V̄ ⊗_{k} (k[t]/(h_{i}(t))) and thus v(α) is unimodular in V̄_{k′} if and only if the elements v(t) ∈ V̄ [t] does not vanish modulo any of h_{i}(t) (i = 1, …, r).

The polynomial h_{i}(t) divides h(t) and the polynomial h(t) divides ϕ^{(n)}(v(t)) (see (∗)). Thus h_{i}(t) divides ϕ^{(n)}(v(t)). Since h_{i}(t) is irreducible and ϕ^{(n)}(v(t)) is separable Lemma 1.4 proves that v(t) does not vanish modulo h_{i}(t). Thus indeed the element v(α) is a unimodular vector in V̄_{k′} and the element v(A) is a unimodular vector in V_{S′}. This completes the proof of the theorem.

3 The proof of Proposition 1.1

3.1. Since the quadratic space (W_{l}, ϕ_{l}) contains a hyperbolic plane ℍ_{l} as a direct summand one can find a unimodular isotropic vector w ∈ W_{l}. Choose and fix such a vector w ∈ W_{l}.

Now set X(l) = {v ∈ W_{l} | ϕ(v) = 0} and consider a map ρ_{w} : X(l) → l taking v into the scalar product ⟨w, v⟩ ∈ l. Consider then a diagram of sets and their polynomial maps

\xymatrix{l & X(l) \ar[l]_{ρ_{w}} \ar@{^{(}->}[d]^{i} & Y \ar[l] \ar@{^{(}->}[d]^{j} \\ & W_{l} & W × … × W = W [t]^{(n)} \ar[l]_{ev} \ar[r]^{ϕ^{(n)}} & k[t]^{(2n−1)} \ar[r]^{disc} & k} ,

where ev is the map defined in the beginning of Section 1 (this map is clearly an isomorphism) and where Y = ev^{−1}(X(l)) and maps i, j are the inclusions and ϕ^{(n)} is defined in the beginning of Section 1. The map disc takes a polynomial g(t) ∈ k[t]^{(2n−1)} into its discriminant. It is well-known that disc(g(t)) has a polynomial expression in terms of coefficients of the polynomial g(t).

This diagram is the diagram of k-rational points and their maps induced by the following one

\xymatrix{R_{l/k}(𝔸_{l}^{1}) & R_{l/k}(X_{l}) \ar[l]_{ℰ_{w}} \ar@{^{(}->}[d]^{i} & 𝕐 \ar[l]_{≅} \ar@{^{(}->}[d]^{j} \\  & R_{l/k}(𝕎_{l}) & 𝕎 × … × 𝕎 \ar[l]_{ev}^{≅} \ar[r]^{ϕ^{(n)}} & 𝔸_{k}^{1} × … × 𝔸_{k}^{1} \ar[r]^{disc} & 𝔸_{k}^{1}}

of algebraic varieties. Here R_{l/k} is the Weil restriction functor. Here 𝕎 (resp. 𝕎_{l}) is the k-vector space (resp. the l-vector space) considered as the k-variety (resp. as the l-variety) and X_{l} is the affine quadric in 𝕎_{l} given by the equation ϕ_{l} = 0 and ℰ_{w} is the Weil restriction of the morphism ρ_{w} : X_{l} → 𝔸_{l}^{1} taking v into ⟨w, v⟩ and the variety 𝕐 is the preimage of R_{l/k}(X_{l}) under the evaluation isomorphism ev : 𝕎 × … × 𝕎 → R_{l/k}(𝕎_{l}) and morphisms i and j are the closed embeddings. The product 𝕎 × … × 𝕎 consists of n factors and the product 𝔸_{k}^{1} × … × 𝔸_{k}^{1} consists of 2n − 1 factors.

Define open subsets of the variety 𝕐 as follows

U_{1} = j^{−1}((ϕ^{(n)})^{−1}(𝔸_{k}^{1} × … × 𝔸_{k}^{1} × (𝔸_{k}^{1} − {0})))

U_{2} = j^{−1}((ϕ^{(n)})^{−1}(disc^{−1}(𝔸_{k}^{1} − {0})))

U_{3} = ev^{−1}(ℰ_{w}^{−1}(R_{l/k}(𝔸_{l}^{1} − {0}_{l})))

We need in the following Lemmas

3.2 Lemma. 𝕐 is k-rational variety.

3.3 Lemma. U_{1} ∩ U_{2} ≠ ∅

3.4 Lemma. U_{3} ≠ ∅

Having these three Lemmas one can prove Proposition 1.1 as follows. Since 𝕐 is k-rational variety any its non-empty open subset has a k-rational point (even infinitely many). Thus by Lemma 3.3 and Lemma 3.4 one can find a point

v ∈ (U_{1} ∩ U_{2} ∩ U_{3})(k) ⊂ 𝕐(k) ⊂ (𝕎 × … × 𝕎)(k).

Set v(t) = j(v). Then the element v(t) satisfies the condition (1), (2) and (3) of Proposition 1.1 by the very definition of the open sets U_{1}, U_{2} and U_{3}. Since v(θ) =^{def} ev(v(t)) thus v(θ) is a k-rational point of the variety R_{l/k}(X_{l}), i.e., v(θ) ∈ X(l). Thus ϕ_{l}(v(θ)) = 0 and the condition (4) is satisfied as well. To complete the proof of Proposition 1.1 it remains to prove Lemmas 3.2 – 3.4.

3.5 (Proof of Lemma 3.2). Since 𝕐 is isomorphic to the variety R_{l/k}(X_{l}) it suffices to check the k-rationality of the last one. The quadric X_{l} has an l-rational point and thus X_{l} is an l-rational variety. This implies that there exist non-empty open subvarieties V_{1} in X_{l} and V_{2} in 𝔸_{l}^{r−1} (r = dim_{k}W) and an isomorphism α : V_{1} ⟶^{≅} V_{2} of l-varieties. Consider now the diagram of l-varieties

X_{l} ⊃ V_{1} ⟶^{≅} V_{2} ⊂ 𝔸_{l}^{r−1}

and apply the Weil restriction functor to this diagram. One obtains a diagram

R_{l/k}(X_{l}) ⊃ R_{l/k}(V_{1}) ⟶^{≅} R_{l/k}(V_{2}) ⊂ R_{l/k}(𝔸_{l}^{r−1})

of k-varieties. Since the left and the right hand side inclusions are open imbeddings and since the variety R_{l/k}(𝔸_{l}^{r−1}) is an affine space over k one concludes that R_{l/k}(X_{l}) is k-rational. Lemma is proven.

3.6 (Proof of Lemma 3.4). The morphism ev : 𝕐 → R_{l/k}(X_{l}) is an isomorphism by the very definition of the variety 𝕐. Therefore it suffices to verify that the variety ℰ_{w}^{−1}(R_{l/k}(𝔸_{l}^{1} − {0}_{l})) is non-empty. We show now that the last variety has k-rational points. For this recall that the morphism ℰ_{w} : R_{l/k}(X_{l}) → R_{l/k}(𝔸_{l}^{1}) induces a map of k-rational points which coincides with the one

ρ_{w} : X(l) → l (v ↦ ⟨v, w⟩).

Take any v ∈ X(l) with ⟨w, v⟩ ∈ l^{∗} and observe that for this element v one has ρ_{w}(v) ∈ l^{∗} and moreover the group l^{∗} coincides with the set of k-rational points of the variety R_{l/k}(𝔸_{l}^{1}−{0}_{l}). Thus the element v is a k-rational point of the variety ℰ_{w}^{−1}(R_{l/k}(𝔸_{l}^{1}−{0}_{l})). Lemma is proved.

3.7 (Proof of Lemma 3.3). We show that already (U_{1} ∩ U_{2})(l) ≠ ∅. For this consider an l-basis e_{1}, e_{2}, …, e_{r} of the free l-module W_{l} such that e_{1}^{2} = 0, e_{2}^{2} = 0, (e_{1}, e_{2}) = 1 and (e_{1}, e_{i}) = (e_{2}, e_{i}) = 0 for i ≥ 3. For any polynomial h(t) ∈ l[t]^{(n)} and for any element e ∈ W_{l} we set h(t)·e = (a_{0} ·e)·1+(a_{1} ·e)·t+…+(a_{n−1} ·e)·t^{n−1}) ∈ l ⊗_{k} W [t]^{(n)} = W_{l}[t]^{(n)}. Since l = k[t]/(f(t)) and θ = t mod f(t) (see page 3 for the definition of θ) one has a unique decomposition f(t) = f^{(1)}(t)(t − θ) in l[t]. Let g(t) ∈ k[t] be a separable of degree n − 2 polynomial which is coprime with f(t). Consider now the element

v(t) = f^{(1)}(t) · e_{1} + (t − θ)g(t) · e_{2} ∈ l ⊗_{k} W [t]^{(n)} = W_{l}[t]^{(n)}.

3.8 Claim. v(t) ∈ (U_{1} ∩ U_{2})(l)

To check this claim observe first that

ϕ^{(n)}(v(t)) = 2 · f^{(1)}(t) · (t − θ) · g(t) · ⟨e_{1}, e_{2}⟩ = 2f(t) · g(t)  (∗)

Thus ϕ(v(θ)) = 2 · f(θ) · g(θ) = 0 in l and therefore v(t) ∈ 𝕐(l). Further (∗) shows that ϕ^{(n)}(v(t)) ∈ k[t] ⊂ l[t] and it has degree 2n − 2. Therefore v(t) ∈ U_{1}(l).

Finally, ϕ^{(n)}(v(t)) is separable because f(t), g(t) are separable and coprime polynomials in k[t]. Thus disc(ϕ^{(n)}(v(t))) ∈ k^{∗} ↪ l^{∗} and, therefore, v(t) ∈ U_{2}(l). So (U_{1} ∩ U_{2})(l) ∋ v(t) and, hence, (U_{1} ∩ U_{2})(l) ≠ ∅. Lemma is proved.

References

[La] T.Y.Lam. The algebraic theory of quadratic forms, Math. Lect. Notes Ser., W.A.Benjamin, Inc., 1973.

[P] I.Panin. Rationally isotropic quadratic spaces are locally isotropic,

       http://www.mathematik.uni-bielefeld.de/LAG/man/127.html