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Pfister forms and a conjecture due to Colliot–Thélène in the mixed characteristic case I. A. Panina, D. N. Tyurinab a St. Petersburg Department of Steklov Mathematical Institute of Russian Academy of Sciences b Leonard Euler International Mathematical Institute at Saint Petersburg (SPB LEIMI), St. Petersburg
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     § 1. IntroductionLet $R$ be a local regular ring with fraction field $K$, let $Q$ be a non-degenerate quadratic form over $R$, and $c$ in $R$ be an invertible element. In 1977, Colliot–Thélène [2] conjectured that the equation $Q=c$ has a solution over the ring $R$ if and only if it has a solution over the field $K$. Panin [4] proved this conjecture for the case when $R$ contains the field of rational numbers. His proof was based on the moving lemma for Levine–Morel’s algebraic cobordism. Later, together with K. I. Pimenov (see [7], [8]), he extended this result to the case when $R$ contains an arbitrary infinite field. The proof in [7] used the sophisticated tool – namely, the Gabber version of the de Jong theorem on alteration, whereas the proof of [8] was based upon quite “elementary” arguments, which could be discovered right after the appearance of the paper [3]. Finally, Scully [9] proved the conjecture for all regular local ring containing a subfield (particularly, a finite field). Nevertheless, the mixed characteristic case remained nearly uninvestigated until recently. In this way, we note the recent paper by Panin (see [5], § 7) which was concerned with the case where $R$ is a local ring of an $A$-smooth scheme $X$ satisfying the $A$-relative Noether Normalization Lemma. In [6], Corollary 3.3, the conjecture was settled for any unramified $R$ under the assumption that the form $Q$ is a Pfister form of dimension 4. The goal of the present paper is to extend the aforementioned Corollary 3.3 in [6] to arbitrary Pfister forms (for the case of mixed characteristic). Our main result is as follows. Let $R$ be a regular local ring of mixed characteristic $(0,p)$, where $p\neq 2$ is a prime number. Suppose that the quotient ring $R/pR$ is also regular. We fix a non-degenerate Pfister form $Q(T_{1},\dots,T_{2^{m}})$ over $R$ and some invertible element $c$ in $R$. Then the equation $Q(T_{1},\ldots,T_{2^{m}})=c$ has a solution over $R$ if and only if it has a solution over the fraction field $K$. Originally, this theorem was proved under the assumption that the residue field of the ring $R$ is infinite. However, due to Scully’s variant of Artin–Springer’s theorem [9], we were also able to provide the proof for the case of a finite residue field. The sketch of the proof is as follows: suppose first that the residue field of $R$ is infinite. Assume that the equation $Q=c$ has a solution over $K$. This means that a certain auxiliary Pfister form $\widetilde{Q}$ over $R$, which is constructed from the pair $(Q,c)$, is isotropic over $K$. Since $\widetilde{Q}_{K}$ is a Pfister form, it is hyperbolic over $K$. According to the recent result of Cesnavicius (see [1], Corollary 9.6), the form $\widetilde{Q}$ is hyperbolic also over $R$. Using now the key geometric theorem (see below), we conclude the proof in the case of infinite residue fields The case of finite residue field follows from the previous result and Scully’s variant of Artin–Springer’s theorem (Theorem 4.1 in [9]). Let us now recall the aforementioned key geometric theorem. Let $R$ be a local ring with the infinite residue field $R/\mathfrak{m}R$, let $M$ be a free module over $R$ of rank $2n$ with fixed hyperbolic quadratic form $Q$, and let $N$ be its free direct summand of rank $n+1$ such that the restriction $Q|_{N}$ is also non-degenerate. Then $N$ contains a unimodular isotropic vector. The most basic case of this theorem is when $R$ coincides with some algebraically closed field $k$. Generally, by $k$ we denote the residue field of $R$. In what follows, we assume that the characteristic of $k$ is always different from 2. The paper is organized as follows. In § 2, we prove the key geometric theorem. The proof of the main result is given in § 3. § 2. The key geometric theoremIn what follows, by $R$ and $k$ we will denote, respectively, a regular local ring and its residue field. Recall that $\operatorname{char}(k)\neq2$. Let us now prove the key geometric theorem for the basic case of a vector space over an algebraically closed field $k$. Proposition 2.1. Let $k$ be an algebraically closed field, $V$ be a $2n$-dimensional $k$-vector space equipped with a hyperbolic quadratic form $Q$, and $P\subset V$ be an $(n+ 1)$-dimensional subspace such that the restriction $Q|_{P}$ is non-degenerate. Then the space $V$ contains a totally isotropic $n$-dimensional subspace $W\subset V$ such that $\operatorname{dim}(P\cap W)= 1$. The following simplified version of the proof was proposed by the referee. The original proof is given in § 4 for the reader’s convenience. Proof. Let $W_{0}\subset V$ be some totally isotropic subspace of dimension $n$. Consider the subset of the Grassmannian variety $\mathrm{Gr}_{n+1}(V)$. We claim that this set is open in $\mathrm{Gr}_{n+1}(V)$. Note that, in our case, the condition $ \operatorname{dim}(P'\cap W_{0})=1$ can be replaced with the equality $P'+W_{0}=V$. Since $Q$ is a non-degenerate form, there is an isomorphism $\mathrm{Gr}_{n+1}(V)\cong \mathrm{Gr}_{n-1}(V)$ that sends an arbitrary $(n+1)$-dimensional subspace $P'\subset V$ to its orthogonal subspace $(P')^{\perp}$. The image of $\mathrm{Gr}^{W_{0}}_{n+1}(V)$ under this isomorphism is the open set Indeed, since $W_{0}^{\perp}=W_{0}$, any element $v\in ((P')^{\perp}\cap W_{0})$ is orthogonal to both $P'$ and $W_{0}$. Therefore, it is also orthogonal to the sum $P'+W_{0}=V$. The form $Q$ is non-degenerate, and hence $v$ is not zero. On the other hand, if $U$ is some $(n-1)$-dimensional subspace of $V$ such that $U\cap W_{0}=\{0\}$, we have the equality $(U\cap W_{0})^{\perp}=U^{\perp}+W_{0}^{\perp}=V$. Since $W_{0}^{\perp}=W_{0}$, we have $U^{\perp}\in \mathrm{Gr}^{W_{0}}_{n+1}(V)$. Applying the inverse isomorphism, we find that $\mathrm{Gr}^{W_{0}}_{n+1}(V)$ is also open in $\mathrm{Gr}_{n+1}(V)$.
Let us now show that the set
$$
\begin{equation*}
\mathrm{Gr}^Q_{n+1}(V):=\{P'\subset V\mid \operatorname{dim}(P')=n+1,\, Q|_{P'}\text{ is a non-degenerate form}\}
\end{equation*}
\notag
$$
is also open in $\mathrm{Gr}_{n+1}(V)$. Let $F_{n+1}(V)$ be the variety of all $(n+1)$-frames on $V$. Consider the function which maps an $(n+1)$-frame to the determinant of its Gram matrix. This function is regular, and hence $\mathrm{Gram}^{-1}(\mathbb{A}^1\setminus\{0\})$ is an open subset of $F_{n+1}(V)$. At the same time, it is clear that the image of $\mathrm{Gram}^{-1}(\mathbb{A}^1\setminus \{0\})$ under the natural smooth morphism $F_{n+1}(V)\to\mathrm{Gr}_{n+1}(V)$, which sends an arbitrary $(n+1)$-frame to the vector space it spans, coincides with $\mathrm{Gr}^Q_{n+1}(V)$. Hence $\mathrm{Gr}^Q_{n+1}(V)$ is also open in $\mathrm{Gr}_{n+1}(V)$, since a smooth morphism is open.
The sets $\mathrm{Gr}^{W_0}_{n+1}(V)$ and $\mathrm{Gr}^Q_{n+1}(V)$ are non-empty (the later set is non-empty by the assumptions of Proposition 2.1), Hence their intersection $\mathrm{Gr}^{W_0}_{n+1}(V)\cap \mathrm{Gr}^Q_{n+1}(V)$ is also non-empty. In particular, since $k$ is algebraically closed, this intersection contains a point. Therefore, there exists an $(n+1)$-dimensional subspace $P^{'}\subset V$ such that the form $Q|_{P'}$ is non-degenerate and $\operatorname{dim}(P'\cap W_{0})=1$. Since both $Q|_{P}$ and $Q|_{P'}$ are non-degenerate and $k$ is algebraically closed, there exists an isometry between $P'$ and $P$. By Witt’s theorem, this isometry can be extended to an isometry $\rho:V\to V$. Setting $W=\rho(W_{0})$, we see that $W$ obviously satisfies the conditions of Proposition 2.1, which completes the proof. Let us now assume that $k$ is not algebraically closed. We let $\mathrm{GO}(n,V)$ denote the Grassmannian of all totally isotropic $n$-dimensional subspaces of $V$. It is known that $\mathrm{GO}(n,V)=\mathrm{GO}(n,V)_{+}\sqcup \mathrm{GO}(n,V)_{-}$, where $\mathrm{GO}(n,V)_{\pm}$ are the irreducible components of $\mathrm{GO}(n,V)$. We also consider the subset
$$
\begin{equation*}
\mathrm{GO}(n,V)^{0}= \{W\in \mathrm{GO}(n,V)\mid \operatorname{dim}(W\cap P)=1\}\subset \mathrm{GO}(n,V).
\end{equation*}
\notag
$$
We already know that the condition $\operatorname{dim}(W\cap P)=1$ is equivalent to saying that $W\cap P^{\perp}=\{0\}$. Therefore, the set $\mathrm{GO}(n,V)^{0}$ is open in $\mathrm{GO}(n,V)$. Correspondingly, the subsets $\mathrm{GO}(n,V)_{\pm}^{0}:=\mathrm{GO}(n,V)_{\pm}\cap \mathrm{GO}(n,V)^{0}$ are also open in $\mathrm{GO}(n,V)_{\pm}$. Corollary 2.2. The conclusion of Proposition 2.1 holds if $k$ is an infinite field. Proof. It suffices to show that the open subset $\mathrm{GO}(n,V)^{0}$ of $\mathrm{GO}(n,V)$ contains a $k$-rational point. From Proposition 2.1 we already know that $\mathrm{GO}(n,V)^{0}(\overline{k})\neq\varnothing$, where $\overline{k}$ is an algebraic closure of $k$. Hence at least one of the sets $\mathrm{GO}(n,V)^{0}_{+}(\overline{k})$ or $\mathrm{GO}(n,V)^{0}_{-}(\overline{k})$ is also non-empty. Since $Q$ is a hyperbolic form, both $\mathrm{GO}(n,V)_{\pm}$ are $k$-rational. Hence $\mathrm{GO}(n,V)^{0}_{+}(k)\neq\varnothing$ or $\mathrm{GO}(n,V)^{0}_{-}(k)\neq\varnothing$. This concludes the proof. Let us now proceed with the general case of the key geometric theorem. We first give some important technical facts. Lemma 2.3. A matrix $A\in \mathrm{Mat}_{n\times n}(R)$ is invertible if and only if so is the quotient matrix $\overline{A}\in \mathrm{Mat}_{n\times n}(k)$. Proof. One implication is trivial. Let $\overline{A}$ be invertible. Hence its determinant $\det(\overline{A})$ is non-zero in $k$. This means that $\det(A)$ does not belong to the maximal ideal of the ring $R$. Therefore, $\det(A)$ is invertible in $R$, and hence so is $A$. Lemma 2.4. Let $M$ be a free $R$-module of rank $n$ and let $\overline{u}_{1},\dots,\overline{u}_{r}$ be a set of linearly independent elements of the $k$-linear vector space $\overline{M}$. We fix some arbitrary liftings $u_{i}\in M$ for each $\overline{u}_{i}$. Then the submodule $U:=\langle u_{1},\dots,u_{r}\rangle_{R}$ of $M$ is a free direct summand of rank $r$ in $M$. Proof. We augment the family $\overline{u}_1,\dots,\overline{u}_r$ to a basis $\overline{u}_1,\dots,\overline{u}_r,\overline{v}_1,\dots,\overline{v}_{n-r}$ in $\overline{M}$. We next fix some liftings $v_{j}\in M$ for each $\overline{v}_{j}$. By Nakayama’s lemma, the elements $\{u_{i},v_{j}\}$ generate $M$ as an $R$-module. Let $A$ be the corresponding transformation matrix. Since $\{\overline{u}_{i},\overline{v}_{j}\}$ is a basis for $\overline{M}$, the quotient matrix $\overline{A}$ is invertible. Now by Lemma 2.3, the matrix $A$ is also invertible. Hence $\{u_{i},v_{j}\}$ is also a basis for $M$. Corollary 2.5. Any direct summand $U$ of a free module $R^{n}$ is also free. Moreover if $\{u_{1},\dots,u_{r}\}$ is a basis for $U$, then $\{\overline{u}_{1},\dots,\overline{u}_{r}\}$ is a basis for $\overline{U}$, and vice versa. Proof. We choose a projection $R^{n}\twoheadrightarrow U$. By definition, it is surjective, and its composition $U\hookrightarrow R^{n}\twoheadrightarrow U$ with the natural embedding $U\hookrightarrow R^{n}$ coincides with the identity automorphism $\mathrm{Id}_U$. Reducing modulo maximal ideal $\mathfrak{m}\subset R$, we choose a basis $\{\overline{u}_{1},\dots,\overline{u}_{r}\}$ in the corresponding subspace $\overline{U}$ of $k^{n}$, and fix some lifting $\{u_{1},\dots,u_{r}\}$ that lies in the image of $U\hookrightarrow R^{n}$. Denote by $U'$ the submodule $\langle u_{1},\dots,u_{r}\rangle_{R}$ in $R^{n}$ By Nakayama’s lemma, the morphism $U'\longrightarrow U$ induced by the projection $R^{n}\twoheadrightarrow U$ is surjective, which, in turn, means that $U$ and $U'$ actually coincide as $R$-submodules in $R^{n}$. Now by Lemma 2.4, $U'$ is a free module with basis $\{u_{1},\dots,u_{r}\}$.
Next, if $U$ is a direct summand of $R^{n}$ and $\{u_{1},\dots,u_{r}\}$ is its basis, then there exists a surjective $R$-linear morphism $R^{n}\twoheadrightarrow R^{r}$ that takes $u_{i}$ to the $i$th element of the standard basis of $R^{r}$. The associated quotient morphism $k^{n}\twoheadrightarrow k^{r}$ of $k$-linear vector spaces stays surjective, and hence $\{\overline{u}_{1},\dots,\overline{u}_{r}\}$ is obviously a basis for $\overline{U}$. Lemma 2.6. Let $M$ be a free $R$-module with fixed non-degenerate quadratic form $Q$. We denote by $O_{R}(M)$ the group of $R$-linear orthogonal automorphisms of $M$. Then the natural group homomorphism $O_{R}(M)\to O_{k}(\overline{M})$ is surjective. Proof. For our convenience, we will also denote by $Q$ the corresponding bilinear form. The characteristic of $k$ is not $2$, and hence, by Cartan–Dieudonne’s theorem, the group $O_{k}(\overline{M})$ is generated by reflections of the type where $\overline{u}$ goes over all elements of $\overline{M}$ such that $\overline{Q}(\overline{u},\overline{u})\neq0$. In particular, for any such $\overline{u}$ and any of its liftings $u\mapsto\overline{u}$ in $M$, the corresponding element $Q(u,u)$ does not belong to the maximal ideal of $R$, and, therefore, is invertible. Hence the reflection which is a lifting in $O_R(M)$ of the reflection $r_{\overline{u}}$, is well defined. This proves the lemma. Now suppose that the residue field $k$ is infinite. Let $M$ be a free $R$-module of rank $2n$ equipped with a hyperbolic form $Q$. Let $N\subset M$ be a free direct summand of $M$ such that $rk(N)=n+1$ and the restriction $Q|_{N}$ is non-degenerate. Then the corresponding $k$-linear vector spaces $\overline{M}$ and $\overline{N}$ satisfy the conditions of Corollary 2.2. Indeed, if $\{x_{1},\ldots,x_{n+1}\}$ is a basis for $N$, then $\{\overline{x}_{1},\dots,\overline{x}_{n+1}\}$ is a basis for $\overline{N}\subset\overline{M}$ by Corollary 2.5. The restriction of the form $Q$ to $N$ is non-degenerate, and hence the corresponding morphism $Q|_{N}^{\vee}:N\to N^{\vee}$ of $R$-modules is an isomorphism. Therefore, the restriction $\overline{Q}|_{\overline{N}}$ is non-degenerate according to Lemma 2.3. Finally, it is obvious that the form $\overline{Q}$ on $\overline{M}$ is also hyperbolic. Hence by Corollary 2.2, there exists a totally isotropic $n$-dimensional subspace $\overline{W}\subset\overline{M}$ such that $\operatorname{dim}(\overline{N}\cap\overline{W})=1$. The above $\overline{W}$ is used in the next result. Proposition 2.7. Let $M$ be a free $R$-module of rank $2n$ with fixed hyperbolic form $Q$. Suppose that $N\subset M$ is a free direct summand of $M$ such that $rk(N)=n+ 1$ and the restriction $Q|_{N}$ is non-degenerate. Then there exists an $R$-submodule $W$ in $M$ such that: (1) $W$ is a free direct summand of $M$ of rank $n$; (2) the quotient morphism $M\twoheadrightarrow\overline{M}$ maps $W$ surjectively onto $\overline{W}$; (3) $Q|_{W}\equiv0$. In addition, the submodule $W\cap N$ of $N$ is also a free direct summand of rank $1$. Proof. Let $\mathbb{H}_{R}$ be the hyperbolic plane with standard basis $\{e,f\}$. Since $Q$ is a hyperbolic form, we have $M=\mathbb{H}_{R}\oplus\dots\oplus\mathbb{H}_{R}$. Let $\{e_{1},f_{1},\dots,e_{n},f_{n}\}$ be the corresponding basis for $M$. Then the $R$-submodule $\langle e_{1},\dots,e_{n}\rangle_{R}\subset M$ is obviously a free totally isotropic summand of $M$ of rank $n$. One similarly verifies that the set $\{\overline{e}_{1},\overline{f}_{1},\dots,\overline{e}_{n},\overline{f}_{n}\}$ forms a basis for the $k$-linear vector space $\overline{M}=\mathbb{H}_{k}\oplus\dots\oplus\mathbb{H}_{k}$ and the subspace $\langle\overline{e}_{1},\dots,\overline{e}_{n}\rangle_{k}\subset \overline{M}$ is totally isotropic and has dimension $n$. We now choose some basis $\{\overline{w}_{1},\dots,\overline{w}_{n}\}$ for $\overline{W}$. By Witt’s theorem, for some $\overline{A}\in O_{k}(\overline{M})$ we have $\overline{A}(\overline{e_{i}})=\overline{w_{i}}$ for any $1\leqslant i\leqslant n$. By Lemma 2.6, $\overline{A}$ has some lifting $A\in O_{R}(M)$. We set $w_{i}:=Ae_{i}$ for $1\leqslant i\leqslant n$. Then the submodule $W:=\langle w_{1},\dots,w_{n}\rangle_{R}\subset M$ obviously satisfies the conditions (1)–(3) of Proposition 2.7.
Let us now prove the last assertion of the proposition. Let $\pi:N\rightarrow M/W$ be the composition of the natural embedding $N\hookrightarrow M$ and the quotient morphism $M\twoheadrightarrow M/N$. Reducing modulo maximal ideal $\mathfrak{m}\subset R$, we get the $k$-linear morphism $\overline{\pi}:\overline{N}\rightarrow\overline{M}/\overline{W}$. Its kernel coincides with the subspace $\overline{N}\cap\overline{W}\subset\overline{N}$. From the dimension theorem for vector spaces it is immediate that $\overline{\pi}$ is surjective. Now by Nakayama’s lemma, $\pi$ is surjective as well. Next, by Lemma 2.4, the $R$-module $M/W$ is free of rank $n$. Thus, $N\cap W$ is the kernel of the surjective $R$-linear morphism $\pi$ from the free $R$-module $N$ of rank $n+1$ to the free $R$-module $M/W$ of rank $n$. Therefore, $N\cap W$ is a direct summand of $N$. Now last assertion of Proposition 2.7 is secured by Corollary 2.5. Corollary 2.8. Under the conditions of Proposition 2.7, any generating element $w$ of the free $R$-module $W\cap N$ of rank $1$ is unimodular in $M$. Proof. By Proposition 2.7, $W\cap N$ is a direct summand of $M$. Therefore, by Corollary 2.5, the element $\overline{w}$ of the $k$-linear vector space $\overline{M}$ is non-zero. We fix some basis for $\overline{M}$ that contains $\overline{w}$, and then lift it to $M$ so that $w$ is the lift of $\overline{w}$. Now Corollary 2.8 is a direct consequence of Lemma 2.4. This completes the proof of the key geometric theorem. § 3. The main resultTheorem 3.1. Let $R$ be a regular local ring of a mixed characteristic $(0,p)$ where $p\ne 2 $ is a prime number. Let the quotient ring $R/pR$ be also regular. We fix some non-degenerate Pfister form $Q(T_{1},\dots,T_{2^m})$ over the ring $R$ together with some invertible element $c$ in $R$. Then the equation $Q(T_{1},\dots,T_{2^m})=c$ has a solution over $R$ if and only if it has a solution over the fraction field $K$ of $R$. The proof of Theorem 3.1 depends substantially on the result of Cesnavicius [1]. We will also need the following technical lemma, which was originally proved in [2]. We give its proof for completeness. Lemma 3.2. Let $R$ be a local ring with residue field $k$ of characteristic other than $2$, and let $\alpha_{1},\dots,\alpha_{n}$ be invertible elements of $R$ with $n>2$. Then the equation has a unimodular solution in $R^{n}$ if and only if the equation has a solution in $R^{n-1}$. Proof. The first implication is trivial: if $(v_{1},\dots,v_{n-1})$ is a solution of equation (3.2), then it is clear that $(v_{1},\dots,v_{n-1},1)$ is a unimodular solution of (3.1).
Now let $v:=(v_{1},\dots,v_{n})$ be some unimodular solution of (3.1). Note that if $v_{n}$ is invertible (that is, it does not belong to the maximal ideal of $R$), then
$$
\begin{equation*}
\biggl(\frac{v_1}{v_n},\dots,\frac{v_{n-1}}{v_n}\biggr)
\end{equation*}
\notag
$$
is obviously a solution of (3.2).
Suppose now that $v_{n}$ is not invertible. As earlier, let $Q$ be the diagonal quadratic form corresponding to (3.1); the associated bilinear form is also denoted by $Q$. Note that, for any $u:=(u_{1},\dots,u_{n})$ such that $Q(u,u)$ is invertible in $R$, an application of the reflection $r_{u}$ to $v$ defines another solution of (3.1). Next, if we assume that $u_{n}$ and $Q(u,v)$ are both invertible in $R$, then the $n$th coordinate of $r_{u}(v)$ is also invertible in $R$. Therefore,
$$
\begin{equation*}
\biggl(\frac{(r_u(v))_1}{(r_u(v))_n},\dots,\frac{(r_u(v))_{n-1}}{(r_u(v))_n}\biggr)
\end{equation*}
\notag
$$
is a solution to (3.2). Note that $u$ satisfies these assumptions if and only if the corresponding vector $\overline{u}:=(\overline{u}_{1},\dots,\overline{u}_{n})\in k^n$ satisfies
$$
\begin{equation*}
\overline{Q}(\overline{u},\overline{v})\neq0, \qquad \overline{Q}(\overline{u},\overline{u})\neq0, \qquad \overline{u}_n\neq0.
\end{equation*}
\notag
$$
Moreover, if there exists a vector $\overline{u}$ in $k^{n}$ satisfying these inequalities, then one can just lift it to $R^{n}$ to get a required $u$.
Since $v$ is unimodular, there exists at least one $i<n$ such that the corresponding coordinate $\overline{v}_{i}$ is non-zero. Let us fix such index $i$ together with some $j<n$, $j\ne i$. Assume that all the coordinates of $\overline{u}$ with indices other than $i$, $j$, and $n$ are zeros. Then the system of above inequalities takes the form (for simplicity we will omit the overline symbol). Let $u_i/u_n$ and $u_j/u_n$ be denoted, respectively, by $x$ and $y$. Then the above system simplifies to read Note that if $\alpha_{i}\neq-\alpha_{n}$, then $x=1$ and $y=0$ obviously satisfy this system. On the other hand, if $\alpha_{i}=-\alpha_{n}$, then it is enough to put $x=1$ and find some non-zero $y$ that satisfies $\alpha_{j}v_{j}y\neq-\alpha_{i}v_{i}$. One can always find such an $y$, since the residue field $k$ contains at least two non-zero elements. This proves the lemma.Now we are ready to prove Theorem 3.1. Proof of Theorem 3.1. We first assume that $k$ is infinite. The first implication is trivial. Assume that the equation $Q(T_1,\dots,T_{2^m})=c$ has a solution over $K$. Then the equation $Q(T_{1},\dots,T_{2^{m}})-cT^2_{2^{m}+1}=0$ also has a non-trivial solution $(a_{1},\dots,a_{2^{m}+1})$ over $K$. In view of Lemma 3.2, it suffices to find some unimodular solution of the equation $Q(T_{1},\dots,T_{2^{m}})-cT^{2}_{2^{m}+1}=0$ over $R$. Let $\widetilde{Q}(T_{1},\dots,T_{2^{m+1}})$ be the tensor product of $Q$ and the Pfister form $\langle1,-c\rangle$. It is clear that the vector $(a_{1},\dots,a_{2^{m}+1},0,\dots,0)$ satisfies the equation $\widetilde{Q}(T_{1},\dots,T_{2^{m+1}})=0$ over $K$. Since $\widetilde{Q}$ is a Pfister form this means that it is hyperbolic over $K$. According to Cesnavicius (see Corollary 9.6 in [1]), the form $\widetilde{Q}$ is hyperbolic also over $R$. We set $M=R^{2^{m+1}}$, and take a free $R$-submodule $N\subset M$ generated by the first $2^{m}+1$ basis elements of $M$. Then the pair $(M,N)$ clearly satisfies the conditions of Proposition 2.7. In particular, the unimodular isotropic element $w$ from Corollary 2.8 is a unimodular solution for the equation $Q(T_{1},\dots,T_{2^{m}})-cT^{2}_{2^{m}+1}=0$ over $R$.
Let us now assume that the field $k$ is finite. The first implication is also trivial. Let $q$ be an odd prime number, $q\ne p$. Then we can construct a tower of finite etale extensions
$$
\begin{equation}
R\subset R_1\subset R_{2}\subset\dots\subset R_i\subset\cdots,
\end{equation}
\tag{*}
$$
where all $R_{i}$ are regular local rings of mixed characteristic $(0,p)$, and, for each $i$, the extension $R_{i}\subset R_{i+1}$ is finite etale of degree $q$. This can easily be done by induction by taking some separable polynomial $g_{i}(t)$ of degree $ q$ with coefficients in the corresponding residue field $k_{i}:=R_{i}/\mathfrak{m}_{i}R_{i}$, and putting $R_{i+1}:=R_{i}[t]/\widetilde{g}(t)$ for some arbitrary lifting $\widetilde{g}(t)$ of $g(t)$ to $R_{i}[t]$. Let $R_{\infty}$ and $K_{\infty}$ be, respectively, the corresponding direct limit of this tower of extensions and its quotient field. The equation $Q=c$ has a solution over $K_{\infty}$. Therefore, it has a solution over $R_{\infty}$ according to the first part of the proof, since the residue field of $R_{\infty}$ is obviously infinite. By the construction, there exists some $i\gg 0$ such that the equation $Q=c$ has a solution in $R_{i}$. By Scully’s variant of Artin–Springer’s theorem (see Theorem 4.1 in [9]), the equation $Q=c$ also has a solution over $R$. This concludes the proof of Theorem 3.1. Corollary 3.3. The set $S_{Q}$ of all the invertible elements $c\in R^{*}$ representable by the Pfister form from Theorem 3.1 forms a multiplicative subgroup in $R^{*}$. Proof. Suppose that the elements $c_{1,2}\in R^{*}$ are represented by $Q$, that is, the equations $Q=c_{1}$ and $Q=c_{2}$ have some solutions over $R$. Since Pfister forms are multiplicative over fields, the equation $Q=c_{1}c_{2}$ has some solution over $K$. According to Theorem 3.1, it also has a solution over $R$. Therefore, $S_{Q}$ is closed under multiplication. If $v$ is a solution for the equation $Q=c$, then $Q(c^{-1}v)=c^{-2}Q(v)=c^{-1}$. Therefore, $S_{Q}$ is closed under taking an inverse. Finally, since $Q$ is a Pfister form, the equation $Q=1$ always has a solution. This proves the corollary.
§ 4. AppendixIn this section, we will give another proof of the Proposition 2.1. As above, we suppose that $k$ is an algebraically closed field of characteristic different than $2$. Let $V$ be a vector space over $k$ of dimension $m$. For each $j\leqslant m$, we let $X^{V}_{j}$ denote the projective variety of complete flags of length $j$ in $V$. In particular, for $j=1$, this variety coincides with the projective space $\mathbb{P}(V)$, and therefore, has the dimension $m-1$. For every $j\leqslant m$, the natural surjective morphism
$$
\begin{equation*}
\pi_j=X^V_j\twoheadrightarrow X^V_{j-1}, \qquad \{L_j\supset\dots\supset L_1\} \to \{L_{j-1}\supset\dots\supset L_1\},
\end{equation*}
\notag
$$
is well defined. Note that its fibre over an arbitrary point
$$
\begin{equation*}
\widetilde{L}:=\{L_{j-1}\supset\dots\supset L_1\}\in X^V_{j-1}
\end{equation*}
\notag
$$
is naturally isomorphic to the projectivization $\mathbb{P}(V/L_{j-1})$ of the quotient vector space $V/L_{j-1}$ of dimension $m-j+1$. Hence the dimension of $\pi^{-1}(\widetilde{L})$ is $m- j$. Using the theorem of dimension of fibres and inducting on $j$, we get the equality
$$
\begin{equation}
\operatorname{dim}(X^V_j)=(m-1)+\dots+(m-j)=mj-\frac{j(j+1)}{2}.
\end{equation}
\tag{4.1}
$$
Now suppose that $m\geqslant2$. We fix some non-degenerate quadratic form $Q$ on $V$. A flag will be said to be totally isotropic if each of its subspaces is a totally isotropic subspace over $Q$ (note that the dimension of such a subspace cannot exceed $[m/2]$). We let $\mathring{X}^{V}_{j}$ denote the projective variety of complete totally isotropic flags of length $j$. There is an obvious closed embedding $\mathring{X}^{V}_{j}\hookrightarrow X^{V}_{j}$, which, in the case $j=1$, identifies $\mathring{X}^{V}_{1}$ with the hypersurface in $X^{V}_{1}\cong\mathbb{P}(V)$ defined by the equation $Q(v)=0$. Consequently $\mathring{X}^{V}_{1}$ is of dimension $m-2$. As in case of usual flags, for every $j\leqslant[m/2]$, the morphism
$$
\begin{equation*}
\tau_j=\mathring{X}^V_j\twoheadrightarrow\mathring{X}^V_{j-1}, \qquad \{L_j\supset\dots\supset L_1\}\to\{L_{j-1}\supset\dots\supset L_1\},
\end{equation*}
\notag
$$
is well defined, and this morphism is also surjective by Witt’s theorem. Its fibre over an arbitrary point $\widetilde{L}:=\{L_{j-1}\supset\dots\supset L_{1}\}$ is isomorphic to the projective variety of all totally isotropic $j$-dimensional subspaces of $V$ that contain $L_{j-1}$. By definition, each such a subspace lies in $L_{j-1}^{\perp}$. Therefore, one can consider $\tau_{j}^{-1}(L)$ as the hypersurface in $\mathbb{P}(L_{j-1}^{\perp}/L_{j-1})$ defined by the equation $\overline{Q}(v)=0$. In particular, the dimension of $\tau_{j}^{-1}(L)$ is $m-2(j-1)-2=m-2j$. Using the fibre dimension theorem and inducting on $j$, we get the equality
$$
\begin{equation}
\operatorname{dim}(\mathring{X}^V_j)=(m-2)+\dots+(m-2j)=mj-j(j+1).
\end{equation}
\tag{4.2}
$$
We next denote by $\mathring{Y}^{V}_{j}$ the projective variety of all totally isotropic $j$-dimensional subspaces in $V$. For every $j\leqslant[m/2]$, the surjective morphism
$$
\begin{equation*}
\gamma_j=\mathring{X}^V_j\twoheadrightarrow\mathring{Y}^V_j, \qquad \{L_j\supset\dots\supset L_1\}\to L_j,
\end{equation*}
\notag
$$
is well defined. The fibre of this morphism over a totally isotropic subspace $L$ in $V$ is naturally isomorphic to $X^{L}_{j-1}$. Hence by (4.1),
$$
\begin{equation*}
\operatorname{dim}(X^{L}_{j-1})=j(j-1)-\frac{j(j-1)}{2}=\frac{j(j-1)}{2}.
\end{equation*}
\notag
$$
Next, applying the theorem of dimension of fibres we get the equality
$$
\begin{equation}
\operatorname{dim}(\mathring{Y}^V_j)=mj-j(j+1)-\frac{j(j-1)}{2}.
\end{equation}
\tag{4.3}
$$
Now consider an arbitrary tuple $(V,Q,P)$ satisfying the hypotheses of Proposition 2.1. We have $m=2n$, and now an appeal to (4.3) shows that the dimension of the projective variety $\mathring{Y}^V_n$ of all $n$-dimensional totally isotropic $W\subset V$ is
$$
\begin{equation}
\begin{aligned} \, \operatorname{dim}(\mathring{Y}^V_n) &=2n^2-n(n+1)- \frac{n(n-1)}{2} \nonumber \\ &=n^2-n-\frac{n^2-n}{2} =\frac{n^2-n}{2}. \end{aligned}
\end{equation}
\tag{4.4}
$$
For each $l\leqslant n$, let $Z_l$ denote the quasi-projective subvariety in $\mathring{Y}^{V}_{n}$ of all totally isotropic $n$-dimensional subspaces $W\subset V$ such that $\operatorname{dim}(W\cap P)=l$. The following disjoint decomposition holds: Let us show that the variety $Z_{1}$ is non-empty. Note that $Z_1\ne\varnothing$ is open in $\mathring{Y}^V_n$, since the set $\{W\in\mathring{Y}^V_n\mid \operatorname{dim}(W\cap P)\geqslant2\}$ is closed.For each $l\leqslant[(n+1)/2]$, we have the natural morphism
$$
\begin{equation*}
\lambda_l\colon Z_l\to\mathring{Y}^{P}_l,\qquad W\to P\cap W.
\end{equation*}
\notag
$$
If $U\subset P$ is an arbitrary point in $\lambda_{l}(Z_{l})\subset\mathring{Y}^{P}_{l}$, then its fibre $\lambda_{l}^{-1}(U)$ coincides with the set
$$
\begin{equation*}
\{W\subset V\mid \operatorname{dim}(W)=n,\, Q|_{W}\equiv0,\, W\cap P=U\}.
\end{equation*}
\notag
$$
Note that the restriction of the form $\overline{Q}$ is still non-degenerate on $U^{\perp}/U$. Indeed, if a vector $\overline{v}\in U^{\perp}/U$ belongs to $\ker(\overline{Q}|_{U^{\perp}/U})$, them $\overline{Q}(\overline{u},\overline{v})=Q(u,v)=0$ for each $u\in U^{\perp}$. Hence $v\in (U^{\perp})^{\perp}=U$, and, therefore, $\overline{v}=0$. Hence $\lambda_l^{-1}(U)$ can be looked upon as an open subset of the projective variety
$$
\begin{equation*}
\mathring{Y}^{U^{\perp}/U}_{n-l}=\{W\subset V\mid \operatorname{dim}(W)=n,\, Q|_{W}\equiv0,\, U\subset W\},
\end{equation*}
\notag
$$
which consists of all $\overline{W}\subset U^{\perp}/U$ such that $\overline{W}\cap\overline{P\cap U^{\perp}}=\{0\}$. We have $\operatorname{dim}(U^{\perp}/U)=2(n-l)$, and hence, applying formula (4.3) to the case $({U^{\perp}/U},\overline{Q}|_{U^{\perp}/U},\,j=n-l)$, we get the equality
$$
\begin{equation*}
\operatorname{dim}\big(\mathring{Y}^{U^{\perp}/U}_{n-l}\big)= \frac{(n-l)^2-(n-l)}{2}.
\end{equation*}
\notag
$$
In particular, all fibres of $\lambda_{l}$ have the same dimension. Applying the theorem of dimension of fibres, we get the estimate
$$
\begin{equation*}
\operatorname{dim}(Z_l)\leqslant \operatorname{dim}(\mathring{Y}^{P}_l)+\frac{(n-l)^2-(n-l)}{2}.
\end{equation*}
\notag
$$
Next, an application of (4.3) to the case $(P,Q|_P,\, j=l)$ transforms the above inequality to
$$
\begin{equation*}
\operatorname{dim}(Z_l)\leqslant(n+1)l-l(l+1)-\frac{l(l-1)}{2}+\frac{(n-l)^2-(n-l)}{2}.
\end{equation*}
\notag
$$
The right-hand side of this inequality is equal to
$$
\begin{equation*}
\frac{n^2-n}{2}-(l^2-l)=\operatorname{dim}(\mathring{Y}^V_n)-(l^2-l).
\end{equation*}
\notag
$$
So, if $l\geqslant2$, then we have the strict inequality $\operatorname{dim}(Z_l)<\operatorname{dim}(\mathring{Y}^V_n)$. In particular, the subset
$$
\begin{equation*}
Z_1=\mathring{Y}^V_n\setminus(Z_{2}\sqcup\dots\sqcup Z_{[(n+1)/2]})
\end{equation*}
\notag
$$
is open and non-empty in $\mathring{Y}^V_n$. Since $k$ is algebraically closed this means that $Z_{1}$ contains some point. This proves Proposition 2.1.
Citation in format AMSBIB
\Bibitem{PanTyu24} |
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