Symmetric square-central elements in products of orthogonal involutions in characteristic two

TL;DR

This paper provides criteria to determine when symmetric square-central elements in certain algebraic structures with involution are contained within invariant quaternion subalgebras, specifically in characteristic two.

Contribution

It introduces new criteria for symmetric square-central elements in totally decomposable algebras with orthogonal involution in characteristic two.

Findings

Criteria for symmetric square-central elements containment in quaternion subalgebras

Conditions specific to characteristic two

Advances understanding of algebraic involutions

Abstract

We obtain some criteria for a symmetric square-central element of a totally decomposable algebra with orthogonal involution in characteristic two, to be contained in an invariant quaternion subalgebra.

Equations

$$\mathrm{Sym}(A,\sigma )=\{x\in A\mid \sigma (x)=x\}\mathrm{and}\mathrm{Alt}(A,\sigma )=\{\sigma (x)-x\mid x\in A\}.$$

$$\mathrm{Alt}(A,\sigma {)}^{+}$$

$$\mathrm{Sym}(A,\sigma {)}^{+}$$

$$\sigma (w)w$$

$$=xyxy+yxyx+\alpha \beta +\alpha \beta =\sigma (yxyx)-yxyx\in \mathrm{Alt}(A,\sigma ).$$

$$v_1^{\prime }$$

$$v_2^{\prime }$$

$$(Q_{n-2},{\sigma }_{n-2})\otimes (Q_{n-1},{\sigma }_{n-1})\otimes (M_2(F),t),$$

$$(xy{)}^{-r}=(y^{-1}{x}^{-1}{)}^r=({\beta }^{-1}y{\alpha }^{-1}x{)}^r={\alpha }^{-r}{\beta }^{-r}(yx{)}^r.$$

$$(xy{)}^{r-s}y(xy{)}^{s-r}y=y(xy{)}^{r-s}y(xy{)}^{s-r}.$$

$$\lambda (xy{)}^{r-s}y(yx{)}^{r-s}y=\lambda y(xy{)}^{r-s}y(yx{)}^{r-s}.$$

$$(xy{)}^s={\alpha }^s(xy{)}^{-s}={\alpha }^s{\alpha }^{-s}(yx{)}^s=(yx{)}^s.$$

$${\sigma }_K(y)y$$

$$={\sigma }_K(x\otimes 1)\cdot (x\otimes 1)+\alpha \otimes 1+(1\otimes \sqrt{\alpha })(x\otimes 1+{\sigma }_K(x\otimes 1)).$$

$$\alpha +\sigma (y)y$$

$$=(\alpha +\sum _{i=1}^m\sigma (x_i)x_i)+\sum _{i<j}(\sigma (x_j)x_i+\sigma (x_i)x_j)$$

$$=(\alpha +\sum _{i=1}^m\sigma (x_i)x_i)+\sum _{i<j}(\sigma (\sigma (x_i)x_j)-\sigma (x_i)x_j).$$

$$\mathrm{Alt}(A,\sigma )$$

$$\mathrm{Sym}(A,\sigma )$$

$$A_i^t{A}_i=\left(\begin{array}{cc}{a}_i^2+c_i^2& a_i{b}_i+c_i{d}_i\\ a_i{b}_i+c_i{d}_i& b_i^2+d_i^2\end{array}\right).$$

$$\sigma (x_i)x_i=\left(\begin{array}{cc}\rho (a_i)& \rho (c_i)\\ \rho (b_i)& \rho (d_i)\end{array}\right)\cdot \left(\begin{array}{cc}{a}_i& b_i\\ c_i& d_i\end{array}\right).$$

$$\alpha +\sum _{i=1}^m(\rho (a_i)a_i+\rho (c_i)c_i)\in \mathrm{Alt}(B,\rho ).$$

$$f:(B,\rho {)}_K\simeq (C,\sigma {|}_C).$$

$$(Q_3,{\sigma }_3)\otimes (B,\rho )\simeq (M_2(F),t)\otimes (Q_1,{\sigma }_1)\otimes (Q_2,{\sigma }_2).$$

$$(M_2(L),{\tau }_3)\otimes (B,\rho {)}_L\simeq (M_2(L),t)\otimes (M_2(L),{\tau }_1)\otimes (M_2(L),{\tau }_2).$$

$$(M_2(L),{\tau }_3)\simeq (Q_3,{\sigma }_3{)}_L\simeq \mathrm{Ad}(⟨⟨\alpha +{\lambda }^2⟩{⟩}_L),$$

$$x=\left(\begin{array}{cc}w& w+u\\ w+u& w\end{array}\right)\in A.$$

$$(C,\sigma {|}_C){\simeq }_L(B,\rho {)}_L,$$

Full text

Mathematics Subject Classification: 16W10, 16K20, 11E39.

1 Introduction

A classical question concerning central simple algebras is to identify under which conditions a square-central element lies in a quaternion subalgebra. This question is only solved for certain special cases in the literature. Let $A$ be a central simple algebra of exponent two over a field $F$. In [2, (3.2)] it was shown that if $\mathop{\mathrm{char}}F\neq 2$ and $F$ is of cohomological dimension less than or equal $2$, then every square-central element of $A$ is contained in a quaternion subalgebra (see also [4, (4.2)]). In [3, (4.1)], it was shown that if $\mathop{\mathrm{char}}F\neq 2$, then an element $x\in A$ with $x^{2}=\lambda^{2}\in F^{\times 2}$ lies in a (split) quaternion subalgebra if and only if $\dim_{F}(x-\lambda)A=\frac{1}{2}\dim_{F}A$. This result was generalized in [16, (3.2)] to arbitrary characteristics, including $\lambda=0$. On the other hand, in [19] it was shown that there is an indecomposable algebra of degree $8$ and exponent $2$, containing a square-central element (see [9, (5.6.10)]). Using similar methods, it was shown in [3] that for $n\geqslant 3$ there exists a tensor product of $n$ quaternion algebras containing a square-central element which does not lie in any quaternion subalgebra.

A similar question for a central simple algebra with involution $(A,\sigma)$ over $F$ is whether a symmetric or skew-symmetric square-central element of $A$ lies in a $\sigma$-invariant quaternion subalgebra. In the case where $\mathop{\mathrm{char}}F\neq 2$, $\deg_{F}A=8$, $\sigma$ has trivial discriminant, the index of $A$ is not $2$ and one of the components of the Clifford algebra $C(A,\sigma)$ splits, it was shown in [18, (3.14)] that every skew-symmetric square-central element of $A$ lies in a $\sigma$-invariant quaternion subalgebra. In [15], some criteria were obtained for symmetric and skew-symmetric elements whose squares lie in $F^{2}$, to be contained in a $\sigma$-invariant quaternion subalgebra. Also, a sufficient condition was obtained in [14, (6.3)] for symmetric square-central elements in a totally decomposable algebra with orthogonal involution in characteristic two, to be contained in a stable quaternion subalgebra.

In this work we study the aforementioned question for totally decomposable algebras with orthogonal involution in characteristic two. Let $(A,\sigma)$ be a totally decomposable algebra with orthogonal involution over a field $F$ of characteristic two and let $x\in A\setminus F$ be a symmetric element with $\alpha:=x^{2}\in F$. Since the case where $\alpha\in F^{2}$ was settled in [15], we assume that $\alpha\in F^{\times}\setminus F^{\times 2}$. We first study in §3 some properties of inseparable subalgebras, introduced in [14]. It is shown in Theorem 3.10 that $(A,\sigma)$ has a unique inseparable subalgebra if and only if either $\deg_{F}A\leqslant 4$ or $\sigma$ is anisotropic. In §4, the notion of the representation of an element of $F$ by $\sigma$ is defined. Theorem 4.10 shows that the set of elements in $F$ represented by $\sigma$ coincides with the set of elements in $F$ represented by the Pfister invariant of $(A,\sigma)$ (a bilinear Pfister form associated to $(A,\sigma)$ and defined in [6]). In particular, if $0\neq x\in A$ is symmetric and square-central, then the element $x^{2}\in F$ is represented by the Pfister invariant (see Corollary 4.9).

We then study our main problem in §5 and §6. For the case where $\sigma$ is anisotropic or $A$ has degree $4$, it is shown that every symmetric square-central element of $A$ lies in a $\sigma$-invariant quaternion subalgebra (see Theorem 5.1 and Proposition 5.2). However, we will see in Proposition 6.2 that if $\sigma$ is isotropic, $\deg_{F}A\geqslant 8$ and $(A,\sigma)\not\simeq(M_{2^{n}}(F),t)$, there always exists a symmetric square-central of $(A,\sigma)$ which is not contained in any $\sigma$-invariant quaternion subalgebra of $A$. If $A$ has degree $8$ or $\sigma$ satisfies certain isotropy condition, it is shown in Proposition 5.7 and Theorem 5.10 that a symmetric square-central element of $A$ lies in a $\sigma$-invariant quaternion subalgebra if and only if it is contained in an inseparable subalgebra of $(A,\sigma)$. Finally, in Example 6.3 we shall see that this criterion cannot be applied to arbitrary involutions.

2 Preliminaries

Throughout this paper, $F$ denote a field of characteristic two.

Let $A$ be a central simple algebra over $F$. An involution on $A$ is an antiautomorphism $\sigma:A\rightarrow A$ of order two. If $\sigma|_{F}=\mathop{\mathrm{id}}$, we say that $\sigma$ is of the first kind. The sets of alternating and symmetric elements of $(A,\sigma)$ are defined as

[TABLE]

For a field extension $K/F$ we use the notation $A_{K}=A\otimes K$, $\sigma_{K}=\sigma\otimes\mathop{\mathrm{id}}$ and $(A,\sigma)_{K}=(A_{K},\sigma_{K})$. An extension $K/F$ is called a splitting field of $A$ if $A_{K}$ splits, i.e., $A_{K}$ is isomorphic to the matrix algebra $M_{n}(K)$, where $n=\deg_{F}A$ is the degree of $A$ over $F$. If $(V,\mathfrak{b})$ is a symmetric bilinear space over $F$, the pair $(\mathop{\mathrm{End}}_{F}(V),\sigma_{\mathfrak{b}})$ is denoted by $\mathop{\mathrm{Ad}}(\mathfrak{b})$, where $\sigma_{\mathfrak{b}}$ is the adjoint involution of $\mathop{\mathrm{End}}_{F}(V)$ with respect to $\mathfrak{b}$ (see [10, p. 2]). According to [10, (2.1)], if $K$ is a splitting field of $A$, then $(A,\sigma)_{K}$ is adjoint to a symmetric bilinear form. We say that $\sigma$ is symplectic, if this form is alternating. Otherwise, $\sigma$ is called orthogonal. By [10, (2.6)], $\sigma$ is symplectic if and only $1\in\mathop{\mathrm{Alt}}(A,\sigma)$. If $\sigma$ is an orthogonal involution, the discriminant of $\sigma$ is denoted by $\mathop{\mathrm{disc}}\sigma$ (see [10, (7.2)]).

Let $(V,\mathfrak{b})$ be a bilinear space over $F$ and let $\alpha\in F$. We say that $\mathfrak{b}$ represents $\alpha$ if $\mathfrak{b}(v,v)=\alpha$ for some nonzero vector $v\in V$. The set of elements in $F$ represented by $\mathfrak{b}$ is denoted by $D(\mathfrak{b})$. We also set $Q(\mathfrak{b})=D(\mathfrak{b})\cup\{0\}$. Observe that $Q(\mathfrak{b})$ is an $F^{2}$-subspace of $F$. If $K/F$ is a field extension, the scalar extension of $\mathfrak{b}$ to $K$ is denoted by $\mathfrak{b}_{K}$. For $\alpha_{1},\cdots,\alpha_{n}\in F^{\times}$, the diagonal bilinear form $\sum_{i=1}^{n}\alpha_{i}x_{i}y_{i}$ is denoted by $\langle\alpha_{1},\cdots,\alpha_{n}\rangle$. The form $\langle\!\langle\alpha_{1},\cdots,\alpha_{n}\rangle\!\rangle:=\langle 1,\alpha_{1}\rangle\otimes\cdots\otimes\langle 1,\alpha_{n}\rangle$ is called a bilinear ($n$-fold) Pfister form. If $\mathfrak{b}$ is a bilinear Pfister form over $F$, then there exists a bilinear form $\mathfrak{b}^{\prime}$, called the pure subform of $\mathfrak{b}$, such that $\mathfrak{b}\simeq\langle 1\rangle\perp\mathfrak{b}^{\prime}$. The form $\mathfrak{b}^{\prime}$ is uniquely determined, up to isometry (see [1, p. 906]).

3 The inseparable subalgebra

Definition 3.1**.**

An involution $\sigma$ on a central simple algebra $A$ is called isotropic if $\sigma(x)x=0$ for some nonzero element $x\in A$. Otherwise, $\sigma$ is called anisotropic.

The following definition was given in [5, p. 1629].

Definition 3.2**.**

A central simple algebra with involution $(A,\sigma)$ is called direct if $\sigma(x)x\in\mathop{\mathrm{Alt}}(A,\sigma)$ for $x\in A$ implies that $x=0$.

It is clear that every direct algebra with involution is anisotropic.

Notation 3.3**.**

For an algebra with involution $(A,\sigma)$ over $F$ we use the following notation:

[TABLE]

Note that we have $\mathop{\mathrm{Alt}}(A,\sigma)^{+}\oplus F\subseteq\mathop{\mathrm{Sym}}(A,\sigma)^{+}$.

Proposition 3.4**.**

If $(A,\sigma)$ is a direct algebra with orthogonal involution over $F$, then $\mathop{\mathrm{Sym}}(A,\sigma)^{+}$ is a subfield of $A$.

Proof.

Let $x,y\in\mathop{\mathrm{Sym}}(A,\sigma)^{+}$. Set $\alpha=x^{2}\in F$ and $\beta=y^{2}\in F$. Consider the element $w:=xy+yx\in\mathop{\mathrm{Sym}}(A,\sigma)$. We have

[TABLE]

Hence $w=0$, i.e., $xy=yx$. It follows that $\sigma(xy)=xy$ and $(xy)^{2}\in F$, i.e., $xy\in\mathop{\mathrm{Sym}}(A,\sigma)^{+}$. We also have $(x+y)^{2}=\alpha+\beta+xy+yx=\alpha+\beta\in F$, hence $x+y\in\mathop{\mathrm{Sym}}(A,\sigma)^{+}$. The set $\mathop{\mathrm{Sym}}(A,\sigma)^{+}$ is therefore a commutative subalgebra of $A$ with $u^{2}\in F$ for every $u\in\mathop{\mathrm{Sym}}(A,\sigma)^{+}$. Since $\sigma$ is direct, if $0\neq u\in\mathop{\mathrm{Sym}}(A,\sigma)^{+}$, then $u^{2}\neq 0$, i.e., $u$ is a unit. Hence, $\mathop{\mathrm{Sym}}(A,\sigma)^{+}$ is a field. ∎

A central simple algebra with involution is called totally decomposable if it decomposes into tensor products of quaternion algebras with involution. Let $(A,\sigma)\simeq\bigotimes_{i=1}^{n}(Q_{i},\sigma_{i})$ be a totally decomposable algebra of degree $2^{n}$ with orthogonal involution over $F$. By [10, (2.23)] the involution $\sigma_{i}$ is necessarily orthogonal for $i=1,\cdots,n$. Also, according to [14, (4.6)] there exists a $2^{n}$-dimensional subalgebra $S\subseteq\mathop{\mathrm{Sym}}(A,\sigma)^{+}$, called an inseparable subalgebra of $(A,\sigma)$, satisfying (i) $C_{A}(S)=S$, where $C_{A}(S)$ is the centralizer of $S$ in $A$; and (ii) $S$ is generated, as an $F$-algebra, by $n$ elements. Furthermore, for every inseparable subalgebra $S$ of $(A,\sigma)$ we have necessarily $S\subseteq\mathop{\mathrm{Alt}}(A,\sigma)^{+}\oplus F$. According to [14, (5.10)], if $S_{1}$ and $S_{2}$ are two inseparable subalgebras of $(A,\sigma)$, then $S_{1}\simeq S_{2}$ as $F$-algebras. Note that if $v_{i}\in\mathop{\mathrm{Sym}}(Q_{i},\sigma_{i})^{+}\setminus F$ is a unit for $i=1,\cdots,n$, then $F[v_{1},\cdots,v_{n}]$ is an inseparable subalgebra of $(A,\sigma)$.

Theorem 3.5**.**

Let $(A,\sigma)$ be a totally decomposable algebra with anisotropic orthogonal involution over $F$ and let $S$ be an inseparable subalgebra of $(A,\sigma)$. Then $S=\mathop{\mathrm{Alt}}(A,\sigma)^{+}\oplus F=\mathop{\mathrm{Sym}}(A,\sigma)^{+}$ is a maximal subfield of $A$. In particular, the inseparable subalgebra $S$ is uniquely determined.

Proof.

We already know that $S\subseteq\mathop{\mathrm{Alt}}(A,\sigma)^{+}\oplus F\subseteq\mathop{\mathrm{Sym}}(A,\sigma)^{+}$. By [6, (6.1)], $(A,\sigma)$ is direct. Hence, $\mathop{\mathrm{Sym}}(A,\sigma)^{+}$ is a field by Proposition 3.4. Since $S$ is maximal commutative, we obtain $\mathop{\mathrm{Sym}}(A,\sigma)^{+}=S$. Finally, as $\dim_{F}S=\deg_{F}A$, $S$ is a maximal subfield of $A$. ∎

The following definition was given in [6].

Definition 3.6**.**

Let $(A,\sigma)\simeq\bigotimes_{i=1}^{n}(Q_{i},\sigma_{i})$ be a totally decomposable algebra with orthogonal involution over $F$. The Pfister invariant of $(A,\sigma)$ is defined as $\mathfrak{Pf}(A,\sigma):=\langle\!\langle\alpha_{1},\cdots,\alpha_{n}\rangle\!\rangle$, where $\alpha_{i}\in F^{\times}$ is a representative of the class $\mathop{\mathrm{disc}}\sigma_{i}\in F^{\times}/F^{\times 2}$, $i=1,\cdots,n$.

According to [6, (7.2)], the isometry class of the Pfister invariant is independent of the decomposition of $(A,\sigma)$. Moreover, if $S$ is an inseparable subalgebra of $(A,\sigma)$, then $S$ may be considered as an underlying vector space of $\mathfrak{Pf}(A,\sigma)$ such that $\mathfrak{Pf}(A,\sigma)(x,x)=x^{2}$ for $x\in S$ (see [14, (5.5)]).

Notation 3.7**.**

For a positive integer $n$, we denote the bilinear $n$-fold Pfister form $\langle\!\langle 1,\cdots,1\rangle\!\rangle$ by $\langle\!\langle 1\rangle\!\rangle^{n}$. We also set $\langle\!\langle 1\rangle\!\rangle^{0}=\langle 1\rangle$.

Let $\mathfrak{b}$ be a bilinear Pfister form over $F$. In view of [1, A.5], one can find a non-negative integer $r$ and an anisotropic bilinear Pfister form $\mathfrak{b}^{\prime}$ such that $\mathfrak{b}\simeq\langle\!\langle 1\rangle\!\rangle^{r}\otimes\mathfrak{b}^{\prime}$. As in [15], we denote the integer $r$ by $\mathfrak{i}(\mathfrak{b})$. If $(A,\sigma)$ is a totally decomposable $F$-algebra with orthogonal involution, we simply denote $\mathfrak{i}(\mathfrak{Pf}(A,\sigma))$ by $\mathfrak{i}(A,\sigma)$. By [6, (5.7)], $(A,\sigma)$ is anisotropic if and only if $\mathfrak{i}(A,\sigma)=0$. Hence, if $\sigma$ is isotropic, then the pure subform of $\mathfrak{Pf}(A,\sigma)$ represents $1$. In other words, if $S$ is an inseparable subalgebra of $(A,\sigma)$, then there exists $x\in S\setminus F$ with $x^{2}=1$. Also, if $r=\mathfrak{i}(A,\sigma)>0$, there exists a totally decomposable algebra with anisotropic orthogonal involution $(B,\rho)$ over $F$ such that $(A,\sigma)\simeq(M_{2^{r}}(F),t)\otimes(B,\rho)$, where $t$ is the transpose involution (see [15, p. 7]). In particular, if $A$ is of degree $2^{n}$ then $\mathfrak{i}(A,\sigma)=n$ if and only if $(A,\sigma)\simeq(M_{2^{n}}(F),t)$.

We recall that every quaternion algebra $Q$ over $F$ has a quaternion basis, i.e., a basis $(1,u,v,w)$ satisfying $u^{2}+u\in F$, $v^{2}\in F^{\times}$ and $w=uv=vu+v$ (see [10, p. 25]). In this case, $Q$ is denoted by $[\alpha,\beta)_{F}$, where $\alpha=u^{2}+u\in F$ and $\beta=v^{2}\in F^{\times}$.

Lemma 3.8**.**

If $(Q,\sigma)$ is a quaternion algebra with orthogonal involution over $F$, then there is a quaternion basis $(1,u,v,w)$ of $Q$ such that $u,v\in\mathop{\mathrm{Sym}}(Q,\sigma)$.

Proof.

Let $v\in\mathop{\mathrm{Alt}}(Q,\sigma)$ be a unit. Since $v\notin F$ and $v^{2}\in F^{\times}$, it is easily seen that $v$ extends to a quaternion basis $(1,u,v,w)$ of $Q$. By [15, (4.5)], we have $\sigma(u)=u$. ∎

Lemma 3.9**.**

Let $(A,\sigma)$ be a totally decomposable algebra of degree $8$ with orthogonal involution over $F$. If $\sigma$ is isotropic, then there are two inseparable subalgebras $S_{1}$ and $S_{2}$ of $(A,\sigma)$ with $S_{1}\neq S_{2}$.

Proof.

Since $\mathfrak{i}(A,\sigma)>0$, we may identify $(A,\sigma)=(B,\rho)\otimes(M_{2}(F),t)$, where $(B,\rho)\simeq(Q_{1},\sigma_{1})\otimes(Q_{2},\sigma_{2})$ is a tensor product of two quaternion algebras with orthogonal involution. By Lemma 3.8 there exists a quaternion basis $(1,u_{i},v_{i},w_{i})$ of $Q_{i}$ over $F$ such that $u_{i},v_{i}\in\mathop{\mathrm{Sym}}(Q_{i},\sigma_{i})$, $i=1,2$. Let $v_{3}\in\mathop{\mathrm{Alt}}(M_{2}(F),t)$ be a unit. By scaling we may assume that $v_{3}^{2}=1$, because $\mathop{\mathrm{disc}}t$ is trivial (see [10, p. 82]). Then $S_{1}=F[v_{1}\otimes 1,v_{2}\otimes 1,1\otimes v_{3}]$ is an inseparable subalgebra of $(A,\sigma)$ (we have identified $Q_{1}$ and $Q_{2}$ with subalgebras of $B$). Let

[TABLE]

Computation shows that $S_{2}=F[v^{\prime}_{1},v^{\prime}_{2},1\otimes v_{3}]$ is another inseparable subalgebra of $(A,\sigma)$. Clearly, we have $S_{1}\neq S_{2}$. ∎

Theorem 3.10**.**

A totally decomposable algebra with orthogonal involution $(A,\sigma)$ over $F$ has a unique inseparable subalgebra if and only if either $\deg_{F}A\leqslant 4$ or $\sigma$ is anisotropic.

Proof.

Let $S$ be an inseparable subalgebra of $(A,\sigma)$. If $A$ is a quaternion algebra, then $S=\mathop{\mathrm{Alt}}(A,\sigma)\oplus F$ by dimension count. If $\deg_{F}A=4$, then $S=\mathop{\mathrm{Alt}}(A,\sigma)^{+}\oplus F$ by [17, (4.4)]. Also, if $\sigma$ is anisotropic, then $S$ is uniquely determined by Theorem 3.5. This proves the ‘if’ implication. To prove the converse, let $\deg_{F}A=2^{n}$. Suppose that $\sigma$ is isotropic and $\deg_{F}A\geqslant 8$, i.e., $n\geqslant 3$. As $\mathfrak{i}(A,\sigma)>0$, we may identify $(A,\sigma)=\bigotimes_{i=1}^{n-1}(Q_{i},\sigma_{i})\otimes(M_{2}(F),t)$, where $(Q_{i},\sigma_{i})$ is a quaternion algebra with orthogonal involution over $F$ for $i=1,\cdots,n-1$. By Lemma 3.9 the algebra with involution

[TABLE]

admits two inseparable subalgebras $S_{1}$ and $S_{2}$ with $S_{1}\neq S_{2}$. Let $S$ be an inseparable subalgebra of $\bigotimes_{i=1}^{n-3}(Q_{i},\sigma_{i})$. Then $S\otimes S_{1}$ and $S\otimes S_{2}$ are two inseparable subalgebras of $(A,\sigma)$ with $S\otimes S_{1}\neq S\otimes S_{2}$, proving the result. ∎

Lemma 3.11**.**

Let $(A,\sigma)$ be a totally decomposable algebra of degree $2^{n}$ with orthogonal involution over $F$ and let $x\in\mathop{\mathrm{Sym}}(A,\sigma)^{+}$ be a unit. If $S$ is an inseparable subalgebra of $(A,\sigma)$, then for every unit $y\in S$, there exists a positive integer $k$ such that $(xy)^{k}\in\mathop{\mathrm{Sym}}(A,\sigma)^{+}$. In addition, for such an integer $k$ we have $(xy)^{k}x=x(xy)^{k}$.

Proof.

Since $x$ and $y$ are units, the element $(xy)^{r}$ is a unit for every integer $r$. For $r\geqslant 0$ let $S_{r}=(xy)^{r}S(xy)^{-r}$. Then $S_{r}$ is a $2^{n}$-dimensional commutative subalgebra of $A$, which is generated by $n$ elements and satisfies $u^{2}\in F$ for every $u\in S_{r}$. Set $\alpha=x^{2}\in F^{\times}$ and $\beta=y^{2}\in F^{\times}$. Then

[TABLE]

Hence, $S_{r}=\alpha^{-r}\beta^{-r}(xy)^{r}S(yx)^{r}\subseteq\mathop{\mathrm{Sym}}(A,\sigma)$, i.e., $S_{r}$ is an inseparable subalgebra of $(A,\sigma)$. However, there exists a finite number of inseparable subalgebras of $(A,\sigma)$, so $S_{r}=S_{s}$ for some non-negative integers $r,s$ with $r>s$. It follows that $S_{r-s}=S_{0}=S$. In particular, we have $(xy)^{r-s}y(xy)^{s-r}\in S$ and

[TABLE]

Set $\lambda=\alpha^{s-r}\beta^{s-r}$, so that $(xy)^{s-r}=\lambda(yx)^{r-s}$. Replacing in (1) we obtain

[TABLE]

It follows that $\lambda y^{2}(xy)^{2(r-s)}=\lambda y^{2}(yx)^{2(r-s)}$, because $y^{2}\in F^{\times}$. Hence, $(xy)^{k}=(yx)^{k}$, where $k=2(r-s)$. We also have $\sigma((xy)^{k})=(yx)^{k}=(xy)^{k}$ and $((xy)^{k})^{2}=(xy)^{k}(yx)^{k}\in F^{\times}$, hence $(xy)^{k}\in\mathop{\mathrm{Sym}}(A,\sigma)^{+}$. Finally, we have $(xy)^{k}x=x(yx)^{k}=x(xy)^{k}$, completing the proof. ∎

Proposition 3.12**.**

Let $(A,\sigma)$ be a totally decomposable algebra with orthogonal involution over $F$ and let $x\in\mathop{\mathrm{Sym}}(A,\sigma)^{+}$ with $x^{2}\notin F^{2}$. Then $\sigma$ is isotropic if and only if $\sigma|_{C_{A}(x)}$ is isotropic.

Proof.

Observe first that since $x^{2}\notin F^{2}$, $C_{A}(x)$ is a central simple algebra over $F(x)=F(\sqrt{\alpha})$, where $\alpha=x^{2}\in F^{\times}$. If $\sigma|_{C_{A}(x)}$ is isotropic, then $\sigma$ is clearly isotropic. To prove the converse, let $S$ be an inseparable subalgebra of $(A,\sigma)$. Since $\sigma$ is isotopic, there exists $y\in S\setminus F$ with $y^{2}=1$. By Lemma 3.11, there is a positive integer $k$ such that $(xy)^{k}\in\mathop{\mathrm{Sym}}(A,\sigma)^{+}$. Let $r$ be the minimum positive integer with $(xy)^{r}\in\mathop{\mathrm{Sym}}(A,\sigma)^{+}$, hence $(xy)^{r}=(yx)^{r}$. We claim that $(xy)^{r}\neq x^{r}$. Suppose that $(xy)^{r}=x^{r}$. If $r$ is odd, write $r=2s+1$ for some non-negative integer $s$. The equality $(xy)^{r}=x^{r}$ then implies that $(yx)^{s}y(xy)^{s}=x^{2s}=\alpha^{s}$. As $(xy)^{s}=\alpha^{s}(yx)^{-s}$, we get $\alpha^{s}(yx)^{s}y(yx)^{-s}=\alpha^{s}$. Hence, $y=1\in F$, which contradicts the assumption. If $r$ is even, write $r=2s$ for some positive integer $s$, so that $(xy)^{r}=x^{r}=\alpha^{s}$. Multiplying with $(xy)^{-s}$ we obtain

[TABLE]

It follows that $(xy)^{s}\in\mathop{\mathrm{Sym}}(A,\sigma)^{+}$, contradicting the minimality of $r$. This proves the claim. According to Lemma 3.11, we have $(xy)^{r}\in C_{A}(x)$. Set $z=(xy)^{r}+x^{r}\in C_{A}(x)$. Then $z\neq 0$ and $\sigma(z)z=\alpha^{r}+\alpha^{r}=0$, i.e., $\sigma|_{C_{A}(x)}$ is isotropic. ∎

4 The set of elements represented by $\sigma$

Definition 4.1**.**

Let $(A,\sigma)$ be a totally decomposable algebra with orthogonal involution over $F$ and let $\alpha\in F$. We say that $\sigma$ represents $\alpha$ if there exists a nonzero element $x\in A$ such that $\sigma(x)x+\alpha\in\mathop{\mathrm{Alt}}(A,\sigma)$. The set of all elements in $F$ represented by $\sigma$ is denoted by $D(A,\sigma)$.

The following result is a restatement of [6, (6.1)].

Lemma 4.2**.**

Let $(A,\sigma)$ be a totally decomposable algebra with orthogonal involution over $F$. Then $\sigma$ is anisotropic if and only if $0\notin D(A,\sigma)$.

Lemma 4.3**.**

Let $\mathfrak{b}$ be an anisotropic bilinear Pfister form over $F$ and let $\alpha\in F$. If $\alpha\notin D(\mathfrak{b})$, then $\mathfrak{b}_{F(\sqrt{\alpha})}$ is anisotropic.

Proof.

Since $\alpha\notin D(\mathfrak{b})$, we have $\alpha\notin F^{\times 2}$, hence $F(\sqrt{\alpha})$ is a field. By [7, (6.1)], $\mathfrak{b}\otimes\langle\!\langle\alpha\rangle\!\rangle$ is anisotropic, hence $\mathfrak{b}_{F(\sqrt{\alpha})}$ is also anisotropic by [8, (4.2)]. ∎

Lemma 4.4**.**

Let $(A,\sigma)$ be a totally decomposable algebra with orthogonal involution over $F$. If $\sigma$ is anisotropic, then $D(A,\sigma)\subseteq D(\mathfrak{Pf}(A,\sigma))$.

Proof.

Let $\alpha\in D(A,\sigma)$. Then there exists $0\neq x\in A$ such that $\sigma(x)x+\alpha\in\mathop{\mathrm{Alt}}(A,\sigma)$. Suppose that $\alpha\notin D(\mathfrak{Pf}(A,\sigma))$ and set $K=F(\sqrt{\alpha})$. Since $\mathfrak{Pf}(A,\sigma)$ is a Pfister form, [6, (7.5 (1))] implies that $\mathfrak{Pf}(A,\sigma)$ is anisotropic. Hence $\mathfrak{Pf}(A,\sigma)_{K}$ is also anisotropic by Lemma 4.3. By [6, (7.5 (2))], $(A,\sigma)_{K}$ is anisotropic. As $\mathop{\mathrm{Alt}}((A,\sigma)_{K})=\mathop{\mathrm{Alt}}(A,\sigma)\otimes K$, we have $\sigma_{K}(x\otimes 1)\cdot(x\otimes 1)+\alpha\otimes 1\in\mathop{\mathrm{Alt}}((A,\sigma)_{K})$. Set $y=x\otimes 1+1\otimes\sqrt{\alpha}\in A_{K}$. Then $y\neq 0$ and

[TABLE]

We therefore obtain $\sigma_{K}(y)y\in\mathop{\mathrm{Alt}}((A,\sigma)_{K})$, i.e., $0\in D((A,\sigma)_{K})$. This contradicts Lemma 4.2, because $(A,\sigma)_{K}$ is anisotropic. ∎

Corollary 4.5**.**

Let $(A,\sigma)$ be a totally decomposable algebra with anisotropic orthogonal involution over $F$. If $\alpha+\sum_{i=1}^{m}\sigma(x_{i})x_{i}\in\mathop{\mathrm{Alt}}(A,\sigma)$ for some $\alpha\in F$ and $x_{1},\cdots,x_{m}\in A$, then $\alpha\in Q(\mathfrak{Pf}(A,\sigma))$.

Proof.

Set $y=\sum_{i=1}^{m}x_{i}\in A$. Then

[TABLE]

Hence, $\alpha+\sigma(y)y\in\mathop{\mathrm{Alt}}(A,\sigma)$. If $y=0$, then $\alpha\in\mathop{\mathrm{Alt}}(A,\sigma)$ and the orthogonality of $\sigma$ implies that $\alpha=0\in Q(\mathfrak{Pf}(A,\sigma))$. Otherwise, we have $\alpha\in D(A,\sigma)$ and the result follows from Lemma 4.4. ∎

Remark 4.6**.**

Let $(B,\rho)$ be a central simple algebra with involution over $F$ and set $(A,\sigma)=(B,\rho)\otimes(M_{2}(F),t)$. Then every element $x\in A$ can be written as $\left(\begin{smallmatrix}a&b\\ c&d\end{smallmatrix}\right)$, where $a,b,c,d\in B$. The involution $\sigma$ maps $x$ to $\left(\begin{smallmatrix}\rho(a)&\rho(c)\\ \rho(b)&\rho(d)\end{smallmatrix}\right)$. It follows that

[TABLE]

Lemma 4.7**.**

If $\alpha I+\sum_{i=1}^{m}A_{i}^{t}A_{i}\in\mathop{\mathrm{Alt}}(M_{2}(F),t)$, where $A_{1},\cdots,A_{m}\in M_{2}(F)$ and $I\in M_{2}(F)$ is the identity matrix, then $\alpha\in F^{2}$.

Proof.

Write $A_{i}=\left(\begin{smallmatrix}a_{i}&b_{i}\\ c_{i}&d_{i}\end{smallmatrix}\right)$ for some $a_{i},b_{i},c_{i},d_{i}\in F$, so that

[TABLE]

Since $\mathop{\mathrm{Alt}}(M_{2}(F),t)$ is the set of symmetric matrices with zero diagonal, the assumption implies that $\alpha+\sum_{i=1}^{m}a_{i}^{2}+c_{i}^{2}=0$, hence $\alpha\in F^{2}$. ∎

Lemma 4.8**.**

Let $(A,\sigma)$ be a totally decomposable algebra with orthogonal involution over $F$. If $\alpha+\sum_{i=1}^{m}\sigma(x_{i})x_{i}\in\mathop{\mathrm{Alt}}(A,\sigma)$ for some $\alpha\in F$ and $x_{1},\cdots,x_{m}\in A$, then $\alpha\in Q(\mathfrak{Pf}(A,\sigma))$.

Proof.

Let $\deg_{F}A=2^{n}$. We use induction on $n$. If $\sigma$ is anisotropic, the conclusion follows from Corollary 4.5. Suppose that $\sigma$ is isotropic. If $n=1$, then $(A,\sigma)\simeq(M_{2}(F),t)$ and $\mathfrak{Pf}(A,\sigma)\simeq\langle\!\langle 1\rangle\!\rangle$ and the result follows from Lemma 4.7. Let $n\geqslant 2$. Since $\mathfrak{i}(A,\sigma)>0$, we may identify $(A,\sigma)=(B,\rho)\otimes(M_{2}(F),t)$, where $(B,\rho)$ is a totally decomposable algebra with orthogonal involution over $F$. Then $\mathfrak{Pf}(A,\sigma)\simeq\mathfrak{Pf}(B,\rho)\otimes\langle\!\langle 1\rangle\!\rangle$, which implies that $Q(\mathfrak{Pf}(A,\sigma))=Q(\mathfrak{Pf}(B,\rho))$. Write $x_{i}=\left(\begin{smallmatrix}a_{i}&b_{i}\\ c_{i}&d_{i}\end{smallmatrix}\right)$, where $a_{i},b_{i},c_{i},d_{i}\in B$. Using Remark 4.6 we get

[TABLE]

Hence the $(1,1)$-entry of $\sum_{i=1}^{m}\sigma(x_{i})x_{i}$ is $\sum_{i=1}^{m}(\rho(a_{i})a_{i}+\rho(c_{i})c_{i})$. Since $\alpha+\sum_{i=1}^{m}\sigma(x_{i})x_{i}\in\mathop{\mathrm{Alt}}(A,\sigma)$, Remark 4.6 implies that

[TABLE]

By induction hypothesis we get $\alpha\in Q(\mathfrak{Pf}(B,\rho))=Q(\mathfrak{Pf}(A,\sigma))$. ∎

Corollary 4.9**.**

Let $(A,\sigma)$ be a totally decomposable algebra with orthogonal involution over $F$. If $x\in\mathop{\mathrm{Sym}}(A,\sigma)^{+}$, then $x^{2}\in Q(\mathfrak{Pf}(A,\sigma))$.

Proof.

Set $\alpha=x^{2}\in F$. Then $\alpha+\sigma(x)x=0\in\mathop{\mathrm{Alt}}(A,\sigma)$ and the result follows from Lemma 4.8. ∎

Theorem 4.10**.**

If $(A,\sigma)$ is a totally decomposable algebra with orthogonal involution over $F$, then $D(A,\sigma)=D(\mathfrak{Pf}(A,\sigma))=\{x^{2}\mid 0\neq x\in S\}$, where $S$ is any inseparable subalgebra of $(A,\sigma)$.

Proof.

If $\alpha\in D(\mathfrak{Pf}(A,\sigma))$, then there exists $0\neq x\in S$ such that $x^{2}=\mathfrak{Pf}(A,\sigma)(x,x)=\alpha$. Hence, $\sigma(x)x+\alpha=0\in\mathop{\mathrm{Alt}}(A,\sigma)$, i.e., $\alpha\in D(A,\sigma)$. Conversely, let $\alpha\in D(A,\sigma)$. If $\alpha=0$, then $\sigma$ is isotropic by Lemma 4.2. Hence, $\mathfrak{Pf}(A,\sigma)$ is isotopic by [6, (7.5) and (6.2)], i.e, $0\in D(\mathfrak{Pf}(A,\sigma))$. Otherwise, as $Q(\mathfrak{Pf}(A,\sigma))=D(\mathfrak{Pf}(A,\sigma))\cup\{0\}$, the result follows from Lemma 4.8. The equality $D(\mathfrak{Pf}(A,\sigma))=\{x^{2}\mid 0\neq x\in S\}$ follows from the relation $\mathfrak{Pf}(A,\sigma)(x,x)=x^{2}$ for $x\in S$. ∎

Lemma 4.11**.**

Let $\mathfrak{b}$ be a bilinear $n$-fold Pfister form over $F$. If $\alpha\in Q(\mathfrak{b})\setminus F^{2}$, then $\mathfrak{i}(\mathfrak{b}_{F(\sqrt{\alpha})})=\mathfrak{i}(\mathfrak{b})+1$.

Proof.

Set $K=F(\sqrt{\alpha})$ and $r=\mathfrak{i}(\mathfrak{b})$. As $Q(\langle\!\langle 1\rangle\!\rangle)=F^{2}$ and $\alpha\in Q(\mathfrak{b})\setminus F^{2}$ we have $\mathfrak{b}\not\simeq\langle\!\langle 1\rangle\!\rangle^{n}$, i.e., $r<n$. Write $\mathfrak{b}\simeq\langle\!\langle 1\rangle\!\rangle^{r}\otimes\mathfrak{b}^{\prime}$ for some anisotropic bilinear Pfister form $\mathfrak{b}^{\prime}$ over $F$. Since $Q(\mathfrak{b})=Q(\mathfrak{b}^{\prime})$, we have $\alpha\in Q(\mathfrak{b}^{\prime})$. Hence, the pure subform of $\mathfrak{b}^{\prime}$ represents $\alpha+\lambda^{2}$ for some $\lambda\in F$. By [1, A.2], there exist $\alpha_{2},\cdots,\alpha_{s}\in F$ such that $\mathfrak{b}^{\prime}\simeq\langle\!\langle\alpha+\lambda^{2},\alpha_{2},\cdots,\alpha_{s}\rangle\!\rangle$. Note that we have $\alpha+\lambda^{2}\in K^{\times 2}$, hence $\mathfrak{b}^{\prime}_{K}\simeq\langle\!\langle 1,\alpha_{2},\cdots,\alpha_{s}\rangle\!\rangle_{K}$. Since $\mathfrak{b}^{\prime}=\langle\!\langle\alpha+\lambda^{2}\rangle\!\rangle\otimes\langle\!\langle\alpha_{2},\cdots,\alpha_{s}\rangle\!\rangle$ is anisotropic and $K=F(\sqrt{\alpha+\lambda^{2}})$, by [8, (4.2)] the form $\langle\!\langle\alpha_{2},\cdots,\alpha_{s}\rangle\!\rangle_{K}$ is anisotropic. Hence $\mathfrak{i}(\mathfrak{b}^{\prime}_{K})=1$, which implies that $\mathfrak{i}(\mathfrak{b}_{K})=r+1=\mathfrak{i}(\mathfrak{b})+1$. ∎

Corollary 4.12**.**

Let $(A,\sigma)$ be a totally decomposable algebra with orthogonal involution over $F$. If $x\in\mathop{\mathrm{Sym}}(A,\sigma)^{+}$ with $\alpha:=x^{2}\notin F^{\times 2}$, then $\mathfrak{i}((A,\sigma)_{F(\sqrt{\alpha})})=\mathfrak{i}(A,\sigma)+1$. In particular, $(A,\sigma)\not\simeq(M_{2^{n}}(F),t)$.

Proof.

By Corollary 4.9, we have $\alpha\in Q(\mathfrak{Pf}(A,\sigma))$. Hence, the result follows from Lemma 4.11. ∎

5 Stable quaternion subalgebras

In this section we study some conditions under which a symmetric square-central element of a totally decomposable algebra with orthogonal involution is contained in a stable quaternion subalgebra. We start with anisotropic involutions.

Theorem 5.1**.**

Let $(A,\sigma)$ be a totally decomposable algebra with anisotropic orthogonal involution over $F$. Then every $x\in\mathop{\mathrm{Sym}}(A,\sigma)^{+}$ is contained in a $\sigma$-invariant quaternion subalgebra of $A$.

Proof.

Since $\sigma$ is anisotropic, Theorem 3.5 shows that $x$ is contained in the unique inseparable subalgebra of $(A,\sigma)$. If $x^{2}=\lambda^{2}$ for some $\lambda\in F$, then $(x+\lambda)^{2}=0$. Hence, $x=\lambda$ by [6, (6.1)] and the result is trivial. Otherwise, $x^{2}\notin F^{2}$ and the conclusion follows from [14, (6.3 (ii))]. ∎

We next consider algebras of degree $4$ and $8$.

Proposition 5.2**.**

Let $(A,\sigma)$ be a totally decomposable algebra of degree $4$ with orthogonal involution over $F$. If $x\in\mathop{\mathrm{Sym}}(A,\sigma)^{+}$ with $x^{2}\notin F^{2}$, then $x$ is contained in a $\sigma$-invariant quaternion subalgebra of $A$.

Proof.

By Corollary 4.12 we have either $\mathfrak{i}(A,\sigma)=0$ or $\mathfrak{i}(A,\sigma)=1$. In the first case, the result follows from Theorem 5.1. Suppose that $\mathfrak{i}(A,\sigma)=1$. Set $C=C_{A}(x)$ and $K=F(x)=F(\sqrt{\alpha})$. By Proposition 3.12, $(C,\sigma|_{C})$ is isotropic, i.e., $(C,\sigma|_{C})\simeq(M_{2}(K),t)\simeq(M_{2}(F),t)_{K}$. Hence, the algebra $M_{2}(F)$ may be identified with a subalgebra of $C\subseteq A$. Let $Q=C_{A}(M_{2}(F))$. Then $(Q,\sigma|_{Q})$ is a $\sigma$-invariant quaternion subalgebra of $A$ containing $x$. ∎

Lemma 5.3**.**

Let $\mathfrak{b}$ be a $2$-dimensional anisotropic symmetric bilinear form over $F$ and let $K=F(\sqrt{\alpha})$ for some $\alpha\in F^{\times}\setminus F^{\times 2}$. If $\mathfrak{b}_{K}$ is isotropic, then $\mathfrak{b}\simeq\langle a,sa\rangle$ for some $a\in F^{\times}$ and $s\in K^{\times 2}$.

Proof.

The result follows from [12, (3.5)]. ∎

We recall that a symmetric bilinear space $(V,\mathfrak{b})$ over $F$ is called metabolic if there exists a subspace $W$ of $V$ with $\dim_{F}W=\frac{1}{2}\dim_{F}V$ such that $\mathfrak{b}|_{W\times W}=0$.

Lemma 5.4**.**

Let $\mathfrak{b}$ be a $4$-dimensional isotropic symmetric bilinear form over $F$. Let $K=F(\sqrt{\alpha})$ for some $\alpha\in F^{\times}\setminus F^{\times 2}$. If $\mathfrak{b}\otimes\langle\!\langle\alpha\rangle\!\rangle$ is a Pfister form, then $\mathfrak{b}_{K}$ is similar to a Pfister form.

Proof.

Since $\mathfrak{b}\otimes\langle\!\langle\alpha\rangle\!\rangle$ is a Pfister form, the form $\mathfrak{b}$ is not alternating. If $\mathfrak{b}$ is metabolic, then by [7, (1.24) and (1.22 (3))] either $\mathfrak{b}\simeq\langle a,a,b,b\rangle$ or $\mathfrak{b}\simeq\langle a,a\rangle\perp\mathbb{H}$, where $a,b\in F^{\times}$ and $\mathbb{H}$ is the hyperbolic plane. In the first case $\mathfrak{b}$ is similar to $\langle 1,1,ab,ab\rangle=\langle\!\langle 1,ab\rangle\!\rangle$. In the second case, using the isometry $\langle a,a,a\rangle\simeq\langle a\rangle\perp\mathbb{H}$ in [7, (1.16)], we get $\mathfrak{b}\simeq\langle a,a,a,a\rangle$. Hence, $\mathfrak{b}$ is similar to $\langle\!\langle 1,1\rangle\!\rangle$.

Suppose that $\mathfrak{b}$ is not metabolic. By Witt decomposition theorem [7, (1.27)] and the isometry $\langle a,a,a\rangle\simeq\langle a\rangle\perp\mathbb{H}$ for $a\in F^{\times}$, the form $\mathfrak{b}$ is similar to $\mathfrak{b}^{\prime}\perp\langle 1,1\rangle$ for some $2$-dimensional anisotropic bilinear form $\mathfrak{b}^{\prime}$ over $F$. Since $\mathfrak{b}\otimes\langle 1,\alpha\rangle$ is an isotropic Pfister form, it is metabolic by [7, (6.3)]. Hence, $\mathfrak{b}^{\prime}\otimes\langle 1,\alpha\rangle$ is also metabolic. By [8, (4.2)] the form $\mathfrak{b}^{\prime}_{K}$ is metabolic. Using Lemma 5.3, one can write $\mathfrak{b}^{\prime}\simeq\langle a,sa\rangle$, where $a\in F^{\times}$ and $s\in K^{\times 2}$. The form $\mathfrak{b}_{K}$ is therefore similar to $\langle 1,1,a,sa\rangle_{K}\simeq\langle 1,1,a,a\rangle_{K}\simeq\langle\!\langle 1,a\rangle\!\rangle_{K}$, proving the result ∎

Lemma 5.5**.**

Let $(A,\sigma)$ be a central simple algebra of degree $4$ with orthogonal involution over $F$ and let $K/F$ be a separable quadratic extension. If $(A,\sigma)_{K}$ is totally decomposable, then $(A,\sigma)$ is also totally decomposable.

Proof.

The result follows from [11, (7.3)] and the relation $K^{\times 2}\cap F^{\times}=F^{\times 2}$. ∎

Lemma 5.6**.**

Let $(A,\sigma)$ be a totally decomposable algebra of degree $2^{n}$ with orthogonal involution over $F$. Let $x\in\mathop{\mathrm{Sym}}(A,\sigma)^{+}$ with $x^{2}\notin F^{2}$ and set $C=C_{A}(x)$. If $(C,\sigma|_{C})$ is totally decomposable and $x$ is contained in a $\sigma$-invariant quaternion subalgebra $Q$, then there exists an inseparable subalgebra of $(A,\sigma)$ containing $x$.

Proof.

Let $\alpha=x^{2}\in F^{\times}\setminus F^{\times 2}$, $K=F(x)=F(\sqrt{\alpha})$, $B=C_{A}(Q)$ and $\rho=\sigma|_{B}$. Then there exists a natural isomorphism of $K$-algebras with involution

[TABLE]

In particular, $(B,\rho)_{K}$ is totally decomposable. By [15, (3.6)], there exists a totally decomposable algebra of degree $2^{n-1}$ with orthogonal involution $(B^{\prime},\rho^{\prime})$ over $F$ such that $(B,\rho)_{K}\simeq(B^{\prime},\rho^{\prime})_{K}$. Let $S=F[v_{1},\cdots,v_{n-1}]$ be an inseparable subalgebra of $(B^{\prime},\rho^{\prime})$. The algebra $K[v_{1},\cdots,v_{n-1}]\simeq S_{K}$ may be identified with an inseparable subalgebra of $(B,\rho)_{K}$. Set $v^{\prime}_{i}=f(v_{i})\in C$, $i=1,\cdots,n-1$. Then $S^{\prime}:=K[v^{\prime}_{1},\cdots,v^{\prime}_{n-1}]$ is an inseparable subalgebra of $(C,\sigma|_{C})$ satisfying $u^{2}\in F$ for $u\in S^{\prime}$. Since $(C,\sigma|_{C})\subseteq(A,\sigma)$, we have $S^{\prime}\subseteq\mathop{\mathrm{Sym}}(A,\sigma)^{+}$. By [14, (3.11)], $S^{\prime}$ is a Frobenius subalgebra of $A$, because $\dim_{F}S^{\prime}=2^{n}$ and $S^{\prime}$ is generated, as an $F$-algebra, by $x,v^{\prime}_{1},\cdots,v^{\prime}_{n-1}$. It follows from [9, (2.2.3)] that $C_{A}(S^{\prime})=S^{\prime}$. Hence, $S^{\prime}$ is an inseparable subalgebra of $(A,\sigma)$ containing $x$. ∎

Proposition 5.7**.**

Let $(A,\sigma)$ be a totally decomposable algebra of degree $8$ over $F$. For an element $x\in\mathop{\mathrm{Sym}}(A,\sigma)^{+}$ with $x^{2}\notin F^{2}$, the following conditions are equivalent:

• (1)

There exists a $\sigma$-invariant quaternion subalgebra of $A$ containing $x$.

• (2)

There exists an inseparable subalgebra $S$ of $(A,\sigma)$ such that $x\in S$.

Proof.

If $\mathfrak{i}(A,\sigma)=0$, by Theorem 5.1 and Theorem 3.5 both conditions are satisfied. Let $\mathfrak{i}(A,\sigma)>0$. Then $(A,\sigma)\simeq(M_{2}(F),t)\otimes(Q_{1},\sigma_{1})\otimes(Q_{2},\sigma_{2})$, where $(Q_{i},\sigma_{i})$, $i=1,2$, is a quaternion algebra with orthogonal involution over $F$. Suppose first that $x$ is contained in a $\sigma$-invariant quaternion subalgebra $Q_{3}$ of $A$. Set $B=C_{A}(Q_{3})$, $\sigma_{3}=\sigma|_{Q_{3}}$ and $\rho=\sigma|_{B}$, so that $(A,\sigma)\simeq(Q_{3},\sigma_{3})\otimes(B,\rho)$. It follows that

[TABLE]

Let $C=C_{A}(x)$ and $K=F(x)=F(\sqrt{\alpha})$, where $\alpha=x^{2}\in F^{\times}\setminus F^{\times 2}$. Then $(C,\sigma|_{C})\simeq(B,\rho)_{K}$ as $K$-algebras. By Proposition 3.12, $(C,\sigma|_{C})$ is isotropic, hence $(B,\rho)_{K}$ is also isotropic. We claim that $(B,\rho)_{K}$ is totally decomposable. The result then follows from Lemma 5.6.

By Lemma 3.8, for $i=1,2,3$ there exists a quaternion basis $(1,u_{i},v_{i},w_{i})$ of $Q_{i}$ such that $u_{i}\in\mathop{\mathrm{Sym}}(Q_{i},\sigma_{i})$. Let $\beta_{i}=u_{i}^{2}+u_{i}\in F$. For $i=0,1,2,3$, define a field $L_{i}$ inductively as follows: set $L_{0}=F$. For $i\geqslant 1$ set $L_{i}=L_{i-1}(u_{i})$ if $\beta_{i}\notin\wp(L_{i-1}):=\{y^{2}+y\mid y\in L_{i-1}\}$ and $L_{i}=L_{i-1}$ otherwise. In other words, either $L_{i}=L_{i-1}$ or $L_{i}/L_{i-1}$ is a separable quadratic extension. Note that we have $L_{i}^{\times 2}\cap F^{\times}=F^{\times 2}$, hence either $L_{i}(\sqrt{\alpha})=L_{i-1}(\sqrt{\alpha})$ or $L_{i}(\sqrt{\alpha})/L_{i-1}(\sqrt{\alpha})$ is a separable quadratic extension. We show that $\rho_{L_{3}(\sqrt{\alpha})}$ is totally decomposable, which implies that $\rho_{L_{i}(\sqrt{\alpha})}$ is totally decomposable for $i=0,1,2$, thanks to Lemma 5.5. In particular, $\rho_{K}=\rho_{F(\sqrt{\alpha})}$ is also totally decomposable, as required.

Set $L=L_{3}$. For $i=1,2,3$, the algebra ${Q_{i}}_{L}$ splits. Hence we may identify $(Q_{i},\sigma_{i})_{L}=(M_{2}(L),\tau_{i})$, where $\tau_{i}$ is an orthogonal involution on $M_{2}(L)$. Using (2) we obtain

[TABLE]

Hence, $B_{L}$ splits and we may identify $(B,\rho)_{L}=\mathop{\mathrm{Ad}}(\mathfrak{b})$ for some symmetric bilinear form $\mathfrak{b}$ over $L$. As $(B,\rho)_{L}$ is isotropic, the form $\mathfrak{b}$ is isotropic by [6, (4.3)]. Since $x\in\mathop{\mathrm{Sym}}(Q_{3},\sigma_{3})^{+}\setminus F$, by [16, (5.4)] there exists $\lambda\in F$ for which $0\neq x+\lambda\in\mathop{\mathrm{Alt}}(Q_{3},\sigma_{3})$. Hence, $\mathop{\mathrm{disc}}\sigma_{3}=(\alpha+\lambda^{2})F^{\times 2}$, which implies that

[TABLE]

by [10, (7.4)]. The right side of (3) is the adjoint involution of a $3$-fold bilinear Pfister form over $L$. Hence, using (4) and [10, (4.2)] it follows that $\mathfrak{b}\otimes\langle\!\langle\alpha+\lambda^{2}\rangle\!\rangle$ is similar to a Pfister form over $L$. By Lemma 5.4, $\mathfrak{b}_{L(\sqrt{\alpha})}$ is similar to a Pfister form. Hence, $\rho_{L(\sqrt{\alpha})}$ is totally decomposable, as required. This proves the implication $(1)\Rightarrow(2)$. The converse follows from [14, (6.3 (ii))]. ∎

Lemma 5.8**.**

If $\mathfrak{b}$ is a $2^{n}$-dimensional symmetric bilinear form over $F$, then $\mathop{\mathrm{Ad}}(\mathfrak{b})\simeq(M_{2^{n}}(F),t)$ if and only if $\mathfrak{b}\simeq\lambda\cdot\langle\!\langle 1\rangle\!\rangle^{n}$ for some $\lambda\in F^{\times}$.

Proof.

The result follows from [14, (5.7)], [6, (7.5 (3))] and [10, (4.2)]. ∎

Lemma 5.9**.**

Let $(A,\sigma)$ be a central simple algebra with orthogonal involution over $F$. If $(A,\sigma)\otimes(M_{2}(F),\tau)\simeq(M_{2^{n}}(F),t)$ for some orthogonal involution $\tau$ on $M_{2}(F)$, then $(A,\sigma)\simeq(M_{2^{n-1}}(F),t)$.

Proof.

Observe first that $A$ splits, hence we may identify $(A,\sigma)=\mathop{\mathrm{Ad}}(\mathfrak{b})$ for some symmetric bilinear form $\mathfrak{b}$ over $F$. By [10, (7.3 (3)) and (7.4)], we have $(M_{2}(F),\tau)\simeq\mathop{\mathrm{Ad}}(\langle\!\langle\alpha\rangle\!\rangle)$, where $\alpha\in F^{\times}$ is a representative of the class $\mathop{\mathrm{disc}}\tau\in F^{\times}/F^{\times 2}$. Also, $(M_{2^{n}}(F),t)\simeq\mathop{\mathrm{Ad}}(\langle\!\langle 1\rangle\!\rangle^{n})$ by Lemma 5.8. The assumption implies that $\mathop{\mathrm{Ad}}(\mathfrak{b}\otimes\langle\!\langle\alpha\rangle\!\rangle)\simeq\mathop{\mathrm{Ad}}(\langle\!\langle 1\rangle\!\rangle^{n})$. Hence, the forms $\mathfrak{b}\otimes\langle\!\langle\alpha\rangle\!\rangle$ and $\langle\!\langle 1\rangle\!\rangle^{n}$ are similar by [10, (4.2)], i.e., there exists $\lambda\in F^{\times}$ for which $\mathfrak{b}\otimes\langle\!\langle\alpha\rangle\!\rangle\simeq\lambda\cdot\langle\!\langle 1\rangle\!\rangle^{n}$. As $Q(\langle\!\langle 1\rangle\!\rangle^{n})=F^{2}$, we obtain $Q(\mathfrak{b})\subseteq\lambda\cdot F^{2}$, i.e., $\mathfrak{b}$ is similar to $\langle\!\langle 1\rangle\!\rangle^{n-1}$. Using Lemma 5.8 we get $(A,\sigma)\simeq(M_{2^{n-1}}(F),t)$. ∎

Theorem 5.10**.**

Let $(A,\sigma)$ be a totally decomposable algebra of degree $2^{n}$ with orthogonal involution over $F$ and let $x\in\mathop{\mathrm{Sym}}(A,\sigma)^{+}$ with $x^{2}\notin F^{2}$. If $\mathfrak{i}(A,\sigma)=n-1$, then the following statements are equivalent:

• (1)

There exists a $\sigma$-invariant quaternion subalgebra $Q$ of $A$ containing $x$.

• (2)

There exists an inseparable subalgebra $S$ of $(A,\sigma)$ such that $x\in S$.

Proof.

The implication $(2)\Rightarrow(1)$ follows from [14, (6.3 (ii))]. To prove the converse, let $C=C_{A}(x)$. In view of Lemma 5.6, it suffices to show that $(C,\sigma|_{C})$ is totally decomposable. Let $\tau=\sigma|_{Q}$, $B=C_{A}(Q)$ and $\rho=\sigma|_{B}$. Then $(A,\sigma)\simeq(B,\rho)\otimes(Q,\tau)$. Set $K=F(x)=F(\sqrt{\alpha})$, where $\alpha=x^{2}\in F^{\times}\setminus F^{\times 2}$. We have $(C,\sigma|_{C})\simeq_{K}(B,\rho)_{K}$. Hence, it is enough to show that $(B,\rho)_{K}$ is totally decomposable. By Corollary 4.12 we have $\mathfrak{i}(A,\sigma)_{K}=n$, so $(A,\sigma)_{K}\simeq(M_{2^{n}}(K),t)$. It follows that $(B,\rho)_{K}\otimes_{K}(Q,\tau)_{K}\simeq(M_{2^{n}}(K),t)$. Since $x\in Q$ and $x^{2}=\alpha\in K^{2}$, the algebra $Q_{K}$ splits. Hence, using Lemma 5.9 we get $(B,\rho)_{K}\simeq_{K}(M_{2^{n-1}}(K),t)$. In particular, $(B,\rho)_{K}$ is totally decomposable, proving the result. ∎

6 Examples for isotropic involutions

In this section we show that the criteria obtained in §5 do not necessarily apply to arbitrary involutions.

Lemma 6.1**.**

Let $(A,\sigma)$ be a totally decomposable algebra of degree $2^{n}$ with orthogonal involution over $F$. If $n\geqslant 2$ and $(A,\sigma)\not\simeq(M_{2^{n}}(F),t)$, then there exist an element $w\in\mathop{\mathrm{Sym}}(A,\sigma)\setminus(\mathop{\mathrm{Alt}}(A,\sigma)\oplus F)$ and a unit $u\in\mathop{\mathrm{Alt}}(A,\sigma)$ such that $u^{2}\in F^{\times}\setminus F^{\times 2}$ and $uw=wu$.

Proof.

Let $(A,\sigma)\simeq\bigotimes_{i=1}^{n}(Q_{i},\sigma_{i})$ be a decomposition of $(A,\sigma)$. As $(A,\sigma)\not\simeq(M_{2^{n}}(F),t)$, (by re-indexing) we may assume that $(Q_{1},\sigma_{1})\not\simeq(M_{2}(F),t)$. Let $u\in\mathop{\mathrm{Alt}}(Q_{1},\sigma_{1})$ be a unit, so that $u^{2}\in F^{\times}$. Note that $u^{2}\notin F^{\times 2}$, since otherwise $Q_{1}$ splits and $\mathop{\mathrm{disc}}\sigma_{1}$ is trivial. As $\mathop{\mathrm{disc}}t$ is also trivial (see [10, p. 82]), we get $(Q_{1},\sigma_{1})\simeq(M_{2}(F),t)$ by [10, (7.4)], contradicting the assumption. Hence, $u^{2}\in F^{\times}\setminus F^{\times 2}$. By [10, (2.6)] we have $\dim_{F}\mathop{\mathrm{Sym}}(Q_{2},\sigma_{2})=3$ and $\dim_{F}\mathop{\mathrm{Alt}}(Q_{2},\sigma_{2})=1$. Hence there exists an element $w\in\mathop{\mathrm{Sym}}(Q_{2},\sigma_{2})\setminus(\mathop{\mathrm{Alt}}(Q_{2},\sigma_{2})\oplus F)$. Clearly, we have $uw=wu$. The elements $u$ and $w$ may be identified with elements of $A$, so that $w\in\mathop{\mathrm{Sym}}(A,\sigma)$ and $u\in\mathop{\mathrm{Alt}}(A,\sigma)$. Since $\alpha+w\notin\mathop{\mathrm{Alt}}(Q_{2},\sigma_{2})$ for every $\alpha\in F$, by [13, (3.5)] we have $\alpha+w\notin\mathop{\mathrm{Alt}}(A,\sigma)$ for all $\alpha\in F$, i.e., $w\in\mathop{\mathrm{Sym}}(A,\sigma)\setminus(\mathop{\mathrm{Alt}}(A,\sigma)\oplus F)$, as required. ∎

The next result shows that Theorem 5.1 does not hold for isotropic involutions of degree $\geqslant 8$ (see also Proposition 5.2).

Proposition 6.2**.**

Let $(A,\sigma)$ be a totally decomposable algebra of degree $2^{n}$ with isotropic orthogonal involution over $F$. If $n\geqslant 3$ and $(A,\sigma)\not\simeq(M_{2^{n}}(F),t)$, then there exists an element $x\in\mathop{\mathrm{Sym}}(A,\sigma)^{+}$ with $x^{2}\notin F^{2}$ which is not contained in any $\sigma$-invariant quaternion subalgebra of $A$.

Proof.

Since $\mathfrak{i}(A,\sigma)>0$, we may identify $(A,\sigma)=(B,\rho)\otimes(M_{2}(F),t)$, where $(B,\rho)$ is a totally decomposable algebra with orthogonal involution over $F$. The assumptions $n\geqslant 3$ and $(A,\sigma)\not\simeq(M_{2^{n}}(F),t)$ imply that $\deg_{F}B\geqslant 4$ and $(B,\rho)\not\simeq(M_{2^{n-1}}(F),t)$. By Lemma 6.1, there exists an element $w\in\mathop{\mathrm{Sym}}(B,\rho)\setminus(\mathop{\mathrm{Alt}}(B,\rho)\oplus F)$ and a unit $u\in\mathop{\mathrm{Alt}}(B,\rho)$ for which $uw=wu$. Set

[TABLE]

By Remark 4.6 we have $x\in\mathop{\mathrm{Sym}}(A,\sigma)\setminus(\mathop{\mathrm{Alt}}(A,\sigma)\oplus F)$. As $u^{2}\in F^{\times}\setminus F^{\times 2}$ we have $x^{2}\in F^{\times}\setminus F^{\times 2}$. By [16, (5.4)], the element $x$ is not contained in any $\sigma$-invariant quaternion subalgebra of $A$, because $x+\alpha\notin\mathop{\mathrm{Alt}}(A,\sigma)$ for every $\alpha\in F$. ∎

We conclude by showing that the implication $(1)\Rightarrow(2)$ in Proposition 5.10 and Proposition 5.7 does not hold for arbitrary involutions. We use the ideas of [6, (9.4)]. Recall that the canonical involution $\gamma$ on a quaternion $F$-algebra $Q$ is defined as $\gamma(x)=x-\mathop{\mathrm{Trd}}_{Q}(x)$ for $x\in Q$, where $\mathop{\mathrm{Trd}}_{Q}(x)$ is the reduced trace of $x$ in $Q$. Also, for a division algebra with involution $(D,\theta)$ over $F$ and $\alpha_{1},\cdots,\alpha_{n}\in D^{\times}\cap\mathop{\mathrm{Sym}}(D,\theta)$, the diagonal hermitian form $h$ on $D^{n}$ defined by $h(x,y)=\sum_{i=1}^{n}\theta(x_{i})\alpha_{i}y_{i}$ is denoted by $\langle\alpha_{1},\cdots,\alpha_{n}\rangle_{\theta}$.

Example 6.3**.**

Let $F\neq F^{2}$ and let $K=F(X,Y,Z)$, where $X$, $Y$ and $Z$ are indeterminates. Let $Q=[X,Y)_{K}$ and let $\gamma$ be the canonical involution on $Q$. By [6, (9.3)], $Q$ is a division algebra over $K$. Choose an element $s\in\mathop{\mathrm{Sym}}(Q,\gamma)$ with $s^{2}=Y$. Let $\psi$ be the diagonal hermitian form $\langle 1,Z,s,s\rangle_{\gamma}$ over $(Q,\gamma)$ and set $(B,\rho)=\mathop{\mathrm{Ad}}(\psi)$. By [6, (9.4)], $(B,\rho)$ is not totally decomposable, but $(B,\rho)_{L}$ is totally decomposable for every splitting field $L$ of $A$.

Now, choose $\alpha\in F^{\times}\setminus F^{\times 2}$ and let $Q^{\prime}=[X,\alpha)_{K}$ with a quaternion basis $(1,u,v,w)$. Let $\tau$ be the involution on $Q^{\prime}$ induced by $\tau(u)=u$ and $\tau(v)=v$. Then $\tau$ is an orthogonal involution and $v=\tau(uv)-uv\in\mathop{\mathrm{Alt}}(Q^{\prime},\tau)$. Set $(A,\sigma)=(B,\rho)\otimes_{K}(Q^{\prime},\tau)$. Then $(A,\sigma)$ is a central simple algebra with orthogonal involution over $K$. We claim that $(A,\sigma)$ is totally decomposable. Let $L=K(u)\subseteq Q^{\prime}$ and set $C=C_{A}(1\otimes u)$. Then $L/K$ is a separable quadratic extension and

[TABLE]

is a central simple $L$-algebra with orthogonal involution. Since $u^{2}+u=X$, $Q_{L}\simeq[X,Y)_{L}$ splits, which implies that $B_{L}$ is also split. It follows that $(B,\rho)_{L}$ is totally decomposable, i.e., $(C,\sigma|_{C})$ is totally decomposable by (5). Using [15, (7.3)] and the isomorphism (5), one can find a totally decomposable algebra with orthogonal involution $(C^{\prime},\sigma^{\prime})$ over $K$ such that $(C,\sigma|_{C})\simeq(C^{\prime},\sigma^{\prime})_{L}$. As $C\subseteq A$, the algebra $C^{\prime}$ may be identified with a subalgebra of $A$. Let $Q^{\prime\prime}=C_{A}(C^{\prime})$. Then $Q^{\prime\prime}$ is a quaternion $K$-subalgebra of $A$ and $(A,\sigma)\simeq_{K}(C^{\prime},\sigma^{\prime})\otimes_{K}(Q^{\prime\prime},\sigma|_{Q^{\prime\prime}})$ is totally decomposable, proving the claim.

The element $1\otimes v\in\mathop{\mathrm{Alt}}(A,\sigma)^{+}$ is contained in the copy of $Q^{\prime}$ in $A$, which is a $\sigma$-invariant quaternion subalgebra of $A$. Note that $(C_{A}(1\otimes v),\sigma|_{C_{A}(1\otimes v)})\simeq(B,\rho)_{K(v)}$ as $K(v)$-algebras. We show that $(B,\rho)_{K(v)}$ is not totally decomposable, which implies that $1\otimes v$ is not contained in any inseparable subalgebra of $(A,\sigma)$, thanks to [14, (6.3)]. Since $v^{2}=\alpha\in F^{\times}\setminus F^{\times 2}$, we have $K(v)\simeq F(\sqrt{\alpha})(X,Y,Z)$. Hence, $Q_{K(v)}$ is still a division algebra by [6, (9.3)]. Using [6, (9.4)] it follows that $(B,\rho)_{K(v)}$ is not totally decomposable.