\titlea{\S20}{Strongly Multiplicative Forms}
\leftrunningtitle{Chapter III. Some Applications}
%\leftrunningtitle{Specialization of Quadratic and Symmetric
%Bilinear Forms}
\rightrunningtitle{\S20 Strongly Multiplicative Forms}

Let $\vf$ be a (symmetric) bilinear or \qf\ over a field $k$ in
$n$ variables and let $t = (t_1, \dots, t_n)$ be a tuple of $n$
unknowns  over $k$.

\begdefinition{1.} The form $\vf$ is called {\it strongly
multiplicative}\Idx{strongly multiplicative form} 
when $\vf(t)$ is a similarity norm of the form $\varphi_{k(t)} = \varphi \otimes
k(t)$,\fonote{In the bilinear case we defined $\varphi(t):=\varphi(t,t)$. Thus $\varphi(t)=(\varphi\otimes [1])(t)$.} 
in other words when
$$\langle \varphi (t) \rangle \otimes \varphi_{k(t)} \; \cong \; \varphi_{k(t)}.$$
\enddefinition

\begexamples{} (1) Clearly every hyperbolic \qf\ $r
\times H$ is strongly multiplicative.

(2) If $\tau$ is a bilinear or \q\ Pfister form over $k$, then
$\tau_{k(t)}$ is likewise such a form over $k(t)$. Thus \S16,
Theorem~1 tells us that $N(\tau_{k(t)}) = D (\tau_{k(t)})$. In
particular we have $\tau (t) \in N (\tau_{k(t)})$. Thus every
Pfister form is \sm.

(3) If $\vf$ is a \sm\ bilinear form, then $\varphi
\otimes [1]$ is clearly a \sm\ quasi-linear \qf. \Ip, for every bilinear Pfister form
$\tau$, the \qf\ $\tau \otimes[1]$ is \sm.
\endexamples

The main aim of this section is to determine all \sm\ bilinear and
\qf s over $k$. We start with two very simple theorems whose
proofs only require the Substitution Principle.

As before we use $D(\varphi)$ to denote the set of elements of
$k^*$ which are represented by $\vf$, i.e. the set of all $a \in k^*$
with $\langle a \rangle < \varphi$ in the bilinear case and
$[a] < \varphi$ in the \q\ case. This set consists of cosets of
the norm group $N(\varphi)$ in the group $k^*$.

\begtheorem{1.} For a bilinear or \qf\ $\vf$ over $k$ the
following statements are equivalent:
\smallskip
\setitemindent{$(a)$}
\item{$(a)$} $\vf$ is \sm.
\smallskip
\item{$(b)$} For every field extension $L$ of $k$ we have $D(\varphi_L) \subset
N(\varphi_L)$, and so $D(\varphi_L) = N(\varphi_L)$.
\endtheorem

\begproof{} $(a) \Rightarrow (b)$: We have $\varphi_{k(t)} \cong \langle
\varphi (t) \rangle \otimes \varphi_{k(t)}$. This implies
$\varphi_{L(t)}
\cong \langle \varphi (t) \rangle \otimes \varphi_{L(t)}$
and so, by the Substitution Principle,
$\varphi_L \cong \langle \varphi (c) \rangle \otimes
\varphi_L$ for every $c \in L^n$ with $\varphi (c) \ne 0$.

$(b) \Rightarrow (a)$: Trivial.\qed
\endproof

\begtheorem{2.} If $\vf$ is a \qf\ with associated \nd\ \qf\
$\widehat{\varphi}$, i.e. (cf. \S6, Definition~9)
$$\varphi \; = \; \widehat{\varphi} \perp \delta (\varphi) \; = \; \widehat{\varphi}
\perp s \times [0]$$
for some $s \in {\Bbb N}_0$, then $\vf$ is \sm\ \iff\
$\widehat{\varphi}$ is \sm.
\endtheorem

\begproof{} $\varphi (t_1, \dots, t_n) = \widehat{\varphi} (t_1, \dots,
t_{n-s})$ is a similarity norm of $\varphi_{k(t)}$ \iff\ this
polynomial is a similarity norm of $\widehat{\varphi}_{k(t)}$.
This will be the case exactly when $\widehat{\varphi} (t_1, \dots,
t_{n-s})$ is a similarity norm of
$\widehat{\varphi}_{k (t_1, \dots, t_{n-s})}$, as can be seen by
the Substitution Principle for instance.\qed
\endproof

In the study of \sm\ \qf s we may thus restrict ourselves to
\nd\ forms $\vf$. By an analogous argument we may also do this in
the bilinear case. Thus we always assume that $\vf$ is \nd\ in
what follows. Let us commence our study with \sm\ {\it quadratic} forms.

\begtheorem{3.} Let $\vf$ be a \nd\ \sm\ \qf\ over $k$. Then $\vf$
is either quasi-linear or strictly regular.
\endtheorem

\begproof{} We recall that the quasi-linear part $QL(\vf)$ of
$\vf$ is uniquely determined by $\vf$ up to isometry. Thus
$$\langle \varphi (t) \rangle \otimes \varphi_{k(t)} \; \cong \; \varphi_{k(t)}$$
implies that
$$\langle \varphi (t) \rangle \otimes QL (\varphi)_{k(t)} \; \cong \; QL (\varphi)_{k(t)}.$$
Assume that $QL(\varphi) \ne 0$. We choose an element $b \in D
(QL(\varphi))$. Then $b \varphi (t) \in D (QL(\varphi)_{k(t)})$.
The form $QL(\varphi)$ is anisotropic. Thus $b \varphi$ is a
subform of $QL(\varphi)$ by the Subform Theorem of \S17. Since $\dim \varphi \geq
\dim QL(\varphi)$ it follows that $\varphi = QL(\varphi)$.\qed
\endproof

\begtheorem{4.} Let $\vf$ be a strictly regular \sm\ \qf\ over $k$
which is not hyperbolic. Assume that $\dim \varphi> 1$.
\smallskip
\setitemindent{$(ii)$}
\item{$(i)$} $\vf$ is an anisotropic Pfister form.
\smallskip
\item{$(ii)$} If $\tau$ is another \q\ Pfister form over $k$ with $\tau <
\varphi$, then there exists a bilinear Pfister form  $\sigma$ over
$k$ with $\varphi \cong \sigma \otimes \tau$.
\endtheorem

\begproof{} $(i)$ Let $\varphi_0$ be the kernel form of $\vf$,
i.e. $\varphi \; \cong \; \varphi_0 \perp r \times H$ for some $r \geq
0$. We have $\varphi_0 \ne 0$. Now
$\langle \varphi (t) \rangle \otimes \varphi_{k(t)} \cong \varphi_{k(t)}$
implies that
$$\langle \varphi (t) \rangle \otimes (\varphi_0)_{k(t)} \; \cong \;
(\varphi_0)_{k(t)}$$
and then $b \varphi (t) \in D(\varphi_0 \otimes k(t))$ for an
arbitrarily chosen $b \in D(\varphi_0)$. By the Subform Theorem of
\S17, $b \varphi$ is a subform of $\varphi_0$. Therefore $\dim \varphi \leq \dim
\varphi_0$ and so $r = 0$. Thus $\vf$ is anisotropic.

$(ii)$ By Theorem~1 we have $N(\varphi) = D(\varphi)$. \Ip\ $1 \in D
(\varphi)$. Since $\vf$ is \sr, there exists a $2$-dimensional
form $\left[ {1 \atop 1} {1 \atop a} \right] < \varphi$. This
form is a \Pf.

Now let some \Pf\ $\tau < \varphi$ be given. We choose a bilinear
\Pf\ $\sigma$ of maximal dimension such that $\sigma \otimes \tau <
\varphi$. $\{$Note that possibly $\sigma = \langle 1 \rangle$.$\}$
The theorem will be proved if we can show that $\varphi \cong \sigma \otimes
\tau$.

By way of contradiction, assume that $\varphi \not\cong \sigma \otimes
\tau$. Then we have a decomposition
$$\varphi \; \cong \; \sigma \otimes \tau \perp \psi$$
with $\psi \ne 0$. Assume that the \Pf\  $\rho \colon = \sigma \otimes \tau$
has dimension $m$ (a power of $2$). Let $t = (t_1, \dots, t_m)$ be
a tuple of $m$ unknowns  over $k$. Since $\vf$ is \sm, we have
by Theorem~1 that
$$\rho (t) \rho_{k(t)} \; \perp \; \rho (t) \psi_{k(t)} \; \cong \;
\rho_{k(t)} \; \perp \;
\psi_{k(t)}.$$
Now $\rho (t) \rho_{k(t)} \cong \rho_{k(t)}$. By the Cancellation
Theorem (\S6, Theorem~2) it follows that
$$\rho (t) \psi_{k(t)} \; \cong \; \psi_{k(t)}.$$
We choose a $b \in D (\psi)$ and then have $b \rho (t) \in
D(\psi_{k(t)})$. By the Subform Theorem we get $b \rho < \psi$
and so
$$(\langle 1, b \rangle \otimes \sigma) \otimes \tau \; = \; \langle 1, b \rangle
\otimes \rho \; < \; \varphi.$$
This contradicts the maximality of $\dim \sigma$. Thus $\sigma \otimes
\tau \cong \varphi$.\qed
\endproof

In \S16 we determined that every anisotropic \q\ \Pf\ $\tau$ has
height~$1$. The same is therefore true for $a \tau$ where $a \in
k^*$ is an arbitrary scalar. We are now in the position to prove a
converse of this fact.

\begtheorem{5.} Let $\vf$ be an anisotropic \sr\ \qf\ over $k$ of
height~$1$. Choose $a \in D (\varphi)$. Then $\varphi
\cong a \tau$ for some \Pf\ $\tau$.
\endtheorem

\begproof{} Upon replacing $\varphi$  by $a \varphi$ we may assume
without loss of generality that $a=1$. As before let $n \colon = \dim \varphi$ and let $t = (t_1, \dots,
t_n)$ be a tuple of $n$ unknowns  over $k$. We have $n \geq 2$.

The polynomial $\varphi (t)$ over $k$ is irreducible. Moreover, after an
appropriate linear transformation of the variables $t_1, \dots, t_n$
over $k$, we have that $\varphi (t)$ is monic.

Since the height of $\vf$ is $1$, $\varphi \otimes k (\varphi) \sim
0$. From the Norm Theorem of \S19 it follows that $\varphi (t)$ is a
similarity norm of $\vf$, thus  $\vf$ is \sm. And so, by
Theorem~4, $\vf$ is a \Pf.\fonote{More precisely we should say ``$\vf$ is isometric
to a Pfister form''. In the following we frequently allow inaccuracies like this one,
in order to simplify the language.}\qed
\endproof

We now move on to {\it quasi-linear} \sm\ (\q) forms. Let $\Char k =
2$ and let us contemplate arbitrary anisotropic quasi-linear forms
over $k$. We will utilize the geometric language to this end.

Thus, let $E = (E, q)$ be an anisotropic quasi-linear \qs\ over
$k$. As usual, we denote by $k^{\frac{1}{2}}$ the subfield of the
algebraic closure $\tilde{k}$, consisting of all $\lambda \in \tilde{k}$
with $\lambda^2 \in k$. Recall that for every $a \in k$ there is a
unique element $\sqrt{a} \in k^{\frac{1}{2}}$ such that $(\sqrt{a})^2 = a$
and also that the map $a \mapsto \sqrt{a}$ is an isomorphism from
the field $k$ to $k^{\frac{1}{2}}$.

Clearly $f \colon E \to k^{\frac{1}{2}}$, $f(x) \colon =
\sqrt{q(x)}$ is a $k$-linear map from $E$ to $k^{\frac{1}{2}}$.
This map is injective since $(E, q)$ is anisotropic. We equip the
$k$-vector space $k^{\frac{1}{2}}$ with the quadratic form
$\sq \colon k^{\frac{1}{2}} \to k$,
$\sq (x) = x^2$. $\{$sq as in ``square''.$\}$ Then
$k^{\frac{1}{2}}$ becomes an -- in general infinite-dimensional --
quasi-linear quadratic $k$-module and $f$ an isometry from $(E,q)$
to a submodule $U$ of ($k^{\frac{1}{2}}, \sq$). The following
stronger statement is now obvious.

\begtheorem{6.}
For every anisotropic quasi-linear \qs\ $(E,q)$ over $k$ there
exists {\rm exactly one} isometry $f \colon E \buildrel \sim \over
\longrightarrow U$, where $U$ is a finite-dimensional quadratic
submodule of $(k^{\frac{1}{2}}, \sq)$.\qed
\endtheorem

Now it is easy to classify the strongly multiplicative anisotropic quasi-linear
forms over $k$.

\begtheorem{7.}
Let $\vf$ be an anisotropic quasi-linear \qf\ over $k$. The
following statements are equivalent:
\smallskip
\setitemindent{$(4)$}
\item{$(1)$} $\vf$ is \sm.
\smallskip
\item{$(2)$} $D(\varphi) = N(\varphi)$.
\smallskip
\item{$(3)$} There exists exactly one subfield $K$ of $k^{\frac{1}{2}}$ with $k
\subset K$, $[K \colon k]$ finite, such that the \qf\ $(K, \sq
\vert K)$ is isometric to $\varphi$.
\smallskip
\item{$(4)$} $\varphi \cong \tau \otimes [1]$ where $\tau$ is an
anisotropic bilinear Pfister form.
\endtheorem

\begproof{}
The implication (1) $\Rightarrow$ (2) is clear by Theorem~2.

(2) $\Rightarrow$ (3): By Theorem~6 there exists exactly one
finite-dimensional $k$-subspace $U$ of $k^{\frac{1}{2}}$ such that
the form $\sq \vert U$ is isometric to $\varphi$. By assumption
$(2)$ we certainly have $1 \in D(\varphi)$. Hence $U$ contains an
element $z$ with $z^2=1$. We must have $z=1$ and thus $k \subset
U$. Furthermore, for every two elements $x,y$ in $U$  there exists
an element $z$ in $U$ such that $z^2 = x^2 \cdot y^2$, and so $z =
xy$. Finally, if $x \ne 0$ is any element in $U$, then $x^{-1} = (x^{-1})^2
\cdot x$ is also in $U$. We conclude that $U$ is a subfield of $k^{\frac{1}{2}}$.

(3) $\Rightarrow$ (4): We have $K = k (\sqrt{a_1}, \dots,
\sqrt{a_d})$ with finitely many elements $a_1, \dots, a_d$ in $k$
which are linearly independent over the field
$k^{(2)} = \{ x^2 | x \in k \}$. The associated \qf\
$\vf$ is the orthogonal sum of the one-dimensional forms
$[a_{i_1} a_{i_2} \dots a_{i_r} ]$ with $1 \leq i_1 <
i_2 < \cdots < i_r \leq d$, i.e.
$$\varphi = \langle 1, a_1 \rangle \otimes \cdots \otimes \langle 1, a_d
\rangle \otimes [1].$$

(4) $\Rightarrow$ (1): We already established above (Example~3)
that for every bilinear Pfister form $\tau$, the form $\tau \otimes
[1]$ is strongly multiplicative.\qed
\endproof

\begdefinition{2.} Following established terminology (e.g. [Lo])
we call a qua\-dratic or bilinear form $\vf$ over a field $k$ {\it
round}\Idx{round form} 
whenever $N(\varphi) = D(\varphi)$. Analogously we call a
\q\ or bilinear module $E$ over $k$ round whenever $N(E) = D(E)$,
thus, in finite dimensions, when the (up to isometry unique)
associated form $\vf$ is round.
\enddefinition

By Theorem~2 it is clear that a form $\vf$ over $k$ is \sm\ if and
only if for every field extension $L$ of $k$ the form $\vf_L$ is
round. For this to hold, it is sufficient to know that
$\varphi_{k(t)}$ is round where, as before, $t = (t_1, \dots,
t_n)$ is a tuple of $n = \dim \varphi$ unknowns  over $k$.

Theorem~7 tells us, among other things, that a round quasi-linear
qua\-dratic form $\vf$ over a field of characteristic~$2$ is already
strongly multiplicative. In Theorem~7 we supposed that $\vf$ is
anisotropic, to be sure, but this assumption can subsequently
be dropped.

We remark that in general there are many more round forms
than  strongly multiplicative forms among
the non quasi-linear quadratic forms. For example if $k={\Bbb R}$, then
for every $n \in {\Bbb N}$ the form $n \times [1]$ is round. This
form is nonetheless \sm\ only when $n$ is a power of $2$.

Now let us assume again that $\Char k=2$. Among other things,
Theorem~7 tells us that every anisotropic quasi-linear round \qs\
over $k$ can be considered as a finite field extension $K$ of $k$
with $K \subset k^{\frac{1}{2}}$ in a unique way. We want to give
a second, more structural, proof of this fact. This will guide us
in the right direction for a classification of the anisotropic
round bilinear forms over $k$.

Thus, let $(E,q)$ be an anisotropic round quasi-linear \qm\ over
$k$. Every vector $x\in E$ is uniquely determined by the value $q(x) \in
k$. For if $q(x) = q(y)$, then $q(x-y) = 0$ and so $x - y = 0$.
$\{$Note: this holds for {\it every} \an\ \ql\ space.$\}$
Since $D(E)$ is a group, there exists a vector $e_0$ with $q(e_0) =
1$. Furthermore there exists for every $x \in E$, $x \ne 0$ {\it
exactly one} similarity transformation $\sigma_x \colon E \to E$
with norm $q(x)$. Hence we have for every vector $y \in E$,
$$q(\sigma_x y) = q(x) q(y).$$
We define a multiplication $(x, y) \mapsto x \cdot y$ on $E$
by $x \cdot y \colon = \sigma_x y$ for $x \ne 0$, $0 \cdot y\colon = 0$.
As usual we write $xy$ instead of $x \cdot y$ most of the time.
Clearly we then have for arbitrary $x, y \in E$ that
$$q(xy) = q(x) q(y).$$
Since $q(yx) = q(y) q(x) = q(xy)$ it follows that $yx = xy$. If
$z$ is a further element in $E$ we have
$$q((xy)z) = q(x) q(y) q(z) = q(x(yz)),$$
so that $(xy)z = x(yz)$. Finally,
$$q(e_0 z) = q(e_0) q(z) = q(z),$$
and so $e_0 z = z$.
Our multiplication is thus commutative and associative and has
$e_0$ as identity element. By definition it is $k$-linear in the
second argument and thus also in the first argument. Hence $E$ is
a commutative $k$-algebra. If $x \in E$, $x \ne 0$, then $q(x) \ne 0$
and furthermore $q(x \cdot x) = q(x)^2$, so that
$$q(q(x)^{-1} x \cdot x) = 1.$$
It follows that $q(x)^{-1} x \cdot x = e_0$. Thus $x$ has an
inverse $x^{-1} = q(x)^{-1} x$. The ring $E$ is therefore a field.
As usual we denote its identity element $e_0$ from now on by $1$
and identify $k$ with the subfield $k \cdot 1$ of $E$. In this way
$E$ becomes a field extension of $k$ and for every $x\in E$ we
have
$$x^2 = q(x).$$
If $k^{\frac{1}{2}}$ is the set of all square roots $\sqrt{a}$ of elements
$a \in k$ in a given fixed algebraic closure $\tilde{k}$ of $k$,
then we have exactly one field embedding $E \hookrightarrow k^{\frac{1}{2}}$ of $E$ in
the field $k^{\frac{1}{2}}$ over $k$. Thus we have derived the
implication (2) $\Rightarrow$ (3) in Theorem~7 for the second
time. In fact we obtained an analogous result for $E$ of
infinite dimension over $k$, which we actually will not need in
what follows.

Now let $(E, B)$ be a round \an\ {\it bilinear} module over the
field $k$ of characteristic~$2$. The \qf\ $n(x)
\colon = B(x, x)$ on $E$, associated to $B$, is round, \ql\ and
\an. Thus $E$ carries the structure of a field extension of $k$
with $x^2 = n(x)$ for all $x\in E$.

\begdefinition{3.}
We call the extension field $U$ of $k$ in $k^{\frac{1}{2}}$ which
is isomorphic to $E$ over $k$, the {\it inseparable field
extension of $k$ associated to} $(E, B)$.
\enddefinition

The field $U \subset k^{\frac{1}{2}}$ should be considered as a
first invariant of the space $(E, B)$. What else do we need in
order to describe $(E, B)$ completely up to isometry?

We endow the $k$-vector space $E$ with the linear form
$$t \colon \; E \to k, \quad t(x) \colon = B(1, x).$$
Clearly, $t(1) = n(1) = 1$. Now $B$ is completely determined by
$t$ since for every $x \in E$, $x \ne 0$ there is exactly one
similarity transformation $\sigma_x \colon E \buildrel \sim \over
\longrightarrow E$ of norm $n(x)$ and for $y \in E$ we have $\sigma_x (y) =
xy$, all of this being justified by our analysis of the round \ql\ space $(E,n)$
above. Hence we have for arbitrary $y, z \in E$
$$B(xy, xz) = n(x) B(y, z). \leqno(\ast)$$
This equality remains valid when $x=0$. Let $y
\ne 0$, $x = y^{-1}$, then ($\ast$) implies that
$n(y)^{-1} B(y, z) = B (1, y^{-1} z) = t(y^{-1} z),$
and so
$B(y, z) = n(y) t(y^{-1} z) = t(n(y) \cdot y^{-1} z) = t(y^2 y^{-1} z) =
t(yz).$
The equality
$$B(y,z)=t(yz)$$
remains valid when $y=0$.

Conversely, let $U$ be an extension field of $k$ with $k \subset U \subset
k^{\frac{1}{2}}$ and let $t \colon U \to k$ be a linear form on the
$k$-vector space $U$ with $t(1) = 1$. On $U$ we define a symmetric
bilinear form
$$\beta_t \colon\, U \times U \to k, \quad \beta_t (x, y) \colon = t(xy).$$
We have $\beta_t (x, x) = t(x^2) = x^2t (1) = x^2$. The quadratic
standard form $x \mapsto x^2$ on $U$ is thus associated to $\beta_t$.
In particular, the bilinear $k$-module $(U, \beta_t)$ is \an. For
every $x \in U$, $x \ne 0$ we define the $k$-linear transformation
$$\sigma_x \colon U \longrightarrow U, \quad y \longmapsto xy.$$
We have
$$\beta_t (\sigma_x y, \sigma_x z) = \beta_t (xy, xz) = t(x^2 yz) =
x^2t(yz) = x^2 \beta_t (y, z).$$
Thus $\sigma_x$ is a similarity transformation on $U$ of norm
$x^2$. This shows that $(U, \beta_t)$ is round.

We summarize these considerations in the following theorem.

\begtheorem{8.}
The round \an\ bilinear modules over $k$ are up to isometry the
pairs $(U, \beta_t)$ with $U$ an extension field of $k$ in
$k^{\frac{1}{2}}$ and $t$ a $k$-linear form on $U$ with $t(1) = 1$.
For any round \an\ bilinear module $(E,B)$ over $k$ there is
exactly one such pair $(U, \beta_t)$ with $(E, B) \cong (U,
\beta_t)$ and furthermore exactly one isometry from $(E, B)$
to $(U, \beta_t)$.\qed
\endtheorem

\begtheorem{9.}
Let $\vf$ be an \an\ round bilinear form over a field $k$ of
characteristic~$2$.
\smallskip
\setitemindent{$(ii)$}
\item{$(i)$} $\vf$ is a Pfister form.
\smallskip
\item{$(ii)$} If $\tau$ is a bilinear Pfister form  with $\tau <
\varphi$, then there exists a further bilinear Pfister form $\sigma$
with $\varphi \cong \sigma \otimes \tau$.
\endtheorem

\begproof{}
We have $\langle 1 \rangle < \varphi$.  Let $\tau < \varphi$ be an
arbitrary Pfister form. We choose a Pfister form $\sigma$ over $k$
of maximal dimension with $\sigma \otimes \tau < \varphi$.
$\{$Note: it is possible that $\sigma = \langle 1 \rangle$.$\}$ We
will show that $\sigma \otimes \tau = \varphi$, which will prove
the theorem.

We will use the geometric language. Let $E = (E, B)$ be a bilinear
space which arises from $\vf$ and let $F$ be a subspace of $E$
arising from $\sigma \otimes \tau$. $E$ carries the structure of a
field extension of $k$ with $x^2 = B(x, x)$ for all $x \in E$.
Furthermore there is a linear form $t
\colon E \to k$ with $B(x, y) = t(xy)$ for all $x, y \in E$.
Finally, $F$ is a subfield of $E$ with $k \subset E$. $\{$If $\sigma \otimes
\tau \cong \langle 1 \otimes a_1 \rangle \otimes \cdots \otimes \langle 1, a_m
\rangle$, then $F = k(\sqrt{a_1}, \dots, \sqrt{a_m})$ and $t(\sqrt{a_{i_1}
\dots a_{i_r}}) = 0$ for $r > 0$, $1 \leq i_1 < \cdots < i_r \leq m.\}$

Assume for the sake of contradiction that $F \ne E$. We choose
$z \in E$ with $z \ne 0$, $B(z, F) = 0$. For $x,y\in F$ we then
have
$$B(x, zy) = t(xyz) = B(xy, z) = 0.$$
Thus $zF$ is a subspace of $E$, perpendicular to $F$. Finally we
have for $x, y \in F$ with $b \colon = B(z, z)$ the equality
$$B(zx, zy) = bB (x, y).$$
All this shows that $F \, \perp \, zF$ is a subspace of $E$, with
associated bilinear form the Pfister form $\langle 1, b \rangle \otimes \sigma \otimes
\tau$. This contradicts the maximality of $\dim \sigma$. Thus
$F=E$.\qed
\endproof

By Theorems~4 and 9 it is now clear that over {\it every} field
the \an\ \sm\ bilinear forms are exactly the \an\ Pfister forms.
