\titlea{\S21}{Divisibility by Pfister Forms}
\leftrunningtitle{Chapter III. Some Applications}
%\leftrunningtitle{Specialization of Quadratic and Symmetric
%Bilinear Forms}
\rightrunningtitle{\S21 Divisibility by Pfister Forms}

In this section we will deal with the divisibility of a \qf\ by a
given \q\ or bilinear Pfister form in the sense of the following
general definition.

\begdefinition{1.} Let $\varphi$ and $\tau$ be \qf s and $\sigma$
a bilinear form over $k$. We say that $\vf$ is {\it divisible by}\Idx{divisibility}
$\tau$ if there exists a bilinear form $\psi$ over $k$ with
$\varphi \cong \psi \otimes \tau$. Similarly we say that $\vf$ is
{\it divisible by} $\sigma$ if there exists a \qf\ $\chi$ over $k$
with $\varphi \cong \sigma \otimes \chi$. We then write $\tau
\vert \varphi$ resp. $\sigma \vert \varphi$.\Idx{111taudivfi@$\tau
\vert \varphi$}
\enddefinition

First we will study the divisibility by a given (anisotropic) {\it
quadratic} Pfister form.

\begtheorem{1.} Let $\tau$ be a \q\ \Pf\ over a field $k$ and
$\vf$ an \an\ \sr\ \qf\ over $k$. The following statements are
equivalent.
\smallskip
\setitemindent{$(iii)$}
\item{$(i)$} $\tau \vert \varphi$.
\smallskip
\item{$(ii)$} There exists a \nd\ bilinear form $\psi$ over $k$
with $\varphi \sim \psi \otimes \tau$.
\smallskip
\item{$(iii)$} $\varphi \otimes k(\tau) \sim 0$.
\endtheorem

\begproof{} The implications $(i)$ $\Rightarrow$ $(ii)$
$\Rightarrow$ $(iii)$ are trivial.

$(iii)$ $\Rightarrow$ $(i)$: Let $\dim \tau = n$ (a $2$-power) and
let $t = (t_1, \dots, t_n)$ be a tuple of $n$ unknowns  over
$k$. We can think of $\tau (t) \in k[t]$ as a monic polynomial. By
the Norm Theorem (cf. \S 19) our assumption $\varphi \otimes
k(\tau) \sim 0$ implies that $\tau (t)$ is a similarity norm of
$\varphi_{k(t)}$.  We choose an element $a \in D(\varphi)$.
Then $a \tau (t) \in D(\varphi_{k(t)})$ and thus, by the Subform
Theorem (cf. \S17),
$$\varphi \cong a \tau \perp \varphi_1$$
where $\vf_1$ is a further \qf\ over $k$. It now follows that
$\varphi_1 \otimes k(\tau) \sim 0$. By induction on $\dim \varphi$
we obtain $\varphi \cong \chi \otimes \tau$ for a bilinear form
$\chi = \langle a, a_2, \dots, a_r \rangle$.\qed
\endproof

The following theorem is a variation of Theorem~1 in the case where
$\vf$ is also a \Pf.

\begtheorem{2.} Let $\tau$ and $\rho$ be \q\ \Pf s over a field
$k$. Assume that the form $\rho$ is \an. The following statements
are equivalent.
\smallskip
\setitemindent{$(iii)$}
\item{$(i)$} There exists a bilinear \Pf\ $\sigma$ over $k$ with $\rho
\cong \sigma \otimes \tau$.
\smallskip
\item{$(ii)$} $\tau \vert \rho$.
\smallskip
\item{$(iii)$} There exists an element $a \in k^*$ with $a \tau < \rho$.
\smallskip
\item{$(iv)$} $\rho \otimes k(\tau) \sim 0$.
\smallskip
\item{$(v)$} There exists a place $\lambda \colon \; k(\rho) \to
k(\tau) \cup \infty$ over $k$.
\endtheorem

\begproof{} The implications $(i)$ $\Rightarrow$ $(ii)$  and
$(ii)$ $\Rightarrow$ $(iii)$ are trivial. The implications $(iii)$
$\Rightarrow$ $(v)$ and $(iv)$ $\Leftrightarrow$ $(v)$ both follow
from the fact that $k(\rho)$ is a generic zero field of $\rho$
and that every isotropic \q\ \Pf\ is already hyperbolic. The
implication $(iv)$ $\Rightarrow$ $(ii)$ is clear by Theorem~1. To
finish the proof we verify the implication $(iii)$ $\Rightarrow$
$(i)$. Thus let $a \tau < \rho$ for some $a \in k^*$. Then $a \in
D(\rho) = N(\rho)$. Hence $\tau < a \rho \cong \rho$. Now \S20,
Theorem~4 tells us that there exists a bilinear \Pf\ $\sigma$ over
$k$ with $\rho \cong \sigma \otimes \tau$.\qed
\endproof

\begcorollary{} Let $\tau_1$ and $\tau_2$ be \q\ \Pf s over $k$
such that\break $k(\tau_1) \sim_k k (\tau_2)$. Then $\tau_1 \cong
\tau_2$.\qed
\endcorollary

Note that in the proof of Theorem~1 we showed that for a given
$a \in D(\varphi)$ we could choose the factor $\psi$ in such a way
that $a \in D(\psi)$. Similarly one could ask whether in Theorem~2
the \Pf\ $\sigma$ can be chosen so as to be divisible by $\langle
1, a \rangle$, for an appropriately given binary form $\langle 1,
a \rangle$. This leads us to the next lemma which is very useful
for what follows.

\beglemma{1.} Let $\sigma$ be a bilinear and $\tau$ a \q\ \Pf\
over $k$. Let $\dim \sigma > 1$ and $a \in D(\sigma' \otimes
\tau)$, where -- as before $(\S4, \S16)$ -- $\sigma'$ denotes the pure
part of $\sigma$. Then there exists a bilinear \Pf\ $\gamma$ over
$k$ with $$\sigma \otimes \tau \; \cong \; \gamma \otimes \langle
1, a \rangle \otimes \tau.$$
\endlemma

\begproof{} (a) If $\sigma \otimes \tau$ is isotropic, then
$\sigma \otimes \tau$ is hyperbolic and the statement can easily
be verified. Namely, if $\dim \sigma > 2$, then we can choose
$\gamma$ to be the hyperbolic \Pf\ of the correct dimension. If
$\sigma = \langle 1, c \rangle$ however, then $a = cu$ for some $u
\in D(\tau)$, so that $\langle 1, c \rangle \otimes \tau \; \cong
\; \langle 1, a \rangle \otimes \tau$, and $\gamma = \langle 1
\rangle$ will do.

(b) Suppose next that $\sigma \otimes \tau$ is anisotropic. Let
$\mu \colon = \langle 1, a \rangle \otimes \tau$ and $K \colon = k
(\mu)$. By Theorem~2 it is sufficient to show that $(\sigma
\otimes \tau)_K \sim 0$.

Since $\mu_K \sim 0$ we have that $\tau_K \cong - a \tau_K$ and
$$(\sigma \otimes \tau)_K \; = \; \tau_K \; \perp \; (\sigma'
\otimes \tau)_K \; \cong \; (- a \tau \; \perp \; \sigma' \otimes
\tau)_K.$$ By assumption the element $a$ is represented by
$\sigma' \otimes \tau$. Thus the form\break $- a \tau  \perp
\sigma' \otimes \tau$ is isotropic. Hence $(\sigma \otimes
\tau)_K$ is isotropic. Since $\sigma \otimes \tau$ is a \Pf\ we
have $(\sigma \otimes \tau)_K \sim 0$, as required.\qed
\endproof

Now suppose that $\rho_1, \rho_2$ are two \q\ \Pf s over $k$,
possibly of different dimension. We are looking for a \q\ \Pf\
$\tau$, perhaps of higher dimension, which divides both $\rho_1$
and $\rho_2$. In this situation a concept emerges which was coined
by Elman and Lam [EL, \S 4] in characteristic $\ne 2$:

\begdefinition{2.} (a) Let $r$ be a natural number. We call
$\rho_1$ and $\rho_2$ {\it $r$-linked},\Idx{r-linked Pfister forms@$r$-linked Pfister forms}\Idx{quadratically r-linked Pfister forms@quadratically $r$-linked Pfister forms}  
more precisely,
 {\it quadratically $r$-linked}, if over $k$
there exist an $r$-fold \q\ \Pf\ $\tau$ and bilinear \Pf s
$\sigma_1, \sigma_2$ such that $$\rho_1 \cong \sigma_1 \otimes
\tau,\quad \rho_2 \cong \sigma_2 \otimes \tau.$$

(b) We define the {\it quadratic linkage number}\Idx{quadratic linkage number}\Idx{111jr1r2@$j(\rho_1, \rho_2)$} 
$j(\rho_1,
\rho_2)$ as follows: if $\rho_1$ and $\rho_2$ are $1$-linked, then
$j(\rho_1, \rho_2)$ is the largest number $r$ such that $\rho_1$
and $\rho_2$ are $r$-linked. Otherwise we set $j(\rho_1, \rho_2) =
0$.
\enddefinition

\begremark{} Elman and Lam only use the term ``$r$-linked''. In
our context the adjective ``quadratic'' is necessary since in
$\Char k=2$ we also can, and should, inquire about {\it bilinear}
\Pf s which divide both $\rho_1$ and $\rho_2$, see below.
\endremark

\begtheorem{3.}{\rm (cf. [EL, Prop.4.4] for $\Char k \ne
2$).}\fonote{Elman and Lam do not require the assumption that
$j(\rho_1, \rho_2) \geq 1$, cf. our Theorem~6 below.}
Assume
that the \q\ \Pf s $\rho_1, \rho_2$ are anisotropic and
$1$-linked. Then the form $\rho_1  \perp  -\rho_2$ has Witt index
$$\ind (\rho_1 \; \perp \; -\rho_2) \; = \; 2^r$$
with $r \colon = j (\rho_1, \rho_2) \geq 1$. Furthermore, if $\tau$
is a \q\ $s$-fold \Pf\ with $1 \leq s < r$, dividing both $\rho_1$ and $\rho_2$,
then there exists an
$(r-s)$-fold bilinear \Pf\ $\gamma$ and bilinear \Pf s $\sigma_1,
\sigma_2$ such that
$$\rho_1
\; \cong \; \sigma_1 \otimes \gamma \otimes \tau, \qquad
\rho_2 \; \cong \; \sigma_2 \otimes \gamma \otimes \tau.$$
\endtheorem

\begproof{}
We assume that we have factorizations $\rho_1
\cong \sigma_1 \otimes \tau$, $\rho_2 \cong \sigma_2 \otimes \tau$
where $\tau$ is a \q\ \Pf\ and $\sigma_1$ and $\sigma_2$ are
bilinear \Pf s. We verify the following two statements from which
the theorem readily follows:
\smallskip
\setitemitemindent{$(ii)$}
\itemitem{$(i)$} $\ind (\rho_1 \perp \rho_2) \geq \dim \tau$.
\smallskip

\itemitem{$(ii)$} If $\ind (\rho_1 \perp - \rho_2) > \dim \tau$,
then there exists an $a \in k^*$ and bilinear Pfister forms $\gamma_1, \gamma_2$
such that $$\rho_1 \; \cong \; \gamma_1 \otimes \langle 1, a \rangle
\otimes \tau, \qquad \rho_2 \; \cong \; \gamma_2 \otimes
\langle 1, a \rangle \otimes \tau.$$

\noi $(i)$: We have $\rho_1 \, \cong \; \tau \; \perp \; \sigma'_1
\otimes \tau$, $\rho_2 \; \cong \; \tau \; \perp \; \sigma'_2
\otimes \tau$, and thus $$\rho_1 \; \perp \; - \rho_2 \; \cong \;
\tau \otimes \langle 1, - 1 \rangle \; \perp \; \sigma'_1 \otimes
\tau \; \perp \; (- \sigma'_2 \otimes \tau).$$

\noi $(ii)$: Now let $\ind (\rho_1 \perp \rho_2) > \dim \tau$. By
our calculation this shows that the form $\sigma'_1 \otimes \tau
\perp  (- \sigma'_2 \otimes \tau)$ is isotropic. Hence there
exists an $a \in k^*$ which is represented by both of the forms $\sigma'_1 \otimes
\tau$ and $\sigma'_2 \otimes \tau$. By Lemma~1 we have
factorizations
$$\sigma_1 \otimes \tau \; \cong \; \gamma_1
\otimes \langle 1, a \rangle \otimes \tau, \qquad  \sigma_2
\otimes \tau \; \cong \; \gamma_2 \otimes \langle 1, a \rangle
\otimes \tau$$
with bilinear \Pf s $\gamma_1, \gamma_2$, as required.\qed
\endproof


Let us now deal with the problems of divisibility by a given {\it
bilinear} \Pf. Unfortunately there apparently does not (yet) exist
a good counterpart to Theorem~1 above. There is one for Theorem~2
however. To derive it we need a further lemma.

\beglemma{2.} Let $\rho$ be a \q\ and $\sigma$ a bilinear \Pf\
over $k$. Assume that the form $\rho$ is \an. Let $\tau = \left[
{1 \atop 1} \; {1 \atop b} \right]$ be a binary \Pf\ over $k$ with
$$\tau \; \perp \; \sigma' \otimes [1] \; < \; \rho.$$
Then there exists a bilinear \Pf\ $\alpha$ over $k$ such that
$$\rho \; \cong \; \alpha \otimes \sigma \otimes \tau.$$
\endlemma

\begproof{} Assume that the dimension of $\sigma$ is $2^d$. Then
the \Pf\ $\mu \colon = \sigma \otimes \tau$ is of dimension
$2^{d+1}$. Let $E = (E, q)$ be a \qs\ corresponding to $\mu$ and
let $K \colon = k(\mu)$. The space $K \otimes E$ is hyperbolic.
Hence it contains a $2^d$-dimensional vector subspace $U$ with
$q(U) = 0$. Let $V$ be a subspace of $K \otimes E$ corresponding
to the subform $(\tau \perp \sigma' \otimes [1])_K$ of $\mu_K$. It
has dimension $2^d +1$. From the obvious inequality $$\dim (V \cap
U) + \dim E \; \geq \; \dim V + \dim U$$ we see that $\dim (V \cap
U) \geq 1$. Hence the quadratic space $V$ is isotropic. Thus $\rho
\otimes K$ is isotropic and so $\rho \otimes K \sim 0$. Now
Theorem~2 tells us that $\rho \cong \alpha \otimes \mu$ for some
bilinear \Pf\ $\alpha$.\qed
\endproof

\begtheorem{4.} Let $\sigma$ be a bilinear \Pf\ and $\rho$ an
anisotropic \q\ \Pf\ over $k$ with $\sigma \otimes [1] < \rho$.
Then there exists a \q\ \Pf\ $\tau$ over $k$ such that $\rho \cong
\sigma \otimes \tau$.
\endtheorem

\begproof{} Let $E = (E, q)$ be a \qs\ over $k$ which corresponds
to the form $\rho$, and let $V$ be a subspace of $E$ which
corresponds to the subform $\sigma \otimes [1]$ of $\rho$. We have
an orthogonal decomposition $V = ke_0 \perp V'$ with $q(e_0) = 1$
and $V'$ a \ql\ space corresponding to the form $\sigma' \otimes
[1]$. The bilinear form $B = B_q$ vanishes on $V \times V$.
There is a linear form $s$ on $V$ with $s(e_0) = 1$, $s(V') = 0$.
Furthermore there is a vector $f_0 \in E$ with $s(v) = B(v, f_0)$
for all $v \in V$. The subspace $V + k f_0$ of $E$ corresponds to
the \qf\ $\left[ {1 \atop 1} \; {1 \atop b} \right] \perp \sigma'
\otimes [1]$. This form is thus a subform of $\rho$. The theorem
now follows from Lemma~2.\qed
\endproof

\begdefinition{3.} If $\vf$ is an \an\ bilinear form over $k$ and
$\dim \varphi > 1$, then we define $k(\varphi) \colon = k(\varphi
\otimes [1])$. If $\varphi = \langle a_1, \dots, a_n \rangle$,
then $k(\varphi)$ is the quotient field of the ring $k[t_1, \dots,
t_n] /(a_1 t^2_1 + \cdots + a_n t_n^2)$. Note that the polynomial
$\varphi (t) = a_1 t_1^2 + \cdots + a_n t_n^2$ is irreducible
since $\vf$ is \an.
\enddefinition

\begtheorem{5.} Let $\sigma$ be an \an\ bilinear \Pf\ and $\rho$
an an\-iso\-trop\-ic \q\ \Pf\ over $k$. Assume that $\sigma \ne
\langle 1 \rangle$. The following statements are equivalent.
\smallskip
\setitemindent{$(iii)$}
\item{$(i)$} There exists a \q\ \Pf\ $\tau$ over $k$ with
$\rho \cong \sigma \otimes \tau$.
\smallskip
\item{$(ii)$} $\sigma \vert \rho$.
\smallskip
\item{$(iii)$} There exists an $a \in k^*$ with $\sigma \otimes [a] <
\rho$.
\smallskip
\item{$(iv)$} $\rho \otimes k (\sigma) \sim 0$.
\smallskip
\item{$(v)$} There exists a place $\lambda \colon \; k(\rho) \to
k(\sigma) \cup \infty$.
\endtheorem

\begproof{} The implications $(i)$ $\Rightarrow$ $(ii)$
$\Rightarrow$ $(iii)$ $\Rightarrow$ $(iv)$ and $(iv)$
$\Leftrightarrow$ $(v)$ are obvious.

$(iv)$ $\Rightarrow$ $(iii)$: Let $t = (t_1, \dots, t_n)$ be a
tuple of $n \colon = \dim \sigma$ unknowns  over $k$. We can
treat $\sigma (t) \in k[t]$ as a monic polynomial. By the Norm
Theorem, $(iv)$ implies that $\sigma(t)$ is a similarity norm of
$\rho_{k(t)}$ and thus that $\sigma (t)$ is represented
by$\rho_{k(t)}$. The Subform Theorem then gives us $\sigma \otimes
[1] < \rho$.

$(iii)$ $\Rightarrow$ $(i)$: We have $a \in D(\rho)$, hence $\rho
\cong a \rho$, and so $\sigma \otimes [1] < \rho$. By Theorem~4 we
then get $\rho \cong \sigma \otimes \tau$ for some \q\ \Pf\
$\tau$.\qed
\endproof

Starting from Theorem~4 we would also like to establish a
counterpart to Theorem~3, dealing with bilinear \Pf s  that
simultaneously divide two given \q\ \Pf s. We need a further lemma
for this purpose.

\beglemma{3.} Let $\sigma$ be a bilinear \Pf\ and $\rho$ an \an\
\q\ \Pf\ over $k$. Furthermore let $a \ne 0$ be an element of $k$
with
$$\sigma \otimes [1] \; \perp \; [a] < \rho.$$
Then there exists a
\q\ \Pf\ $\tau$ over $k$ such that
$$\rho \; \cong \; \sigma
\otimes \langle 1, a \rangle \otimes \tau.$$
\endlemma

\begproof{} Let $n \colon = \dim \sigma$ (a $2$-power), and let
$(t, u) = (t_1, \dots, t_n, u_1, \dots, u_n)$ be a tuple of $2n$
unknowns  over $k$. The form $\sigma_{k(t, u)}$ is round
(\S16, Theorem~1). Hence there exists an $n$-tuple $v \in k(t,
u)^n$ with
$$\sigma (t) + a \sigma (u) \; = \; \sigma (u) (\sigma (v) + a).$$
By our assumption the elements $\sigma
(u)$ and $\sigma (v) + a$ are represented by $\rho_{k(t, u)}$.
Since $\rho_{k(t, u)}$ is round as well, $\sigma (t) + a \sigma
(u)$ is represented by  $\rho_{k(t, u)}$. By the Subform Theorem
we then get
$$\sigma \otimes \langle 1, a \rangle \otimes [1] \; < \;
\rho\ ,$$
and we can conclude the proof by applying Theorem~4.\qed
\endproof

\begdefinition{4.} (a) Let $r \in {\Bbb N}_0$. We call two \q\ \Pf
s $\rho_1$, $\rho_2$ over $k$ {\it bilinearly $r$-linked}\Idx{bilinearly r-linked Pfister forms@bilinearly $r$-linked Pfister forms}  
 if there
exist a bilinear $r$-fold \Pf\ $\sigma$ and \q\ \Pf s $\tau_1$,
$\tau_2$ over $k$ such that $$\rho_1 \; \cong \; \sigma \otimes
\tau_1, \qquad  \rho_2 \; \cong \; \sigma \otimes \tau_2.$$

(b) We define the {\it bilinear linkage number\/}\Idx{bilinear linkage number}\Idx{111ir1r2@$i(\rho_1, \rho_2)$}  
$i (\rho_1,
\rho_2)$ of $\rho_1$ and $\rho_2$ as follows: If $\rho_1$ is
neither divisible by $\rho_2$, nor $\rho_2$ divisible by $\rho_1$,
then $i (\rho_1, \rho_2)$ is the largest number $r$ such that
$\rho_1$ and $\rho_2$ are bilinearly $r$-linked. In the cases
$\rho_1 \vert \rho_2$ and $\rho_2 \vert \rho_1$ we let $i(\rho_1,
\rho_2)$ be the number $d$ such that $2^d = \min (\dim \rho_1,
\dim \rho_2)$.
\enddefinition

\begremark{} If $\Char k\ne 2$, it follows immediately from
Definitions~2 and 4 that $i (\rho_1, \rho_2)$ coincides with the
quadratic linkage number $j(\rho_1, \rho_2)$, which we then simply
call the {\it linkage number} of $\rho_1$ and $\rho_2$.
\endremark

\begtheorem{6.} {\rm([EL, Prop.4.4] for $\Char k \ne 2$).} Let
$\rho_1$ and $\rho_2$ be two \an\ \q\ \Pf s over $k$.
\smallskip
\setitemindent{$(ii)$}
\item{$(i)$} The Witt index $\ind (\rho_1 \; \perp \; -\rho_2)$ is
a $2$-power $2^r$ with $r \geq j (\rho_1, \rho_2)$ and $r \geq i
(\rho_1, \rho_2)$. If $j (\rho_1, \rho_2) > 0$, then $r =
j(\rho_1, \rho_2)$. Otherwise $r = i (\rho_1, \rho_2)$.
\smallskip
\item{$(ii)$} Let $\sigma$ be a bilinear $s$-fold \Pf\ with $0
\leq s < r$ that divides $\rho_1$ and $\rho_2$, i.e.
$$\rho_1 \cong \sigma \otimes \tau_1,\qquad \rho_2 \cong \sigma \otimes \tau_2$$
with \q\ \Pf s $\tau_1, \tau_2$ $($cf. Theorem~$2)$. Then there exists
either an $(r-s)$-fold bilinear \Pf\ $\gamma$ or an $(r-s)$-fold
\q\ \Pf\ $\gamma$ such that $\sigma \otimes \gamma$ divides both
\Pf s $\rho_1$ and $\rho_2$, i.e. either
$$\rho_1 \cong \sigma
\otimes \gamma \otimes \mu_1,\qquad \rho_2 \cong \sigma \otimes \gamma
\otimes \mu_2$$
with \q\ \Pf s $\mu_1, \mu_2$, or
$$\rho_1 \cong
\alpha_1 \otimes \sigma \otimes \gamma,\qquad \rho_2 \cong \alpha_2
\otimes \sigma \otimes \gamma$$
with bilinear \Pf s $\alpha_1, \alpha_2$.
\endtheorem

\begproof{} We proceed as in the proof of Theorem~3 and assume
that we have factorizations $\rho_1 \cong \sigma \otimes \tau_1$,
$\rho_2 \cong \sigma \otimes \tau_2$ with \q\ \Pf s $\tau_1,
\tau_2$ and a bilinear \Pf\ $\sigma$. $\{$ We allow $\sigma =
\langle 1 \rangle$.$\}$ We now want to verify the following two
statements which, together with Theorem~4, will establish the
proof.
\smallskip
\setitemindent{$(b)$}
\item{$(a)$} $\ind (\rho_1 \perp -\rho_2) \geq \dim \sigma$.
\smallskip
\item{$(b)$} If $\ind (\rho_1 \perp \rho_2) > \dim \sigma$, then
either there exists an $a \in k^*$ and quadratic Pfister forms
$\mu_1, \mu_2$ such that
$$\rho_1 \; \cong \; \sigma \otimes \langle 1,
a \rangle \otimes \mu_1, \qquad \rho_2 \; \cong \; \sigma
\otimes \langle 1, a \rangle \otimes \mu_2, \leqno(1)$$
or there exist over $k$ a quadratic Pfister form $\mu$ of
dimension $2$ and bilinear Pfister forms $\alpha_1, \alpha_2$ such
that
$$\rho_1 \; \cong \; \alpha_1 \otimes \sigma \otimes \mu, \qquad
\rho_2 \; \cong \; \alpha_2 \otimes \sigma \otimes \mu.
\leqno(2)$$

\noi Let $(E_1, q_1)$ and $(E_2, q_2)$ be \qs s associated to
$\rho_1$ and $\rho_2$, and let $(E, q) \colon = (E_1, q_1) \perp
(E_2, - q_2)$. The corresponding vector space decomposition $E =
E_1 \oplus E_2$ gives rise to projections $p_1 \colon E
\longrightarrow E_1$, $p_2 \colon E \longrightarrow E_2$.

We have $[1] < \tau_1$ and $[1] < \tau_2$, and thus
$$\sigma
\otimes [1, -1] < \sigma \otimes \tau_1 \; \perp \; \sigma \otimes (-
\tau_2) \; \cong \; \rho_1 \; \perp \; - \rho_2.$$
Hence $E$ contains a subspace $U$ with
$q(U) = 0$, $\dim U = \dim \sigma$, such that the form $\sigma \otimes
[1]$ corresponds to $p_1 (U)$ as a subspace of $(E_1, q_1)$ as
well as to $p_2(U)$ as a subspace of $(E_2, q_2)$. To see this,
note that $p_i$ $(i = 1, 2)$ maps the vector space $U$ linearly
isomorphic to $p_i (U)$ since $E_1$ and $E_2$ are anisotropic as
subspaces of $(E, q)$.

Since $E$ contains the totally isotropic subspace $U$, it is
already clear that $\ind (q_1 \perp - q_2) \geq \dim \sigma$.

Assume now that $\ind (q_1 \perp -q_2) > \dim \sigma$. Then $(E,
q)$ contains a totally isotropic subspace $V = U + kv$ of
dimension $1 + \dim \sigma$. Again we know that $p_i$ maps the
vector space $V$ linearly isomorphic to $p_i (V)$ $(i = 1, 2)$,
and for arbitrary $x \in V$ we have:
$$q_1 (p_1 (x))\; = \; q_2 (p_2 (x)),$$
and thus also for $x, y \in V$:
$$B_1 (p_1 (x), p_1 (y)) \; = \;
B_2 (p_2 (x), p_2 (y)),$$
where $B_i$ denotes the bilinear form $B_{q_i}$ corresponding to
$q_i$. We distinguish two cases:

{\it Case 1:} $B_1 (p_1 (v), p_1 (u)) = 0$. Now $\sigma \perp [a]$
corresponds to $p_1 (V)$ with respect to $q_1$ and also to  $p_2
(V)$ with respect to $q_2$ where $a \colon = q_1 (p_1 (v)) = q_2
(p_2 (v))$. Thus $\sigma \perp [a]$ is a subform of both $\rho_1$
and $\rho_2$. By Lemma~3 we obtain (1) with \q\ \Pf s $\mu_1$ and
$\mu_2$.

{\it Case 2:} $B_1 (p_1 (v), p_1 (u)) \ne 0$. We choose an
orthogonal basis $e_1, \dots, e_n$ of $p_1 (U)$ \wrt\  $q_1$ with
$B_1 (e_1, p_1 (v)) = 1$, $B_1 (e_i, p_1 (v)) = 0$ for $2 \leq i
\leq n$, which is certainly possible.  Then, corresponding to $p_1
(V)$ \wrt\ $q_1$ and also to $p_2 (V)$ \wrt\ $q_2$, we have the
form
$$\left[ {a_1 \atop
1} \; {1 \atop b_1} \right] \; \perp \; [a_2, \dots, a_n]$$
with $b_1 \colon = q_1 (p_1 (v))$ and $a_i = q_1 (e_i)$, and
we have
$$\sigma \otimes [1] \; \cong \; [a_1, a_2, \dots, a_n].$$
Now by \S 6, Lemma 2,
$$\left[ {a_1 \atop 1} \; {1 \atop
b_1} \right] \; \cong \; \langle a_1 \rangle \otimes \left[ {1
\atop 1} \; {1 \atop b} \right]$$
with $b \colon = a_1b_1$. Since the three forms $\sigma$,
$\rho_1$, $\rho_2$ are round, it follows that
$$\left[ {1 \atop 1} \; {1
\atop b} \right] \; \perp \; [a_2 a_1^{-1}, \dots, a_n a_1^{-1} ]
\; < \; \rho_1$$
for $i = 1$ and $i = 2$, and also that
$$[1, a_2 a_1^{-1} , \dots,
a_n a_1^{-1}] \; \cong \; \sigma \otimes [1].$$
Hence
$$\left[ {1 \atop 1} \; {1 \atop b} \right] \;
\perp \; \sigma' \otimes [1] \; < \; \rho_i$$
for $i = 1$ and $i = 2$.
Now Lemma~2 tells us that the factorizations~(2) above hold with
$\mu = \left[ {1 \atop 1} \; {1
\atop b} \right]$ and bilinear Pfister forms $\alpha_1$ and
$\alpha_2$.\qed
\endproof

\begproblem{} Let $\Char k = 2$, $a \in k^*$, and let $\rho_1,
\rho_2$ be $2$-fold anisotropic \q\ \Pf s with $\langle 1, a
\rangle \vert \rho_1$ and $\langle 1, a \rangle \vert \rho_2$.
Does there always exist a \qf\ $\tau = \left[ {1 \atop 1} \; {1
\atop b} \right]$ over $k$ with $\tau \vert \rho_1$ and $\tau
\vert \rho_2$?
\endproblem

The author has not yet been able to give either a positive or a negative
answer to this question. If the answer is ``yes'', then by
Theorem~6 it is immediate that for two \an\ \q\ \Pf s $\rho_1,
\rho_2$ over $k$ we {\it always} have that
$$\ind (\rho_1 \; \perp
\; - \rho_2) \; = \; 2^{j (\rho_1, \rho_2)}.$$

Finally we remark that R.~Baeza proved several theorems in this
section, or parts of them, more generally over semi-local rings
instead of fields, see in particular [Ba$_1$, Chap.IV, \S 4].
As our proofs rely on the Norm Theorem, the Subform Theorem, and
elements of generic splitting, it is till this day not possible to
preserve them over semi-local rings. Thus Baeza's proofs
differ from the ones we presented here. They are substantially
more ``constructive'' than ours. Therefore they deserve also over
fields  an interest for their own sake, just as the proofs of Elman and
Lam in [EL] when the characteristic is different from two.
