\titlea{\S17}{The Subform Theorem}
\leftrunningtitle{Chapter III. Some Applications}
%\leftrunningtitle{Specialization of Quadratic and Symmetric
%Bilinear Forms}
\rightrunningtitle{\S17 The Subform Theorem}


The so-called Subform Theorem of Cassels and Pfister (e.g. [S,
Th.3.7]) is one of the most important tools in the algebraic
theory of quadratic forms over fields of characteristic not two.
In 1969/70, Pfister's Egyptian student Mahmud Amer proved
the Subform Theorem in characteristic two in his
dissertation in a manner
which does not leave anything to be desired. Unaware of this fact,
R.~Baeza proved a slightly more special version of the Subform
Theorem in characteristic two  in 1974 [Ba$_3$]. Unfortunately Amer's proof
was never published. In Pfister's book [P] one can find a sketch
of Amer's proof which leaves a gap however. Since the Subform
Theorem will play an important role in what follows,
and  will furnish us with a description of all strictly
regular forms of height one in
particular, we will now present Amer's proof in its completeness.
Apart from this,  the paper [Ba$_3$] on this topic
is also worth reading in our opinion.

\begtheorem{1}{\rm (``Cassels's Lemma'')\bf{.}}
Let $k$ be a field of arbitrary characteristic, $x$ an
indeterminate over $k$ and $p \in k [x]$ a polynomial. Let $\vf$
be a quadratic form in $n$ variables over $k$ which represents the
polynomial $p$ over the field $k(x)$. Then $\vf$ already represents
the polynomial $p$ over the ring $k[x]$, i.e. there exist
polynomials $f_1, \dots, f_n \in k [x]$ with $\varphi (f_1, \dots, f_n) = p$.
\endtheorem

\begproof{} 1) We may assume that $p\not=0$. We may further assume
that $\vf$ is \nd\ upon replacing $\vf$ by $\widehat\vf$ (cf. \S6,
Definition 9). If $\vf$ is isotropic, then
$\left[ {0 \atop 1} \; {1 \atop 0} \right]$ is a subform of $\vf$
(cf. \S6, Lemma~4). Clearly $\left[ {0 \atop 1} \; {1 \atop 0}
\right]$ already represents every polynomial over $k[x]$. Thus we
assume from now on that $\vf$ is anisotropic.

2) We choose a representation
$$p = \varphi \left( \frac{f_1}{f_0}, \frac{f_2}{f_0}, \dots, \frac{f_n}{f_0}
\right) \leqno(1)$$
with $f_i \in k [x] \; (0 \leq i \leq n)$, $f_0 \ne 0$ and $\deg f_0$ minimal.
(``$\deg$'' denotes the degree of a polynomial.) We assume that $\deg f_0
> 0$. This should lead to a contradiction.

Consider the form $\psi \colon = \langle - p \rangle \, \bot \, \varphi$
in $n+1$ variables over $k(x)$. For every polynomial $f_i$ with $1 \leq i
\leq n$, let
$$f_i = f_0 g_i + r_i,$$
the division with remainder of $f_i$ by $f_0$, thus with $g_i, r_i \in k[x]$, $\deg
r_i < \deg f_0$.

It can happen that $r_i = 0$, i.e. that $\deg r_i = - \infty$.
However, for at least one $i \in \{ 1, \dots, n \}$ we have $r_i \ne
0$. We introduce the vectors $f \colon
= (f_0, f_1, \dots, f_n)$ and $g \colon = (g_0, g_1, \dots, g_n)$
with $g_0 \colon = 1$.  By (1) we have $\psi (f) = 0$. By the
minimality of $\deg f_0$ we certainly have $\psi (g) \ne 0$
however. In particular, the vectors $f$ and $g$ are linearly
independent over $k(x)$.

In what follows we will denote the ``inner product'' $B_{\psi} (f,
g)$ simply by $fg$ and we will use such a  notation (as in Amer) also for forms
different from $\psi$. Let
$$u \colon = \psi (g) f - (fg)g.$$
We have
$$\psi (u) = \psi (g)^2 \psi (f) + (fg)^2 \psi (g) - (fg)^2 \psi (g) =
0,$$
since $\psi (f) = 0$. From the linear independence of $f$ and $g$
it follows that $u\neq 0$.

The form $\psi$ is also isotropic over $k(x)$. Hence the first
coordinate $u_0$ of $u$ is likewise different from zero. We have
$$u_0 = \psi (g) f_0 - fg =
               \frac{1}{f_0} [ \psi (g) f_0^2 - (fg) f_0 ] =
               \frac{1}{f_0} \psi (f-f_0 g).$$
Every coordinate of $f - f_0 g = (0, r_1, \dots, r_n)$ is of lower
degree than $f_0$. Hence $\deg \psi (f-f_0 g) \leq 2 \deg f_0 - 2$
and so $\deg u_0 \leq \deg f_0 - 2$. Since $\psi (u) = 0$ we thus
obtained a representation of $p$ of the form (1) in which the
denominator polynomial $u_0$ is of lower degree than $f_0$,
contradiction!\qed
\endproof

\begaddendum {\rm{(Amer)}\bf.} In the situation of Theorem~1, let
$\vf$ be anisotropic and $p\neq 0$. Furthermore, let $M$ be a
subset of $\{ 1, \dots, n \}$ and $p =
\varphi (h_1, \dots, h_n)$ a representation with $h_i \in k (x)$ for $1 \leq i
\leq n$ and even $h_i \in k [x]$ for $i \in M$. Then there exist
polynomials $g_1, \dots, g_n \in k [x]$ with $p = \varphi (g_1, \dots, g_n)$
and $g_i = h_i$ for $i \in M$.
\endaddendum

\begproof{} We choose a representation
$$p = \varphi \left( \frac{f_1}{f_0}, \dots, \frac{f_n}{f_0} \right)\leqno(2)$$
with
$f_i \in k [x] \quad (0 \leq i \leq n)$, $f_0 \ne 0$, $\frac{f_i}{f_0} = h_i$
for $i \in M$ and $\deg f_0$ minimal. By way of contradiction,
suppose again that $\deg f_0 > 0$. Using these polynomials $f_i$, we go through the proof above
again, starting from step~2. This time we have $r_i = 0$ and $g_i = h_i$
for every $i \in M$. For $i\in M$, the vector $u$ above has $i$-th
coordinate
$$\eqalign{u_i &= \psi (g) f_i - (fg) g_i = \psi (g) f_0 h_i - (fg) h_i
\cr
               &= [\psi (g) f_0 - (fg)]h_i = u_0 h_i.}$$
Thus $\frac{u_i}{u_0} = h_i$ for every $i \in M$. By the above
considerations we get $\deg u_0 < \deg f_0$, contradiction!\qed
\endproof

\begremark{} If $\Char k\ne 2$, then the Addendum is dispensable
since it can easily be obtained from Theorem~1 as follows: as
usual we interpret $\vf$ as a quadratic form over the $k$-vector
space $V:=k^n$ with standard basis $e_1, \dots, e_n$. Suppose
without loss of generality that $M = \{ 1, \dots, m
\}$ for an $m \leq n$ and $W \colon = \sum\limits_{i = 1}^m k e_i$.
The form $\varphi_1 \colon = \varphi \vert W$ is anisotropic and
thus strictly regular. We have an orthogonal decomposition $\varphi \cong \varphi_1 \, \bot \,
\varphi_2$. Now $\varphi_2$ represents $f_1 \colon = p - \varphi_1 (h_1, \dots,
h_m)$ over $k(x)$ and thus over $k[x]$ by Theorem~1. This yields
the representation of $p$ by $\vf$, as given in the Addendum.
\endremark

\begtheorem{2}{\rm (Subform Theorem)\bf{.}}\Idx{Subform Theorem}
Let $\psi$ be an anisotropic \qf\ over $k$ and let $\vf$ be a \nd\
\qf\ over $k$ in $m$ variables. Let $x_1, \dots, x_m$ be
indeterminates over $k$. Suppose that the form $\psi$ represents
the element $\varphi (x_1, \dots, x_m)$ over the field $k
(x_1, \dots, x_m)$. Then $\vf$ is a subform of $\psi$.
\endtheorem

The proof involves several lemmata. We assume all the time that
$\psi$ is an $n$-dimensional anisotropic \qf\ over $k$ and that
{\it $k$ has characteristic two.} For $\Char k\neq 2$ one can find
the proof of the Subform Theorem in many books, in particular in
[P], [S].

\beglemma{1.} Let $a, b, c \in k$, $b \ne 0$ and let $x$ be an
indeterminate over $k$. Suppose that $\psi$ represents the element
$ax^2 + bx + c$ over $k(x)$. Then
$\left[ {a \atop b} \; {b \atop c} \right] < \psi$.
\endlemma

\begproof{} By Theorem~1, there exists a vector $f \in k [x]^n$ with $\psi (f)
= ax^2 + bx + c$. Since $\psi$ is anisotropic, every coordinate of
$f$ has to be a polynomial of degree $\leq 1$. Hence $f = xu + v$,
for certain vectors $u, v \in k^n$. Again using the notation $uv \colon = B_{\psi} (u,
v)$, we then have the equation
$$\psi (u) x^2 + (uv) x + \psi (v) = ax^2 + bx + c.$$
Therefore, $\psi (u) = a$, $\psi (v) = c$ and $uv = b$. Hence the value matrix of
$\psi \vert ku + kv$ is  $\left[ {a \atop b} \; {b \atop c}
\right]$.\qed
\endproof

\beglemma{2.} Let $E = (E, q)$ be an anisotropic \qm\ over $k$ and
let $u,v$ be vectors in $E$ with $q(u) = q(v) \ne 0$. Then there
exists an isometry $\sigma \colon E \buildrel \sim \over \longrightarrow E$
with $\sigma (u) = v$ and $\sigma (v) = u$.
\endlemma

\begproof{} Suppose without loss of generality that $u \ne v$. Then $u + v \ne 0$, so that  $q(u+v) = 2 q
(u) + uv = uv \ne 0$. $\{$As before, we write $uv$ instead of $B_q (u, v)$.$\}$
The quadratic submodule $ku+kv$ of $E$ is therefore strictly
regular and it follows that
$$E = (ku + kv) \, \bot \, F$$
with $F \colon = (ku + kv)^{\bot}$ (cf. \S 5, Lemma 1). The linear
map $\sigma \colon E \to E$ with $\sigma (u) = v$, $\sigma (v) = u$, $\sigma (w) =
w$ for all $w \in F$ is the one we are looking for.\qed
\endproof

\begremark{} The map $\sigma$ is nothing else than the reflection
in the hyperplane $(u + v)^{\bot}$, perpendicular to the vector $u + v \ne 0$
(cf. [Bo$_1$, \S 6, No. 4]),
$$\sigma (z) = z - \frac{(u+v)z}{q(u+v)} (u+v),\qquad \hbox{\rm for all } z \in E.$$
\endremark

\beglemma{3.} Let $\psi = \varphi \, \bot \, \chi$ be an
orthogonal decomposition with $\vf$ strictly regular. Let $\gamma$
be a further \qf\ over $k$ with $\varphi \, \bot \, \gamma <
\psi$. Then $\gamma < \chi$.
\endlemma

This lemma is a special case of a much more general theorem, which
deserves attention in its own right. We will prove this theorem at the
end of this section.

\beglemma{4.} Let $\psi = \varphi \, \bot \, \chi$ with $\dim \varphi =
2$ and $\varphi$ strictly regular. Let $x_1, x_2$ be
indeterminates over $k$ and $d$ an element $\ne 0$ of $k$. Suppose
that $\psi$ represents the polynomial $\varphi (x_1, x_2) + d$
over $k(x_1, x_2)$. Then $d \in D(\chi)$.\fonote{One recalls
the notation in \S16, Definition~2.}
\endlemma

\begproof{} Let $\varphi = \left[ {a \atop b} \; {b \atop c} \right]$ and
$\dim \psi = n$. We have $b \ne 0$. Therefore, evaluating $\psi$
at the first two standard vectors $e_1 = (1, 0, \dots, 0)$,
$e_2 = (0, 1, 0, \dots, 0)$ of the vector space $k^n$ yields the
values $\psi (e_1) = a$, $\psi (e_2) = c$, $e_1 e_2 = b$. Let us
now work in the vector space $V = K^n$ over the field $K \colon =
k(x_2)$ which we also equip with the form $\psi$. (More precisely,
we are dealing with the form $\psi \otimes K$ here, but we simply
write $\psi$.) By our assumption, $\psi$ represents the element
$$\varphi (x_1, x_2) + d = ax_1^2 + (b x_2) x_1 + (cx_2^2 + d)$$
over $K (x_1)$. Lemma~1 tells us now that there exist vectors $e'_1,
h'$ in $V$ with
$\psi (e'_1) = a$, $\psi (h') = cx_2^2 + d$, $e'_1 h' = bx_2$. By
Lemma~2 there is an isometry $\sigma \colon V \buildrel \sim \over
\longrightarrow V$ \wrt\  $\psi$ with $\sigma (e'_1) = e_1$. Letting $h \colon =
\sigma (h') \in V$, we then have
$$\psi (e_1) = a, \quad \psi (h) = cx^2_2 + d, \quad e_1 h = bx_2. $$
If $h_1, \dots, h_n$ are the coordinates of $h$, i.e. $h = \sum\limits_{i =
1}^n h_i e_i$, then the last equation gives $e_1 h = h_2 b = x_2 b$,
and so $h_2 = x_2$. $\{$Bear in mind that $e_i e_j = 0$ for $i \leq 2$, $j > 2$,
and $uu = 0$ for every $u \in V$.$\}$ By the Addendum
there exists a vector $g = (g_1, \dots, g_n) \in k[x_2]^n$ with $g_2 = h_2 = x_2$
and $\psi(g) = cx_2^2 + d$. Since $\psi$ is anisotropic, all
polynomials $g_i$ have degree $\leq 1$. Thus, $g = x_2 u + v$ with vectors $u
= (u_1, \dots, u_n)$, $v = (v_1, \dots, v_n)$ in $k^n$ and $u_2 = 1$, $v_2 =
0$. We have
$$cx_2^2 + d = \psi (u) x^2_2 + (uv) x_2 + \psi (v),$$
so that $\psi (u) = c$, $uv = 0$ and $\psi (v) = d$. Furthermore,
$e_1 u = u_2
b = b$, $e_1 v = v_2 b = 0$. Therefore $\psi \vert k e_1 + ku\; \cong
\left[ {a \atop b} \; {b \atop c} \right]$ and $v \in (k e_1 + ku)^{\bot}$.
This shows that
$$\left[ {a \atop b} \; {b \atop c} \right] \, \bot \, [d] \quad < \quad \psi =
\left[ {a \atop b} \; {b \atop c} \right] \, \bot \, \chi.$$
By Lemma~3 we get $[d] < \chi$.\qed
\endproof

\beglemma{5.} Let $\psi = \varphi \, \bot \, \chi$ with $\vf$
strictly regular, $\dim \varphi = 2r$ and let $x_1, \dots, x_{2r}$
be indeterminates over $k$. Further, let $d$ be an element $\ne 0$ of
$k$ such that $\psi$ represents the element $\varphi (x_1, \dots, x_{2r}) +
d$ over $k(x_1, \dots, x_{2r})$. Then $d \in D(\chi)$.
\endlemma

\begproof{} After a linear coordinate transformation of the
variables $x_1, \dots, x_{2r}$ we may assume that
$$\varphi = \left[ {a_1 \atop b_1} \; {b_1 \atop c_1} \right] \, \bot \dots
\bot \, \left[ {a_r \atop b_r} \; {b_r \atop c_r} \right]$$
with elements $a_i, b_i, c_i \in k$, $b_i \ne 0$. The form $\psi$
represents the element $\sum\limits_{i = 1}^r (a_i x^2_{2i-1} +
b_i x_{2i-1} x_{2i} + c_i x_{2i}^2) + d$ over
$k(x_1, \dots, x_{2r})$. The statement now follows from Lemma~4 by
induction on $r$.\qed
\endproof

\beglemma{6.} Let $\varphi = [d_1, \dots, d_m]$ be an anisotropic
quasi-linear form over $k$. Assume that there is a form $\psi$
which represents the element $\varphi
(x_1, \dots, x_m) = d_1 x_1^2 + \dots + d_m x_m^2$ over the field
$k (x_1, \dots, x_m)$ with indeterminates $x_1, \dots, x_m$ over
$k$. Then $\varphi < \psi$.
\endlemma

\begproof{} By induction on $m$. Let $m=1$. Then $\psi$ represents
the element $d_1$ over $k(x_1)$. Using the Substitution Principle
\S11, Theorem~3, it follows that $\psi$ represents the element
$d_1$ over $k$ as well.

Suppose now that $m > 1$. By the Substitution Principle we
immediately get that $d_m \in D (\psi)$. We choose a vector $u\in
k^n$ with $\psi (u) = d_m$. Let $d \colon = d_1 x_1^2 + \dots + d_{m-1}
x^2_{m-1}$. By Theorem~1, $\psi$ represents the element $d + d_m
x_m^2$ over $K [x_m]$ with $K \colon = k (x_1, \dots,
x_{m-1})$. Since $\psi$ is anisotropic, we have vectors $w', u' \in K^n$
with $\psi (x_m u' + w') = d_m x_m^2 + d$. This implies
$$\psi (u') = d_m, \; \psi (w') = d, \; u'w' = 0.$$
By Lemma~2, the vector space $K^n$ has an isometry
$\sigma \colon K^n \buildrel \sim \over \longrightarrow K^n$ \wrt\
$\psi$ with $\sigma (u') = u$. Letting $w \colon = \sigma (w') \in
K^n$, we then have
$$\psi (u) = d_m, \; \psi (w) = d, \; uw = 0.$$
Thus, the form $\psi_1 \colon = \psi \vert (ku)^{\bot}$ represents
the element $d$ over $K$, and so $[d_1, \dots, d_{m-1}] <
\psi_1$, by our induction hypothesis. Hence there is a system of
linearly independent vectors $v_1,
\dots, v_{m-1}$ in $(ku)^{\bot}$ with $\psi (v_i) = d_i \quad (1 \leq i \leq
m-1)$ and $v_i v_j = 0$ for $i \ne j$.
The vector $u$ cannot be a linear combination of $v_1, \dots, v_{m-1}$,
for if there was a relation $u = a_1 v_1 + \dots
+ a_{m-1} v_{m-1}$ with coefficients $a_i \in k$, applying $\psi$
to this relation would give $d_m = a_1^2 d_1 + \dots + a_{m-1}^2
d_{m-1}$, contradicting the anisotropy of $\vf$. Thus
$W \colon = kv_1 + \dots + k v_{m-1} + ku$  is  an $m$-dimensional
subspace of $k^n$ and $\psi \vert W \cong [d_1, \dots, d_m]$.\qed
\endproof

Theorem~2 now follows in a straightforward way from Lemmas~5 and
6.
We  conclude this section with the generalization of Lemma~3, announced above.

\begtheorem{3.} Let $E$ and $F$ be free quadratic modules over a
valuation ring ${\nam o}$ and let $G$ be a strictly regular
quadratic space over ${\nam o}$. If $G \, \bot \, F < G \, \bot \, E$,
then $F <E$.
\endtheorem

\begproof{} We work in the qua\-drat\-ic module $U \colon = G \bot
E$ with associated quadratic form $q$. By our assumption, $U$
contains a submodule $V$, which is a direct summand of $U$ (as an ${\nam
o}$-module), so that $V = (V, q \vert V)$ has an orthogonal
decomposition $V = G' \, \bot \, F'$ with $G \cong G'$, $F \cong
F'$. Since $G'$ is strictly regular, we have a further orthogonal
decomposition $U = G' \, \bot \, E'$. This gives $E \cong E'$ by
the Cancellation Theorem \S6, Theorem~2. Hence, there is an
isometry
$\sigma \colon U \buildrel \sim \over \longrightarrow U$ with $\sigma (G') =
G$ and $\sigma (E') = E$. Since $F'$ is orthogonal to $G'$, $F'' \colon =
\sigma (F')$ has to be orthogonal to $G$. Therefore, $F'' \subset
E$. The ${\nam o}$-module $\sigma (V) = G \, \bot \, F''$ is a
direct summand of $U = G \, \bot \, E$, and so the ${\nam
o}$-module $E/F'' \cong U/\sigma (V)$ is free. Thus $F''$ is a
direct summand of $E$, and so $F < E$.\qed
\endproof

\begremark{} The statement of Theorem~3 remains valid for a local
ring ${\nam o}$ because the Cancellation Theorem \S6, Theorem~2
holds more generally over local rings.
\endremark
