\titlea{\S19}{A Norm Theorem}
\leftrunningtitle{Chapter III. Some Applications}
%\leftrunningtitle{Specialization of Quadratic and Symmetric
%Bilinear Forms}
\rightrunningtitle{\S19 A Norm Theorem}

Let $\vf$ be a (symmetric) bilinear or \qf\ over a field $k$. We
are interested in the {\it norm group} $N(\vf)$ of $\vf$, which is
the group of similarity norms of $\vf$. It consists of all $a\in
k^*$ with $\varphi \cong a \varphi$ and was already introduced in
$\S16$.

If $\vf$ is a \qf, then we have a decomposition
$$\varphi = \widehat{\varphi}
\perp r \times [0]$$
with $\widehat{\varphi}$ \nd\ and $r \in {\Bbb N}_0$ (cf. \S
6). Clearly, $N(\varphi) = N (\widehat{\varphi})$. Therefore we
may assume without loss of generality that $\vf$ is \nd. By an analogous argument we
may also assume this  in the bilinear case.

Thus we always assume that $\vf$ is \nd\ in what follows. In the
quadratic case we have a Witt decomposition
$$\varphi \; \cong \; \varphi_0 \perp r \times H$$
with $\varphi_0$ anisotropic, $r \in {\Bbb N}_0$. Clearly, $N(\varphi) =
N(\varphi_0)$. Thus we may always assume that $\vf$ is
anisotropic in the quadratic case. In the bilinear case and when
$\Char k=2$ we only have a stable isometry
$$\varphi \; \approx \; \varphi_0 \perp r \times \tilde{H},$$
with  $\varphi_0$ anisotropic, $r \in {\Bbb N}_0$ and $\vf_0$ is up
to isometry uniquely determined by $\vf$. Now
$N(\varphi)$ is a subgroup of $N(\varphi_0)$ and is possibly
different from $N(\varphi_0)$. In what follows we will almost
always assume that $\vf$ is anisotropic.

Let $K \colon = k(t)$ be the rational function field over $k$ for
a set of indeterminates $t = (t_1, \dots, t_r)$. We want to
compare the norm groups of $\varphi$ and $\varphi \otimes K$.
Clearly, $N(\varphi)$ is a subgroup of $N(\varphi \otimes k(t))$.
Furthermore, $N(\varphi \otimes k(t))$ contains the group $K^{*2}$
of all squares $f^2$ with $f \in K$, $f \ne 0$. Thus $N(\varphi \otimes
K)$ is obtained from the group $N(\varphi) K^{*2}$ by adjoining
certain square-free polynomials $f(t) \in k[t]$, i.e. polynomials $f(t) \ne
0$ which are products of pairwise  different irreducible
polynomials. Our task will now be to determine these polynomials
$f(t)$.

We endow the monomials $t_1^{\alpha_1} \dots t_r^{\alpha_r} \quad
(\alpha_i \in {\Bbb N}_0)$ with the lexicographic ordering \wrt\
the sequence of variables $t_1 \geq t_2 \geq \dots \geq t_r$.
Thus, if $(\alpha_1, \dots, \alpha_r) \in {\Bbb N}_0^r$ and $(\beta_1, \dots, \beta_r)
\in {\Bbb N}_0^r$ are two multi-indices, then the monomial $t_1^{\alpha_1} \dots
t_r^{\alpha_r}$ precedes the monomial $t_1^{\beta_1} \dots
t_r^{\beta_r}$ if there exists a $j \in \{ 1, \dots, r \}$ with
$\alpha_i = \beta_i$ for $i <j$, but $\alpha_j < \beta_j$.

For every polynomial $f \in k[t]$ with $f \ne 0$, let $f^* \in
k^*$ denote the leading coefficient of $f$, i.e. the coefficient
of the highest monomial occurring in $f$. Clearly we have $(fg)^* = f^*
g^*$. We call the polynomial $f$ {\it monic} if $f \ne 0$ and $f^* =
1$.

The following theorem tells us, among other things, that we may
restrict ourselves to {\it monic} square-free polynomials in the
task above.

\begtheorem{1.} Let $\vf$ be anisotropic. Let $f(t)\in k[t]$ be a
similarity norm of $\varphi \otimes K$ and let $t_1^{m_1} \dots
t_r^{m_r}$ be the highest monomial occurring in $f$. Then the
exponents $m_i$ are all even and $f^*$ is a similarity norm of
$\vf$.
\endtheorem

\begproof{} By induction on $r$. Let $r=1$. Write $t_1=t$. Let $m \colon = \deg
f$. We use the place $\lambda \colon k (t)
\to k \cup \infty$ over $k$ with $\lambda (t) = \infty$. $\{$Note:
the associated valuation ring is $k[t^{-1}]_{(t^{-1})}.\}$

We choose an element $a \in D (\varphi)$. Then $a f \in D (\varphi
\otimes k(t))$. The form $\varphi \otimes k (t)$ has good
reduction \wrtl\ and $\lambda_* (\varphi \otimes k (t)) = \varphi$
is anisotropic. By \S15, Theorem 3, there exists a $\xi \in
k(t)^*$ with $\lambda (af \xi^2)$ finite and $\ne 0$. This shows
that the degree $m$ of $f$ is even. It follows that
$$\varphi \otimes k(t) \; \cong \; \langle t^{-m} f \rangle \otimes (\varphi
\otimes k(t)). \leqno(\dagger)$$
Now $\lambda (t^{-m} f) = f^*$. In particular, $\langle t^{-m} f
\rangle$ has good reduction \wrtl. Applying $\lambda_*$ to
($\dagger$) gives
$$\varphi \; \cong \; \langle f^* \rangle \otimes \varphi.$$
This proves the theorem for $r=1$. Assume that $r>1$. We write
$t= (t_1, t')$ with $t' = (t_2, \dots, t_r)$ and then have $K = K'(t_1)$
with $K' \colon = k(t_2, \dots, t_r)$. Applying the result for
$r=1$ to the anisotropic form $\varphi \otimes K'$, we get that
$m_1$ is even and that the highest coefficient $h \in k[t']$ of
$f$, as a polynomial in $t_1$, is a similarity norm of $\varphi \otimes
K'$. By the induction hypothesis, $m_2, \dots, m_r$ are also even
and $h^* =f^*$ is a similarity norm of $\vf$.\qed
\endproof

If $p(t) \in k[t]$ is an {\it irreducible} polynomial, then $k(p)$
denotes the quotient field of $k[t]/(p)$, i.e. the function field
of the affine hyperplane $p(t) = 0$.

\begtheorem{2}{{\rm (Norm Theorem, [K$_4$] for bilinear $\vf$,
[Ba$_4$] for quadratic $\vf$)}{\bf.}}\Idx{Norm Theorem} 
Assume that the form $\vf$ is
anisotropic. In the quadratic case, assume moreover that $\vf$ is
strictly regular. Further, let $p(t) \in k[t] = k[t_1, \dots,
t_r]$ be a monic irreducible polynomial. Then the following are equivalent.
\smallskip
\setitemindent{$(iii)$}
\item{$(i)$} $p(t) \in N (\varphi \otimes k(t))$.
\smallskip
\item{$(ii)$} $p(t)$ divides a square-free polynomial $f(t) \in k[t]$ with $f(t)
\in N (\varphi \otimes k(t))$.
\smallskip
\item{$(iii)$} $\varphi \otimes k(p) \sim 0$.
\endtheorem


Under the same assumptions as in the theorem, we immediately
obtain the following

\begcorollary{} A polynomial $f(t) \in k[t]$, $f(t) \ne 0$ is a
similarity norm of $\varphi \otimes k(t)$ if and only if $f^* \in N (\varphi)$
and $\varphi \otimes k(p) \sim 0$ for every monic irreducible
polynomial $p \in k[t]$ such that the exponent of $p$ in $f$ is
odd.
\endcorollary

The rest of this section will be devoted to the proof of
Theorem~2. In the main part of the proof (showing that $(iii) \Rightarrow
(i)$) we will use induction on $r$. A priori we note that if the
theorem holds for a given fixed $r \in {\Bbb N}$, then the
corollary also holds for this $r$.

The implication $(i) \Rightarrow (ii)$ is trivial. In order to
prove $(ii)\Rightarrow (iii)$, we use the place $\lambda_p \colon k(t) \to k(p)
\cup \infty$ over $k$ with $\lambda_p (p) = 0$ (cf. \S~18; there we write
$k_p$ instead of $k(p)$). By assumption we have
$$\varphi \otimes k(t) \; \cong \; \langle f \rangle \otimes (\varphi \otimes
k(t)).$$
To this isometry we apply the map $(\lambda_p)\lower5pt\hbox{$\scriptstyle
W$}$, i.e. weak specialization \wrt\ $\lambda_p$, as introduced in \S3 in the
bilinear case and in \S7 in the quadratic case. We obtain
$\varphi \otimes k(p) \sim 0$, as desired.

Now we come to the prove of the implication $(iii)\Rightarrow
(i)$, which will keep us busy for a longer time. Suppose that
$r=1$ and let $\vf$ be an anisotropic {\it bilinear} form. As
before we write $t \colon = t_1$. We use the exact sequence from
\S18,
$$0 \longrightarrow W(k) \buildrel i \over \longrightarrow W(k(t)) \buildrel
\partial \over \longrightarrow \bigoplus\limits_{q \in {\Cal P}} W(k(q))
\longrightarrow 0,$$
with retraction
$$(\lambda_{\infty})_* \colon \; W(k(t)) \longrightarrow W(k)$$
\wrt\ the place $\lambda_{\infty} \colon \; k(t) \to k \cup \infty$
over $k$, which maps $t$ to $\infty$. Let\fonote{Now we will often
write $\varphi_{k(t)}$ instead of $\vf \otimes k(t)$ etc.}
$$z \colon \; = \; \langle 1, - p \rangle \otimes
\varphi_{k(t)} \; \in \; W(k(t)).$$
Clearly, $\partial_q (z) = 0$ for every monic irreducible
polynomial $q \ne p$. Furthermore, $\partial_p (z)$ is the Witt
class of $- \varphi \otimes k(p)$ which is also zero by
assumption.

Assume that the polynomial $p$ has degree $m$. By a well-known
theorem of Springer (e.g. [S,~Th.5.3]) which also holds for
bilinear forms in characteristic two by \S10, Theorem~1\fonote{For $\vf$ one
considers the quadratic form $\vf\otimes [1]$.},
$\varphi \otimes
E$ is anisotropic for every finite extension $E/k$ of odd degree.
Since $\varphi \otimes k(p) \sim 0$ by assumption, $m$ has to be
even. Hence
$(\lambda_{\infty})\lower5pt\hbox{$\scriptstyle W$} (z) = 0$ and
it follows from \S18 that $z=0$. Since $\vf$ is anisotropic, this
gives
$$\varphi_{k(t)} \; \cong \; \langle p \rangle \otimes \varphi_{k(t)} \; ,$$
as required.

Since we never proved Springer's Theorem for $\Char k\not=2$, we
will show again that $m$ has to be even, using a different
argument.

We have seen that $\partial (z) = 0$. Hence, by \S18, there exists an
anisotropic form $\psi$ over $k$ with $z \sim \psi_{k(t)}$. Assume
that $m$ is odd. Then $(\lambda_{\infty})\lower5pt\hbox{$\scriptstyle W$} (z) =
(\lambda_{\infty})_* (\varphi_{k(t)}) = \varphi$. On the other
hand, $(\lambda_{\infty})\lower5pt\hbox{$\scriptstyle W$} (z) =
(\lambda_{\infty})_* (\psi_{k(t)}) = \psi$. Therefore, $\psi =
\varphi$. We obtain
$$\langle 1, - p \rangle \otimes \varphi_{k(t)} \; \sim \; \varphi_{k(t)},$$
and so $\langle - p \rangle \otimes \varphi_{k(t)} \sim 0$. This
is absurd since $\varphi_{k(t)}$ is anisotropic. Thus $m$ has to
be even. This proves the theorem for $r=1$ in the bilinear case.

Let us now consider the case where $r=1$ and $\vf$ is an
anisotropic strictly regular \qf\ over $k$. By assumption we have $\varphi
\otimes k(p) \sim 0$. From this we can deduce that $p$ is a
similarity norm of $\varphi \otimes k(t)$. In case $k$ has
characteristic $\neq 2$, we already know this since $\vf$ can then
be interpreted as  a bilinear form.

So, suppose that $\Char k=2$. We want to ``lift'' $\vf$ to a form over a
field of characteristic zero in the easiest way possible, cf.
[Ba$_4$]. For this purpose we choose a henselian discrete
valuation ring $A$ with maximal ideal ${\nam m} = 2 A \ne 0$,
generated by $2$, and residue class field $A/{\nam m} = k$. That
this is possible is well-known, cf.~[E], [EP]. The quotient field
$K \colon = \Quot\, A$ has characteristic zero.

We choose elements $a_1, \dots, a_n$ in $A$ such that
$$p(t) \colon \; = \; t^n + \overline{a}_1 t^{n-1} + \dots + \overline{a}_n,$$
where $\overline{a}_i$ of course denotes the image of $a_i$ in $A/{\nam m} =
k$. The polynomial
$$\tilde{p} (t) \colon \; = \; t^n + a_1 t^{n-1} + \dots + a_n$$
is irreducible in $A[t]$ and thus also irreducible in $K[t]$. Let
$$A(\tilde{p}) \colon \; = \; A[t]/(\tilde{p} (t)).$$
This is a henselian discrete valuation ring with maximal ideal $2 A
(\tilde{p})$ and residue class field
$$A(\tilde{p})/2A (\tilde{p}) \; = \; k[t]/(p(t)) \; = \; k(p).$$
The valuation ring $A(\tilde{p})$ has $K(\tilde{p})$ as its
quotient field.

Let us use the geometric language now. Suppose that the form $\vf$
corresponds to the \qs\ $(E, q)$ over $k$. Here we may choose $E =
M/{\nam m} M$ for some free $A$-module $M$ of rank $m$. We equip
$M$ with a strictly regular \qf\ $\tilde{q} \colon
M \to A$ in such a way that $(E, q)$ is the reduction of $(M, \tilde{q})$ modulo ${\nam
m}$, which is possible. Let $\tilde{E} \colon = K \otimes_A M$.
Our form $\tilde{q}$ extends to a strictly regular \qf\
$\tilde{q}_K \colon \tilde{E} \to K$. This gives us a strictly
regular \qs\ $\tilde{E} \colon = (\tilde{E}, \tilde{q}_K)$ over
$K$ which has good reduction \wrt\ the canonical place $\lambda \colon K \to k \cup
\infty$ associated to $A$, and whose specialization $\lambda_*
(\tilde{E})$ is equal to the given space $E = (E, q)$ over $k$.
By assumption we have $k(p) \otimes E \sim 0$.

We may interpret the canonical place $\mu \colon \tilde{K} (p) \to k(p) \cup
\infty$ associated to the valuation ring $A(\tilde{p})$ as an
extension of the place $\lambda$. By \S8, Theorem~8, the space
$\tilde{K}(p) \otimes_{\tilde{K}}
\tilde{E}$ again has good reduction \wrt\ $\mu$ and
$$\mu_* (K(p) \otimes \tilde{E} ) \; = \; \lambda_*
(k(p) \otimes \tilde{E}) \; = \; k(p)
\otimes E \; \sim \; 0.$$
From the \qs\ $M = (M, \tilde{q})$ over $A$ we obtain a \qs\
$A(\tilde{p}) \otimes_A M$ over $A(\tilde{p})$. The statement
$\mu_* (k(p) \otimes \tilde{E}) \sim 0$ tells us that the
reduction of $A(\tilde{p}) \otimes_A M$ to the maximal ideal of
$A(\tilde{p})$ is hyperbolic. Since $A(\tilde{p})$ is henselian,
$A(\tilde{p}) \otimes_A M$  itself is hyperbolic and then
$$K(\tilde{p}) \otimes_K \tilde{E} \; = \; K(\tilde{p}) \otimes_{A(\tilde{p})} (A
(\tilde{p}) \otimes_A M)\; \sim \; 0.$$
Since $\Char K\neq 2$, it follows from what we proved above (the
bilinear case, $r=1$) that $\tilde{p} (t)$ is a similarity norm of
$K(t)\otimes_K \tilde{E}$, i.e.
$$K(t) \otimes_K \tilde{E}\; \cong \; \langle \tilde{p} \rangle \otimes (K(t)
\otimes_K \tilde{E}). \leqno(\ast)$$
We want to specialize this isometry \wrt\ a suitable place from
$K(t)$ to $k(t)$. Let $S \subset A[t]$ be the multiplicatively
closed set of all polynomials $f \in A[t]$ for which at least one
coefficient is not in the maximal ideal ${\nam m} = 2 A$
(``unimodular'' polynomials). We construct the localization
$A(t) \colon = S^{-1} A[t]$ of $A[t]$ at this set. One can easily
verify that $A(t)$ is a discrete valuation ring with maximal ideal
$2 A (t)$, residue class field $A(t)/2A(t) = k(t)$ and quotient
field $\Quot\, A(t) = K(t)$.
Let $\rho \colon K(t) \to k(t) \cup\infty$ be the canonical place
associated to the valuation ring $A(t)$. It extends the canonical
place $\lambda \colon K \to k \cup \infty$ associated to the
valuation ring $A$. Clearly the \qs\ $K(t) \otimes_K \tilde{E}$
has good reduction \wrt\ $\rho$ and
$$\rho_* (K(t) \otimes_K \tilde{E}) \; = \; k(t) \otimes_k \lambda_*
(\tilde{E}) \; = \;
k(t) \otimes_k E.$$
Applying $\rho_W$ to the isometry ($\ast$) above gives
$$k(t) \otimes_k E \; \cong \; \langle p \rangle \otimes (k(t) \otimes_k E).$$
This shows that $p$ is a similarity norm of $\varphi \otimes
k(t)$ and we have completely proved the theorem for $r=1$.

To conclude, we prove the implication $(iii) \Rightarrow (i)$ for
$r>1$, simultaneously for $\vf$ bilinear or quadratic. We may suppose
that the theorem, and thus also the corollary, is valid for
$r-1$ variables. Also, we already established that the
implications $(i) \Rightarrow (ii) \Rightarrow (iii)$ hold for $r$
variables. Next we will prove the implication $(ii) \Rightarrow
(i)$ for $r$ variables and only then the implication $(iii)
\Rightarrow (i)$.

Thus, let $t = (t_1, t_2, \dots, t_r) = (t_1, t')$ with $t' = (t_2, \dots,
t_r)$. Let $p \in k[t]$ be monic and irreducible and let $f \in k[t] \setminus \{ 0
\}$. Assume that the exponent of $p$ in $f$ is odd.
We have to show that $p$ is a similarity norm of $\varphi_{k(t)}$.

Suppose first that the field $k$ contains infinitely many
elements. Let $n \colon = \deg_{t_1} p$ be the degree of $p$, as a
polynomial over $k[t']$. We distinguish  the cases $n = 0$, $n >
0$.

Let $n=0$. We have a decomposition
$$f(t) = p(t') h(t)$$
in $k[t]$. We choose an element $c$ in $k$ such that $p(t')$ in
$k[t']$ does not divide the polynomial $h(c, t')$. This is
possible since $\vert k \vert = \infty$. $\{$One considers the
image of $h(t)$ in $(k [t']/(p)) [t_1]$.$\}$ By the Substitution
Principle (\S3, Theorem~5 in the bilinear case, \S11, Theorem~3 in
the quadratic case) we have $f(c, t') \in N (\varphi_{k(t')})$. By
the implication $(ii) \Rightarrow (i)$ for $r-1$ variables, it
follows that
$p = p(t') \in N (\varphi_{k(t')})$ and thus also $p \in N
(\varphi_{k(t)})$.

Now let $n>0$. Let $a(t')$ be the highest coefficient of $p(t)$, as
a polynomial in $t_1$. The polynomial $a(t') \in k[t']$ is monic
and $\tilde{p} \colon = a^{-1} p \in k(t') [t_1]$ is a monic
polynomial in the variable $t_1$. In the ring $k(t') [t_1]$,
the exponent of $\tilde{p}$ in $f$ is odd. Thus, by the
established case $r=1$, $\tilde{p}$ is a similarity norm of
$\varphi_{k(t')}$, and hence also of $\varphi_{k(t)}$. Therefore,
$$\langle p \rangle \otimes \varphi_{k(t)} \; \cong \; \langle a \rangle \otimes
\varphi_{k(t)}. \leqno(\ddag)$$
Thus, it suffices to show that $a$ is a similarity norm of
$\varphi_{k(t')}$. Let $\pi (t') \in k[t']$ be a monic irreducible
polynomial whose exponent in $a$ in the ring $k[t']$ is odd (if
such a polynomial $\pi$ exists at all). We will show that $\pi$ is
a similarity norm of $\varphi_{k(t')}$ and know then that  $\pi$ is
indeed a similarity norm of $\varphi_{k(t)}$.

$\pi (t')$ does not divide any of the coefficients of $p(t)$ as a
polynomial in $t_1$. Since $\vert k \vert = \infty$, there exists
an element $c$ in $k$ with $\pi (t') \nmid p(c, t')$ in the ring
$k[t']$. By the Substitution Principle, we obtain from ($\ddag$)
that
$$\langle p(c, t') \rangle \otimes \varphi_{k(t')} \; \cong \; \langle a(t')
\rangle \otimes \varphi_{k(t')},$$
and so
$$a(t') p(c, t') \; \in \; N (\varphi_{k(t')}).$$
The exponent of $\pi (t')$ in $a(t') p(c, t')$ is odd. From the
implication $(ii) \Rightarrow (i)$ for $r-1$ variables it follows
that $\pi \in N (\varphi_{k(t')})$ and we are finished.

Now suppose that $k$ is a finite field. We consider the field
$k(u)$ with $u$ an indeterminate. Upon applying what we proved
before to $\varphi_{k(u)}$ and $p, f$, considered as polynomials
over $k(u)$, we see that $p$ is a similarity norm of
$\varphi_{k(u)}$. By the Substitution Principle we see that $p$ is
a similarity norm  of $\vf$, upon specializing $u$ to $0$. This completely
proves the implication $(ii)\Rightarrow (i)$ for $r$ variables.

Finally, we come to the proof of the implication $(iii)\Rightarrow
(i)$ for $r$ variables. As before, let $a(t')$ be the highest
coefficient of $p(t)$ as a polynomial in $t_1$ and let
$\tilde{p} \colon = a^{-1} p$. Then $k(p) = k(t') (\tilde{p})$ and
by our assumption we have $\varphi \otimes k(p) \sim 0$. The
implication $(iii) \Rightarrow (i)$ in the case $r=1$ tells us
that $\tilde{p}$ is a similarity norm of $\varphi_{k(t)}$. Hence,
$a(t') p(t)\in N (\varphi_{k(t)})$. By the implication $(ii)
\Rightarrow (i)$ for $r$ variables, proved above, $p(t)$ is a
similarity norm of $\varphi_{k(t)}$. The proof of Theorem~2 is now
complete.
