\titlea{\S30}{Realization of Quadratic Places; Generic Splitting of Specialized Forms  in Characteristic $\not=2$}
\leftrunningtitle{Chapter IV. Specialization with Respect to Quadratic Places}
%\leftrunningtitle{Specialization of Quadratic and Symmetric
%Bilinear Forms}
\rightrunningtitle{\S30 Realization of Quadratic Places $\ldots$}%; Generic Splitting of Specialized Forms  in Characteristic $\not=2$}

We consider fields of arbitrary characteristic and want to investigate ``realizations'' of a quadratic place in the sense of the following definition. 

\begdefinition{1.} Let  $\Lambda \colon K \to L \cup \infty$ be a quadratic place.
A {\it realization of}\Idx{realization of a Q-place@realization of a $Q$-place} 
$\Lambda$ (as an ordinary place) is a pair $(i, \mu)$ consisting of a field extension  $i \colon K \hookrightarrow E$ and a place
$\mu \colon E \to L \cup \infty$ with the property $\Lambda
\subset \widehat{\mu} \vert K$. $\{$Here $\widehat{\mu}$ denotes the quadratic place determined by  $\mu$, cf.  \S 28.$\}$ The realization is called
{\it strict} when actually $\Lambda =
\widehat{\mu} \vert K$.\Idx{strict realization of a Q-place@strict realization of a $Q$-place}
\enddefinition

Strict realizations are not easy to find, but in general the following theorem holds which suffices for our generic splitting theory. 

\begtheorem{1.}  For every quadratic place $\Lambda \colon K \to L
 \cup  \infty$ there exists a realization   $(i \colon K
\hookrightarrow E, \mu)$ with $E$ purely transcendental over $K$.
\endtheorem

For the proof we need a lemma, which follows easily from general principles pertaining to the extension of valuations, cf.  [Bo$_2$, \S 2, n$^{\circ}$ 4, Prop.3 and \S 8, n$^{\circ}$ 3, Th.1].

\beglemma{1.} Let $K$ be a quadratic field extension of a field
$E$, $K = E (\alpha)$ with $\alpha^2 = a \in E$, $\alpha
\not\in E$. Let $\rho \colon E \to L \cup \infty$ be a place with
$\rho (ac^2) = 0$ or $\infty$ for all $c \in E$. Then 
$\rho$ extends uniquely to a place  $\lambda
\colon K \to L \cup \infty$.\qed
\endlemma

\begproof{of Theorem 1.} Let  $\Lambda = (\lambda, H, \chi)$, ${\nam
o} \colon = {\nam o}_{\lambda}$ and ${\nam m}$ the maximal ideal of the valuation ring ${\nam o}$. We choose a subgroup $P$ of $Q(K)$ with $Q(K) = P \times Q ({\nam o})$ and a family $(\pi_i \; \vert \; i \in I)$ in ${\nam m}$ such that
the square classes  $\langle \pi_i \rangle$, $i \in I$, form a basis
of the ${\Bbb F}_2$-vector space $P$. Finally we choose families of indeterminates
$(t_i \; \vert \; i \in I)$ and $(u_i \; \vert \; i \in I)$ over the fields $K$ and $L$ respectively. 

Let  $E'\colon = K (t_i \; \vert \; i \in I)$ and $F \colon = L (u_i \;
\vert \; i \in I)$. The place $\lambda \colon K \to L \cup \infty$
extends uniquely to a place
$\tilde{\lambda} \colon E' \to F \cup \infty$ with $\tilde{\lambda}
(u_i) = t_i$ for every $i \in I$, cf. [Bo$_2$, \S 10, Prop.2].
$\{$Note:  with $\tilde{{\nam o}} \colon = {\nam o}_{\tilde{\lambda}}$
and $\tilde{{\nam m}}$ the maximal ideal of $\tilde{{\nam o}}$
we have $\tilde{{\nam o}}/\tilde{{\nam m}} \cong ({\nam o}/{\nam m})
(\overline{t_i} \; \vert \; i \in I)$, where $\overline{t_i}$ denotes the image 
of $t_i$ in $\tilde{{\nam o}}/\tilde{{\nam m}}$.
The family $(\overline{t_i} \; \vert \; i \in I)$
is algebraically free over ${\nam o}/{\nam m}$. The natural
map $K^*/{\nam o}^* \to E^{\prime *}/\tilde{{\nam o}}^*$ is an
isomorphism.$\}$

For every $i \in I$ and $f \in E'$ we have $\tilde{\lambda} (\pi_i
f^2) = 0$ or $\infty$. Let $E \colon = E' (\sqrt{\pi_i t_i^{-1}}
\; \vert \; i \in I)$. This field is again purely transcendental over
$K$. It follows from Lemma~1  (possibly with Zorn's lemma)
that $\tilde{\lambda}$ extends uniquely to a place
$\rho \colon E \to F \cup
\infty$. (We have $\rho (\sqrt{\pi_i t_i^{-1}}) = 0$ for every $i
\in I$.) Now $\langle \pi_i \rangle_E = \langle t_i \rangle_E$.
Thus $\langle \pi_i \rangle_E$ has GR with respect to $\rho$ and $\rho_*
(\langle \pi_i \rangle_E) = \langle u_i \rangle$ for every $i \in
I$.

Finally we choose for every $i \in I$ an element $a_i$
of $L$ with $\chi ( \langle \pi_i \rangle) = \langle a_i \rangle$.
There exists a place $\sigma \colon F \to L \cup \infty$ over $L$,
in general not uniquely determined, with $\sigma (u_i) = a_i$ for every
$i$. For every such place $\sigma$ we have 
%$(\sigma \circ\rho)^{\wedge} \; \vert \; K \supset \Lambda$, 
$\widehat{\,\sigma\circ\rho\,}\vert K\supset \Lambda$,
which can be verified easily.\qed
\endproof

In what follows we require that {\it all fields have characteristic $\ne 2$}. As before, a ``form'' is a nondegenerate (symmetric) bilinear form, which is now the same as  a nondegenerate quadratic form.

As indicated at the end of \S 29, we want to study the generic splitting of quadratically specialized forms $\Lambda_*
(\varphi)$ of  forms with almost good reduction in a manner, similar to what was done in \S4 for  specialized forms $\lambda_* (\varphi)$ of  forms with good reduction. For this purpose we need a supplement to Theorem~2 in \S 4 (which itself is a special case of \S 9, Theorem~4) about generic zero fields.

\begtheorem{3}{\rm [K$_5$, Th.3.3.ii].}  Let $\gamma \colon k \to L \cup
\infty$ be a place and $\varphi$ a form over $k$ with bad $($$=$ not good\/$)$
reduction with respect to $\gamma$. Assume there exists an element $c
\in D (\varphi)$ with $\gamma (c) \ne 0, \infty$. Then
$\gamma$ extends to a place $\lambda \colon k (\varphi)
\to L \cup \infty$.
\endtheorem

\begproof{} We may assume that  $\varphi = \langle a_1,
\dots, a_n \rangle$ with $n \geq 2$, $\gamma (a_i) \ne \infty$
for all $i$, $\gamma (a_1) \ne 0$, $\gamma (a_n c^2) = 0$ or
$\infty$ for every $c \in k$. We can write $k(\varphi) =
k (x_1, \dots, x_n)$ with the only relation $\sum\limits_1^n a_i
x_i^2 = 0$. The elements $x_1, \dots, x_{n-1}$ are algebraically independent over
$k$ and $k(\varphi)$ is a quadratic extension
$E(x_n)$ of the purely transcendental extension $E = k (x_1, \dots,
x_{n-1})$ of $k$.

Let $u_1, \dots, u_{n-1}$ be indeterminates over $L$ and $F \colon = L
(u_1, \dots, u_{n-1})$. Then $\gamma$ has a unique extension $\mu
\colon E \to F$ with $\mu (x_i) = u_i$ $(i = 1, \dots, n-1)$. We
want to show that $\mu (x_n^2 z^2) = 0$ or $\infty$ for every
$z \in E$. By  Lemma~1 above we then know that 
$\mu$ extends to a place $\rho \colon k (\varphi) \to F \cup
\infty$. By composing  $\rho$ with a place $\sigma \colon F \to L \cup
\infty$ over $L$, we obtain a place $\lambda =
\sigma \circ \rho \colon k (\varphi) \to L \cup \infty$, which extends
$\gamma$.

Let $z \in E^*$.  We write $z = dfg^{-1}$ with $d \in k$
and polynomials $f, g \in k [x_1, \dots, x_{n-1}]$ whose 
coefficients are all in the valuation ring ${\nam o}$ of
$\gamma$, but not all in the maximal ideal of 
${\nam o}$. Then $\mu (f)$ and $\mu (g)$ are finite and nonzero.
We obtain
$$\mu (x_n^2 z^2)  =  - [\gamma (a_1) u_1^2 + \dots + \gamma (a_{n-1})
u^2_{n-1}] \mu (f)^2 \mu (g)^{-2} \gamma (a_n^{-1} d^2).$$
On the right hand side all factors, except the last one, are finite and nonzero. In contrast,  $\gamma (a^{-1}_n d^2) = 0$ or $\infty$. Therefore the same is true for
$\mu (x_n^2 z^2)$. \qed
\endproof

\begtheorem{4.} Let $\gamma \colon k \to L \cup \infty$ be a place and
$\varphi$ a form over $k$ which has AGR with respect to $\widehat{\gamma}$.
Assume that the form  $\widehat{\gamma}_* (\varphi)$ is isotropic. Then there exists
a place $\lambda \colon k (\varphi) \to L \cup \infty$, which extends
$\gamma$.
\endtheorem

\begproof{} Let  $\Gamma \colon = \widehat{\gamma}$. If $\varphi$ has
GR with respect to $\gamma$ then our statement is covered by 
Theorem~2 from \S 4, since  $\widehat{\gamma}_* (\varphi) = \gamma_*
(\varphi)$ in this case. Assume now that $\varphi$ has bad reduction with respect to
$\gamma$. We choose a subgroup $P$
of $Q(K)$ with $Q(k) = Q ({\nam o}_{\gamma}) \times P$. Let
$$\varphi  \cong  \raise5pt\hbox{$\newbot\limits_{\alpha \in P}$} \; \alpha
\; \varphi_{\alpha}$$
be a $\Gamma$-modular decomposition of  $\varphi$. This implies that
all forms $\varphi_{\alpha}$ are  $\gamma$-unimodular. If
$\varphi_1 \ne 0$ then  Theorem~3 tells us that $\gamma$ extends to a place from $k(\varphi)$ to $L$. 

Finally assume that $\varphi_1 = 0$. We have $\dim
\varphi = \dim \Gamma_* (\varphi) \geq 2$. Thus there certainly exists an
$\alpha_0 \in P$ with $\varphi_{\alpha_0} \ne 0$. Since $\varphi$
has AGR, we have $\gamma_* (\varphi_{\alpha_0}) \sim 0$. The form
$\tilde{\varphi} \colon = \alpha_0 \varphi$ has  AGR with respect to $\Gamma$,
since  $\varphi_1 = 0$ and thus $\gamma_* (\varphi_1) = 0$.
($\gamma_* (\varphi_1) \sim 0$ would suffice.)
We have  $k(\varphi) = k (\tilde{\varphi})$. $\Gamma_* (\tilde{\varphi})$ has
an  orthogonal summand $\gamma_* (\varphi_{\alpha_0})$ and is thus isotropic. 
By the above $\gamma$ extends to a place
from $k(\tilde{\varphi})$ to $L$.\qed
\endproof

Now we can prove a theorem, crucial for a generic splitting theory of specialized forms. It is the converse of  \S 29, Theorem~3 in characteristic $\ne 2$.

\begtheorem{5.} Let $\Gamma \colon k \to L \cup \infty$ be a 
$Q$-place and  $\varphi$ a form over $k$ which has AGR with respect to
$\Gamma$. Assume that the form  $\Gamma_* (\varphi)$ is isotropic. Let
$K \supset k$ be a generic zero field. Then there exists a  $Q$-place
$\Lambda \colon K \to L \cup \infty$, which extends $\Gamma$ $\{$i.e., with $\Lambda\vert k \supset \Gamma$, cf. \S $29$, Definition~$3$$\}$.
\endtheorem

\begproof{} Following Theorem~1 we choose a purely transcendental extension
$E \supset k$ together with a place
$\mu \colon E \to L \cup \infty$ such that $\widehat{\mu} \vert k
\supset \Gamma$. By  \S 29, Theorem~2,  $\varphi \otimes E$ has AGR
with respect to $\widehat{\mu}$ and $\widehat{\mu}_* (\varphi \otimes E) =
\Gamma_* (\varphi)$. Therefore $\widehat{\mu}_* (\varphi \otimes E)$ is
isotropic. The free compositum $K \cdot E$ of $K$ and $E$ over $k$
is specialization equivalent to $k(\varphi) \cdot E =
E (\varphi \otimes E)$ over $E$. By  Theorem~4 we can extend $\mu$ to a
place $\rho \colon K \cdot E \to L \cup \infty$. Let $\Lambda
\colon K \to L \cup \infty$ be the restriction of the  $Q$-place
$\widehat{\rho}$ to $K$. Clearly,
$$\Lambda \vert k  =  (\widehat{\rho}\, \vert K) \vert k  =  \widehat{\rho}\, \vert k
 =  (\widehat{\rho}\, \vert E) \vert k,$$
and also $\widehat{\rho}\, \vert E \supset \widehat{\mu}$. Therefore
$$\Lambda \vert k  \supset  \widehat{\mu} \vert k  \supset  \Gamma.\eqno\square$$
\endproof

With this theorem and \S 29, Theorem~3, we have all the ingredients at our disposal  to replicate the arguments of \S4 about generic splitting of a specialized form.
If we traverse the proof of \S 4, Theorem~3 and replace the words ``place'' and ``good reduction'' everywhere by ``$Q$-place'' and ``almost good reduction'' respectively, we obtain the following theorem about the Witt decomposition of a specialized form.

\begtheorem{6.} Let  $\varphi$ be a form over a field $k$. Let
$(K_r \; \vert \; 0 \leq r \leq h)$ be a generic splitting tower of
$\varphi$ with associated higher kernel forms
$\varphi_r$ and indices $i_r$. Let $\Gamma \colon k \to L
\cup \infty$ be a  $Q$-place, with respect to which $\varphi$ has almost good
reduction. Finally, consider a $Q$-place $\Lambda \colon
K_m \to L \cup \infty$ for some $m$, $0 \leq m \leq h$, which extends
$\Gamma$ and which in the case  $m < h$ cannot be extended to 
$K_{m+1}$. Then $\varphi_m$ has good reduction with respect to
$\Lambda$. The form $\Gamma_* (\varphi)$ has kernel form
$\Lambda_* (\varphi)$ and Witt index $i_0 + \dots + i_m$.\qed
\endtheorem 
 
A subtle point to make here is that $\varphi_m$ has  GR with respect to $\Lambda$, not just AGR. Namely, if $\varphi_m$ only had AGR, then 
$\Lambda_* (\varphi_m)$ would certainly be isotropic.

We now also obtain a detailed proposition about the generic splitting of
$\Gamma_* (\varphi)$, which parallels word for word Scholium~3 of
 \S 4, and whose formulation we may thus leave to the Reader.

