\titlea{\S29}{Almost Good Reduction with Respect to Extensions of Quadratic Places}

\leftrunningtitle{Chapter IV. Specialization with Respect to Quadratic Places}
%\leftrunningtitle{Specialization of Quadratic and Symmetric
%Bilinear Forms}
\rightrunningtitle{\S29 Almost Good Reduction with Respect to Extensions of $Q$-Places}
%\rightrunningtitle{\S29 Almost Good Reduction with Respect to Extensions of %Quadratic Places}

In this section a ``form'' will again always mean a nondegenerate (symmetric) bilinear form.

\begdefinition{1.} Let  $\Lambda \colon K \to L \cup \infty$ and
$\Lambda' \colon K \to L \cup \infty$ be two quadratic places from
$K$ to $L$, $\Lambda = (\lambda, H, \chi)$, $\Lambda' = (\lambda',
H', \chi')$. We call $\Lambda'$ an {\it expansion}\Idx{expansion} 
of
$\Lambda$, and write $\Lambda \subset \Lambda'$, when
$\lambda = \lambda'$, $H \subset H'$, and $\chi' \vert H = \chi$.
\enddefinition

Note that every $Q$-place  $\Lambda = (\lambda, H, \chi)$ is an
expansion of the  $Q$-place $\widehat{\lambda} = (\lambda, Q ({\nam
o}), \; \lambda_W \vert Q ({\nam o}))$ with ${\nam o} = {\nam
o}_{\lambda}$.

\begremark{} If  $\Lambda \subset \Lambda'$ and if some form
$\varphi$ over $K$ has good reduction   (respectively, almost good reduction) with respect to
$\Lambda$, then it  obviously has
 GR (respectively,  AGR) with respect to $\Lambda'$ and we have
 $\Lambda'_* (\varphi) = \Lambda_* (\varphi)$. 
 \endremark

Consider a  $Q$-place $\Lambda = (\lambda, H, \chi)$ from $K$
to $L$. If $k$ is a subfield of  $K$, we obtain from $\Lambda$   a $Q$-place $\Gamma = (\gamma, D, \psi)$ from $k$ to $L$ as follows:
let $\gamma$ be the restriction $\lambda \vert k$ of $\lambda$. The 
group $D \subset Q(k)$ is the preimage of $H$ under the
homomorphism $j \colon Q(k) \to Q(K)$, which maps a square class
$\langle a \rangle = ak^{*2}$ of $k$ to the square class $\langle
a \rangle_K = aK^{*2}$.  (Note that $\langle a \rangle_K =
\langle a \rangle \otimes K$ when we interpret square classes as one-dimensional forms.) Finally  $\psi$ is the character  $\chi \circ j \colon D \to Q(L)$.

\begdefinition{2.} We call $\Gamma$ the {\it restriction}\Idx{restriction of a Q-place@restriction of a $Q$-place}\Idx{111lwk@$\Lambda\vert k$}
of the
$Q$-place $\Lambda$ to $k$ and write $\Gamma = \Lambda
\vert k$.
\enddefinition

\noi {\it Caution!\/} If $\Lambda = \widehat{\lambda}$ for a place
$\lambda \colon K \to L \cup \infty$ and $\gamma = \lambda \vert k$,
then $\Lambda \vert k \supset \widehat{\gamma}$, but $\Lambda
\vert k$  is in general  different from  $\widehat{\gamma}$.
\medskip


The following important theorem for quadratic places has no counterpart for places. 

\begtheorem{1.} Let $k$ be a subfield of $K$ and $\varphi$ a form over $k$. Let $\Lambda \colon K \to L \cup \infty$ be a 
$Q$-place and $\Gamma = \Lambda \vert k$.
The form $\varphi \otimes K$ has AGR with respect to $\Lambda$
if and only if $\varphi$ has  AGR with respect to $\Gamma$, and then
$\Lambda_* (\varphi \otimes K) = \Gamma_* (\varphi)$.
\endtheorem

\begproof{} As above we write $\Lambda = (\lambda, H, \chi)$,
$\Gamma = (\gamma, D, \psi)$ with $\gamma = \lambda \vert k$ etc. We choose
a subgroup $S$ of $Q(k)$ with $Q(k) = S \times D$.
The homomorphism $j \colon Q (k) \to Q(K)$ maps $S$ isomorphically to
a subgroup of $Q(K)$, which we denote by $S_K$. We further choose
a subgroup $T \supset S_K$ of $Q(K)$ such that
$Q(K) = T \times H$. It is always possible to do this.
We now start with a
$\Gamma$-modular decomposition of $\varphi$,
$$\varphi  \approx  \raise5pt\hbox{$\newbot\limits_{s \in S}$} \; s \; \varphi_s,
$$
for  $\Gamma$-unimodular forms $\varphi_s$ over $k$. We
can interpret
$$\varphi \otimes K  \approx  \raise5pt\hbox{$\newbot\limits_{s \in S}$} \; s_K
(\varphi_s \otimes K)$$
as a  $\Lambda$-modular decomposition of $\varphi \otimes K$
for the decomposition $Q(K) = T \times H$, in which the components
for $t \in T \setminus S_K$ are all zero. 
By definition $\varphi$ has AGR with respect to $\Gamma$ if and only if
$\Gamma_* (\varphi_s) \sim 0$ for all $s \ne 1$ in $S$, and
$\varphi \otimes K$ has AGR with respect to $\Lambda$ if and only if
$\Lambda_* (\varphi_s \otimes K) \sim 0$ for all $s \ne 1$ in
$S$. But for every $s \in S$ we have $\Lambda_* (\varphi_s \otimes
K) \approx \Gamma_* (\varphi_s)$. Thus $\varphi \otimes K$ has AGR
with respect to $\Lambda$ if and only if $\varphi$ has AGR with respect to $\Gamma$, and in this case we have, with $r \colon = \frac{1}{2} (\dim \varphi -
\dim \varphi_1)$, that
$$\Lambda_* (\varphi \otimes K)  =  \Lambda_* (\varphi_1 \otimes K) \perp r
\times \tilde{H}  \approx  \Gamma_* (\varphi_1) \perp r \times
\tilde{H}  =   \Gamma_* (\varphi).\eqno{\square}$$
\endproof

\begdefinition{3.} As before, let $K$ be a field and $k$ a
subfield of $K$. Consider $Q$-places $\Lambda \colon K \to L
\cup \infty$ and $\Gamma \colon k \to L \cup \infty$. We call
$\Lambda$ an {\it extension of}\Idx{extension} 
$\Gamma$ when $\Lambda
\vert k \supset \Gamma$, and a {\it strict extension of}\Idx{strict extension}
$\Gamma$ when actually $\Lambda \vert k = \Gamma$.
\enddefinition

\begtheorem{2.} Let  $\Lambda \colon K \to L \cup \infty$ be an
extension of $\Gamma \colon k \to L \cup \infty$. Let $\varphi$
be a form over $k$ which has AGR with respect to $\Gamma$. Then $\varphi \otimes
K$ has AGR with respect to $\Lambda$ and $\Lambda_* (\varphi \otimes K) =
\Gamma_* (\varphi)$.
\endtheorem

\begproof{} By Theorem~1 the statement holds when 
$\Lambda$ is a strict extension of 
$\Gamma$.  Thus it suffices to consider the case where
$K = k$ and $\Lambda = (\lambda, H, \chi)$ is an expansion of
$\Gamma = (\gamma, D, \psi)$.  Thus, let $K = k$, $D
\subset H$, $\chi \vert D = \psi$.

We choose subgroups  $S_1$ of $H$ and $T$ of $Q(k)$ with $H =
S_1 \times D$, $Q(k) = T \times H$. Then $S \colon = T \times
S_1$ is a subgroup of $Q(k)$ with $Q(k) = S \times D$. Let
$$\varphi  \approx  \raise5pt\hbox{$\newbot\limits_{(\alpha, \beta) \in T
\times S_1}$} \; \alpha \beta \; \varphi_{\alpha, \beta}$$
be a  $\Gamma$-modular decomposition of $\varphi$. We let
$$\varphi_{\alpha} \colon  =  \raise5pt\hbox{$\newbot\limits_{\beta \in S_1}$}
\; \beta \; \varphi_{\alpha, \beta} \qquad (\alpha \in T).$$

For every  $\alpha \in T$ the form $\varphi_{\alpha}$ is
$\Lambda$-unimodular and
$$\Lambda_* (\varphi_{\alpha})  \approx 
\raise5pt\hbox{$\newbot\limits_{\beta \in S_1}$} \; \chi (\beta) \;
\Gamma_* (\varphi_{\alpha, \beta}). \leqno(1)$$
It follows that
$$\varphi  \approx  \raise5pt\hbox{$\newbot\limits_{\alpha \in T}$} \; \alpha
\; \varphi_{\alpha}$$
is a $\Lambda$-modular decomposition of $\varphi$. If
$\alpha \ne 1$, then $\alpha \beta \ne 1$ for every $\beta \in
S_1$, and so $\Gamma_* (\varphi_{\alpha \beta}) \sim 0$ for every
$\beta \in S_1$. From (1) we then see that  $\Lambda_*
(\varphi_{\alpha}) \sim 0$.  Therefore $\varphi$ has AGR with respect to
$\Lambda$.

Furthermore,
$$\Lambda_* (\varphi)  \approx  \Lambda_* (\varphi_1) \perp \frac{\dim \varphi -
\dim \varphi_1}{2} \times \tilde{H}. \leqno(2)$$
If $\beta \in S_1$, $\beta \ne 1$, then $\Gamma_*
(\varphi_{1, \beta}) \sim 0$, and we see from the expression (1) with
$\alpha = 1$ that
$$\Lambda_* (\varphi_1)  \approx  \Gamma_* (\varphi_{1, 1}) \perp \frac{\dim
\varphi_1 - \dim \varphi_{1, 1}}{2} \times \tilde{H}. \leqno(3)$$
From (2) and (3) we obtain
$$\Lambda_* (\varphi)  \approx  \Gamma_* (\varphi_{1, 1}) \perp \frac{\dim
\varphi - \dim \varphi_{1, 1}}{2} \times \tilde{H}  \approx 
\Gamma_* (\varphi).\eqno\square $$ 
\endproof

\begtheorem{3.} 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 there exists a field extension
$K \supset k$ such that $\varphi
\otimes K$ is isotropic  and $\Gamma$ has an extension to a 
$Q$-place $\Lambda \colon K \to L \cup \infty$. Then
$\Gamma_* (\varphi)$ is isotropic. More precisely,
$$\ind\, \Gamma_* (\varphi)  \geq  \ind (\varphi \otimes K).$$
\endtheorem

\begproof{} We have $\varphi \otimes K \approx \psi_0 \perp r
\times \tilde{H}$ with $\psi_0$ anisotropic, $r = \ind (\varphi \otimes
K) > 0$. By Theorem~2,  $\varphi \otimes K$ has  AGR with respect to $\Lambda$ and
$\Gamma_* (\varphi) \approx \Lambda_* (\varphi \otimes K)$. Theorem~4
from \S 28 tells us that $\psi_0$ has AGR with respect to $\Lambda$. 
Therefore
$$\Gamma_* (\varphi)  \approx  \Lambda_* (\psi_0) \perp r \times \tilde{H}.\eqno\square$$
\endproof

Theorem~3 inspires the hope that for almost good reduction  there are theorems about the generic splitting of  $\Gamma_* (\varphi)$, analogous to those in \S4. As in \S4 one will have to assume that the characteristic of $L$ is different from two though. 




