\titlea{\S8}{Good Reduction}
\leftrunningtitle{Chapter I. Fundamentals of Specialization Theory}
%\leftrunningtitle{Specialization of Quadratic and Symmetric
%Bilinear Forms}
\rightrunningtitle{\S8 Good Reduction}

As before, we let $\lambda \colon K \to L \cup \infty$ be a place,
${\nam o} = {\nam o}_{{\lambda}}$ the valuation ring of $K$,
${\nam m}$ its maximal ideal, $k = {\nam o}/{\nam m}$ its residue
class field and $v$ a valuation on $K$, associated to $\o$,  with
value group $\Gamma$. Also, let $\Sigma \subset Q(K)$ be a
complement of $Q({\nam o})$ in $Q(K)$ and $S$ a system of
representatives of $\Sigma$ in $K^*$ (with $1 \in S$), as
introduced in \S7.

From now on, we call a nondegenerate quadratic $\o$-module a {\bf
quadratic space over} $\o$. In case $\o=K$, we already used this
terminology in \S7. Recall further the concept of a {\it bilinear
space} over $\o$ (\S3, Definition~4).

If $E$ is a quadratic (resp. bilinear) space over $K$ then,
according to Definitions~1 and 8 in \S7, $E$ has {\it good
reduction \wrt\ } $\lambda$ (or: \wrt\ $\o$) if there exists a
quadratic (resp. bilinear) space $M$ over $\o$ with $E \; \cong \;
L \otimes_{{\nam o}} M$.

\begtheorem{1.} Let $E$ be a quadratic or bilinear space over $K$,
which has good reduction \wrt\ $\lambda$, and let $M$, $M'$ be
quadratic $($resp. bilinear$)$ spaces over $\o$ with $E \; \cong \; K
\otimes_{{\nam o}} M \; \cong \; K \otimes_{{\nam o}} M'$. Then
the $k$-spaces $M/{\nam m} M$ and $M' /{\nam m} M'$ are isometric
in the quadratic case and stably isometric\fonote{See \S2 for the
term ``stably isometric''.}
in the bilinear case.
Therefore we also have, $L \otimes_{\lambda} M \; \cong \; L
\otimes_{\lambda} M'$ resp. $L
\otimes_{\lambda} M \approx L \otimes_{\lambda} M'$.
\endtheorem

\begproof{} By the theory of \S7 (Theorems~4 and 7), $M/{\nam
m} M$ and $M'/{\nam m} M'$ are Witt equivalent. Since these spaces
have the same dimension, namely $\dim E$, they are isometric resp.
stably isometric.\qed
\endproof

In the following, the words ``good reduction'' are used so often,
that it is appropriate to introduce an abbreviation. {\it From now
on, we mostly write {\rm GR} instead of ``good reduction''.}

\begdefinition{1.} Let $E$ be a quadratic or bilinear space over
$K$, which has GR \wrt\ $\lambda$, and let $M$ be a quadratic, resp.
bilinear, space over $\o$ with $E \; \cong \; K \otimes_{{\nam o}}
M$. Then we denote the quadratic, resp. bilinear, module $L \otimes_{\lambda} M$
by $\lambda_* (E)$ and call it {\it ``the'' specialization of $E$ \wrt\
$\lambda$}.
\enddefinition

In the bilinear case, $\lambda_* (E)$ is nondegenerate, and thus a
space. This is true in the quadratic case as well, as long as
$\Char K\ne 2$. If $\Char K=2$ however,  $\lambda_* (E)$ can
be degenerate (cf. our discussion about $\lambda_W (E)$ in \S 7
after Definition~6).

This terminology is convenient, but sloppy. According to
Theorem~1, $\lambda_* (E)$ is only determined by $E$ up to
isometry in the quadratic case, and in the bilinear case even only
up to stable isometry. Anyway, in what follows, we are
interested in quadratic spaces (or modules) only up to isometry,
and in bilinear spaces almost always only up to stable isometry.

Note that in the bilinear case, Definition~1 is only a translation
of \S3, Definition~3 in geometric language.

\begremark{} Let $F$ and $G$ be bilinear spaces over $K$, which
have
GR \wrt\ $\lambda$. Then $F \, \perp \, G$ clearly also has
GR \wrt\ $\lambda$, and
$$\lambda_* (F \, \perp \, G) \approx \lambda_* (F) \, \perp \, \lambda_*
(G).$$
For quadratic spaces, we have to realize  that (when $\Char K=2$)
the orthogonal sum of two spaces over $\o$ can possibly be
degenerate and thus need not again be a space. However, if $F$ and $G$ are
quadratic spaces, which have GR \wrt\ $\lambda$, {\it and if $F$ is
strictly regular}, then $F \, \perp \, G$ has GR \wrt\ $\lambda$
and
$$\lambda_* (F \, \perp \, G) \; \cong \; \lambda_* (F) \, \perp \, \lambda_*
(G).\eqno{\square}$$
\endremark

For further applications, the following theorem is of the utmost
importance.

\begtheorem{2.} $(a)$ Let $F$ and $G$ be bilinear spaces over $K$.
If the spaces $F$ and $F \, \perp \, G$ have {\rm GR} \wrt\
$\lambda$, then $G$ also has {\rm GR} \wrt\ $\lambda$.

$(b)$ Let $F$ and $G$ be quadratic spaces over $K$. Suppose that
$F$ is strictly regular. If $F$ and $F \, \perp \, G$ have
\GR\ \wrtl, then $G$ also has \GR\ \wrtl.
\endtheorem

\begproof{} Part (a) has already been proved in \S3, Theorem~4.
Upon replacing every occurrence of  ``metabolic'' by
``hyperbolic'', the argument there also yields a proof of part
(b). (Use Theorem~3 and Lemma~3 in \S6.)\qed
\endproof

\begcorollary{} Let $E$ and $F$ be bilinear $($resp. quadratic$)$
spaces with $E \sim F$. If $E$ has \GR\ \wrtl, then so has $F$,
and $\lambda_* (E) \sim \lambda_* (F)$. Besides, if $E \approx F$
$($resp. $E \, \cong \; F$$)$, then $\lambda_* (E) \approx \lambda_* (F)$
$($resp. $\lambda_* (E) \; \cong \; \lambda_* (F)$$)$.
\endcorollary

This was already established in \S3 for the bilinear case. The
quadratic case can be proved similarly.

In part (b) of Theorem~2, the assumed strict regularity of $F$ is
essential.

\begexample{} Let $\Char K=2$ and let $E$ be a 3-dimensional
quadratic space over $K$ with basis $e,f,g$ and corresponding
value matrix
$\left[ {1
\atop 1} \; {1 \atop a} \right] \, \perp \, [ \varepsilon ]$, where $a \in {\nam
o}$ and $\varepsilon \in {\nam o}^*$. Let
$F \colon = E^{\perp} = Kg$ and
$G \colon = Ke + K(f + cg) \; \cong \;
\left[ {1
\atop 1} \; {1 \atop a+c^2 \varepsilon} \right]$ for some $c \in K$.
We have $E = F \, \perp \, G$. The quadratic spaces $E$ and $F$
have good reduction, but $G$ can have bad reduction because the
possibility exists that the Arf-invariant
$a + c^2 \varepsilon + \wp K$, $\wp K \colon = \{ x^2 + x \; \vert \; x \in K
\}$ does not contain an element of $\o$.

For instance, let $K = k(t)$ where $k$ is an imperfect field of
characteristic $2$ and $t$ is an indeterminate. As in the example
of a disobedient space, \S7, let ${\nam o} = k [t]_{(t)}$. We
choose $\varepsilon \in k^*$ such that $\varepsilon$ is not a
square in $k$. Furthermore, we choose $a = 0$, $c = t^{-2}$. We
have $\varepsilon t^{-2} + x^2 + x \not \in {\nam o}$ for {\it
every} $x\in K$. This is immediately clear when we look at the power
series expansions in $k((t))$: Let $x = \sum\limits_{n \geq d} c_n
t^n$ for some $d\in{\Bbb Z}$, all $c_n \in k$ and $c_d \ne 0$. If
it was true that $\varepsilon t^{-2} + x^2 + x \in {\nam o}$,
then we would have
$$\varepsilon t^{-2} + \sum\limits_{n \geq d} c^2_n t^{2n} + \sum\limits_{n
\geq d} c_n t^n \in k [[t]].$$
But then $d < 0$, and so $d = -1$. It would follow that $\varepsilon =
c^2_d$, an impossibility.

\begtheorem{3.} Let $E$ be a quadratic space over $K$. $E$ has
\GR\ \wrtl\ if and only if $QL(E)$ has \GR\ \wrtl\ and  there exists a
decomposition $E = F \, \perp \, QL(E)$ such that $F$ has
\GR\ \wrtl. In this case $\lambda_* (E) \; \cong \; \lambda_* (F) \, \perp \,
\lambda_* (QL(E))$.
\endtheorem

\begproof{} Suppose first that we have a decomposition $E = F \, \perp \,
QL(E)$, in which both $F$ and $QL(E)$ have \GR\ \wrtl. Then, by  the
remark preceding Theorem~2, $E$ has \GR\ \wrtl\ and
$\lambda_* (E) \; \cong \; \lambda_* (F) \, \perp \, \lambda_*
(QL(E))$,
because $F$ is strictly regular.

Next, suppose that $E$ has \GR. We choose a quadratic space $M$
over $\o$ with $E \; \cong \; K \otimes_{{\nam o}} M$ and a
decomposition $M = N \, \perp \, QL(M)$. From the definition of
spaces over $\o$, in other words of nondegenerate quadratic
$\o$-modules (\S6, Definition~3), follows immediately that $QL(M)$
is nondegenerate and even that $N$ is strictly regular. Now
$E \; \cong \; K \otimes_{{\nam o}} N \, \perp \,
K \otimes_{{\nam o}} QL(M)$, so that $QL(E) \; \cong \; K
\otimes_{{\nam o}} QL(M)$. Therefore $QL(E)$ has \GR.
Furthermore $F \colon = K \otimes_{{\nam o}} N$ has \GR\
and is strictly regular.\qed
\endproof

If $F$ and $G$ are bilinear spaces over $K$, which have \GR\ \wrtl, then
we know from \S3 that $F \otimes G$ also has \GR\ \wrtl\ and that
$$\lambda_* (F \otimes G) \approx \lambda_* (F) \otimes \lambda_* (G),$$
see \S3, Theorem~6. As explained in \S5, given a bilinear space
$F$ and a quadratic space $G$ over $K$ (or $\o$), we can also
construct a quadratic module $F \otimes G$ over $K$ (resp. $\o$),
which can however be degenerate when $\Char K=2$.
For $F \otimes G$ to be again a space, we must require that $G$ is
strictly regular. Analogous to the above statement, we have

\begtheorem{4.} Let $F$ be a bilinear space over $K$, which has
\GR\ \wrtl\ and let $G$ be a quadratic space, having \GR\ \wrtl. In
case $\Char K=2$, suppose furthermore that $G$ is strictly
regular. Then $F \otimes G$ als has \GR\ \wrtl\ and
$$\lambda_* (F \otimes G) \; \cong \; \lambda_* (F) \otimes \lambda_* (G).$$
\endtheorem

\begproof{} We choose spaces $M$ and $N$ over $\o$ with $F \; \cong \;
K \otimes_{{\nam o}} M$, $G \; \cong \; K \otimes_{{\nam o}} N$.
Then $N$ is strictly regular. Therefore $M \otimes_{{\nam o}} N$
is a strictly regular quadratic space over $\o$. Furthermore,
$F \otimes G \; \cong \; K
\otimes_{{\nam o}} (M \otimes_{{\nam o}} N)$.
Hence, $F \otimes G$ has \GR\ and
$$\lambda_* (F \otimes G) \; \cong \; L \otimes_{\lambda} (M
\otimes_{\o} N) \; \cong \; (L \otimes_{\lambda} M) \otimes_L (L
\otimes_{\lambda} N) \; \cong \; \lambda_* (F) \otimes \lambda_* (G).$$
\qed
\endproof

\begremark{}$\lambda_* (F)$ is only determined up to stable
isometry and $\lambda_* (F \otimes G)$ only up to isometry. The
theorem is valid -- as the proof shows -- for every choice of $\lambda_*
(F)$. More generally we have: if $E$ and $E'$ are bilinear space
over $K$ with $E \approx E'$, and if $F$ is a strictly regular
quadratic space over $K$, then $E
\otimes F \; \cong \; E' \otimes F$.
This follows from the fact that there exists a bilinear space $U$
over $K$ with $E \, \perp \, U \; \cong \; E' \perp U$, implying
that $(E
\otimes F) \, \perp \, (U \otimes F) \; \cong \; (E' \otimes F) \, \perp \,
(U \otimes F)$, and so $E \otimes F \; \cong \; E' \otimes
F$ by the Cancellation Theorem.
\endremark


The currently developed notion of the specialization
$\lambda_* (E)$ of a space $E$ which has \GR, allows us
to complete our understanding of the
results from \S7 about obedience and weak specialization.

\begtheorem{5.} Let $E$ be a quadratic module over $K$. $E$ is
nondegenerate and obedient \wrt\ $\lambda$ if and only if
$$E \; \cong \; \raise6pt\hbox{$\newbot\limits_{s \in
S}$} \; \langle s \rangle \otimes F_s, \leqno(\ast)$$
where the $F_s$ are spaces over $K$, which have \GR\ \wrtl, only
finitely many of them being nonzero. For every decomposition of
the form $(\ast)$, we have
$\lambda_W (E) = \{ \lambda_* (F_1) \}$.
\endtheorem

\begproof{} If $E$ is a space and is obedient \wrtl, we clearly have
a decomposition as above. The equality $\lambda_W (E) = \{
\lambda_* (F_1) \}$ follows from the definition of $\lambda_W (E)$
in \S7 and the definition of $\lambda_* (F_1)$.

Now let $(F_s \; \vert \; s \in S)$ be a family of spaces, having
\GR\ \wrtl\ and $F_s = 0$ for almost all $s \in S$. Let
$E \colon = \raise6pt\hbox{$\newbot\limits_{s \in
S}$} \; \langle s \rangle \otimes F_s$. Then, the quadratic module
$E$ over $K$ has finite dimension and
$E^{\perp} = \raise6pt\hbox{$\newbot\limits_{s \in
S}$} \; \langle s \rangle \otimes F_s^{\perp}$.
According to the Final Consideration of \S7, $E^{\perp}$ is
anisotropic and therefore nondegenerate. It follows immediately
from \S7, Definition~3 that $E$ is obedient \wrtl.\qed
\endproof

We call every decomposition of the form ($\ast$), having the
properties of Theorem~5, a $\lambda$-{\it modular decomposition} of $E$.
This terminology is a bit sloppier than in \S7, in the sense that
we no longer discriminate between internal and external orthogonal
sums. We talk about $\lambda$-modular decompositions of bilinear
spaces in a similar fashion.

\begtheorem{6.} Let $F$ be a bilinear space over $K$, which has \GR\
\wrtl.
\medskip

\setitemindent{(ii)}
\item{\rm (i)} If $G$ is a bilinear space over $K$, then
$$\lambda_W (F \otimes G) = \lambda_W (F) \lambda_W (G) = \{ \lambda_* (F) \}
\lambda_W (G). \leqno(\dagger)$$
\medskip

\item{\rm (ii)} If $G$ is a strictly regular quadratic space over
$K$, obedient \wrtl, then $F \otimes G$ is also obedient \wrtl,
and $(\dagger)$ holds again.
\endtheorem

\begproof{} (ii) Let $G \; \cong \;
\raise6pt\hbox{$\newbot\limits_{s \in
S}$} \; \langle s \rangle \otimes G_s$ be a $\lambda$-modular
decomposition of $G$. Every $G_s$ has \GR\ \wrtl\ and is strictly
regular. By Theorem~4, $F \otimes G_s$ has \GR\ \wrtl\ and
$\lambda_* (F \otimes G_s) \;
\cong \; \lambda_* (F) \otimes \lambda_* (G_s)$.
Hence,
$$F \otimes G \; \cong \;
\raise6pt\hbox{$\newbot\limits_{s \in
S}$} \; \langle s \rangle \otimes (F \otimes G_s)$$
is a $\lambda$-modular decomposition of $F \otimes G$ and
$\lambda_W
(F \otimes G) = \{ \lambda_* (F \otimes G_1) \} = \{ \lambda_* (F) \otimes
\lambda_* (G_1) \} = \{ \lambda_* (F) \} \{
\lambda_* (G_1) \} = \lambda_W (F) \lambda_W (G)$.
\medskip

The proof of (i) is analogous. Less care is needed here than for
(ii). \qed
\endproof



Similarly we can show,

\begtheorem{7.} Let $F$ be a bilinear space and $G$ a strictly
regular quadratic space over $K$. Suppose that the space $G$ has
\GR\ \wrtl. Then $F \otimes G$ is obedient \wrtl, and
$$\lambda_W (F \otimes G) = \lambda_W (F) \lambda_W (G) = \lambda_W (F) \{
\lambda_* (G) \}.\eqno{\square}$$
\endtheorem

To conclude this section, we have a look at the reduction
behaviour of quadratic spaces under basis extensions. The
following -- almost banal -- theorem, together with Theorem~2(b)
above, lays the foundations upon which we will build the
generic splitting theory  of regular quadratic spaces in the next section.

\begtheorem{8.} Let $K' \supset K$ be a field extension and $\mu \colon K'
\to L \cup \infty$ an extension of the place $\lambda \colon K \to L \cup
\infty$. Let $E$ be a $\underline{\hbox{regular}}$ quadratic space
over $K$, which has \GR\ \wrtl. The space $K' \otimes_K E$ has \GR\
\wrt\ $\mu$ and
$\mu_* (K' \otimes_K E) \; \cong \; \lambda_* (E)$.
\endtheorem

\begproof{} Let $\o'$ be the valuation ring associated to $\mu$.
Furthermore, let $M$ be a regular quadratic space over $\o$ with
$E \; \cong \; K \otimes_{{\nam o}} M$. Then ${\nam o}' \otimes_{{\nam o}}
M$ is a regular quadratic space over $\o'$ (sic!), and
$K' \otimes_K E \; \cong \; K' \otimes_{{\nam o}} M = K' \otimes_{{\nam o}'}
({\nam o}' \otimes_{{\nam o}} M)$. Therefore, $K' \otimes_K E$ has
\GR\ \wrt\ $\mu$ and
$$\mu_* (K'
\otimes_K E) \; \cong \; L \otimes_{\mu} ({\nam o}' \otimes_{{\nam o}}
M) = L \otimes_{\lambda} M \; \cong \; \lambda_* (E).$$
\qed
\endproof

If we only require that $E$ is nondegenerate instead of regular,
the statement of Theorem~8 becomes false. Looking for a
counterexample, we can restrict ourselves to quasilinear spaces,
by Theorems~3 and 8.

\begexample{} Let $k$ be an imperfect field of characteristic~2
and let $a$ be an element of $k$ which is not a square.
Consider the power series field $K = k((t))$ in one indeterminate
$t$. Let $\lambda_0 \colon K \to k \cup \infty$ be the place with
valuation ring ${\nam o} = k [[t]]$, which maps every power
series $f(t) \in {\nam o}$ to its constant term $f(0)$.
The quasilinear quadratic $\o$-module $M =[1, a +t]$ is
nondegenerate. For, if $f(t) = \sum\limits_{i \geq 0}
b_it^i$ and $g(t) = \sum\limits_{i \geq 0} c_it^i$ are elements of
$\o$, whose constant terms $b_0$, $c_0$ are not both zero, then
$f(t)^2 + (a+t) g(t)^2$ has nonzero constant term $b_0^2 + a
c_0^2$, and is thus a unit in $\o$. Therefore, axiom (QM2) of
\S6, Definition~3 is satisfied.

Hence, the space $E \colon = K \otimes_{{\nam o}} M$ has \GR\
\wrt\ $\lambda_0$, and $(\lambda_0)_* (E) \; \cong \; [1, a ]$.
Consider now $L \colon = k (\sqrt{a})$ and the place
$\lambda \colon = j \circ \lambda_0
\colon K \to L \cup \infty$, obtained by composing $\lambda_0$
with the inclusion $j \colon k \hookrightarrow L$. Then $E$ also
has \GR\ \wrtl, and over $L$ we have,
$$\lambda_* (E) = L \otimes_k (\lambda_0)_*
(E) \; \cong \; [1, a] \; \cong \; [1, 1] \; \cong \; [1, 0].$$

Finally, let $K' \colon = K(\sqrt{a}) = L((t))$ and  let $\mu \colon K' \to L \cup
\infty$ be  the place with valuation ring ${\nam o}' \colon = L
[[t]]$, which again maps every power series $f(t) \in {\nam o}'$
to its constant term $f(0)$. $\mu$ extends the place $\lambda$.
Over $K'$, we have
$$K' \otimes_K E = [1, (\sqrt{a})^2 + t] \; \cong \; [1, t].$$
Therefore, $K' \otimes_K E$ is obedient \wrt\ $\mu$ and
$\mu_W (K' \otimes_K E) = \{ [1] \}$.
It is now clear that $K' \otimes_K E$ has bad reduction \wrt\
$\mu$, since the Witt class $\mu_W (K'
\otimes_K E)$ would contain a two-dimensional module  otherwise.
\endexample

\begremark{} This argumentation shows nicely that it is profitable
to give such an elaborate
definition of Witt equivalence of degenerate quadratic modules over
fields, as done in \S6, Definition~10.
\endremark
