\titlea{\S18}{Milnor's Exact Sequence}
\leftrunningtitle{Chapter III. Some Applications}
%\leftrunningtitle{Specialization of Quadratic and Symmetric
%Bilinear Forms}
\rightrunningtitle{\S18 Milnor's Exact Sequence}

Let $k$ be an arbitrary field. We want to derive an exact
sequence, going back to J. Milnor [M$_2$],  which establishes a
close relationship between the Witt ring $W(k(t))$ of the rational
function field $k(t)$ in one variable $t$ and the Witt rings of
field extensions of $k$, generated by one element. We need this
sequence in order to prove the announced Norm Theorem in \S19. It
should be emphasized that we will deal with \nd\ {\it
bilinear forms} now, although we are mostly interested in quadratic
forms where applications are concerned. Milnor assumes in [M$_2$, \S
5] that $\Char k\neq 2$, but this restriction is not necessary
here.

Let ${\Cal P}$ be the set of monic irreducible polynomials $p(t) \in
k[t]$. For every $p \in {\Cal P}$, let $k_p$ denote the residue
class field $k[t]/(p)$ and $\lambda_p \colon k(t) \to k_p \cup
\infty$ the canonical place with valuation ring ${\nam o}_p \colon =
k[t]_{(p(t))}$. $\{$Note: except for the ring ${\nam o}_{\infty} \colon =
k[t^{-1}]_{(t^{-1})}$, these rings are exactly all nontrivial
valuation rings of $k(t)$ which contain the field $k$. They are
all discrete.$\}$

For every polynomial $p \in {\Cal P}$ and every \nd\ (symmetric)
bilinear form $\vf$ over $k(t)$, let $\partial_p (\varphi)$ denote
the weak specialization of the form
$\langle p \rangle \varphi = \langle p \rangle
\otimes \varphi$,
$$\partial_p (\varphi) \colon \; = \; (\lambda_p)\lower5pt\hbox{$\scriptstyle
W$} (\langle p \rangle \varphi)\; \in \;
W (k_p),$$
(cf. \S3). The Witt class $\partial_p (\varphi)$ can be described
in an elementary way for one-dimensional forms as follows:
Let $f(t) \in k[t]$ be a polynomial $\neq 0$. Consider the
decomposition $f(t) = p(t)^m g(t)$, with $m \in {\Bbb
N}_0$ and $g(t)$ not divisible by $p(t)$. If $m$ is even, then
$\partial_p (\langle f \rangle ) = 0$. If $m$ is odd, then $\partial_p
(\langle f \rangle ) = \langle \overline{g} \rangle$, where
$\overline{g}$ denotes the residue class of $g$ in $k_p$.
$\{$Actually, we should write $[
\langle \overline{g} \rangle ]$ instead of $\langle \overline{g}
\rangle$, but we avoid such intricacies, now and in the future.
From now on we will usually not make a notational distinction
between a form and its associated Witt class.$\}$

Thus we obtain an additive map
$$\partial_p \colon \; W (k(t)) \longrightarrow W(k_p).$$
Furthermore we have a natural ring homomorphism
$$i \colon \; W(k) \longrightarrow W(k(t)), \quad \varphi \longmapsto \varphi
\otimes k(t).$$
$W(k(t))$ is a $W(k)$-algebra by means of this ring homomorphism.
Analogously, we interpret every ring $W(k_p)$, $p \in {\Cal P}$,
as a $W(k)$-algebra. In what follows, only the  structure
of $W(k(t))$ and the $W(k_p)$ as modules over $W(k)$ will be important
however. The maps $\partial_p$ are $W(k)$-linear. We combine them
in a $W(k)$-linear map
$$\partial = (\partial_p \vert p \in {\Cal P}) \colon \; W(k(t)) \longrightarrow
\bigoplus\limits_{p \in {\Cal P}} W(k_p).$$

\begtheorem{\rm (Milnor [M$_2$]){\bf.}} The sequence of $W(k)$-modules
$$0 \longrightarrow W(k) \buildrel i \over \longrightarrow W(k(t)) \buildrel
\partial \over \longrightarrow \bigoplus\limits_{p \in {\Cal P}} W(k_p)
\longrightarrow 0$$
is split exact. If $\lambda \colon k(t) \longrightarrow k \cup
\infty$ is an arbitrarily chosen place over $k$, then
$$\lambda_W \colon W(k(t)) \longrightarrow W(k)$$
is a retraction for $i$, in other words $\lambda_W \circ i = \id_{W(k)}$.
\endtheorem

We will carry out the {\it proof} in several steps, broadly following the
style of Milnor's procedure [M$_2$, p.335ff]. In what follows, a
``form'' is always a \nd\ symmetric bilinear form, which will be
identified with its Witt class.
\medskip

\noindent 1) If $\vf$ is a form over $k$, then $\varphi \otimes k(t)$ has
good reduction \wrt\ every place $\lambda \colon k(t) \to k \cup
\infty$ over $k$ and
$$\lambda_W (\varphi \otimes k(t)) = \lambda_* (\varphi \otimes k(t)) =
\varphi.$$
This proves the last statement of the theorem. In particular, $i$
is injective. In the following, we think of $W(k)$ as a subring of
$W(k(t))$ via the map $i$.
\medskip

\noindent 2) We consider an ascending filtration $(L_d \; \vert \; d \in {\Bbb
N}_0)$ of the $W(k)$-algebra $W(k(t))$. For every $d \in {\Bbb
N}_0$, let $L_d$ be the subring of $W(k(t))$, generated by
one-dimensional forms $\langle f \rangle$ for polynomials $f \in k[t]$ with $f \ne 0$, $\deg
f \leq d$. Clearly we then have
$$W(k) = L_0 \subset L_1 \subset L_2 \subset \dots \quad ,$$
and $W(k(t))$ is the union of all $L_d$. The forms $\langle f_1 \dots f_s
\rangle$ for products of finitely many polynomials
$f_1, \dots, f_s$ with $\deg f_i \leq d \quad (1 \leq i \leq s)$
generate $L_d$ additively. Further, let $(U_d \; \vert \; d \in {\Bbb
N}_0)$ be the filtration of
$\bigoplus\limits_{p \in {\Cal P}} W(k_p)$ with
$$U_0\; \colon = 0, \quad \quad U_d\; \colon  = \;
\bigoplus_{ p \in {\Cal P} \atop\deg p \leq d} W(k_p)\qquad
(d>0).$$
Clearly $\partial (L_d) \subset U_d$ for every $d \in {\Bbb N}_0$
and in particular, $\partial (L_0) = 0$.
Therefore $\partial$ induces for every $d\geq 1$ a $W(k)$-linear homomorphism
$$\overline{\partial}_d \colon \; L_d/L_{d-1} \longrightarrow U_d/U_{d-1}.$$
{\it For every $d \in {\Bbb N}$ we will verify that
$\overline{\partial}_d$ is bijective.} As an immediate
consequence we then obtain that for every $d \in {\Bbb N}$, $\partial$ induces an isomorphism
from $L_d/L_0$ to $U_d$ and thus an isomorphism from $W(k(t))/W(k)$ to $\bigoplus\limits_{p}
W(k_p)$, which concludes the proof.
\medskip

We let ${\Cal P}_d$ denote the set of all $p \in {\Cal P}$ of degree
$d$ and we identify $U_d/U_{d-1}$ with the direct sum of the
$W(k)$-modules $W(k_p)$, $p
\in {\Cal P}_d$. The component of index $p \in {\Cal P}_d$ of the map
$$\overline{\partial}_d \colon L_d/L_{d-1} \longrightarrow
\bigoplus\limits_{p \in {\Cal P}_d} W(k_p)$$
is the $W(k)$-linear map
$$\overline{\partial}_p \colon L_d/L_{d-1} \longrightarrow W(k_p),$$
induced by $\partial_p \colon \;
W(k(t)) \to W(k_p)$.
\medskip

\noindent 3) {\bf Proof of the surjectivity of $\overline{\partial}_d \; (d \geq
1)$.} It suffices to prove that, for  given $p \in {\Cal P}_d$ and $\eta \in
W(k_p)$, there exists a $\xi \in L_d$ with $\partial_p (\xi) = \eta$,
$\partial_p (\xi) = 0$ for $q \in {\Cal P}_d, \; q \ne p$.
Furthermore, we may assume that $\eta$ is the Witt class of a
one-dimensional form, i.e.
$\eta = \langle \overline{f} \rangle$ for an $f \in k[t]$ with $\deg f < d$,
$f \ne 0$, where $\overline{f}$ obviously denotes the image of $f$
in $k_p =k[t]/p(t)$. Letting $\xi \colon = \langle p f \rangle$
clearly accomplishes our goal.
\medskip

\noi 4) For the proof of the injectivity of
$\overline{\partial}_d$ we employ the following lemma (statement
and proof are exactly as in Milnor [M$_2$, Lemma 5.5]).

\beglemma{} Let $d\geq 1$. The additive group  $L_d$ is modulo
$L_{d-1}$ generated by  elements $\langle pg_1 \dots g_s \rangle$ with $s
\in {\Bbb N}_0$, $p \in {\Cal P}_d$ and every  $g_i$ a polynomial of degree $<
d$. Moreover,  if $f$ is the polynomial of degree $< d$ with $f \equiv g_1 \dots g_s
\bmod p$, then
$$\langle pf \rangle \equiv \langle p g_1 \dots g_s \rangle \bmod L_{d-1}.$$
\endlemma

\begproof{of the Lemma.} We recall that the identity
$$\langle a + b \rangle = \langle a \rangle + \langle b \rangle - \langle ab
(a+b) \rangle$$
holds in the Witt ring of any field, cf. \S3. Let
$\langle f_1 \dots f_r \; g_1 \dots g_s \rangle$
be a generator of $L_d$ with pairwise different $f_i \in {\Cal P}_d$ and
polynomials $g_j \ne 0$ of degree $< d$. If $r\geq 2$, then $f_1 = f_2 +
h$ with $h \ne 0$, $\deg h < d$, and we obtain the identity
$$\langle f_1 \rangle = \langle f_2 \rangle + \langle h \rangle - \langle f_1
f_2 h \rangle.$$
Multiplying with $\langle f_2 \dots f_r \; g_1 \dots g_s \rangle$
and omitting quadratic factors gives
$$\langle f_1 \dots f_r \; g_1 \dots g_s \rangle = \langle f_3 \dots g_s
\rangle + \langle h f_2 \dots g_s \rangle - \langle f_1 h f_3 \dots g_s
\rangle.$$
Every term on the right-hand side contains at most $r-1$ factors
of degree $d$. By induction on $r$ we now obtain that $L_d$
modulo $L_{d-1}$ is additively generated by forms $\langle f g_1 \dots g_s
\rangle$, where $f$ is monic of degree $d$, and every $g_j$ is a
nonzero polynomial of degree $<d$. Furthermore, we may assume that
$f$ is irreducible.

Consider such a generator $\langle p g_1 \dots g_s \rangle$
with $p \in {\Cal P}_d$, $\deg g_i < d$. Assume that $s \geq 2$.
Then $g_1 g_2 = p k + h$
with polynomials $k, h$ of degree $< d$ and $h \ne 0$. Hence,
$$\langle g_1 g_2 \rangle = \langle h \rangle + \langle pk \rangle - \langle
pkh g_1 g_2 \rangle.$$
Multiplying with $\langle p g_3 \dots g_s \rangle$ gives
$$\langle p g_1 g_2 \dots g_s \rangle = \langle p h g_3 \dots g_s \rangle +
\langle k g_3 \dots g_s \rangle - \langle k h g_1 \dots g_s \rangle,$$
and so
$$\langle p g_1 \dots g_s \rangle \equiv \langle ph g_3 \dots g_s \rangle
\bmod L_{d-1}.$$
Furthermore,
$$g_1 \dots g_s \equiv h g_3 \dots g_s \bmod p.$$
The proof of the Lemma can now be concluded by means of a simple
induction on $s$.
\bigskip

\noi 5) {\bf Proof of the injectivity of $\overline{\partial}_d \quad (d \geq
1)$.} Assume \ftsoc\ that $\overline{\partial}_d$ is not
injective. Then there exists a $\xi
\in L_d \setminus L_{d-1}$ with $\partial_p (\xi) = 0$ for every $p \in {\Cal
P}_d$. By the Lemma, we have a congruence
$$\xi \; \equiv \; \sum\limits_{i = 1}^s \langle p_i \rangle \varphi_i \qquad \bmod
L_{d-1}$$
with finitely many $p_i \in {\Cal P}_d$ and forms $\varphi_i = \langle
f_{i1}, \dots, f_{i, r_i} \rangle$ for nonzero polynomials
$f_{ij}$ of degree $<d$. So, for every $i \in \{ 1, \dots, s \}$,
$$\partial_{p_i} ( \langle p_i \rangle \varphi_i ) \; = \; \partial_{p_i}
(\xi) \; =0.$$
Furthermore, $\partial_q (\langle p_i \rangle \varphi_i) = 0$ for $q \in {\Cal
P}_d, \; q \ne p_i$. Hence every summand $\langle p_i \rangle
\varphi_i$ of $\xi$ modulo $L_{d-1}$ is already in the kernel of
$\overline{\partial}_d$. Thus we certainly have a
$p \in {\Cal P}_d$ and polynomials $f_1, \dots, f_r \in k [t]
\setminus \{ 0 \}$ of degree $< d$ such that
$$\xi \colon \; = \langle p f_1, \dots , p f_r \rangle \; \not\in \; L_{d-1},$$
but $\partial_p (\xi) = 0$, and so $\partial_q (\xi) = 0$ for every $q
\in {\Cal P}_d$. This should now lead to a contradiction.

We assume that the sequence $f_1, \dots, f_r$ is chosen to be
of minimal length~$r$. We have $\partial_p (\xi) = \langle \overline{f}_1, \dots, \overline{f}_r
\rangle$. Thus, the form $\langle \overline{f}_1, \dots, \overline{f}_r
\rangle$  over $k_p$ is metabolic and so in particular isotropic.
Hence, after permuting the $\overline{f}_i$ if necessary,
there certainly exists an index $s \in \{ 2, \dots, r \}$ and
polynomials $u_1, \dots, u_s \; \in \; k [t] \setminus \{ 0 \}$ of degree $<
d$, such that
$$\overline{u}_1^2 \overline{f}_1 + \dots + \overline{u}_s^2 \overline{f}_s =
0$$
in $k_p$. By the Lemma we have
$$\langle p u_i^2 f_i \rangle \equiv \langle p h_i \rangle \bmod L_{d-1},$$
where $h_i \in k [t]$ is the polynomial of degree $<d$ with $u_i^2 f_i \equiv
h_i \bmod p$.
$\{$Note: every $h_i \ne 0$.$\}$ Certainly $h_1 + h_2 \ne
0$, for otherwise we could write $\xi$ modulo $L_{d-1}$ in a
shorter way, contradicting the minimality of $r$. Applying the
relation
$$\langle h_1, h_2 \rangle = \langle h_1 + h_2, \; h_1 h_2 (h_1 + h_2)
\rangle,$$
while keeping in mind that by the Lemma
$$\langle p h_1 h_2 (h_1 + h_2) \rangle \equiv \langle p g_1 \rangle \bmod
L_{d-1}$$
for a polynomial $g_1 \ne 0$ of degree $<d$, we obtain
$$\xi \equiv \langle p (h_1 + h_2), \; p g_1, p h_3, \dots p h_s, \; p
f_{s+1} , \dots p f_r \rangle \bmod L_{d-1}.$$
Continuing in this way, we finally obtain a congruence
$$\xi \equiv \langle p (h_1 + \dots + h_{s-1}), \; p h_s, p g_1, \dots , p
g_{s-2}, \; p f_{s+1}, \dots, p f_r \rangle \bmod L_{d-1}.$$
The polynomial $h_1 + \dots + h_s$ is of degree $<d$ and is
divisible by $p$. Hence, $h_1 + \dots + h_s = 0$.
It follows that
$$\xi \equiv \langle p g_1, \dots,  p g_{s-2}, \; p f_{s+1} , \dots, p f_r
\rangle \bmod L_{d-1},$$
which contradicts the minimality of $r$. This establishes  the
injectivity of $\overline{\partial}_d$ and concludes the proof of
the Theorem.
