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%%% Date: Sat, 22 Aug 1998 07:02:41 -0700
%%% From: wadswrth@euclid.ucsd.edu (A. Wadsworth)
%%% To: rehmann@mathematik.Uni-Bielefeld.DE
%%% 
%%% %% 1-26-98
%%% %%
%%% %% This is the journal version of the paper
%%% %%
%%% %%  Algebraic Extensions of Graded and 
%%% %%  Valued Fields
%%% %%
%%% %%  by  Y.-S. Hwang  and  A. R. Wadsworth.
%%% %%
%%% %%  It will appear in Communications in Algebra.
%%% %%
%%% %%  It is in AMSTeX.  
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\vskip0.8truein
\centerline{\bf{ALGEBRAIC EXTENSIONS OF GRADED AND VALUED
FIELDS}}
%\topmatter
%\title Algebraic Extensions of Graded and
%Valued Fields
%\endtitle
\vskip 0.4truein
\baselineskip = 10pt
\centerline{Y.-S.~Hwang$^1$}
\bigskip
\centerline{Department of Mathematics} 
\centerline{College of Science}
\centerline{Korea University}
\centerline{5--1, Anam-dong, Sungbuk-ku}
\centerline{Seoul, 136--701}
\centerline{Korea}
\vskip0.1truein 
\centerline {{\it e-mail:} {\tt
yhwang\@semi.korea.ac.kr}}
\bigskip
\centerline{and}
\bigskip
\centerline{A. R. Wadsworth$^2$}
\bigskip
\centerline{Department of Mathematics, 0112}
\centerline{University of California, San Diego} 
\centerline {9500 Gilman Drive}
\centerline {La Jolla, California 92093-0112}
\centerline{USA}
\vskip0.1truein
\centerline{{\it e-mail:} {\tt arwadsworth\@ucsd.edu}}

\vskip 0.4truein
%\author Y.-S.~Hwang$^1$ and A. ~R.~Wadsworth$^2$
%\endauthor
%\endtopmatter
\footnote""{$^1$Supported in part by the Non-directed 
Research Fund, Korea.}
\footnote""{$^2$Supported in part by the
NSF.}
\bigskip
 \baselineskip= 15pt
If $F$ is a field with a (Krull) valuation, then 
the filtration of $F$ induced by the valuation
yields an associated graded ring, which is a
graded field.  Conversely, if $R$ is a graded
field with totally ordered grade group, then
$R$ is an integral domain and there is a
canonically associated valuation on the
quotient field of $R$.  The processes of passing
from valued field to graded field and vice
versa are not quite inverses of each other, but
many properties in one setting are
well-reflected in the other.

The goal of this paper is to describe an
algebraic extension theory for graded fields
analogous to what is known for valued fields, 
and then to spell out the correspondence
between tame extensions of graded fields and
Henselian valued fields.  This has the benefit 
that graded fields are easier to work with for
many purposes than valued fields.  But beyond
this, there is a similar correspondence
between graded division rings and valued
division rings, where the graded objects
seem to be significantly easier to work with
than the valued objects.  We first learned of 
this correspondence from a paper by 
M\. Boulagouaz [B$_2$].  The correspondence
for division rings is actually far more
extensive than what was described by
Boulagouaz, and we pursue that
subject in a sequel to this paper [HW].  
The choice of topics to treat here was 
influenced by the needs of the study of 
division rings.   But, we feel that the 
commutative theory presented here is of
interest in its own right.

\bigskip\smallskip
%\bigskip

\head \S 1 Graded fields (with
totally ordered grade group)
\endhead 

Let $(\Gamma , +)$ be an abelian group and let $R =
\bigoplus\limits_{\gamma
\in \Gamma} R_\gamma$ be a graded ring with respect to $\Gamma$ (i.e., $R$
is a ring with $1$, such that $R_\gamma \cdot R_\delta \subseteq R_{\gamma
+
\delta}$ for all $\gamma, \delta \in \Gamma$).  We set
$$
\Gamma_R
= \supp (R) 
= \{  \gamma \in \Gamma \mid R_\gamma \not= (0) \} .
$$
Also, let $R^h = \bigcup\limits_{\gamma \in \Gamma} R_\gamma$, the set
of homogeneous elements of $R$.  If $r \in R_\gamma$, $r \not= 0$, we
write $\degg (r) = \gamma$.

The graded ring $R$ is said to be a {\it graded field} if $R$ is commutative
with $1 \not= 0$ and every nonzero homogeneous element of $R$ is a unit. 
When this occurs, $\Gamma_R$ is a subgroup of $\Gamma$, and we call it
the {\it grade group} of $R$.  We will be interested exclusively in the case
where $\Gamma_R$ is totally ordered.  We adopt as a standing hypothesis
throughout the paper that all the graded fields $R$ we consider are equipped
with a total ordering on $\Gamma_R$.  (But, note that 
any torsion-free abelian group admits a total ordering. 
Therefore, all the results in \S\S 1-3 below on extensions of
graded fields hold if we merely assume that $\Gamma_R$
is torsion-free abelian.  Only when we wish to build a valuation
ring on the quotient field of $R$ will we need to specify an
ordering on $\Gamma_R$.)
  A graded isomorphism $R\to R'$ of graded
fields consists of an order-preserving group
isomorphism $\alpha\: \Gamma_R \to 
\Gamma_{R'}$ and a ring isomorphism
$\beta\: R\to R'$ such that 
$\beta(R_\gamma) = R'_{\alpha(\gamma)}$ for 
all $\gamma\in \Gamma_R$.  When such a
graded isomorphism exists, we write 
$R\cong_g R'$.

 It
follows  from the total ordering of 
$\Gamma_R$  that a graded field $R$ is
an integral domain and that its group of units
$R^* = R^h - \{ 0\}$.  Let
$$
QR
= \text{quotient field of } R .
$$
Since $\Gamma_R$ is totally ordered, it is a torsion-free abelian group,
and we set
$$
\Delta_R = \Bbb Q \otimes_{\Bbb Z} \Gamma_R , 
\text{ the divisible
hull of } \Gamma_R.
$$  
We identify $\Gamma_R$ with its isomorphic image in
$\Delta_R$ $ ( \gamma \leftrightarrow 1 \otimes \gamma)$.  The total
ordering on $\Gamma_R$ extends uniquely to a total ordering on
$\Delta_R$.  Moreover, if $\Lambda$ is any torsion-free abelian group
containing $\Gamma_R$ as a subgroup with $\Lambda /\Gamma_R$ torsion,
then there is a unique monomorphism $\Lambda \to \Delta_R$ extending
the embedding $\Gamma_R \hookrightarrow  \Delta_R$.  We will
thus routinely view any such $\Lambda$ as a subgroup of $\Delta_R$.  The
ordering on
$\Gamma_R$ extends uniquely to an ordering on
$\Lambda$.

Since $R$ is a graded field, it is clear that $R_0$ is a field and each
$R_\gamma$ is a $1$-dimensional vector space over $R_0$.  For each
$\gamma \in \Gamma_R$, fix some nonzero $t_\gamma \in R_\gamma$. 
Then for each $\gamma , \delta \in \Gamma_R$ there is $c_{\gamma,
\delta} \in R^*_0$, such that $t_\gamma t_\delta = c_{\gamma, \delta }
t_{\gamma + \delta}$.  We call $\{ c_{\gamma, \delta } \mid \gamma , \delta
\in \Gamma_R \}$ a family of structure constants of $R$.  The
commutativity and associativity of $R$ imply
$$
c_{\gamma, \delta} 
= c_{\delta , \gamma} \quad
\text{and} \quad
c_{\gamma , \delta} c_{\gamma + \delta , \rho} 
= c_{\delta , \rho} c_{\gamma, \delta + \rho}
\quad \text{for all } 
\gamma , \delta , \rho \in \Gamma_R .
\tag1.1
$$
Conversely, it is clear that given any field $F$, any totally ordered
abelian group $\Gamma$, and any function $\Gamma \times \Gamma \to
F^*$ $((\gamma, \delta ) \mapsto c_{\gamma, \delta})$ satisfying the
conditions in (1.1), there is a graded field
$R$ with
$R_0 = F$, $\Gamma_R = \Gamma$, and structure constants given by the
specified function.

We say that a graded field $R$ is of {\it group-ring type} if there  is a
family of structure constants of $R$ with $c_{\gamma, \delta} = 1$ for all
$\gamma , \delta \in \Gamma_R$.  Clearly, $R$ is of group-ring type iff $R$
is isomorphic (as a graded ring) to the group ring $R_0 [\Gamma ]$.

\proclaim{Proposition 1.1}  A graded field $R$ is of group-ring type iff the
canonical short exact sequence of abelian groups
$$
0 \to R^*_0 \to R^* \to \Gamma_R \to 0
$$
is split exact.\endproclaim

\demo\nofrills{\smc {Proof.}\usualspace}  Clearly, the exact sequence
splits iff there is a family $\{ t_\gamma \mid \gamma \in \Gamma_R \}$
with each
$t_\gamma \in R_\gamma - \{ 0 \}$ and $t_\gamma t_\delta = t_{\gamma +
\delta}$.  This is exactly what is needed so that the structure constants
relative to $\{ t_\gamma \}$ will all be $1$.  $\square$ \enddemo

Prop.~1.1  shows that if $\Gamma_R \cong \Bbb Z$, or $\Gamma_R$ is
any free abelian group, then the graded field $R$ is of group-ring type (and
hence determined up to isomorphism by $R_0$ and $\Gamma_R$).  But, not
every graded field is of group-ring type, as the following example shows.

\example {Example 1.2} {\sl A graded field $R$ with $R_0 = \Bbb Q$ and
$\Gamma_R =
\Bbb Z \big[ \frac 1 2 \big]$, and $R$ not of group-ring type.}  Let $t$ be an
indeterminate, and let $A = \Bbb Q [ t, t^{-1}]$, the Laurent polynomial ring in
$t$ over the rational numbers $\Bbb Q$.  This $A$ is the group ring $\Bbb Q [
\Bbb Z]$, so a graded field, with $\Gamma_A = \Bbb Z$ and $A_i = \Bbb Q t^i$,
$i \in \Bbb Z$.  Let $p$ be any prime number.  In an algebraic closure of the
quotient field of $A$, there are $y_1 , y_2, \ldots$ satisfying $y^2_1 =
pt$, $y^2_2 = p^2 y_1$, $y_3^2 = p^4 y_2 , \ldots$, $y^2_i =
p^{2^{i-1}} y_{i - 1}, \ldots $.  Let $B_1 = A [ y_1]$, $B_2 = B_1 [ y_2 ] ,
\ldots , B_i = B_{i-1} [ y_i ] , \ldots $, and let $R =
\bigcup\limits^\infty_{i=1} B_i$.  Note that, as an $A$-module,
$B_1 = A \oplus y_1 A$, and $B_1$ is a graded field with $(B_1 )_0
= \Bbb Q$, $\Gamma_{B_1} = \frac 1 2 \Bbb Z$, and for $j \in \Bbb
Z$, $(B_1 )_{j/2} = \Bbb Q y^j_1$.    Also, the gradings on
$A$ and
$B_1$ are compatible, so $A$ is a graded subfield of $B_1$.  Proceeding
inductively, we have for each $i$, $B_i = B_{i-1} \oplus B_{i-1} y_i$, $B_i$ is a
graded field with $(B_i)_0 = \Bbb Q$, $\Gamma_{B_i} = 2^{-i} \Bbb Z$,
$(B_i)_{j/2^i} = \Bbb Q y^j_i$ for all $j \in \Bbb Z$, and $B_{i-1}$ is a graded
subfield of $B_i$.  Hence, $R$ is a graded field with $R_0 = \Bbb Q$ and
$\Gamma_R = \Bbb Z \big[ \frac 1 2 \big]$ (the additive group of the ring
$\Bbb Z \big[ \frac 1 2 \big]$, with the usual ordering).  If $R$ were of
group-ring type, then there would exist nonzero $s_1 \in R_1 = \Bbb Qt$ and
$s_{2^{-i}} \in R_{2^{-i}} = \Bbb Q y_i$ such that $(s_{2^{-i}})^{2^i} = s_1$, for
all integers $i > 0$.  However, observe that $y^{2^i}_i = p^{(4^i - 1)/3} t$. 
Hence,  $s_1 \in \bigcap\limits^\infty_{i=1} (R_{2^{-i}})^{2^i} =
\bigcap\limits^\infty_{i=1} \Bbb Q^{2^i} p^{(4^i - 1)/3} t = (0)$, a
contradiction.  Hence, $R$ is not of group-ring type. 
\endexample
\medskip

\proclaim{Corollary 1.3}  Let $R$ be a graded field.  Then $R$ is integrally
closed, and $R_0$ is algebraically closed in
$QR$.\endproclaim

\demo\nofrills{\smc {Proof.}\usualspace}  If $R$ were not integrally closed,
this would be detected by an equation involving only finitely many elements
of $R$, hence only finitely many homogeneous elements.  Thus, there is a
finitely generated subgroup $\Lambda$ of $\Gamma_R$ such that for the
graded subfield $A = R |_\Lambda = \bigoplus\limits_{\lambda \in \Lambda}
R_\lambda$ of $R$, $A$ is not integrally closed.  Note that $\Gamma_A =
\Lambda$ and $A_0 = R_0$.  As $\Lambda$ is finitely generated and
torsion-free, $\Lambda \cong \Bbb Z^n$ for some $n$.  So, Prop.~1.1
shows $A \cong A_0 [ \Lambda ] \cong A_0 [x_1, x^{-1}_1 , \ldots , x_n,
x^{-1}_n]$, where $x_1, \ldots , x_n$ are independent indeterminates. 
Since $A$ is a localization of a polynomial ring over a field, $A$ is
integrally closed, a contradiction.  Hence, $R$ must be integrally closed. 
That $R_0$ is algebraically closed in
$QR$ is proved in the same way, as $R_0 = A_0$, which is algebraically
closed in the rational function field $QA \cong R_0 (x_1 , \ldots , x_n)$. 
$\square$
\enddemo

Let $R$ be a graded field.  By a {\it graded} $R${\it -module}, we mean an
$R$-module $M$ such that $M$ has a direct sum decomposition as abelian
groups $M = \bigoplus\limits_{\mu \in \Gamma_M} M_\mu$, where
$\Gamma_R$ acts freely on the set $\Gamma_M$, and for $\gamma \in
\Gamma_R$, $\mu \in
\Gamma_M$, we have $R_\gamma \cdot M_\mu \subseteq M_{\gamma +
\mu}$.  (Here $\gamma + \mu$ denotes the image of $\mu$ under the
action of $\gamma$.)  There is no group structure assumed on
$\Gamma_M$.  That $\Gamma_R$ acts freely on $\Gamma_M$ means that
for all $\gamma$, $\delta \in \Gamma_ R$, $\mu \in \Gamma_M$, we have
$\gamma + \mu \not= \delta + \mu$ whenever
$\gamma \not= \delta$.  Observe that every graded module $M$ over any
graded field $R$ is a free $R$-module, with a homogeneous base.   Indeed,
any maximal $R$-linearly independent homogeneous subset of $M$ is a
base (cf\. [B$_1$, Th\.~3]).  We write $\dim_R
(M)$ for the rank of $M$ as a free
$R$-module.  Note that if $N$ is any graded $R$-submodule of $M$ then, as
$M/N$ is a graded, hence free, $R$-module, we have $M \cong N \oplus
(M/N)$; so $\dim_R (M) = \dim_R (N) +   \dim_R (M/N)$.  In particular, if
$N$ is a proper graded $R$-submodule of $M$ and $\dim_R (M) < \infty$
then $\dim_R (N) < \dim_R (M)$.

\bigskip\smallskip
%\bigskip

\head \S 2 Graded algebraic extensions of graded fields\endhead

Let $R \subseteq S$ be graded fields (i.e., $R$ is a graded subfield of
$S$).  We set
$$
[S : R]
= \dim_R (S) .
$$
Clearly, $R_0$ is a subfield of $S_0$, $\Gamma_R$ is a subgroup of
$\Gamma_S$, and $QR$ is a subfield of $QS$.  The following proposition is
easy but important.

\proclaim{Proposition~2.1}  (cf\. [B$_2$, p\.
4278])  Let
$R
\subseteq S$ be graded fields with $[S:R] <
\infty$.  Then,
$$
[S:R] 
= [S_0 : R_0]  | \Gamma_S : \Gamma_R |
= [QS: QR] ,
$$
and $QS \cong QR \otimes_R S$. \endproclaim

\demo\nofrills{\smc {Proof.}\usualspace}  If $\{ s_i \}_{i \in I}$ is an
$R_0$-base of $S_0$, and $\{ t_j \}_{j\in J} \subseteq S^*$ with $\{ \degg
(t_j)\}$ a set of representatives for the cosets of $\Gamma_R$ in
$\Gamma_S$.  It is easy to check that $\{ s_i t_j \}$ is an $R$-base of $S$. 
This gives the first formula for $[S:R]$, and the second follows from $QS
\cong QR \otimes_R S$.  This isomorphism holds as $S$ is an integral
domain finitely generated as an $R$-module; so $S$ (a torsion-free
$R$-module) embeds in its localization $QR \otimes_R S$ which is a field,
since it is an integral domain finite-dimensional over the field $QR$. 
$\square$ \enddemo

We will want to consider infinite degree extensions of graded fields which are
algebraic in an appropriate sense.  For this, we first look at gradings on the
polynomial ring $R[x]$ over a graded field $R$.  For any $\delta \in \Delta_R$,
there is a unique grading on $R[x]$ extending that on $R$, such that $x \in R
[x]_\delta$.  We call this the $\delta$-grading of $R[x]$.  An $f \in R[x]$ is said
to be {\it homogenizable} if there is $\delta \in \Delta_R$ such that $f$ is 
homogeneous with respect to the $\delta$-grading of $R[x]$ (cf\. [B$_1$, \S3],
[vGvO, p\. 274]).  The following proposition is
essentially in [vGvO, pp\. 274--276] for
$\Gamma_R \cong \Bbb Z$.

\proclaim{Proposition~2.2}  Let $R$ be a graded field which is a subring of the
(ungraded) field $F$.  For any $\alpha \in F$, the following are equivalent:
\roster
\item"{(i)}"  $\alpha$ is algebraic over $QR$, the minimal polynomial of
$\alpha$ over $QR$, denoted $m_{QR, \alpha}$, lies in $R[x]$ and $m_{QR,
\alpha}$ is homogenizable.

\item"{(ii)}"  $R[\alpha ]$ is a graded field extension of $R$, with $\alpha \in R
[\alpha]^h$ and $[R [\alpha ] : R ] < \infty$.

\item"{(iii)}"  There is a graded field extension $S$ of $R$, with $R \subseteq S
\subseteq F$, $[S:R] < \infty$, and $\alpha \in S^h$.
\endroster\endproclaim

\demo\nofrills{\smc {Proof.}\usualspace} (i) $\Rightarrow$ (ii)  Suppose
$\alpha$ is algebraic over $QR$ and that $m_{QR, \alpha}$ lies in $R[x]$ and
is homogeneous with respect to the $\delta$-grading on $R[x]$, for some
$\delta \in \Delta_R$.  Then $(m_{QR, \alpha})$ is a homogeneous ideal
$R[x]$, so the ring $R[x] / (m_{QR, \alpha})$ inherits a grading from $R[x]$
which extends the grading on $R$.  The canonical evaluation
homomorphism $R [x]
\to F$ given by $f \mapsto  f ( \alpha )$ has image $R[\alpha ]$ and kernel
$\{ f \in R [x] \mid f (\alpha ) = 0\} = (m_{QR, \alpha}  QR [x]) \cap R[x] =
m_{QR,
\alpha} R[x]$.  (The last equality holds as $m_{QR , \alpha}$ is monic in
$R[x]$.)  Hence, the isomorphism $R[\alpha ] \cong R[x] / (m_{QR, \alpha})$
allows us to define a grading on $R[\alpha ]$ extending that on $R$, with
respect to which $\alpha \in R[ \alpha ]^h$ since $\alpha$ is the image of $x
\in R [x]^h$.  Also, $[R (\alpha ) : R] = [R[x]/(m_{QR, \alpha }) : R] = \degg
(m_{QR , \alpha}) < \infty$.  Because $R[ \alpha ]$ is a graded integral
domain (since it lies in $F$), with $[R [\alpha ] : R] < \infty$, $R[ \alpha ]$ is
actually a graded field.  (For, if $\beta \in R[ \alpha ]^h$ , $\beta \not= 0$,
then the map
$\lambda_\beta : R [ \alpha ] \to R [\alpha ]$ given by $c \mapsto \beta c$
is an injective graded $R$-module homomorphism.  Then, as $\im (
\lambda_\beta)$ is a graded $R$-submodule of $R[\alpha ]$ with $\dim_R
(\im (\lambda_\beta )) = [R(\alpha ) : R] < \infty$, so $\lambda_\beta $ is
surjective by dimension count.  (See the remarks at the end of \S1.) 
Hence $\beta \in R [\alpha ]^*$.  

(ii) $\Rightarrow$ (iii)  Take $S = R [ \alpha ]$.

(iii) $\Rightarrow$ (i)  Assume $\alpha \in S^h$ for a graded field $S
\supseteq R$, with $[S:R] < \infty$.  Then $\alpha$ is integral over $R$, since
$[S:R] < \infty$.  So $\alpha$ is algebraic over $QR$, and $m_{QR, \alpha} \in
R[x]$, as $R$ is integrally closed by Cor.~1.3.  Write $m_{QR, \alpha}
= x^n + c_{n-1} x^{n-1} + \ldots + c_0$.  If we take the $n \,\degg ( \alpha )$
homogeneous component of the equation $\alpha^n + c_{n-1} \alpha^{n-1} +
\ldots + c_0$, we get another monic polynomial in $\alpha$ of degree $n$
with coefficients in $R$, which equals $0$.  The uniqueness of $m_{QR ,
\alpha}$ assures that this polynomial must coincide with $m_{QR, \alpha}$. 
Hence, each $c_i \in R_{(n-i) \text{\eightsl deg} ( \alpha)}$.  So $m_{QR,
\alpha}$ is homogeneous for the
$\degg ( \alpha )$-grading of $R[x]$, and $\degg ( \alpha ) = \frac 1 n
\degg (c_0) \in \Delta_R$. $\square$ \enddemo

\definition {Definition~2.3}  An $\alpha$ satisfying the equivalent conditions
of Prop.~2.2 is said to be {\it gr-algebraic} over the graded field
$R$.  For a graded field extension $S$ of $R$, we say $S$ is {\it
gr-algebraic} over
$R$ if each $\alpha \in S^h$ is gr-algebraic over $R$.  (In particular, by
Prop.~2.2, if
$[S:R] < \infty$, then $S$ is gr-algebraic over $R$.)\enddefinition

\proclaim{Corollary 2.4}  Suppose $\alpha$ is gr-algebraic over a graded field
$R$.  Then, 

\roster
\item"{(a)}"  $m_{QR, \alpha}$ determines $\degg ( \alpha)$.

\item"{(b)}"  $\Gamma_{R[\alpha]} = \langle \degg ( \alpha ) \rangle +
\Gamma_R$. 
 
\item"{(c)}"  $\degg ( \alpha ) = 0$ iff $m_{QR, \alpha} \in R_0 [x]$, iff $\alpha$
is algebraic over $R_0$.
 
\item"{(d)}" $R [\alpha ]_0 = R_0$ iff $\alpha^m \in R$, where $m$ is 
the order of
the image of $\degg (\alpha)$ in $\Delta_R / \Gamma_R$.  This occurs iff
$m_{QR, \alpha} = x^n - d_0$, where $ d_0 \in R^h$ and $\degg (d_0)$ has
order $n$ in $\Gamma_ R /n  \Gamma_R$.  Then $n = m = [R[\alpha ] : R]$.
\endroster \endproclaim 

\demo\nofrills{\smc {Proof.}\usualspace}  (a) The proof of Prop.~2.2
shows
$\degg (\alpha ) = \frac 1 n \degg (c_0)$, where $n = \degg (m_{QR, \alpha
})$ and
$c_0$ is the constant term of $m_{QR, \alpha }$.  Part~(b) holds since $\alpha
\in R [\alpha ]^h$.  For (c), if $\alpha$ is algebraic over $R_0$, then $\alpha \in
R [\alpha ]_0$ by Cor.~1.3; the rest of (c) is clear from the proof of
Prop.~2.2.

(d) By (b), $m = | \Gamma_{R[\alpha ]} / \Gamma_R |$.  If $R [\alpha ]_0 =
R_0$, then $R = \bigoplus\limits_{\gamma \in \Gamma_R} R [\alpha
]_\gamma$, by Prop.~2.1, since these graded fields have the same
grade group and the same degree $0$ component.  Since $\degg ( \alpha^m
) \in \Gamma_R$, this shows $\alpha^m \in R$.  Now, suppose we just know
$\alpha^m \in R$.  Let $c_0 = \alpha^m$, so $m_{QR, \alpha} \mid (x^m -
c_0)$.  But, using Prop.~2.1, $\degg (x^m - c_0) = m = |
\Gamma_{R[\alpha ]} /
\Gamma_R | \le [QR(\alpha ) : QR] = \degg (m_{QR, \alpha })$.  Hence,
$m_{QR,
\alpha } = x^m - c_0$.  Since $\degg ( \alpha )$ has order $m$ in $\Delta_R /
\Gamma_R$, we have $\degg (c_0) = m\, \degg (\alpha )$ has order $m$ in
$\Gamma_R / m \Gamma_R$.  Now, suppose instead that $m_{QR, \alpha} =
x^n - d_0$, where $d_0 \in R^h$ and 
$\degg (d_0 )$ has order $n$ in $\Gamma_R / n \Gamma_R$.  Then $\frac 1 n
\degg (d_0)$ has order $n$ in $\Delta_R / \Gamma_R$.  But $\degg ( \alpha
) =
\frac 1 n \degg (d_0)$.  Hence, $m = n$.  Since $m
= |
\Gamma_{R[\alpha ]} / \Gamma_R |$ and $n = [R [\alpha ] : R]$, the equality
$m = n$ forces $R [ \alpha ]_0 = R_0$, by Prop.~2.1.  $\square$
\enddemo

\proclaim{Corollary 2.5}  Let $R \subseteq S$ be graded fields with $S$
gr-algebraic over $R$.  Then,

\roster
\item"{(a)}"  $S$ is the integral closure of $R$ in $QS$.

\item"{(b)}"  $QR \cap S = R$.

\item"{(c)}"  If $\alpha \in S$ and $a$ is gr-algebraic over $R$, then $\alpha \in
S^h$.

\item"{(d)}"  If $\tau$ is any $QR$-automorphism of $QS$, then $\tau (S) = S$
and $\tau |_S \: S \to S$ is a graded field isomorphism.
\endroster \endproclaim

\demo\nofrills{\smc {Proof.}\usualspace}  (a) $S$ is integral over $R$ by
Prop.~2.2, and is integrally closed by Cor.~1.3.  (b)~is immediate
from (a), as
$R$ is integrally closed.

(c)~By Prop.~2.2 there is a grading on $R[x]$ with $\Gamma_{R [x]}
\subseteq \Delta_R$ and $x \in R [x]^h$, such that
$m_{QR, \alpha } \in R [x]^h$.  This grading extends (uniquely) to a grading on
$S[x]$ which extends the grading on $S$.  Since $\Gamma_{S[x]} \subseteq
\Delta_S$, \,$\Gamma_{S[x]}$ is totally ordered.  Hence, the factor $x -
\alpha$ of $m_{QR, \alpha}$ in $S[x]$ must be homogeneous, as $m_{QR,
\alpha} \in S[x]^h$.  So, $\alpha \in S^h$.

(d)~For any $QR$-automorphism $\tau$ of $QS$, we have $\tau (S) = S$ as
$S$ is the integral closure of $R$ in $QS$.  Take any $\alpha \in S^h$.  Then
$m_{QR, \tau (\alpha )} = m_{QR , \alpha}$.  Since $\alpha$ is gr-algebraic
over $R$, Prop.~2.2 shows $\tau( \alpha)$ is also gr-algebraic over
$R$.  Then
$\tau (
\alpha ) \in S^h$ by (c) above, and $\degg ( \tau (\alpha )) = \degg ( \alpha )$
by Cor.~2.4(a).  So the ring isomorphism $\tau |_S$ maps each
homogeneous component of $S$ onto itself.  $\square$ \enddemo

\proclaim{Corollary 2.6}  Let $R \subseteq S$ be graded fields, and let $K$ be
a field containing $S$.  If $\alpha \in K$ is gr-algebraic over $R$, then $\alpha$
is gr-algebraic over $S$.
\endproclaim  

\demo\nofrills{\smc {Proof.}\usualspace}  There is a grading on $R[x]$
extending that on $R$ with $x \in R [x]^h$, such that $m_{QR, \alpha} \in
R[x]^h$.  This grading extends (uniquely) to a grading on $S[x]$ extending that
on $S$, with $\Gamma_{S[x]} \subseteq \Delta_S$.  Since $\alpha$ is integral
over $S$, we have $m_{QS, \alpha} \in S[x]$; then $m_{QS, \alpha } \mid m_{QR,
\alpha }$ in $S[x]$, since this divisibility holds in $QS[x]$, and $m_{QS,
\alpha}$ is monic.  Since $m_{QR, \alpha } \in S[x]^h$ and $\Gamma_{S[x]}$
is totally ordered, we must have $m_{QS, \alpha} \in S[x]^h$.  Hence,
$\alpha$ is gr-algebraic over $S$.  $\square$ \enddemo

The following corollary can now be proved just as in the ungraded case:

\proclaim{Corollary 2.7}  Let $R$ be a graded field which is a subring of the
ungraded field $F$.  Then,

\roster
\item"{(a)}"  If $\alpha_1 , \ldots , \alpha_n \in F$ and each $\alpha_i$ is
gr-algebraic over $F$, then $R [\alpha_1 , \ldots , \alpha_n]$ is a graded field
graded algebraic over $R$, with 
$$[R[\alpha_1 , \ldots , \alpha_n] : R] \le
\prod\limits^n_{i=1} [R[\alpha_i ] : R] < \infty\, .$$

\item"{(b)}"  If $S$ is a graded field, $R \subseteq S \subseteq F$, with $S$
gr-algebraic over $R$, and if $\alpha \in F$ is gr-algebraic over $S$, then
$\alpha$ is gr-algebraic over $R$. 
%\pagebreak 

\item"{(c)}"  Let $A = R [ \{ \alpha \in F \mid \alpha \text{ is gr-algebraic over
} R\} ]$.  Then $A$ is a graded field which is gr-algebraic over $R$, and for
every graded field $S$ with $R \subseteq S \subseteq F$ and $S$
gr-algebraic over $R$, $S$ is a graded subfield of
$A$.
\endroster \endproclaim

\definition {Definition~2.8}  We call the $A$ of Cor.~2.7(c) the {\it graded
algebraic closure} of $R$ in $F$.\enddefinition

Note that even if $F$ is algebraic over $QR$, the field $QA$ may be a
proper subfield of $F$.  The following example is given in [vGvO,
Ex\. 3.10.2]:  Let $R = K[t, t^{-1}]$, where $K$ is
any field of characteristic not $2$, $t$ is
transcendental over $K$, and $R$ is graded
with $R_0 = K$,
$\Gamma_R =
\Bbb Z$, $R_i = Kt^i$; so $QR = K(t)$.  Let $F = K (t) ( \sqrt{t+1})$.  For
$\alpha \in F-QR $ write $\alpha = r + s \sqrt{t+1}$, with $r,s \in QR$, $s
\not= 0$; so,
$m_{QR, \alpha} = x^2 - 2rx + (r^2 - s^2 (t + 1 ))$.  If $\alpha$ is integral
over
$R$, then $r,s \in R$, but then $m_{QR, \alpha}$ is not
homogenizable, since its constant term is not homogeneous in
$R$.  The gr-algebraic  closure of
$R$ in
$F$ is in this case $R$ itself.

Now, for any graded field $R$, let $QR_{alg}$ denote the algebraic closure of
$QR$.  Let $A$ be the graded algebraic closure of $R$ in $QR_{alg}$.  It is clear
that for any gr-algebraic extension field $S$ of $R$, $S$ is graded
$R$-isomorphic to a graded subfield of $A$.  Also, from
Cor.~2.7(b) it follows easily that $A_0 \cong R_{0 alg}$ and
$\Gamma_A = \Delta_R$.  Also,
$QA$ is normal over $QR$, since for each $\alpha \in A^h$, Prop.~2.2 and
Cor.~2.4 show that each root of $m_{QR, \alpha}$ in $QR_{alg}$
actually lies in $A^h$. However, $QA$ need not be Galois over
$QR$.  We call $A$ the {\it gr-algebraic closure} of $R$, denoted
$R_{gr\text{-}alg}$.
\bigskip\smallskip
%\bigskip

\head \S 3.  Unramified, totally ramified, and tame graded field
extensions \endhead 

We now look at some specific types of gr-algebraic extensions of graded
fields.  Let $R \subseteq S$ be graded fields with $S$ gr-algebraic over $R$. 
The torsion group $\Gamma_S / \Gamma_R$ is called the {\it ramification
group} of $S$ over $R$, and $| \Gamma_S / \Gamma_R |$ the {\it ramification
index} of $S$ over $R$.  We say $S$ is {\it totally ramified} over $R$ if $S_0 =
R_0$.  At the other extreme, we say $S$ is {\it unramified} over $R$ if
$\Gamma_S = \Gamma_R$ and $S_0$ is separable over $R_0$.  (This is
analogous to the terminology for extensions of valued fields.)  Note that
every graded extension of  graded fields $R \subseteq S$ has a unique
subextension $A$ with $\Gamma_A = \Gamma_R$ and $S$ totally ramified
over $A$, namely $A = S_0 \otimes_{R_0} R$.

\remark{Remark 3.1}  If $\Gamma_S = \Gamma_R$, then clearly $S
\cong_g S_0 \otimes_{R_0} R$, and $QS \cong S_0 \otimes_{R_0} QR$. 
For any intermediate graded field $A$, $R \subseteq A \subseteq S$,
necessarily
$\Gamma_A = \Gamma_R$, so $A \cong_g A_0 \otimes_{R_0} R$.  Thus,
there is a one-to-one correspondence between intermediate graded fields
$A$, $R \subseteq A \subseteq S$, and intermediate fields $A_0$, $R_0
\subseteq A_0 \subseteq S_0$.  Further $QS$ is separable (resp\. purely
inseparable, resp\. Galois) over $QR$ iff $S_0$ is separable (resp\.
purely inseparable, resp\. Galois) over $R_0$.  For the Galois case, use
Cor.~2.5; note that then $\Cal G (QS /QR) \cong \Cal G (S_0 /
R_0)$.\endremark
\medskip

For any graded field $R$ and any subgroup $\Lambda$ of
$\Gamma_R$, we write
$R|_\Lambda$ for $\bigoplus\limits_{\gamma \in \Lambda} R_\gamma$, a graded
subfield of~$R$.

\proclaim {Proposition 3.2}  Let $R \subseteq S$ be graded fields with $S$
gr-algebraic over $R$.  Suppose $S$ is totally ramified over $R$.  Then,
\roster
\item"{(a)}"  $R = S|_{\Gamma_R}$.

\item"{(b)}"  Every intermediate graded field $A$, $R \subseteq A \subseteq
S$ has the form $A = S |_\Lambda$, where $\Lambda$ is a group, $\Gamma_R
\subseteq \Lambda \subseteq \Gamma_S$.  Thus, there is a one-to-one
correspondence between subgroups of $\Gamma_S / \Gamma_R$ and
intermediate graded fields.

\item"{(c)}"  Suppose $[S:R] = n < \infty$, so $\Gamma_S /\Gamma_R \cong
\Bbb Z / t_1 \Bbb Z \times \ldots \times \Bbb Z / t_k \Bbb Z$ with $t_1
\ldots t_k = n$.  Pick any $s_i \in S^h - \{ 0 \}$ with $\degg (s_i)$ mapping
to a generator of the $i$-th component in the given cyclic decomposition
of
$\Gamma_S / \Gamma_R$, $1 \le i \le k$.  Then, $s^{t_i}_i \in R^h$, $[R[s_i ]:R] =
t_i$, and $S \cong_g R[s_1] \otimes_R \ldots \otimes_R R[s_k]$.  Likewise,
$QS \cong QR (s_1) \otimes_{QR} \ldots \otimes_{QR} QR(s_k)$, with $[QR
(s_i) : QR] = t_i$ and $s^{t_i}_i \in QR$.
\endroster\endproclaim

\demo\nofrills{\smc {Proof.}\usualspace} (a)~is clear since
$(S|_{\Gamma_R})_0 = S_0 = R_0$ and also $S|_{\Gamma_R}$ and $R$ have the
same grade group.  (b)~is immediate from (a), since for $R \subseteq A \subseteq
S$, $S$ is totally ramified over $A$, since it is totally ramified over $R$. For
(c), take any $s_i$ as described in (c).  Since $t_i \degg (s_i) \in
\Gamma_R$, we have $s^{t_i}_i \in S_{t_i \text{\eightsl deg} (s_i)} = R_{t_i
\text{\eightsl deg} (s_i)}$ (see (a)).  Then
$[R[s_i] : R] = t_i$ by Cor.~2.4(d).  The graded ring 
homomorphism \break 
$\rho \: R [s_i] 
\otimes_R \ldots \otimes_R R [s_k] \to S$ has image
a graded subfield of $S$ with
$\Gamma_{\text{\eightsl im} (\rho)} = \Gamma_S$. 
Hence
$\im (\rho) = S$ by (b).  Because $[S: R] = t_i \ldots t_n = \dim_R (R[s_1]
\otimes_R \ldots
\otimes_R R [s_k])$, $\rho$ must be an isomorphism.  The isomorphism for
$QS$ follows from this by Prop.~2.1.  $\square$ \enddemo

The Galois case deserves special attention:

\proclaim{Proposition 3.3}  Let $R \subseteq S$ be graded fields with $[S:R] <
\infty$ and $S$ totally ramified over $R$.  Let $\ell$ be the exponent of
$\Gamma_S / \Gamma_R$, and let $\mu_\ell$ denote the group of all
$\ell$-th roots of unity in $QR_{alg}$.  Then,
\roster
\item"{(a)}" $QS$ is Galois over $QR$ iff $\chr (R_0) \nmid [S:R]$ and
$\mu_\ell
\subseteq R_0$.

\item"{(b)}" When $QS$ is Galois over $QR$, there is a (well-defined)
perfect pairing
$$
\Gamma_S /\Gamma_R \times \Cal G (QS/QR) \to \mu_\ell
$$
given by $(\gamma + \Gamma_R , \tau ) \mapsto a / \tau (a)$, for any $a \in
S^h$,
$a \not= 0$, with $\degg (a) = \gamma$.  Furthermore, $QS$ is an
$\ell$-Kummer extension of $QR$, and if $B = \{ b \in QS^* \mid b^\ell \in
QR^* \}$, then $B /QR^*
\cong \Gamma_S / \Gamma_R$ and there is a commutative diagram
$$
\CD
\Gamma_S/\Gamma_R \times \Cal G (QS / QR) @>>> \mu_\ell\\
@V \cong VV @|\\
B/QR^* \times \Cal G (QS/QR) @>>> \mu_\ell
\endCD
$$
with the lower row the perfect Kummer pairing, given by $(b\, QR^* , \tau )
\mapsto b / \tau (b)$.\endroster\endproclaim  

\demo\nofrills{\smc {Proof.}\usualspace}  (a) From the description of $QS$ in
Prop.~3.2(c) as a radical extension of $QR$, it is clear that $QS$ is
separable over $QR$ iff $\chr (R_0) \nmid [S:R]$; further, as $\ell = \lcm (t,
\dots , t_k)$ for the $t_i$ of Prop.~3.2(c), $QS$ is Galois over $QR$ iff
additionally
$QR$ contains the required roots of unity, i.e., $\mu_\ell \subseteq QR$. 
When this occurs, $\mu_\ell \subseteq R_0$ by Cor.~1.3, since $\mu_\ell$
is algebraic over $R_0$.  Also, $QS$ is clearly an $\ell$-Kummer extension
of $QR$.

For (b) assume now that $QS$ is Galois over $QR$.  It is easy to check that the map
$\Gamma_S / \Gamma_R \times \Cal G (QS / QR) \to \mu_\ell$ is a well-defined
pairing of finite abelian groups.  Also, there is a well-defined group
homomorphism $\beta \: \Gamma_S /\Gamma_R \to B / QR^*$ given by
$\gamma +
\Gamma_R \mapsto a\, QR^*$ for any $a \in S_\gamma$, $a \not= 0$.  This
$\beta$ is clearly injective, and is an isomorphism since $| \Gamma_S /
\Gamma_R | = [S:R] = [QS:QR] = | B / QR^*|$, the last equality given by
Kummer theory.  The diagram of the proposition is clearly commutative,
and its vertical lines are isomorphisms.  Since the bottom row of the
diagram is the perfect pairing of Kummer theory, the top row must also
be  a perfect pairing. 
$\square$ \enddemo

The graded analogue to a tamely ramified extension of a valued field is of
particular interest here.  

\definition{Definition 3.4}  Let $R \subseteq S$ be graded fields with $S$
gr-algebraic over $R$.  Then $S$ is said to be {\it tame} over $R$ if $\chr
(R_0) = 0$ or $\chr (R_0) = p \not= 0$ and $S_0$ is separable over $R_0$
and
$\Gamma_S / \Gamma_R$ has no $p$-torsion.\enddefinition

Clearly, for graded fields $R \subseteq A \subseteq S$, $S$ is tame over $R$
iff $S$ is tame over $A$ and $A$ is tame over $R$.

\proclaim{Proposition 3.5}  (cf\. [B$_1$, Th\.
4])  Let $R \subseteq S$ be a gr-algebraic
extension of graded fields.  Then $S$ is tame
over $R$ iff $QS$ is separable over
$QR$.\endproclaim

\demo\nofrills{\smc {Proof.}\usualspace}  Since this is clear if $\chr (R_0) =
0$, assume $\chr (R_0) = p \not= 0$.  Let $I = S_0 \otimes_{R_0} R$.  Since $QI
\cong S_0 \otimes_{R_0} QR$, $I_0 = S_0$, and $\Gamma_I = \Gamma_R$, we
have $I$ is tame over $R$ iff $I_0$ is separable over $R_0$ iff $QI$ is separable
over
$QR$.  Since $S$ is totally ramified over $I$, Prop.~3.2 shows that
for every finite degree subextension $S^\prime$ of $S$,
$QS^\prime$ is a radical extension of $QI$ 
obtained by adjoining $t_i$-th
roots of homogeneous elements of
$I$, where $\lcm ( \{ t_i \}) = \exp
(\Gamma_{S^\prime}/\Gamma_I)$.  Thus, 
$QS$ is separable over $QI$ iff each $QS^\prime$ is separable
over 
$QI$ iff each $\Gamma_{S^\prime}  /\Gamma_I$ has no $p$-torsion
iff
$\Gamma_S / \Gamma_I$ has no $p$-torsion iff $S$ is tame over
$I$.  The transitivity of tameness and of separable field
extensions completes the proof.  $\square$ \enddemo

\proclaim{Lemma 3.6}  Let $R \subseteq S$ be a graded algebraic extension of
graded fields with $\chr (R_0) = p$.  There is no proper tame extension of $R$
in $S$ iff $S_0$ is purely inseparable over $R_0$ and $\Gamma_S / \Gamma_R$
is a $p$-primary torsion group, iff $QS$ is purely inseparable over
$QR$.\endproclaim

\demo\nofrills{\smc {Proof.}\usualspace}  In the first equivalence,
$\Leftarrow$ is clear.  For $\Rightarrow$, suppose $R$ has no proper tame
extension in $S$.  Let $A_0$ be the separable closure of $R_0$ in $S_0$.  Since
$A_0 \otimes_{R_0} R$ is a tame extension of $R$ in $S$, we must have $A_0
= R_0$, so $S_0$ is purely inseparable over $R_0$.  Suppose $\Gamma_S
/\Gamma_R$ is not $p$-primary.  Then, there is $\gamma \in \Gamma_S -
\Gamma_R$ with image of order $n$ in $\Gamma_S / \Gamma_R$, where $p
\nmid n$.  Take any $s \in S_\gamma$, $s \not= 0$.  Then $\degg (s^n) \in
\Gamma_R$, so if we take any $r \in R_{n \gamma} $, $r \not= 0$, we have $s^n
/ r \in S_0$.  Because $S_0$ is purely inseparable over $R_0$, $(s^n / r)^{p^k} \in
R_0$ for some $k$.  So, $(s^{p^k})^n = (s^n / r)^{p^k} r^{p^k} \in R_{p^k n
\gamma}$.  Since $\degg (s^{p^k}) = p^k \gamma$, which has order $n$ in
$\Gamma_S / \Gamma_R$, $R [s^{p^k}]$ is a proper tame and totally ramified
extension of $R$, by Cor.~2.4(b) and (d), a contradiction.  Hence,
$\Gamma_S / \Gamma_R$ must be $p$-primary.  This proves the first
equivalence.

Let $A = S_0 \otimes_{R_0} R$.  Then, $QA$ is purely inseparable over $QR$
iff $S_0$ is purely inseparable over $R_0$ (see Remark~3.1).  Also, as $S$ is
totally ramified over $A$ and $\Gamma_A = \Gamma_R$, $QS$ is purely
inseparable over $QA$ iff $\Gamma_S /\Gamma_R$ is $p$-primary torsion,
as $QS$ is built from $QA$ by radical extensions of degrees the orders of
elements in
$\Gamma_S /\Gamma_R$, by Prop.~3.2(a).  $\square$ \enddemo 

\proclaim{Proposition 3.7}  Let $R \subseteq S$ be a gr-algebraic
extension of graded fields. Let $T = L \cap S$, where $L$ is the separable
closure of $QR$ in $QS$.  Then $T$ is a graded field gr-algebraic and tame
over $R$.  Also, for any graded field $A$ with $R \subseteq A
\subseteq S$, $A$ is tame over $R$ iff $A \subseteq T$.  \endproclaim

\demo\nofrills{\smc {Proof.}\usualspace}  Let $U$ be a maximal tame
graded field extension of $R$ in $S$. Such a $U$ exists by Zorn's Lemma,
using the transitivity of the property of tameness.  Then $QU$ is separable
over $QR$, by Prop.~3.5, and $QS$ is purely inseparable over $QU$ by
Lemma~3.6.  Hence, $QU = L$, so that by Cor.~2.5(b), $U = QU \cap
S = L
\cap S = T$.  This shows $T$ is a graded field tame over $R$.  Also,
$QT = QU = L$.  If $A$ is any graded field, $R \subseteq A \subseteq S$ and
$A$ is tame over $R$, then $QA$ is separable over $QR$ by Prop.~3.5. 
Hence, by Cor.~2.5(b) again, $A = QA \cap S \subseteq L \cap S = T$. 
Conversely, if $A \subseteq T$, then $A$ is tame over $R$, since $T$ is tame
over $R$. $\square$ \enddemo

\definition{Definition 3.8}  The $T$ of Prop.~3.7 is called the {\it
tame closure} of $R$ in $S$.  Note that by Lemma~3.6, $T_0$ is the
separable closure of $R_0$ in $S_0$, and (if $\chr (R_0) = p \not= 0$)
$\Gamma_T /
\Gamma_R$ is the prime-to-$p$ part  of $\Gamma_S / \Gamma_R$.  Also,
$QT$ is the separable closure of $QR$ in $QS$, as we saw in the proof of
Prop.~3.7.\enddefinition

\proclaim{Theorem 3.9}  Let $R$ be a graded field.  Let $T$ be the tame
closure of $R$ in $R_{gr\text{-}alg}$.  Then,
\roster
\item"{(a)}"  $QT$ is the separable closure of $QR$ in
$Q(R_{gr\text{-}alg})$, so $QT$ is Galois over $QR$.

\item"{(b)}"  There is a one-to-one correspondence between 
the graded fields
$A$ such that $R \subseteq A \subseteq R_{gr\text{-}alg}$ with
$A$ tame over $R$ and the fields
$L$ with $QR \subseteq L \subseteq QT$.  (The correspondence
is given by $A
\mapsto QA$ and $L \mapsto T \cap L$.)\endroster \endproclaim

\demo\nofrills{\smc {Proof.}\usualspace}  (a)~Since $Q(R_{gr\text{-}alg})$
is normal over $QR$, as noted after Definition~2.8, and $QT$ is the
separable closure of $QR$ in $Q(R_{gr\text{-}alg})$, $QT$ is Galois over
$QR$ (typically of infinite degree).  

(b)~If $A$ is a graded field with $R \subseteq A \subseteq
R_{gr\text{-}alg}$  and
$A$ tame over $R$, then $A \subseteq T$ by Prop.~3.7, so $QA
\subseteq QT$ and $A = T \cap QA$, by Cor.~2.5(b).  On the other hand,
let $L$ be any field with $QR \subseteq L \subseteq QT$.  So $QT$ is Galois
over $L$.  Let $H = \Cal G (QT/L)$.  By Cor.~2.5(d), each $\tau \in H$ maps
$T$ to $T$ by a graded field isomorphism.  Hence, as $T \cap L = T^H$, the
$H$-fixed points of $T$, $T \cap L$ is a graded integral domain, which is a
graded field, since it is gr-algebraic over $R$.  For any $\ell \in L$, since
$\ell$ is algebraic over $QR$, there is $r \in R$, $r \not= 0$ with $r \ell $
integral over $R$.  Then, as $T$ is the integral closure of $R$ in $QT$, $r \ell
\in T \cap L$.  This shows $L = Q (T \cap L)$, which proves the one-to-one
correspondence asserted in the Theorem.  $\square$ \enddemo

Note one somewhat surprising consequence of the theorem just proved:  If
$t \in T$, then $QR (t) \cap T$ is a graded field containing $t$, so all the
homogeneous components of $t$ lie in $QR(t)$.  This requires the tameness
of $R[t]$ over $R$, as the next example illustrates:

\example{Example 3.10}  Let $R_0$ be a field,  $\chr (R_0) = p
\not= 0$, and let $R = R_0 [s, t, s^{-1} , t^{-1}]$, where $s$ and $t$
are algebraically independent over $R_0$.  Then $R$ is a graded
field with $\degg (s) = (1,0)$ and $\degg (t) = (0, 1)$ in
$\Gamma_R =
\Bbb Z \times \Bbb Z$, and $QR$ is the rational function field $R_0
(s,t)$.  Let $S = R [\root \uproot 3 p \of s,  \root \uproot 3 p \of
t]$, which is a totally ramified but not a tame gr-algebraic graded
field extension of $R$.
%which is a gr-algebraic graded field extension which is totally
%ramified but not tame over $R$.  
Let $u = \root \uproot 3 p \of s +
\root \uproot 3 p \of t =
\root \uproot 3 p
\of{s+t} \in S$.  Then,
$\root \uproot 3 p \of s , \root \uproot 3 p \of t \notin QR (u)$, since otherwise $QR (u) = QS$.  This
cannot occur, as $[QR (u) : QR] = p$ while $[QS : QR] = p^2$.  Here, $QR (u)
\cap S$ is not a graded subring of $S$ since it contains $u$ but not its
homogeneous components $\root \uproot 3 p \of s $ and $\root \uproot 3 p \of t$.  In fact, the kind
of one-to-one correspondence described in the theorem for the tame case
fails dramatically here, since there are infinitely many fields $L$ with $QR
\subseteq L \subseteq QS$, but by Prop.~3.2(b) there are only
finitely many graded fields $A$ with $R \subseteq A \subseteq S$.
\endexample

\proclaim{Theorem 3.11}  Let $R \subseteq S$ be graded fields with $[S:R]
< \infty$.  Then, 
\roster
\item"{(a)}"  $S$ is separable over $R$ iff $S$ is tame over $R$ iff $QS$ is
separable over $QR$.

\item"{(b)}"  $S$ is Galois over $R$ iff $QS$ is Galois over $QR$.  When this
occurs, $\Cal G (S/R) \cong \Cal G (QS /QR)$. 

\item"{(c)}"  If $S$ is separable over $R$, there is a graded field $A
\supseteq S$ with $[A : S] < \infty$ and $A$ Galois over $R$.
\endroster \endproclaim

\demo\nofrills{\smc {Proof.}\usualspace}  (b)~If $S$ is Galois over $R$, with
group $N$, then by base extension (cf\. [G, p.~5, Lemma~1.11]) $QS$ is Galois
over $QR$, as $QS \cong QR \otimes_R S$ by Prop.~2.1.  Also, the
general base extension result shows $\Cal G (QS /QR) \cong N$.  Conversely,
suppose $QS$ is Galois over $QR$.  Let $G = \Cal G (QS / QR)$.  By
Cor.~2.5, each $\tau \in G$ restricts to a graded automorphism of $S$. 
Furthermore, $S^G = S \cap QR = R$ by Cor.~2.5(b).  For $\tau \in G$,
$\tau \not= \id_{QS}$ there is $t \in S^h$, $t \not= 0$ with $\tau (t) \not=
t$.  Since $\tau$ preserves degrees $\tau (t) - t \in S^h - \{ 0 \} = S^*$.  So,
there is no maximal ideal $M$ of $S$ containing $\{ \tau (s) - s \mid s \in
S\}$.  This together with $S^G = R$ shows that $S$ is Galois over $R$ with
respect to $G$, by [CHR, Th.~1.3] or [G, pp.~2--3, Th.~1.6].

(a) and (c) \ Now, assume $QS$ is separable over $QR$.  Then $S$ is tame
over $R$ by Prop.~3.5, so we can assume $S$ lies in the tame closure
$T$ of $R$ in $R_{gr\text{-}alg}$.  Since $QT$ is Galois over $QR$ by Th.~3.9,
the normal closure $L$ of $QS$ over $QR$ lies in $QT$.  Furthermore, $L$ is
Galois over $QR$ and $[L : QR] < \infty$.  Let $A = T \cap L$, which is a graded
subfield of $T$ with $QA = L$, by Th.~3.9, and $[A:R] = [L:QR] < \infty$
by Prop.~2.1.  So, $A$ is Galois over $R$ with respect to (the
restriction to $A$ of) $\Cal G (L /QR)$, as we just proved.  Let $H = \Cal G (L /
QS) \subseteq \Cal G (L/ QR)$.  Since $A^H = A \cap QS = S$ by
Cor.~2.5(b), $S$ is separable over $R$ by [CHR, Th\.
2.2] or [G, p.~7, Th.~2.2].  On the other hand, if $S$ is
separable over $R$, then $QS \cong QR
\otimes_R S$ is separable over $QR$ as separability is preserved under
base extension.  By Prop.~3.5 this is equivalent to $S$ being tame
over $R$.  This yields (a), and also (c) along the way.  
$\square$ \enddemo
\bigskip\smallskip

\head \S 4. Tame extensions of valued fields
\endhead 

We now recall some facts about tame extensions of valued fields,
which will be needed for our comparison of valued fields and graded
fields in the next section.  Everything we mention in this section is
known, but a concise summary seems worthwhile.

Let $F$ be a field, $\Gamma$ a totally ordered abelian group, and $v \:
F^* \to \Gamma$ a (Krull) valuation on $F$.  (That is, (i)~$v ( \alpha
\beta) = v(\alpha) + v (\beta)$, and 
(ii)~$v ( \alpha + \beta) \ge \min (v ( \alpha ) , v (
\beta))$ for all
$\alpha, \beta \in F^*$ (with $\beta \not= -\alpha$ in
(ii)).)  Let
$\Gamma_F = \im (v)$, the value group of $v$; let $V_F$ be the
valuation ring of $v$; $M_F$ the unique maximal ideal of $V_F$; and
$\overline F = V_F / M_F$, the residue field of $v$.  The indexing by $F$
will cause no confusion, because we will never consider more than
one valuation at a time on any given field.  Let $\overline p = \chr ( \overline
F)$.

Now, let $F \subseteq K$ be fields with $K$ algebraic over $F$, and let
$v$ be a valuation on $F$ which has a unique extension (also called
$v$) to $K$.  If $[K:F] < \infty$, we say $K$ is {\it tame} over $F$ (with
respect to $v$) if $\overline K$ is separable over $\overline F$, \ $\overline p
\nmid |
\Gamma_K : \Gamma_F |$, and $[K:F] = [\overline K : \overline F] | \Gamma_K :
\Gamma_F|$.  In the terminology of Endler [E, pp\.
178-180],
$K$ is tamely ramified and defectless over $F$.  If
$[K:F] = \infty$, we say that $K$ is tame over $F$ if for
every field $L$ with $F \subseteq L \subseteq K$ and
$[L:F] < \infty$ we have $L$ is tame over $F$ (with
respect to the restriction of $v$ from $K$ to $L$). 
Recall that if $K$ is tame over $F$, then $K$ is
separable over $F$.
This follows at once from the fact that if 
$L$ is purely inseparable over $F$, then
$\overline L$ is purely inseparable over 
$\overline F$ and $\Gamma_L/\Gamma_F$ is 
a $\overline p$-group (cf\. [E, Ex. III-6, pp\. 
228-229]).  Note also that if 
$[K:F] < \infty$ and $N$
is any field with $F \subseteq N \subseteq K$, then
$K$ is tame over $F$ iff
$N$ is tame over $F$ and $K$ is tame over $N$.  This
is immediate from the fundamental inequality
[E, (13.10)].

Still assuming $v$ has a unique extension from 
$F$ to $K$, suppose $[K:F] <\infty$ and
$K$ is Galois over $F$. Let  $G$ be the Galois
group $\Cal G 
(K/F)$.  Let $G^V = \{ \sigma \in G \mid
\sigma (c) / c  \in M_K$ for all $c \in K^*\}$ the {\it ramification subgroup}
($=$ Verzweigungsgruppe) of $v$, and let $F^V$ be
the fixed field
$K^{G^V}$, which is called the {\it ramification field} of $K$ over $F$. 
Then it is known (cf\. [E, (20.11), (19.12), (20.20), 
(19.14)])
 that $G^V$ is a $\overline p$-group (if
$\overline p = 0$ this means  
$|G^V| = 1$), $G^V$ is a normal subgroup of $G$, so
$F^V$ is Galois over $F$, and that 
$F^V$ is tame over $F$
(since $F^V$ is tame over the inertia field 
$F^T$ and $F^T$ is tame over $F$), and that
$\overline K$ purely inseparable over
$\overline{F^V}$ and $\Gamma_K /
\Gamma_{F^V}$ is a $\overline p$-group (so $\Gamma_{F^V} = \Gamma_K$
if $\overline p = 0$).  Hence, $F^V$ can have no
proper tame extension in $K$, so $F^V$ is a maximal
tame extension of $F$ in
$K$.  For any field $L$ with $F \subseteq L 
\subseteq K$, let $H$ be the
corresponding group $\Cal G  (K/L)$.  Since $H^V = G^V \cap H$, we
have
$L^V = L \cdot F^V$.  If $L$ is tame over $F$, then $L^V$ is also tame
over $F$, by the transitivity of tameness, so $L^V = F^V$, and hence
$L \subseteq F^V$.  Thus, $F^V$ is the unique maximal tame extension
of $F$, and an intermediate field $L$ is tame over $F$ iff $L \subseteq
F^V$.  It follows that if intermediate fields $L_1, L_2$ are each tame
over
$F$, then $L_1 \cdot L_2$ is tame over $F$.  Thus, if $N$ is any
intermediate field, then there is a unique maximal tame extension
$N_0$ of $F$ inside $N$, and we have $N_0 = F^V \cap N$.  We call
$N_0$ the {\it tame closure} of $F$ in $N$.

All of the preceding paragraph extends readily to the case of infinite
degree algebraic extensions (see [E, (20.12), 
(20.17), (20.18) ]) and
leads to the following in the Henselian case:

Suppose $F$ is a field with a Henselian valuation $v$.  We work in
some fixed algebraic closure $F_{alg}$ of $F$.  That $v$ is Henselian
means that there is a unique extension of $v$ to $F_{alg}$.
%\pagebreak

\proclaim{Proposition 4.1}  Suppose a field $F$ has a Henselian
valuation $v$.  Then, 

\itemitem{(a)}  There is a unique maximal tame extension $F_t$ of
$F$ in $F_{alg}$, and $F_t$ is the compositum of all the finite degree
tame extensions of $F$ in $F_{alg}$.

\itemitem{(b)}  $\overline{F_t} = ( \overline F )_{sep}$,
$\Gamma_{F_t} = \Gamma_F \otimes_{\Bbb Z} {\Bbb Z} _{( \overline
p)}$ (if $\overline p = 0$, then $\Gamma_{F_t} = \Gamma_F
\otimes_{\Bbb Z} \Bbb Q$).

\itemitem{(c)}  For any field $L$, $F \subseteq L \subseteq F_{alg}$,
we have $L$ is tame over $F$ iff $L \subseteq F_t$.

\itemitem{(d)}  For any field $N$, $F \subseteq N \subseteq F_{alg}$,
we have $N_t = F_t \cdot N$, and $N \cap F_t$ is the unique maximal
tame extension of $F$ in $N$.\endproclaim

\demo\nofrills{\smc {Proof.}\usualspace}  Since any tame extension
of $F$ is separable over $F$, this follows easily from the preceding
discussion, when we take $F_t = F^V$, the ramification field with
respect to the (usually infinite degree) Galois extension $F_{sep}$
over $F$, where $F_{sep}$ is the separable closure of $F$ in $F_{alg}$. 
$\square$ \enddemo

\definition{Definition 4.2}  We call the $F_t$ of Prop.~4.1 the
{\it tame closure} of $F$ and $N \cap F_t$ the tame closure of $F$ in
$N$.  Note that $\overline{N \cap F_t}$ is the separable closure of
$\overline F$ in $\overline N$, and $\Gamma_{N \cap F_t} /
\Gamma_F$ is the prime-to-$\overline p$ part of $\Gamma_N /
\Gamma_F$.  Also, $[N: N \cap F_t]$ is a power of $\overline p$ ($N =
N \cap F_t$ if $\overline p = 0$). \enddefinition
\bigskip\smallskip

\head \S 5 Correspondence between graded fields and valued
fields\endhead 

We now recall how valued fields can be obtained from graded fields, and
vice versa.  We will then prove correspondences for the tame extensions
of each kind of field.

Let $R = \bigoplus\limits_{\gamma \in \Gamma_R} R_\gamma$ be a
graded field (with $\Gamma_R$ totally ordered, as we are always
assuming).  Define a function $v \: R - \{0 \} \to \Gamma_R$ by 
$$
v \big( \tsize\sum\limits_\gamma r_\gamma \big)
= \text{the least $\delta$ such that } r_\delta \not= 0 .
$$
So, for $a, a^\prime \in R - \{ 0 \}$, we clearly have
\roster
\item"{(i)}" $v (aa^\prime) = v (a) + v ( a^\prime)$;

\item"{(ii)}" $v (a + a^\prime ) \ge \minn (v (a) , v(a^\prime))$, if $a^\prime
\not= -a$.
\endroster
This function $v$ extends canonically to a function $v \: QR - \{ 0 \} \to
\Gamma_R$ by $v (rs^{-1}) = v(r) - v (s)$, for $r,s \in R - \{ 0 \}$.  Property
(i) shows that $v$ is well-defined on $QR - \{ 0 \}$.  Properties (i) and (ii)
hold for all $a, a^\prime \in QR - \{0 \}$.  Thus $v$ is a valuation on $QR$, for
which clearly $\Gamma_{QR} = \Gamma_R$.  Also, for the residue field,
$\overline {QR} = R_0$.  For, the canonical injection $R_0 \to
\overline{QR}$ is onto, since if $v (rs^{-1}) = 0$, then $v(r) = v(s)$, so $r =
r_\gamma \, +$ higher degree terms, $s = s_\gamma \, +$ higher
degree terms, for $\gamma = v (r) = v (s))$, yielding $\overline
{rs^{-1}} =
\overline{r_\gamma s^{-1}_\gamma} \in \im (R_0)$.  

Let $HR$ denote the Henselization of $QR$ with respect to $v$ (see, e.g\.
[E, p\. 131] for the Henselization of a valued field).  If
$S$ is a gr-algebraic extension graded field of $R$,
then
$QS \supseteq QR$ canonically, and while
$HS$ is determined only up to isomorphism in $QS_{alg}$, we will assume
$HS$ has been chosen to be $QS \cdot HR$.

Now, suppose instead we start with a field $F$ with a valuation $v$ on $F$. 
For each $\gamma \in \Gamma_F$, define
$$
\align
W^\gamma
&= \{ c \in F^* \mid 
v (c) \ge \gamma \} \cup \{ 0 \} \\
W^{> \gamma}
&= \{ c \in F^* \mid v (c) > \gamma \} 
\cup \{ 0 \} .
\endalign
$$
Then set $R_\gamma = W^\gamma / W^{>\gamma}$.  Define multiplication
$R_\gamma \times R_\delta \to R_{\gamma + \delta}$ by, for $a
\in W^\gamma$ and  $b \in W^\delta$:
$\ (a + W^{> \gamma}) \cdot (b + W^{>
\delta}) = ab + W^{> \gamma + \delta }$.  This is well-defined, and extends
to a multiplication on all of $\bigoplus\limits_{\gamma \in \Gamma_F}
R_\gamma$, making it a graded field.  We denote this graded field by
$GF$.  Clearly, $\Gamma_{GF} = \Gamma_F$ and $GF_0 = \overline F$. 
Note that if $K$ is a field with a valuation extending $v$ on $F$, then $GK$ is
a graded field extension of $GF$.

If we start with a graded field $R$, and build the valuation on $QR$ as
described above, then form the associated graded field $GQR$, then $GQR
\cong_g R$, canonically.  For, each $R_\gamma$ maps bijectively onto
$(GQR)_\gamma$.  Likewise, $GHR \cong_g R$ canonically.  On the other
hand, if we start with a valued field $F$, we need not have $QGF \cong F$,
nor $HGF \cong F$, even if the valuation on $F$ is Henselian.  (For example, 
$\chr (QGF) = \chr ( \overline F)$, which need not equal $\chr (F)$.  Also, if
$\overline F$ and $\Gamma_F$ are countable, then $QGF$ is countable,
though $F$ might be uncountable.)

\proclaim{Proposition 5.1}  Let $R$ be a graded field, and let $S$ be the
tame closure of $R$ in $R_{gr\text{-}alg}$.  Then $HS$ is the maximal tame
extension of $HR$, and $\Cal G (HS / HR) \cong \Cal G (QS / QR)$. 
Hence, there are one-to-one correspondences
$$
\align
&\text{tame gr-algebraic graded field extensions of $R$ in }
R_{gr\text{-}alg}\\
&\leftrightarrow \text{ field extensions of } QR \text{ in } QS\\
&\leftrightarrow \text{ tame field extensions of } HR \text{ in } (HR)_{alg} .
\endalign
$$\endproclaim

\demo\nofrills{\smc {Proof.}\usualspace}  Let $T$ be a graded field with $R
\subseteq T \subseteq S$ and $[T:R] < \infty$.  So, $T$ is tame over $R$ by
Prop.~3.7.  Then, $\overline{QT} = T_0$ is separable over $\overline {QR} =
R_0$, and 
$$
[QT : QR] = [T:R] =
%\linebreak 
[T_0 : R_0] | \Gamma_T : \Gamma_R |
= [
\overline {QT} : \overline {QR} ] | \Gamma_{QT} : \Gamma_{QR}| \, ;
$$
hence $QT$ is a tame valued field extension of $QR$.  Since by
convention $HT = QT
\cdot HR$, we have 
$$
\align
[HT : HR] \,&\le\, [QT : QR] \,=\, [\overline{QT} : \overline {QR}]
| \Gamma_{QT} : \Gamma_{QR} | \\
&=\,
%\linebreak
 [ \overline{HT} : \overline{HR}]
|\Gamma_{HT} : \Gamma_{HR} | \,\le \, [HT : HR]\, ;
\endalign
$$ the last inequality is the
fundamental inequality for valued field extensions 
[E, (13.10)].  So, equality
holds throughout.  Since $\overline {HT} = T_0$ is separable over
$\overline {HR} = R_0$, this shows $HT$ is tame over $HR$.  Also, the
equality just proved shows $QT$ and $HR$ are linearly disjoint over $QR$. 
Since $S$ is the union of such finite-degree extensions as $T$, \, $HS$ is
tame over $HR$, and $QS$ and $HR$ are linearly disjoint over $QR$. 
Furthermore, as $\overline {HS} = S_0$ is separably closed and
$(\Gamma_{HS}
\otimes_{\Bbb Z} \Bbb Q) / \Gamma_{HS} = \Delta_R / \Gamma_S$ is
$p$-primary where $p = \chr (R_0) = \chr (\overline {HS})$, $HS$ can have
no proper tame field extensions.  Hence, $HS$ is the maximal tame
extension of the valued field $HR$.  From the linear disjointness, we have
$\Cal G (HS/ HR) = \Cal G (QS \cdot HR / HR) \cong \Cal G (QS / QR)$.  

We have seen in Th.~3.9(b) the one-to-one correspondence between tame
gr-algebraic graded field extensions of $R$ and field extensions of $QR$
in $QS$.  The isomorphism of Galois groups gives the one-to-one
correspondence between field extensions of $QR$ in $QS$ and the field
extensions of $HR$ in $HS$.  The latter are precisely the tame field
extensions of $HR$ in $(HR)_{alg}$. $\square$ \enddemo

\proclaim{Theorem 5.2}  Let $F$ be a field with Henselian valuation $v$, and
let $K$ be the maximal tame extension of $F$ in $F_{alg}$ re $v$.  Then $GK$
is the tame closure of $GF$ in $GF_{gr \text{-}alg}$, and $\Cal G (QGK / QGF)
\cong \Cal G (K/F)$.  There are one-to-one correspondences:
$$
\align
&\text{tame  field extensions of $F$ in $F_{alg}$ re } v\\
&\leftrightarrow \text{ tame graded field extensions of } GF \text{ in
}(GF)_{gr \text{-}alg}\\ 
&\leftrightarrow \text{ field extensions of } QGF
\text{ in } QGK
 .
\endalign
$$\endproclaim

\demo\nofrills{\smc {Proof.}\usualspace}  Let $\overline p = \chr (
\overline F ) = \chr (GF)$.  Since $GK_0 = \overline K = \overline F_{sep} =
(GF_0)_{sep}$ and $\Gamma_{GK} / \Gamma_{GF} = \Gamma_K /
\Gamma_F$, which is the prime-to-$\overline p$ part of $\Delta_{GF} /
\Gamma_{GF}$, $GK$ is tame over $GF$ and has no proper tame extensions. 
Hence, $GK$ is the tame closure of $GF$.  

We define a homomorphism $\alpha \: \Cal G (K/F) \to \Cal G (QGK / QGF)$
as follows:  For $\sigma \in \Cal G (K/F)$, $\sigma (V_K) = V_K$ since $V_K$
is the unique extension of the Henselian valuation ring $V_F$ to $K$. 
Moreover, as $ \Gamma_K / \Gamma_F$ is a torsion group, $\sigma$ must
induce the identity automorphism on $\Gamma_K$.  Hence, $\sigma
(W^\gamma_K) = W^\gamma_K$ and $\sigma (W^{> \gamma}_K ) = W^{>
\gamma}_K$ for each $\gamma \in\Gamma_K$.  Thus, $\sigma $ induces a
graded ring isomorphism $\sigma^\prime \: GK \to GK$ which is the identity
on $GF$.  This isomorphism extends to an isomorphism $\widetilde \sigma
\: QGK \to QGK$ of the quotient field.  Define $\alpha (\sigma ) = \widetilde
\sigma \in \Cal G (QGK / QGF)$.

To show that $\alpha$ is an isomorphism, we proceed in stages.  First, let
$I$ be the maximal unramified extension of $F$ in $K$, re $v$.  Then
$\overline I = \overline F_{sep}$ and $\Gamma_I = \Gamma_F$, and $I$ is
Galois over $F$ with $\Cal G (I/ F) \cong \Cal G (\overline I / \overline F)$
canonically.  So, $GI$ is the maximal unramified graded field extension of
$GF$, since $GI_0 = \overline I = ( \overline F)_{sep} = (GF_0)_{sep}$ and
$\Gamma_{GI} = \Gamma_I = \Gamma_F = \Gamma_{GF}$.  Hence, $QGI
\cong GI_0 \otimes_{GF_0} QGF$, by Remark~3.1, showing that $QGI$ is
Galois over $QGF$, with 
$$
\Cal G (QGI / QGF) \cong \Cal G (GI_0 /GF_0) \cong
\Cal G ( \overline F_{sep} / \overline F) \cong \Cal G (I/F)\, .
$$ 
 The inverse of
this isomorphism corresponds to the mapping induced by $\alpha$.

The field extension $K/I$ is tame and totally ramified.  Let $\Omega = \{
\omega \in I_{sep} \mid \omega^n = 1 $ for some $n \in \Bbb N$ with $
\overline p \nmid n \}$.  Since $V_I$ is Henselian and $\overline I$ is
separably closed, $\Omega \subseteq I$ and $\Omega$ maps injectively to
$\overline I$, let $\Omega^\prime$ be the image of $\Omega$ in $\overline
I = GI_0$.  Let $L$ be any finite-degree field extension of $I$ in $K$.  Then,
$L$ is tame and totally ramified over $I$.  Because further $V_I$ is
Henselian and $I$ contains all $\ell$-th roots of unity for $\ell = [L : I]$, $L$
is an $\ell$-Kummer extension of $I$ (cf\. [S, p\. 64,
Th\. 3] or [TW, Prop\. 1.4(iii)]).
 Likewise, $QGL$ is an $\ell$-Kummer
extension of $QGI$, by Prop.~3.3.  Moreover, by [E,
(20.11)] or [TW, Prop\. 1.4(i)] there is a perfect
pairing
$\Gamma_L / \Gamma_I \times \Cal G (L/I) \to \Omega^\prime$ given by
$(\gamma +\Gamma_I
, \sigma ) \mapsto \overline{ a/ \sigma (a)}$ for any $a \in L$
with
$v (a) = \gamma$.  With respect to the canonical isomorphism $\Gamma_L
/ \Gamma_I \to \Gamma_{GL} / \Gamma_{GI}$ and the map $\Cal G (L/I) \to
\Cal G (QGL / QGI)$ induced by $\alpha$, the following diagram is evidently
commutative,
$$
\CD
\, \ \Gamma_L / \Gamma_I \times \Cal G (L/I) @>>>\  \Omega^\prime \\
@VVV @| \\
\ \ \ \ \ \ \  \Gamma_{GL} / \Gamma_{GI} \times \Cal G (QGL/QGI) @>>>
\ \Omega^\prime
\endCD
$$
where the bottom row is the perfect pairing of Prop\.~3.3(b).  Because both
rows of the diagram are perfect pairings, the map $\Cal G (L/I) \to G(QGL /
QGI)$ must be an isomorphism.  Since $K$ is the union of fields such as $L$
and $QGK$ is the union of the corresponding fields $QGL$, the restriction of
$\alpha$ mapping $\Cal G (K/I)$ to $\Cal G (QGK/ QGI)$ is an isomorphism. 
Thus, we have a commutative diagram
$$
\CD
1 @>>> \Cal G (K/I) @>>> \Cal G (K/F) @>>> \Cal G (I/F) @>>> 1\\
@. @V \cong VV @V \alpha VV @V \cong VV \\
1 @>>> \Cal G (QGK/QGI) @>>> \Cal G (QGK /QGF) @>>> \Cal G(QGI/QGF) @>>> 1
\endCD
$$
The $5$-lemma shows that $\alpha$ is an isomorphism.  The isomorphism
$\alpha$ of Galois groups gives the one-to-one correspondence between
tame field extensions of $F$ in $F_{alg}$ (i.e., subfields of $K$) and field
extensions of $QGF$ in $QGK$.  The correspondence between these fields
and the tame graded field extensions of $GF$ in $GF_{gr \text{-}alg}$ was
given in Th.~3.9. $\square$ \enddemo

In  the setting of Th\.  5.2, it was shown in [B$_1$, Th\. 5] that if $L$
is a field extension of $F$ with $[L:F] = n < \infty$,  then 
$L$ is tame over $F$ iff $QGL$ is separable over $QGF$ with
$[QGL:QGF] = n$.


\proclaim{Corollary 5.3}  Let $K$ be the maximal tame extension of a
Henselian valued field $F$.  Then $HGK$ is the maximal tame extension of
$HGF$, and $\Cal G (HGK /HGF) \cong \Cal G (K/F)$. \endproclaim

\demo\nofrills{\smc {Proof.}\usualspace}   This is immediate from
Prop.~5.1 and Th.~5.2. $\square$ \enddemo

\bigskip \smallskip
%\bigskip

\head References\endhead

\def\hangbox to #1 #2{\vskip1pt\hangindent #1\noindent 
\hbox to #1{#2}$\!\!$}

\def\myitem#1{\hangbox to 47pt {\hfill#1 \ \ }}

\frenchspacing
\baselineskip = 12pt

\myitem{[B$_1$]}
 M. Boulagouaz,
 The graded and tame extensions, pp\. 27--40 in
{\it Commutative Ring Theory} (F\`es, 1992)
(P.  J.  Cahen, et.~al., eds.), 
Lecture Notes in Pure and Applied Math., No.~153,
 Marcel Dekker,  New York, 1994.
\medskip

\myitem{[B$_2$]}
 M. Boulagouaz, 
 Le gradu\`e d'une alg\`ebre \`a division valu\'ee,
 {\it Comm. Algebra}, {\bf 23} (1995),  4275--4300.
\medskip

\myitem{[CHR]}
S.  U.  Chase, D.  K. Harrison, and A. Rosenberg,
{\it Galois Theory and Galois Cohomology of 
Commutative Rings},
Mem. Amer. Math. Soc., No.~52, 
Amer. Math. Soc., Providence, R.I.,  1965.
\medskip

\myitem{[E]} O. Endler, {\it Valuation Theory}, Springer, 
New York, 1972.
\medskip

\myitem{[G]}
C. Greither,
{\it Cyclic Galois Extensions of Commutative Rings},
Lecture Notes in Math.,  No.~1534, Springer,  Berlin,
1992.
\medskip

\myitem{[HW]}
Y.-S. Hwang and A. R. Wadsworth,
{\it Correspondences between valued
division algebras and graded division
algebras}, preprint, UCSD, 1997.
\medskip

%\myitem{[NvO$_1$]}
%C. N\v ast\v asescu and F. van Oystaeyen,
%{\it Graded and Filtered Rings and Modules},
%Lecture Notes in Math., No.~758,
%Springer, Berlin, 1979.
%\medskip

%\myitem{[NvO$_2$]}
%C. N\v ast\v asescu and F. van Oystaeyen,
%{\it Graded  Ring Theory}, North-Holland,
%Amsterdam,  1982.
%\medskip

\myitem{[S]}
O. F. G. Schilling, {\it The Theory of Valuations}, 
Math. Surveys, No. 4, Amer.  Math. Soc., Providence,
R.I., 1950.
\medskip

\myitem{[TW]}
J.-P. Tignol and A. R. Wadsworth,
Totally ramified valuations on finite-dimensional division algebras,
{\it Trans. Amer. Math. Soc.},
{\bf 302} (1987), 223--249.
\medskip

\myitem{[vGvO]}
J. van Geel and F. van Oystaeyen,
About graded fields, {\it Indag. Math.},
{\bf 43} (1981), 273--286.
\medskip
\end

