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\authorrunning{Heinzer, Rotthaus, Wiegand}
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\begin{document}

\title*{Mixed Polynomial/Power Series Rings
and Relations among their Spectra}
% Use \titlerunning{polynomial/power series} for an abbreviated version of
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\author{ William Heinzer \inst{1}\and
Christel Rotthaus \inst{2} \and Sylvia Wiegand \inst{3} }
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\institute{Purdue University, West Lafayette IN 47907-1395
\texttt{heinzer@math.purdue.edu}\and Michigan State University,
East Lansing Mi 488824-1024 \texttt{rotthaus@math.msu.edu}\and
University of Nebraska, Lincoln NE 68588-0130
\texttt{swiegand@math.unl.edu}
\thanks{The authors are grateful for the hospitality and cooperation
of Michigan State, Nebraska and Purdue where several work sessions
on this research were conducted.
Wiegand thanks the National Security Agency for support.}}%
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\maketitle

\centerline{This article is dedicated to Robert Gilmer, an
outstanding commutative} \centerline{algebraist, scholar and
teacher. It relates to his work in ideal theory.}

\medskip

\section{Introduction and Background}
\label{sec:1}

In this article we study the nested mixed polynomial/power series
rings

\begin{equation} \label{1}
 A:=k[x,y]\hookrightarrow B:=k[[y]]\,[x]\hookrightarrow
C:=k[x]\,[[y]]\hookrightarrow
 E:=k[x,1/x]\,[[y]],
\end{equation}
where $k$ is a field and $x$ and $y$ are indeterminates over $k$.
In Equation \ref{1} the maps  are all flat. Also we consider
\begin{equation} \label{2}
C\hookrightarrow D_1:=k[x]\,[[y/x]]\hookrightarrow  \cdots
\hookrightarrow D_n:= k[x]\,[[y/x^n]] \hookrightarrow \cdots
\hookrightarrow  E.
\end{equation}
With regard to Equation \ref{2}, for $n$ a positive integer, the
map $C \hookrightarrow D_n$ is not flat, but $D_n \hookrightarrow
E$ is a localization followed by an adic completion of a
Noetherian ring and therefore is flat. We discuss the  spectra of
these rings and consider  the maps induced on the spectra by the
inclusion maps on the rings. For example, we determine whether
there exist  nonzero primes of one of the larger rings that
intersect a smaller ring in zero. We were led to consider these
rings by questions that came up in two contexts.

  The first motivation is from the introduction to the
paper \cite{AJL} by Alonzo-Tarrio, Jeremias-Lopez and Lipman: If a
map between  Noetherian formal schemes can be factored as a closed
immersion followed by an open one, can  this  map  also be
factored as an open immersion followed by a closed one?   This  is
not true in general. As mentioned in \cite{AJL},  Brian Conrad
observed that a counterexample can be constructed for  every
triple $(R, x, p)$, where

\begin{description} \item[{(1)}] $R$ is an adic domain, that is,    $R$ is a
Noetherian domain that is separated and complete with respect to
the powers of a proper   ideal $I$.

\item[{(2)}]
 $x$ is  a nonzero  element of $R$ such that the completion
of $R[1/x]$ with respect to the powers of $IR[1/x]$, denoted $S: =
R_{\{x\}}$,  is  an integral domain.

\item[{(3)}] $p$ is a nonzero prime ideal of $S$ that intersects $R$ in
$(0)$.
\end{description}

The composition $R \to S \to S/p$ determines a   map on formal
spectra $\Spf(S/p)\to \Spf(S) \to \Spf(R)$ that is a closed
immersion followed by an open one. To see this, recall that a
surjection such as $S \to S/p$ of adic rings gives rise to a
closed immersion $\Spf(S/p) \to \Spf(S)$ while a localization,
such as that of $R$ with respect to the powers of $x$,   followed
by the completion of $R[1/x]$ with respect to the powers of
$IR[1/x]$ to obtain $S$ gives rise to an open immersion $\Spf(S)
\to \Spf(R)$ \cite{EGA, (10.2.2)}.

 The map $\Spf(S/p) \to \Spf(R)$
cannot be factored, however, as an open immersion followed by a
closed one. This is because a closed immersion into $\Spf(R)$
corresponds to a surjective map of adic rings $R \to R/J$, where
$J$ is an ideal of $R$ \cite{EGA, page 441}. Thus if the immersion
$\Spf(S/p) \to \Spf(R)$ factored as an open immersion followed by
a closed one, we would have $R$-algebra homomorphisms from $R \to
R/J \to S/p$, where $\Spf(S/p) \to \Spf(R/J)$ is an open
immersion. Since  $p \cap R = (0)$, we must have $J = (0)$. This
implies $\Spf(S/p) \to \Spf(R)$ is an open immersion, that is, the
composite map $\Spf(S/p) \to \Spf(S) \to \Spf(R)$, is an open
immersion. But also $\Spf(S) \to \Spf(R)$ is an open immersion. It
follows that $\Spf(S/p) \to \Spf(S)$ is both open and closed.
Since $S$ is an integral domain this implies $\Spf(S/p) \cong
\Spf(S)$. This is a contradiction since $p$ is nonzero.

An example of such a triple $(R,x,p)$ is described in \cite{AJL}:
For $w,x,y,z$ indeterminates over a field $k$, set
$R:=k[w,x,z]\,[[y]]$, $S:= k[w,x,1/x,z]\,[[y]]$.  Notice that $R$
is complete with respect to $yR$ and $S$ is complete with respect
to $yS$. An indirect proof is given in \cite{AJL} that there exist
nonzero primes $p$ of $S$ for which  $p \cap R = (0)$.  In
Proposition 4.5 below we give a direct proof of this fact.

A second motivation is from a question raised by  Mel Hochster:
``Can one describe or somehow classify the local maps $R
\hookrightarrow S$ of complete local domains  $R$ and $S$ such
that every nonzero prime ideal of $S$ has nonzero intersection
with $R$?''  In \cite{HRW2} we study this question and define:

\begin{definition} \label{1.1} {\em For $R$ and $S$ integral domains with
$R$ a subring of $S$  we say that $S$ is a {\it trivial generic
fiber} extension of $R$, or a {\bf TGF} extension of $R$,  if
every nonzero prime ideal of $S$ has nonzero intersection with
$R$.}
\end{definition}

In some correspondence to Lipman regarding closed and open
immersions, Conrad asked: ``Is there a nonzero prime ideal of
$E:=k[x,1/x]\,[[y]]$ that intersects $C = k[x]\,[[y]]$ in zero?''
If there were such a prime ideal, then $C:=k[x]\,[[y]]
\hookrightarrow E:=k[x,1/x]\,[[y]]$ would be  a simpler
counterexample to the assertion that a closed immersion followed
by an open one also has a factorization as an open immersion
followed by  a closed one. In the terminology of Definition
\ref{1.1}, one can ask:

\begin{newquestion} \label{1.2}  Let $x$ and $y$ be indeterminates
over a field $k$.  Is  $C:=k[x]\,[[y]]\hookrightarrow
E:=k[x,1/x]\,[[y]]$  a TGF extension?  \end{newquestion}

We show in Proposition \ref{2.7}.2   below that the answer to
Question \ref{1.2}  is ``yes''. This is part of our analysis of
the prime spectra of $A$, $B$, $C$, $D_n$ and $E$, and the maps
induced on these spectra by the inclusion maps on the rings.

\medskip

The following example is a local map of the type described in
Hochster's question.

\begin{newexample}   Let $x$ and $y$ be indeterminates over a field
$k$ and consider the extension  $R := k[[x,y]]\hookrightarrow S :=
k[[x]]\,[[y/x]]$.
 To see  this extension is TGF, it suffices to
show $P \cap R \ne (0)$ for each $P \in \Spec S$ with $\hgt P =
1$.
  This is clear if $x \in P$, while
if $x \not\in P$, then  $k[[x]] \cap P = (0)$, and so $k[[x]]
\hookrightarrow  R/(P \cap R) \hookrightarrow  S/P$ and $S/P$ is
finite over $k[[x]]$. Therefore $\dim R/(P \cap R) = 1$, and so $P
\cap R \ne (0)$.
\end{newexample}

\begin{newremark}  (1) The extension $k[[x, y]]
\hookrightarrow k[[x, y/x]]$ is, up to isomorphism, the same as
the extension $k[[x, xy]] \hookrightarrow k[[x, y]]$.

(2) We show in \cite{HRW2} that the extension $R := k[[x, y, xz]]
\hookrightarrow S := k[[x, y, z]]$ is not TGF.
\end{newremark}

\section{Trivial generic fiber (TGF) extensions and prime spectra}
\label{sec2}

The following two propositions  about trivial generic fiber
extensions from \cite{HRW2} are useful and straightforward to
check.

\begin{proposition}\label{2.1} Let $R \hookrightarrow S$ be an
injective map  where $R$ and  $S$ are integral domains. Then the
following  are equivalent: \begin{description} \item[{\em (1)}]
$S$ is a TGF extension of $R$.

\item[{\em (2)}] Every nonzero element of $S$ has a nonzero multiple in $R$.

\item[{\em (3)}] For $U=R\,\setminus\,\{0\}$, $U^{-1}S$ is a field.
\end{description}
\end{proposition}

\begin{proposition}\label{2.2} Let $R \hookrightarrow S$ and $S
\hookrightarrow T$ be injective maps where $R$, $S$ and $T$ are
integral domains.
\begin{description}
 \item[{\em (1)}] If $R \hookrightarrow S$ and $S
\hookrightarrow T$ are TGF extensions, then so is $R
\hookrightarrow T$.  Equivalently if $R \hookrightarrow T$ is not
TGF, then at least one of the extensions $R \hookrightarrow S$ or
$S \hookrightarrow T$ is not TGF.
\item[{\em (2)}] If $R \hookrightarrow T$
is TGF, then $S \hookrightarrow T$ is TGF.
\item[{\em (3)}] If the map $\Spec
T \to \Spec S$ is surjective, then $R \hookrightarrow T$ is TGF
implies $R \hookrightarrow S$ is TGF.
\end{description}
\end{proposition}

More information about TGF extensions is in \cite{HRW1} and
\cite{HRW2}.

\medskip

\begin{newremark}\label{2.4}  Let $R$ be a commutative ring and let
$R[[y]]$ denote the formal power series ring in the variable $y$
over $R$. Then  \begin{description}
\item[{(1)}]
Each maximal ideal of $R[[y]]$ is of the form $(\m, y)R[[y]]$
where $\m$ is a maximal ideal of $R$. Thus $y$ is in every maximal
ideal of $R[[y]]$.

\item[{(2)}] If $R$ is Noetherian with $\dim R[[y]] = n$ and $x_1,
\ldots, x_m$ are independent indeterminates over $R[[y]]$, then
$y$ is in every height $n+m$ maximal ideal of the polynomial ring
$R[[y]]\,[x_1, \ldots,x_m]$.
\end{description}\end{newremark}

\begin{proof} Item (1) follows from \cite{N, Theorem 15.1}.
For item (2), let $\m$ be a maximal ideal of $R[[y]]\,[x_1,
\ldots, x_m]$ with $\hgt(\m) = n+m$. By \cite{K, Theorem 39},
 $\hgt(\m \cap R[[y]]) = n$ ;
thus $\m \cap R[[y]]$ is maximal in $R[[y]]$, and so, by item (1),
$y \in \m$. \qed
\end{proof}

\begin{proposition} \label{2.5} Let $n$ be a positive integer, let $R$
be an  $n$-dimensional Noetherian domain, let $y$ be an
indeterminate over $R$,
 and let $\q$ be a prime ideal of height $n$ in the power series ring
$R[[y]]$. If $y \not\in \q$, then $\q$ is contained in a unique
maximal ideal of $R[[y]]$.
 \end{proposition}

\begin{proof} The assertion is clear if $\q$ is maximal. Otherwise
$S := R[[y]]/\q$ has dimension one. Moreover,  $S$ is complete
with respect to the $yS$-adic topology \cite{M, Theorem 8.7} and
every maximal ideal of $S$ is a minimal prime of the principal
ideal $yS$. Hence $S$ is a complete semilocal ring. Since $S$ is
also an integral domain, it must be local by \cite{M, Theorem
8.15}. Therefore $\q$ is contained in a unique maximal ideal of
$R[[y]]$. \qed
\end{proof}

In Section 3 we use the following corollary to Proposition
\ref{2.5}.

\begin{corollary} \label{2.6}  Let  $R$ be a one-dimensional Noetherian
domain
 and let $\q$ be a height-one prime ideal of the power series ring $R[[y]]$.
If $\q \ne yR[[y]]$, then $\q$ is contained in a unique maximal
ideal of $R[[y]]$.
 \end{corollary}

\begin{proposition} \label{2.7}  Consider  the nested mixed
polynomial/power series rings

\begin{eqnarray*}
 A:=k[x,y] \, \hookrightarrow B:=k[[y]]\,[x]\, \hookrightarrow
C:=k[x]\,[[y]] \phantom{xxxxxxx} \\ \hookrightarrow D_1:=
k[x]\,[[y/x]] \, \hookrightarrow D_2:=k[x]\,[[y/x^2]] \,
\hookrightarrow \cdots \phantom{xx} \\ \hookrightarrow D_n:=
k[x]\,[[y/x^n]] \, \hookrightarrow \cdots \, \hookrightarrow
E:=k[x,1/x]\,[[y]],
\end{eqnarray*}
where $k$ is a field  and $x$ and $y$ are indeterminates over $k$.
Then
\begin{description}

\item[{\em (1)}] If $S \in \{ B, C, D_1, D_2, \cdots ,  D_n, \cdots , E \}$, then  $A \hookrightarrow
S$ is not TGF.

\item[{\em(2)}]  If $\{R, S\} \subset \{B, C, D_1, D_2, \cdots ,  D_n \cdots , E \}$ are
such that $R \subseteq S$, then $R \hookrightarrow S$ is TGF.

\item[{\em (3)}] Each of the proper associated maps on spectra fails to be
surjective.
\end{description}
\end{proposition}

\begin{proof}  For item (1), let $\sigma(y) \in yk[[y]]$ be such
that $\sigma(y) $ and $y$ are algebraically independent over $k$.
Then $(x-\sigma(y))S \cap A=(0)$, and so $A \hookrightarrow S$ is
not TGF.

For item (2), observe that every maximal ideal of $C$, $D_n$ or
$E$ is of height two with residue field finite algebraic over $k$.
To show $R \hookrightarrow S$ is TGF, it suffices to show $\q \cap
R \ne (0)$ for each height-one prime ideal $\q$ of $S$. This is
clear if $y \in \q$. If $y \not\in \q$,  then $k[[y]]\cap\q=(0)$,
and so $k[[y]]\hookrightarrow R/(\q\cap R)\hookrightarrow S/q$ are
injections.
 By Corollary \ref{2.6}, $S/\q$ is a one-dimensional local domain.
Since the residue field of $S/\q$ is finite algebraic over $k$, it
follows that $S/\q$ is finite over $k[[y]]$.   Therefore $S/\q$ is
integral over $R/(\q \cap R)$. Hence $\dim(R/(\q\cap R)=1$ and so
$\q\cap R\ne (0)$.

For item (3), observe that $xD_n$ is a prime ideal of  $D_n$ and
$x$ is a unit of $E$. Thus  $\Spec E \to \Spec D_n$ is not
surjective. Now, considering $C = D_0$ and $n > 0$, we have $xD_n
\cap D_{n-1} = (x, y/x^{n-1})D_{n-1}$. Therefore $xD_{n-1}$ is not
in the image of the map $\Spec D_n \to \Spec D_{n-1}$. The  map
from $\Spec C\to \Spec B$ is not onto, because $(1+xy)$ is a prime
ideal of $B$, but $1+xy$ is a unit in $C$. Similarly $\Spec B\to
\Spec A$ is not onto, because $(1+y)$ is a prime ideal of $A$, but
$1+y$ is a unit in $B$. This completes the proof. \qed
\end{proof}

\begin{quesremark} \label{2.8}  Which of the Spec maps of
Proposition \ref{2.7}  are one-to-one and which are finite-to-one?
\begin{description}
\item[{(1)}] For $S \in \{B, C, D_1, D_2, \cdots, D_n, \cdots , E\}$, the generic fiber ring of the
map $A \hookrightarrow  S$ has infinitely many prime ideals and
has dimension one. Every height-two maximal ideal of $S$ contracts
in $A$ to a maximal ideal. Every maximal ideal of $S$ containing
$y$ has height two. Also $yS \cap A = yA$ and  the map $\Spec S/yS
\to \Spec A/yA$ is one-to-one.
\item[{(2)}]
Suppose  $R \hookrightarrow  S$ is as in Proposition \ref{2.7}.2.
Each height-two prime of $S$ contracts in $R$ to a height-two
maximal ideal of $R$. Each height-one prime of $R$ is the
contraction of at most finitely many prime ideals of $S$ and all of
these prime ideals have height one. If $R \hookrightarrow S$ is
flat, which is true if $S \in \{B, C, E \}$, then  ``going-down''
holds for $R \hookrightarrow S$, and so, for $P$ a height-one prime
of $S$, we have $\hgt(P \cap R)  \le 1$.

\item[{(3)}] As mentioned in
\cite{HW, Remark 1.5}, $C/P$ is Henselian for every nonzero prime
ideal $P$ of $C$ other than  $yC$.
\end{description}
\end{quesremark}

\section{Spectra for two-dimensional mixed polynomial/power
series rings} \label{sec3}

 Let $k$ be a field and let $x$ and $y$ be
indeterminates over $k$. We consider the prime spectra, as
partially ordered sets, of the mixed polynomial/power series rings
$A$, $B$, $C$, $D_1, D_2, \cdots, D_n, \cdots$  and $E$ as given
in  Equations \ref{1} and \ref{2} of the introduction.

Even for $k$ a countable field   there are at least two
non-order-isomorphic partially ordered sets that can be the prime
spectrum of the polynomial ring $A := k[x, y]$.  Let $\mathbb Q$ be
the field of rational numbers, let $F$ be a field contained in the
algebraic closure of a finite field and let $\mathbb Z$ denote the
ring of integers. Then, by \cite{rW1} and \cite{rW2}, $\Spec\mathbb
Q[x, y] \not\cong \Spec F[x, y] \cong \Spec \mathbb Z[y]$.

The prime spectra of the rings $B$, $C$, $D_1, D_2, \cdots, D_n,
\cdots$, and $E$ of Equations \ref{1} and \ref{2}  are simpler
since they involve power series in $y$. Remark \ref{2.4}.2 implies
that $y$ is in every maximal ideal of height two of each of these
rings.

The partially ordered set  $\Spec B =\Spec k[[y]]\,[x]$  is similar
to a prime ideal space studied in \cite{HW} and \cite{Shah}. The
difference from \cite{HW} is that here $k[[y]]$ is uncountable, even
if $k$ is countable. It follows that  $\Spec B$ is also uncountable.
As a partially ordered set, $\Spec B$ can be described uniquely up
to isomorphism by the axioms of \cite{Shah} (similar to the CHP
axioms of \cite{HW}), since $k[[y]]$ is Henselian and has
cardinality at least equal to $c$, the cardinality of the real
numbers $\mathbb R$.

 The
following theorem characterizes $U:=\Spec B$ as  a {\it Henselian
affine} partially ordered set (where the ``$\le$" relation is ``set
containment"):

\begin{theorem}\label{3.2} {\em \cite{HW, Theorem 2.7}  \cite{Shah, Theorem
2.4}} Let $B=k[[y]]\,[x]$ be as in Equation \ref{1}, where $k$ is
a field, the cardinality of the set of maximal ideals of $k[x]$ is
$\alpha$ and the cardinality of $k[[y]]$ is $\beta$. Then the
partially ordered set $U:=\Spec B$ is called {\it Henselian affine
of type $(\beta,\alpha)$} and is characterized as a partially
ordered set by the following axioms:
\begin{description}\item[{\em (1)}] $|U|=\beta$. \item[{\em (2)}] $U$ has a
unique minimal element.
\item[{\em (3)}] $\dim(U)=2$ and $|\{$ height-two elements of
$U\,\}|=\alpha$. \item[{\em (4)}] There exists a unique {\it
special} height-one element $u \in U$ such that $u$ is contained
in every height-two element of $U$. \item[{\em (5)}] Every
nonspecial height-one element of $U$ is in at most one height-two
element. \item[{\em (6)}] Every height-two element $t\in U$
contains cardinality $\beta$ many height-one elements that are
{\it only} contained in $t$. If $t_1,t_2\in U$ are distinct
height-two elements, then the special element from (4) is the
unique height-one element less than both.
\item[{\em (7)}]  There are cardinality $\beta$ many height-one elements that
are maximal.
\end{description}
\end{theorem}

 \begin{newremark} \label{3.3} (1) The axioms of Theorem \ref{3.2} are redundant.
We feel this redundancy helps in understanding  the relationships
between the prime ideals.

(2) The theorem applies to the spectrum of $B$ by defining
 the unique minimal element to be  the ideal $(0)$ of $B$  and the
 special height-one element to be  the prime ideal $yB$. Every height-two maximal
 ideal $\m$ of $B$ has  nonzero intersection with $k[[y]]$. Thus  $\m/yB$ is
principal and so
 $\m=(y,f(x))$, for some monic irreducible polynomial $f(x)$  of $k[x]$. Consider
$\{f(x) + ay \, | \, a \in k[[y]] \}$. This set has cardinality
$\beta$ and each $f(x) + ay$ is contained in a nonempty finite set
of height-one primes contained in $\m$.  If $\p$ is a height-one
prime contained in $\m$ with $\p \ne yB$, then $\p \cap k[[y]] =
(0)$, and so $\p k((y))[x]$ is generated by a monic polynomial in
$k((y))[x]$. But for $a, b \in k[[y]]$ with $a \ne b$, we have
$(f(x) + ay, f(x) + by)k((y))[x] = k((y))[x]$. Therefore no
height-one prime contained in $\m$ contains both $f(x) + ay$ and
$f(x) + by$.  Since $B$ is Noetherian and $|B| = \beta$ is an
infinite cardinal, we conclude that the cardinality of the set of
height-one prime ideals contained in $\m$ is $\beta$.
 Examples of height-one maximals are $(1+xyf(x,y)\,)$, for
 various $f(x,y)\in k[[y]]\,[x]$. The set of height-one maximal ideals of $B$ also
has cardinality $\beta$.

 (3) These axioms {\it
characterize} $\Spec B$ in the sense that every two partially
ordered sets satisfying these axioms are order-isomorphic.
\end{newremark}

The picture of $\Spec B$ is shown below: \vskip 10 pt
\setbox4=\vbox{\hbox{%
    \rlap{\kern  .650in\lower 0.916in\hbox to .3in{\hss%
$\boxed{\beta}$\hss}}%
    \rlap{\kern  1.200in\lower 0.916in\hbox to .3in{\hss%
    $(y)$\hss}}%
    \rlap{\kern  1.750in\lower 0.916in\hbox to .3in{\hss%
$\boxed{\beta}$\hss}}%
    \rlap{\kern  2.350in\lower 0.916in\hbox to .3in{\hss%
$\boxed{\beta}$\hss}}%
    \rlap{\kern  2.850in\lower 0.916in\hbox to .3in{\hss $\cdots$\hss}}%
    \rlap{\kern  1.150in\lower 0.166in\hbox to .3in{\hss $\bullet$\hss}}%
    \rlap{\kern  1.750in\lower 0.166in\hbox to .3in{\hss $\bullet$\hss}}%
    \rlap{\kern  2.350in\lower 0.166in\hbox to .3in{\hss $\bullet$\hss}}%
    \rlap{\kern  3.050in\lower 0.166in\hbox to .3in{\hss $\cdots$\hss}}%
    \rlap{\kern 3.550in\lower 0.466in\hbox to .3in{\hss ($\#\{$ bullets$ \}=\alpha$)\hss}}%
    \rlap{\kern  1.850in\lower 1.666in\hbox to .3in{\hss $(0)$\hss}}%
    \rlap{\special{pa 1200 180}    \special{pa 1200 800}    \special{fp}}%
    \rlap{\special{pa 1800 180}    \special{pa 2400 750}    \special{fp}}%
 \rlap{\special{pa 1800 180}    \special{pa 1200 800}    \special{fp}}%
    \rlap{\special{pa 2400 180}    \special{pa 3000 750}    \special{fp}}%
    \rlap{\special{pa 1200 180}    \special{pa 1800 770}    \special{fp}}%
    \rlap{\special{pa 2400 180}    \special{pa 1200 800}    \special{fp}}%
   \rlap{ \special{pa 3000 180}    \special{pa 1200 780}    \special{fp}}%
    \rlap{\special{pa 680 1000}    \special{pa 1850 1500}    \special{fp}}%
    \rlap{\special{pa 1230 970}    \special{pa 1850 1500}    \special{fp}}%
   \rlap{ \special{pa 1800 1000}    \special{pa 1800 1500}    \special{fp}}%
   \rlap{ \special{pa 2400 1000}    \special{pa 1800 1500}    \special{fp}}%
   \rlap{ \special{pa 3000 1000}    \special{pa 1800 1500}    \special{fp}}%
   }}
 \box4 \vskip 10 pt \centerline{$\Spec k[[y]]\,[x]$} \vskip 18 pt

In the diagram, $\beta$ is the cardinality of $k[[y]]$, and
$\alpha$ is the cardinality of the set of maximal ideals of $k[x]$
(and also the cardinality of the set of maximal ideals of
$k[[y]]\,[x]$\,); the boxed $\beta$ means there are cardinality
$\beta$ height-one primes in that position with respect to the
partial ordering.
\medskip

Next we consider $\Spec R[[y]] $, for $R$  a Noetherian
one-dimensional domain. Then $\Spec R[[y]]$ has the following
picture by Theorem \ref{3.4} below:

\vskip 10 pt
\setbox4=\vbox{\hbox{%

    \rlap{\kern  1.200in\lower 0.916in\hbox to .3in{\hss%
    $(y)$\hss}}%
    \rlap{\kern  1.750in\lower 0.916in\hbox to .3in{\hss%
$\boxed{\kappa_i}$\hss}}%
    \rlap{\kern  2.350in\lower 0.916in\hbox to .3in{\hss%
$\boxed{\kappa_j}$\hss}}%
    \rlap{\kern  2.850in\lower 0.916in\hbox to .3in{\hss $\cdots$\hss}}%
    \rlap{\kern  1.150in\lower 0.166in\hbox to .3in{\hss $\bullet$\hss}}%
    \rlap{\kern  1.750in\lower 0.166in\hbox to .3in{\hss $\bullet$\hss}}%
    \rlap{\kern  2.350in\lower 0.166in\hbox to .3in{\hss $\bullet$\hss}}%
    \rlap{\kern  3.050in\lower 0.166in\hbox to .3in{\hss $\cdots$\hss}}%
    \rlap{\kern  3.550in\lower 0.466in\hbox to .3in{\hss ($\#\{$ bullets$ \}=\alpha$)\hss}}%
    \rlap{\kern  1.850in\lower 1.666in\hbox to .3in{\hss $(0)$\hss}}%
    \rlap{\special{pa 1200 180}    \special{pa 1200 800}    \special{fp}}%
    \rlap{\special{pa 1800 180}    \special{pa 2400 750}    \special{fp}}%
 \rlap{\special{pa 1800 180}    \special{pa 1200 800}    \special{fp}}%
    \rlap{\special{pa 2400 180}    \special{pa 3000 750}    \special{fp}}%
    \rlap{\special{pa 1200 180}    \special{pa 1800 770}    \special{fp}}%
    \rlap{\special{pa 2400 180}    \special{pa 1200 800}    \special{fp}}%
   \rlap{ \special{pa 3000 180}    \special{pa 1200 780}    \special{fp}}%
    \rlap{\special{pa 1230 970}    \special{pa 1850 1500}    \special{fp}}%
   \rlap{ \special{pa 1800 1000}    \special{pa 1800 1500}    \special{fp}}%
   \rlap{ \special{pa 2400 1000}    \special{pa 1800 1500}    \special{fp}}%
   \rlap{ \special{pa 3000 1000}    \special{pa 1800 1500}    \special{fp}}%
   }}
 \box4 \vskip 10 pt \centerline{Spec$(R[[y]])$} \vskip 18 pt

Here $\alpha$ is the cardinality of the set of maximal ideals of $R$
(and also the cardinality of the set of maximal ideals of $R[[y]]$
by Remarks \ref{2.4}.1\,); the boxed $\kappa_i$ (one for each
maximal ideal of $R$) means that there are cardinality $\kappa_i$
prime ideals in that position, where each $\kappa_i$ is uncountable.
If $R$ satisfies certain cardinality conditions described in Remark
\ref{3.3.5}.3, for example, if $R=k[x]$, for $k$ a countable field,
then each $\kappa_i$ equals the cardinality of $R[[y]]$. By Remark
\ref{3.3.5}.2 below, each $\kappa_i$ is at least
$\gamma^{\aleph_0}$, where $\gamma$ is the cardinality of $R/\m$ and
$\m$ is the maximal ideal of $R$ such that $(\m, y)$ is the maximal
ideal of $R[[y]]$ above the $\kappa_i$ height-one primes in the
picture above.


 \begin{newremark}\label{3.3.5} Let $\aleph_0$ denote the
cardinality of the set of natural numbers. Suppose that $T$ is a
commutative ring of cardinality $\delta$, that ${\m}$ is a maximal
ideal of $T$ and that $\gamma$ is the cardinality of $T/\m$. Then

(1) The cardinality of $T[[y]]$ is $\delta^{\aleph_0}$, because the
elements of $T[[y]]$ are in one-to-one correspondence with
$\aleph_0$-tuples having entries in $T$. If $T$ is Noetherian, then
$T[[y]]$ is Noetherian, and so every prime ideal of $T[[y]]$ is
finitely generated. Since the cardinality of the finite subsets of
$T[[y]]$ is $\delta^{\aleph_0}$, it follows that $T[[y]]$ has at
most $\delta^{\aleph_0}$ prime ideals.

(2) If $T$ is Noetherian, there are at least $\gamma^{\aleph_0}$
distinct height-one prime ideals (other than $(y)T[[y]]$\,)  of
$T[[y]]$ contained in $(\m,y)T[[y]]$. To see this, choose a set
$C=\{c_i\,|\, i\in I\}$ of elements of $T$ so that $\{c_i +\m\,|\,
i\in I\}$ gives the distinct coset representatives for $T/\m$.
Thus there are $\gamma$ elements of $C$, and for $c_i, c_j\in C$
with $c_i\ne c_j $, we have $c_i-c_j\notin \m$.
 Now also let
$a\in \m, a\ne 0$. Consider the set $$ G=\{ a+\sum_{n\in
\N}d_ny^n\,|\, d_n\in C\, \forall n\in \N \}.$$ Each of the
elements of $G$ is in $(\m, y)T[[y]] \setminus yT[[y]]$ and hence
is contained in a height-one prime contained in $(\m, y)T[[y]]$
distinct from $yT[[y]]$.

Moreover, $|G| = |C|^{\aleph_0} = \gamma^{\aleph_0}$.  Let $P$ be
a height-one prime ideal of $T[[y]]$ contained in $(\m,y)T[[y]]$
but such that $y\notin P$. If two distinct elements of $G$, say
$f=a+\sum_{n\in \N}d_ny^n$ and $g=a+\sum_{n\in \N}e_ny^n$, with
the $ d_n,e_n\in C$, are both in $P$, then so is their difference;
that is

$$f-g =\sum_{n\in \N}d_ny^n- \sum_{n\in \N}e_ny^n=\sum_{n\in
\N}(d_n-e_n)y^n\in P.$$

\noindent Now let $t$ be the smallest power of $y$ so that $d_t\ne
e_t$. Then $(f-g)/y^t\in P$, since $P$ is prime and $y\notin P$,
but the constant term, $d_t-e_t\notin\m$, which contradicts the
fact that $P\subseteq (\m,y)T[[y]]$. Thus there must be at least
$|C|^{\aleph_0}=\gamma^{\aleph_0}$ distinct height-one  primes
contained in $(\m, y)T[[y]]$.

(3) Using (1) and (2), if $T$ is Noetherian and if
$\gamma^{\aleph_0}=\delta^{\aleph_0}$, then there are exactly
$\gamma^{\aleph_0}=\delta^{\aleph_0}$ distinct height-one prime
ideals (other than $yT[[y]]$\,)  of $T[[y]]$ contained in
$(\m,y)T[[y]]$. This is the case, for example, if $T$ is
countable, say $T=k[x]$ where $k$ is a countable field, for then
$|T|=\aleph_0$ and for every $\gamma$ with $1<\gamma\le\aleph_0$,
$\gamma^{\aleph_0}=\aleph_0^{\aleph_0}=c$, the cardinality of the
real numbers.
\end{newremark}

\begin{theorem} \label{3.4} Suppose that $R$ is a one-dimensional
Noetherian domain with cardinality $\delta:=|R|=$ and that the
cardinality of the set of maximal ideals of $R$ is $\alpha$
($\alpha$ can be finite). Let $U=\Spec R[[y]]$, where $y$ is an
indeterminate over $R$. Then

\noindent {\em  (a)}   $U$ as a partially ordered set (where the
``$\le$" relation is ``set containment") satisfies the following
axioms:
\begin{description}\item[{\em (1)}] $|U|\le\delta^{\aleph_0}$.
\item[{\em (2)}] $U$ has a unique minimal element, namely $(0)$.
\item[{\em (3)}]
$\dim(U)=2$ and $|\{$ height-two elements of $U\,\}|=\alpha$.
  \item[{\em (4)}] There exists a unique {\it
special} height-one element $u \in U$ (namely $u=(y)$) such that $u$
is contained in every height-two element of $U$.
\item[{\em (5)}] Every
nonspecial height-one element of $U$ is in exactly  one height-two
element.
\item[{\em (6)}] Every height-two element $t\in U$ contains uncountably
 many height-one elements that are  contained {\it only} in
$t$. (The number of height-one elements contained only in $t$ is
at least $\gamma^{\aleph_0}$, where $\gamma$ is the cardinality of
the residue field of the corresponding maximal ideal of $R$.) If
$t_1,t_2\in U$ are distinct height-two elements, then the special
element from (4) is the unique height-one element less than both.
\item[{\em (7)}] There are no height-one maximal elements in $U$.
Every maximal element has height two.
\end{description}

\noindent {\em  (b)}  If $R$  is countable, or more generally if
$\delta=|R|$ satisfies the condition of Remarks \ref{3.3.5}.3,
that is, $\gamma^{\aleph_0}=\delta^{\aleph_0}$, for every $\gamma$
that occurs as $|R/\m|$, where $\m$ is a maximal ideal of $R$,
then $U=\Spec R[[y]] $ satisfies (1)-(7), with the stronger axioms
(1$'$) and  (6$'$):
\begin{description}
\item[{\em (1$'$)}] $|U|=\delta^{\aleph_0}$. (For $R$ countable, this is $c$, the
cardinality of the real numbers.)
\item[{\em (6$'$)}] Every height-two element $t\in U$ contains
$\delta^{\aleph_0}$ (uncountably many) height-one elements that are
contained {\it only} in $t$.
\end{description}

\noindent {\em  (c)}  With the additional  hypotheses of (b), $U$
is characterized as a
 partially ordered set  by the axioms given in (a) and (b).
Every partially ordered set satisfying the axioms (1)-(7) in (a)
and (b) is order-isomorphic to every other such partially ordered
set.
  \end{theorem}

  \begin{proof}  In part(a), item (1) is from Remark \ref{3.3.5}.1.
Item (2) and the first part of (3) are clear. The second part of
(3) follows immediately from Remark \ref{2.4}.1.

For items (4) and   (5), suppose that $P$ is a height-one prime of
$R[[y]]$. If $P={yR[[y]]}$, then $P$ is contained in each maximal
ideal of $R[[y]]$  by Remark \ref {2.4}.1, and so ${yR[[y]]}$ is the
special element. If $y\notin P$, then, by Corollary \ref{2.6}, $P$
is contained in a unique maximal ideal of $R[[y]]$.

For item (6) and items (1$'$) and (6$'$) of part (b) use Remarks
\ref{3.3.5}.2 and \ref{3.3.5}.3.

For item (c), all partially ordered sets satisfying the axioms of
Theorem \ref{3.2} are order-isomorphic,  and the partially ordered
set $U$ of the present theorem satisfies the same axioms as in
Theorem \ref{3.2} except axiom (7) that involves height-one
maximals. Since $U$ has no height-one maximals, an
order-isomorphism between two partially ordered sets as in item
(c) can be deduced by adding on height-one maximals and then
deleting them. \qed
\end{proof}

\begin{corollary} In the terminology of Equations 1 and 2 of the
introduction, we have $\Spec C \cong\Spec D_n \cong\Spec E$, but
$\Spec B \ncong \Spec C$.
\end{corollary}

\begin{proof} The rings $C,D_n,$ and $E$ are all formal power series rings in
one variable over a one-dimensional Noetherian domain $R$, where
$R$  is either $k[x]$ or $k[x, 1/x]$. Thus the domain $R$
satisfies the hypotheses of Theorem \ref{3.4} with the cardinality
conditions of parts (b) and (c).  If $k$ is finite, then
$|R|=|k[x]|=\aleph_0$ and $\alpha$, the number of maximal ideals
of $R$, is also $\aleph_0$; in this case $|R/\m| = \gamma$ is
finite for each maximal ideal $\m$ of $R$ and $\delta = |R|
=\gamma\cdot\aleph_0 =\alpha$, and so
$\gamma^{\aleph_0}=\delta^{\aleph_0}$. On the other hand, if $k$
is infinite, then $|k|=|k[x]|= |R| = \alpha$, and
 $|k|=|R/\m|=\gamma$ is the
same for every maximal ideal $\m$ of $R$.  Hence also in this case
$\delta = |R| =\gamma\cdot\aleph_0 =\alpha$, and so
$\gamma^{\aleph_0}=\delta^{\aleph_0}$.

Also the number of maximal ideals is the same for $C,D_n,$ and
$E$, because in each case, it is the same as  the number of
maximal ideals of $R$  which is $|k[x]| = |k|\cdot\aleph_0$.

 Thus in the picture of $R[[y]]$ shown above,
for $R[[y]]=C,D_n$ or $E$,  the $\kappa_i$ are all equal to
$|k|^{\aleph_0}$ and  $\alpha=|k|\cdot\aleph_0$, and so the
spectra are isomorphic. The spectrum of $B$ is not isomorphic to
that of $C$, however, because $B$ contains height-one maximal
ideals, such as that generated by $1+xy$, whereas $C$ has no
height-one maximal ideals. \qed
\end{proof}

\begin{newremark}
As mentioned at the beginning of this section, it is shown in
\cite{rW1} and \cite{rW2} that  $\Spec \mathbb Q[x, y] \not\cong
\Spec F[x, y] \cong \Spec \mathbb  Z[y]$, where $F$ is a field
contained in the algebraic closure of a finite field. Corollary
\ref{3.6} shows that the spectra of power series extensions in $y$
behave differently in that $\Spec \mathbb Z[[y]] \cong\Spec
\Q[x]\,[[y]] \cong \Spec F[x]\,[[y]]$.
\end{newremark}

\begin{corollary} \label{3.6}  If  $\mathbb Z$ is the ring of integers, $\Q $
is the rational numbers, $F$ is a field contained in the algebraic
closure of a finite field, and $\mathbb R$ is the real numbers, then
$$\Spec \mathbb Z[[y]] \cong\Spec \Q[x]\,[[y]]
\cong\Spec F[x]\,[[y]]\not\cong\Spec \mathbb R[x]\,[[y]].$$
\end{corollary}

\begin{proof} The rings $ \mathbb Z, \mathbb Q[x]$ and $F[x]$ are all
countable with countably  infinitely many maximal ideals. Thus if
$R=\mathbb Z, \mathbb Q[x]$ or $F[x]$, then  $R$ satisfies the
hypotheses of Theorem \ref{3.4} with the cardinality conditions of
parts (b) and (c). On the other hand,  $\mathbb R[x]$ has
uncountably many maximal ideals; thus $\mathbb R[x]\,[[y]]$ also has
uncountably many maximal ideals. \qed
\end{proof}

\section{Higher dimensional mixed power series/polynomial
rings} \label{sec4}

In analogy to Equation (\ref{1}), we display several embeddings
involving three variables.

\begin{eqnarray*} (4.0) \phantom{xxx}k[x,y,z]\overset\alpha \hookrightarrow k[[z]]\,[x,y]
\overset\beta
 \hookrightarrow k[x]\,[[z]]\,[y] \overset\gamma \hookrightarrow
k[x,y]\,[[z]]\overset\delta \hookrightarrow
k[x]\,[[y,z]], \phantom{xxxx}\\
\phantom{xxxxxxx}k[[z]]\,[x,y]\overset\epsilon \hookrightarrow
k[[y,z]]\,[x] \overset\zeta \hookrightarrow
k[x]\,[[y,z]]\overset\eta \hookrightarrow k[[x,y,z]],
\phantom{xxxxx}
\end{eqnarray*}
 where $k$ is a field and $x$,
$y$ and $z$ are indeterminates over $k$.


\begin{newremark} \label{4.1}
 \begin{description}
\item[{(1)}] By Proposition \ref{2.7}.2  every nonzero prime ideal
of $C=k[x]\,[[y]]$ has nonzero intersection with $B=k[[y]]\,[x]$.
In three or more variables, however, the analogous statements
fail. We show below that the maps $\alpha, \beta,
 \gamma,\delta, \epsilon, \zeta, \eta$ in Equation (4.0) fail to be TGF. Thus,
by Proposition \ref{2.2}.2,  no proper inclusion in (4.0) is TGF.
The dimensions of the generic fiber rings of the maps in the
diagram are either one or two. \item[{(2)}] For those rings in
(4.0) of form $R=S[[z]]$ (ending in a power series variable) where
$S$ is a ring, such as $R=k[x,y][[z]]$, we have some information
concerning the prime spectra. By Proposition \ref{2.5} every
height-two prime ideal not containing $z$ is contained in a unique
maximal ideal. By \cite{N, Theorem 15.1} the maximal ideals of
$S[[z]]$ are of the form $(\m,z)S[[z]]$, where  $\m$ is a maximal
ideal of $S$, and thus the maximal ideals of $S[[z]]$ are in
one-to-one correspondence with the maximal ideals of $S$.  As in
section 3, using Remarks \ref{2.4}, we see that maximal ideals of
$\Spec k[[z]]\,[x,y]$
 can have
height two or three, that $(z)$ is contained in every height-three
prime ideal, and that every height-two prime ideal not containing
$(z)$ is contained in a unique maximal ideal. \item[{(3)}] It
follows by arguments analogous to that in Proposition \ref{2.7}.1,
that $\alpha$, $\delta$, $\epsilon$ are not TGF. For $\alpha$, let
$\sigma(z) \in zk[[z]]$ be transcendental over $k(z)$; then $(x -
\sigma)k[[z]]\,[x, y] \cap k[x, y, z] = (0)$. For $\delta$ and
$\epsilon$:  let $\sigma(y) \in k[[y]]$ be transcendental over
$k(y)$; then  $(x - \sigma)k[x]\,[[z, y] ]\cap k[x]\,[[z]]\,[ y] =
(0)$, and $(x - \sigma)k[[y, z]]\,[x] \cap k[[z]]\,[x, y] = (0)$.
\item[{(4)}] By \cite{HRW1, Theorem 1.1}, $\eta$ is not TGF and
the dimension of the generic fiber ring of $\eta$ is one.


\end{description}




\end{newremark}


In order to show in Proposition \ref{4.3}  below that the map
$\beta$ is not TGF, we first observe:

\begin{proposition} \label{4.01}  The element
$\sigma = \sum_{n=1}^\infty(xz)^{n!} \in k[x]\,[[z]]$ is
transcendental over $k[[z]]\,[x]$.
\end{proposition}

\begin{proof}
Consider an expression

$$
Z := a_\ell \sigma^{\ell} + a_{\ell - 1} \sigma^{\ell - 1} + \cdots
+ a_1\sigma + a_0,
$$
where the $a_i \in k[[z]]\,[x]$ and $a_{\ell} \ne 0$. Let $m$ be
an integer greater than $\ell + 1$ and greater than $\deg_xa_i$
for each $i$  such that  $0 \le i \le \ell$ and $a_i \ne 0$.
Regard each $a_i \sigma^i$ as a power series in $x$ with
coefficients in $k[[z]]$.

For each $i$ with $0 \le i \le \ell$, we have $i(m!) < (m+1)!$. It
follows that the coefficient of $x^{i(m!)}$ in $\sigma^i$ is
nonzero, and the coefficient of $x^j$ in $\sigma^i$ is zero for
every $j$ with $i(m!) < j < (m+1)!$. Thus if $a_i \ne 0$ and $j =
i(m!) + \deg_xa_i$, then the coefficient of $x^j$ in $a_i\sigma^i
$ is nonzero, while for $j$ such that $i(m!) + \deg_xa_i < j <
(m+1)!$, the coefficient of $x^j$ in $a_i\sigma^i$ is zero. By our
choice of $m$, for each $i$ such that  $0 \le i < \ell$ and $a_i
\ne 0$, we have
$$
(m+1)! > \ell(m!) + \deg_xa_{\ell} \ge i(m!) + m! > i(m!) +
\deg_xa_i.
$$
Thus in $Z$,  regarded as a power series in $x$ with coefficients in
$k[[z]]$, the coefficient of $x^j$ is nonzero for $j = \ell(m!) +
\deg_xa_{\ell}$. Therefore $Z \ne 0$. We conclude that $\sigma$ is
transcendental over $k[[z]]\,[x]$. \qed
\end{proof}

\begin{proposition} \label{4.3} $k[[z]]\,[x,y] \overset\beta\hookrightarrow
k[x]\,[[z]]\,[y]$ is not TGF.
\end{proposition}

\begin{proof}  Fix an element $\sigma \in k[x]\,[[z]]$ that is
transcendental over $k[[z]]\,[x]$. We  define $\pi :
k[x]\,[[z]]\,[y] \to k[x]\,[[z]]$ to be the identity map on
$k[x]\,[[z]]$ and $\pi(y)=\sigma z$. Let $\q = \ker \pi$. Then $y
- \sigma z \in \q$. If $h\in \q\cap (k[[z]] \,[x, y])$, then
\begin{eqnarray*}
h=\sum_{j=0}^s \sum_{i=0}^t(\sum_{\ell\in\N}
a_{ij\ell}z^\ell)x^iy^j ,\text{ for some }\,\, s, t \in \N\text{
and } a_{ij\ell}\in k, \text{ and so }
\\ 0=\pi (h)= \sum_{j=0}^s \sum_{i=0}^t(\sum_{\ell\in\N}
a_{ij\ell}z^\ell)x^i(\sigma z)^j = \sum_{j=0}^s
\sum_{i=0}^t(\sum_{\ell\in\N} a_{ij\ell}z^{\ell + j}
)x^i\sigma^j.\end{eqnarray*}


Since $\sigma$ is transcendental over $k[[z]]\,[x]$, we have that
$x$ and $\sigma$ are algebraically independent over $k((z))$. Thus
each of the $a_{ij\ell} = 0$.
 Therefore  $\q\cap (k[[z]] [x, y])=(0)$, and so the
embedding $\beta$ is not TGF. \qed
\end{proof}

\begin{proposition} \label{4.4} $k[[y,z]]\,[x] \overset\zeta\hookrightarrow k[x]\,[[y,
z]]$ and $k[x]\,[[z]]\,[y] \overset\gamma \hookrightarrow
k[x,y]\,[[z]]$ are not TGF.
\end{proposition}

\begin{proof} For $\zeta$, let $t=xy$ and let $\sigma\in k[[t]]$ be
algebraically independent over $k(t)$. Define $\pi : k[x]\,[[y,
z]] \to k[x]\,[[y]]$ as follows. For
$$
f: = \sum_{\ell=0}^\infty \sum_{m+n = \ell}f_{mn}(x)y^mz^n \in
k[x]\,[[y, z]],
$$
where $f_{mn}(x) \in k[x]$, define
$$
\pi(f) : = \sum_{\ell=0}^\infty \sum_{m+n =
\ell}f_{mn}(x)y^m(\sigma y)^n \in k[x]\,[[y]].
$$
In particular, $\pi(z) = \sigma y$. Let $\p := \ker \pi$. Then $z -
\sigma y \in \p$, and so $\p \ne (0)$.  Let $h \in \p \cap k[[y,
z]]\,[x]$. We show $h = 0$. Now $h$ is a polynomial with
coefficients in $k[[y,z]]$, and we define $g\in k[[y, z]]\,[t]$, by,
if $a_i(y,z)\in k[[y,z]]$ and
$$h:=\sum_{i=0}^ra_i(y,z)x^i,\text{ then set }
 g:=y^rh=\sum_{i=0}^r(\sum_{\ell=0}^\infty\,\sum_{m+n=\ell}
b_{imn}y^mz^n)t^i.$$ The coefficients of $g$ are in $k[[y,z]]$,
since $y^rx^i=y^{r-i}t^i$.  Thus
\begin{eqnarray*}
0 =\pi(g)=\sum_{i=0}^r(\sum_{\ell=0}^\infty\,\sum_{m+n=\ell}
b_{imn}y^m(\sigma y)^n)t^i
=\sum_{i=0}^r(\sum_{\ell=0}^\infty\,\sum_{m+n=\ell}
b_{imn}\sigma^n y^\ell)t^i\\
 =\sum_{\ell=0}^\infty({\sum_{m+n=\ell}}
(\sum_{i=0}^rb_{imn}t^i)\sigma^n)y^\ell. \end{eqnarray*} Now $t$
and $y$ are analytically independent over $k$, and so the
coefficient of each $y^\ell$ (in $k[[t]]$) is $0$; since $\sigma$
and $t$ are algebraically independent over $k$, the coefficient of
each $\sigma^n$ is $0$. It follows that each $b_{imn}=0$, that
$g=0$ and hence that $h=0$. Thus the extension $\zeta$ is  not
TGF.

To see that $\gamma$ is not TGF, we switch variables in the proof
for $\zeta$, so that $t=yz$. Again choose $\sigma\in k[[t]]$ to be
algebraically independent over $k(t)$. Define $\psi :
k[x,y]\,[[z]] \to k[y]\,[[z]]$ by $\psi(x)=\sigma z$ and $\psi$ is
the identity on $k[y]\,[[z]]$. Then $\psi $ can be extended to
$\pi: k[y]\,[[x,z]]\to k[y]\,[[z]]$, which is similar to the $\pi$
in the proof above. As above, set $\p := \ker \pi$; then $\p \cap
k[[x, z]]\,[y]=(0)$. Thus $\p \cap k[x]\, [[z]]\,[y]=(0)$ and
$\gamma$ is not TGF. \qed
\end{proof}

\begin{proposition} \label{4.4.5} Let $k$ be a field and let
$x$ and $t$ be indeterminates over $k$. Then $\sigma =
\sum_{n=1}^\infty t^{n!}$ is algebraically independent over $k[[x,
xt]]$.
\end{proposition}

\begin{proof} Let $\ell$ be a positive integer and
consider an expression
$$
\gamma := \gamma_\ell\sigma^{\ell} + \cdots + \gamma_i\sigma^i +
\cdots + \gamma_1\sigma,\text{ where }\gamma_i :=
\sum_{j=0}^\infty f_{ij}(x)(xt)^j\in k[[x,xt]],
$$
that is, each $f_{ij}(x) \in k[[x]]$ and $1\le i\le\ell$. Assume
that $\gamma_{\ell} \ne 0$. Let $a_{\ell}$ be the smallest $j$
such that $f_{\ell j}(x) \ne 0$, and let $m_{\ell}$ be the order
of $f_{\ell {a_{\ell}}}(x)$, that is,
 $f_{\ell a_{\ell}}(x) = x^{m_{\ell}}g_{\ell}(x)$, where $g_{\ell}(0) \ne 0$.
Let $n$ be a positive integer such that
$$
n \ge 2 + \max\{\ell, m_{\ell}, a_{\ell}  \}.
$$
Since $\ell < n$, for each $i$ with $1 \le i \le \ell$,  we have


\begin{equation} \label{eq}
\sigma^i = \sigma_{i1}(t) + c_it^{i(n!)} + t^{(n+1)!}\tau_i(t),
\end{equation}
where $c_i$ is a nonzero element of $k$, $\sigma_{i1}(t)$ is a
polynomial in $k[t]$ of degree at most $(i-1)n! + (n-1)!$ and
$\tau_i(t) \in k[[t]]$.
\end{proof}

\begin{newclaim} \label{4.4.6}
The coefficient of $t^{\ell(n!) + a_{\ell}}$ in
$\sigma^{\ell}\gamma_{\ell} =
\sigma^{\ell}(\sum_{j=a_{\ell}}^\infty f_{\ell j}(x)(xt)^j)$ as a
power series in $k[[x]]$ has order $m_{\ell}  + a_{\ell}$, and
hence, in particular, is nonzero.
\end{newclaim}

\begin{proof}
By the choice of $n$, $(n+1)! > \ell(n!) + a_{\ell}$. Hence by the
expression for $\sigma^{\ell}$ given in Equation \ref{eq}, we see
that all of the terms in $\sigma^{\ell}\gamma_{\ell}$ of the form
$bt^{\ell(n!)+a_{\ell}}$, for some $b \in k[[x]]$, appear in the
product
$$
(\sigma_{\ell 1}(t) + c_{\ell}t^{\ell(n!)})(\sum_{j=
a_{\ell}}^{\ell(n!)+a_{\ell}}f_{\ell j}(x)(xt)^j).
$$
One of the terms of the form $bt^{\ell(n!)+a_{\ell}}$ in this
product is
$$
c_{\ell}t^{\ell(n!)}f_{\ell a_{\ell}}(x)(xt)^{a_{\ell}} =
(c_{\ell}x^{m_{\ell}+a_{\ell}}g_{\ell}(x))t^{\ell(n!) + a_{\ell}}
= (c_{\ell}x^{m_{\ell}+a_{\ell}}g_{\ell}(0) + \hdots )t^{\ell(n!)
+ a_{\ell}}.
$$
Since $c_{\ell}g_{\ell}(0)$ is a nonzero element of $k$,
$c_{\ell}x^{m_{\ell}+a_{\ell}}g_{\ell}(x) \in k[[x]]$ has order
$m_{\ell}  + a_{\ell}$.  The other terms in the product
$\sigma^{\ell}\gamma_{\ell}$ that have  the form
$bt^{\ell(n!)+a_{\ell}}$, for some $b \in k[[x]]$, are in the
product
$$
(\sigma_{\ell 1}(t))(\sum_{j= a_{\ell}}^{\ell(n!)+a_{\ell}}f_{\ell
j}(x)(xt)^j) = \sum_{j= a_{\ell}}^{\ell(n!)+a_{\ell}}f_{\ell
j}(x)(xt)^j \sigma_{\ell 1}(t).
$$
Since $\deg_t\sigma_{\ell 1} \le (\ell -1)n! + (n-1)!$ and since,
for each $j$ with  $f_{\ell j}(x) \ne 0$, we have $\deg_t f_{\ell
j}(x)(xt)^j = j$, we see that  each term in $f_{\ell j}(xt)^j
\sigma_{\ell 1}(t)$ has degree in $t$ less than or equal to $j +
(\ell -1)n! + (n-1)!$.  Thus each nonzero  term in this product of
the form $bt^{\ell(n!) + a_{\ell}}$ has
$$
j \ge \ell (n!) + a_{\ell} - (\ell - 1)(n!) - (n-1)! = a_{\ell} +
(n-1)!(n-1) >
 m_{\ell} + a_{\ell},
$$
by choice of $n$. Moreover,  for $j$ such that $f_{\ell j}(x) \ne
0$, the order in $x$ of $f_{\ell j}(x)(xt)^j$ is bigger than or
equal to $j$. This completes the proof of Claim  \ref{4.4.6}.
\end{proof}

\begin{newclaim} \label{4.4.7} For $i < \ell$, the coefficient of
$t^{\ell(n!) + a_{\ell}}$ in $\sigma^{i}\gamma_{i} $ as a power
series in $k[[x]]$ is either zero or  has order greater than $
m_{\ell}  +  a_{\ell} $.
\end{newclaim}

\begin{proof}
As in the proof of Claim \ref{4.4.6}, all of the terms in
$\sigma^{i}\gamma_{i}$ of the form $bt^{\ell(n!)+a_{\ell}}$, for
some $b \in k[[x]]$, appear in the product
$$
(\sigma_{i 1} + c_{i}t^{i(n!)})(\sum_{j= 0}^{\ell(n!)+a_{\ell}}f_{i
j}(x)(xt)^j)  = \sum_{j= 0}^{\ell(n!)+a_{\ell}}f_{i
j}(x)(xt)^j(\sigma_{i 1} + c_{i}t^{i(n!)}).
$$
Since $\deg_t(\sigma_{i 1} + c_{i}t^{i(n!)}) = i(n!)$, each term
in $ f_{i j}(x)(xt)^j(\sigma_{i 1} + c_{i}t^{i(n!)}) $ has degree
in $t$ at most $j + i(n!)$. Thus each  term in this product of the
form $bt^{\ell(n!) + a_{\ell}}$,  for some nonzero $b \in k[[x]]$,
has
 $$
j \ge \ell (n!) + a_{\ell} - i(n!) \ge n! + a_{\ell}  >
 m_{\ell} + a_{\ell}.
$$
Thus $\ord_x b \ge j >  m_{\ell} + a_{\ell}$. This completes the
proof of Claim \ref{4.4.7}. Hence $\gamma\not\in k[[x,xz]]$ and so
Proposition \ref{4.4.5} is proved.
\end{proof}

\begin{quesremark}
\begin{description}
\item[{(1)}] As we show in Proposition \ref{2.7}, the embeddings
from Equation 1 involving two-dimensional mixed power
 series/polynomial rings over a field  $k$ with
inverted elements are TGF. So far we have not determined whether
the same is true in the three-dimensional case. For example, is
$\theta$ below TGF?
\begin{eqnarray*}
k[x,y]\,[[z]]\overset\theta\hookrightarrow k[x,y, 1/x]\,[[z]]
\end{eqnarray*}
 \item[{(2)}] For the
four dimensional case, as observed in the introduction, it follows
from \cite{HR, p. 364, Theorem 1.12} that the extension
$k[x,y,u]\,[z] \hookrightarrow k[x,y,u,1/x,]\,[[z]]$ is not TGF.
We provide in Proposition \ref{4.9} a direct proof of this fact.
\end{description}\end{quesremark}

 \begin{proposition}\label{4.9} For $k$ a field and
$x, y, u$ and $z$ indeterminates over $k$, the extension
$k[x,y,u]\,[[z]] \hookrightarrow k[x,y,u,1/x,]\,[[z]]$ is not TGF.
\end{proposition}

\begin{proof} Let $t=z/x$ and let  $\sigma\in k[[t]]$
be algebraically independent over $k[[x, z]]$. (By Proposition
\ref{4.4.5}, we may take  $\sigma=\sum_{r=1}^\infty t^{r!}\,$. )

Consider
$$
\pi : k[[x, y, u]]\,[1/x]\,[[z]] \to k[[x, u]]\,[1/x]\,[[z]]
$$
defined by mapping
$$
\sum_{i=0}^\infty a_i(x, y, u,1/x)z^i \mapsto \sum_{i=0}^\infty
a_i(x, \sigma u, u,1/x)z^i,
$$
where $a_i(x, y, u,1/x) \in k[[x, y, u]][1/x]$.  Let $\p = \ker
\pi$. Then $y - \sigma u \in \p$. We show that $\p \cap k[[x, y,
u, z]] = (0)$, and so also $\p \cap k[x, y, u]\,[[ z]] = (0)$. Let
$$
f := \sum_{\ell = 0}^\infty(\sum_{i+j = \ell} d_{ij}u^iy^j) \in
k[[x, y, u, z]],
$$
where $d_{ij} \in k[[x, z]]$. If $f \in \p$, then
$$
0 = \pi(f) = \sum_{\ell = 0}^\infty(\sum_{i+j =
\ell}d_{ij}u^i\sigma^ju^j) = \sum_{\ell = 0}^\infty(\sum_{i+j =
\ell}d_{ij}\sigma^j)u^{\ell}.
$$
This is a power series in $u$, and so, for each $\ell$, $\sum_{i+j =
\ell}d_{ij}\sigma^j = 0$. Since $\sigma$ is algebraically
independent over $k[[x, z]]$, each $d_{ij} = 0$. Thus $f=0$. This
completes the proof of Proposition \ref{4.9}.





 \end{proof}

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