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\newcommand{\heading}[1]{\begin{center}\small\bf #1\end{center}}
\begin{document}

\begin{slide}
\heading{MIXED POLYNOMIAL/POWER SERIES RINGS }

\heading{AND RELATIONS  AMONG THEIR SPECTRA  }

\vspace{0.2in}

\begin{center}
William Heinzer
\end{center}

\begin{center}
Purdue University
\end{center}

\vspace{0.2in}

\begin{center}
joint with Christel Rotthaus and Sylvia Wiegand
\end{center}
\end{slide}




\begin{slide}

\heading{OVERVIEW 1}

At first glance the rings
$$
B:=k[[y]]\,[x] \qquad \text{ and } \qquad C:=k[x]\,[[y]]
$$
look similar. One has
$$
B = k[[y]]\,[x]\hookrightarrow k[x]\,[[y]] = C,
$$
but this is a strict inclusion. For example, $1 - xy$ is a nonunit
of $B$,

and
$$
\frac{1}{1-xy} = \sum_{i=0}^\infty x^ny^n \in C,
$$
so $1-xy$ is a unit of $C$. \quad Indeed, the rings $B$ and $C$ are
not

isomorphic: the intersection of the maximal ideals of $B$ is $(0)$,

while  $y$ is in every maximal ideal of $C$.
\end{slide}

\begin{slide}

\heading{OVERVIEW 2}

\medskip

Consider the mixed polynomial/power series rings

$$
 A:=k[x,y]\hookrightarrow B:=k[[y]]\,[x]\hookrightarrow
C:=k[x]\,[[y]] \hookrightarrow D := k[[x, y]],
$$

where $k$ is  a field. The inclusion maps here are all flat

\medskip

homomorphisms. The prime ideal structure of these rings is

\medskip

well understood. The above inclusions induce maps

$$
\Spec A \quad \leftarrow \quad  \Spec B \quad \leftarrow \quad \Spec
C \quad \leftarrow \quad \Spec D.
$$

We are interested in describing these $\Spec$ maps.

\end{slide}



\begin{slide}

\heading{OVERVIEW 3}

\medskip




Consider
$$
k[x]\,[[y]] = C \hookrightarrow C[1/x] \hookrightarrow
 k[x,1/x]\,[[y]] := E,
$$
At first glance, it appears that $E$ is a localization of $C$, but
it is not.

\medskip

There are elements in $E$ that are not in the fraction field of $C$.

\medskip

However, $E$ is obtained from $C$ by the  localization $C[1/x]$
followed

\medskip

by the $(y)$- adic completion of $C[1/x]$. Thus  $E$ is flat over
$C$.

\medskip
The map $C \hookrightarrow  E$ induces $\Spec C \leftarrow \Spec E$,
and again we are

\medskip
interested in describing this $\Spec$ map.

\end{slide}


\begin{slide}

\heading{OVERVIEW 4}

Also consider

$$
C\hookrightarrow C_1:=k[x]\,[[\frac{y}{x}]]\hookrightarrow  \cdots
\hookrightarrow C_n:= k[x]\,[[\frac{y}{x^n}]] \hookrightarrow \cdots
\hookrightarrow  E.
$$

The maps  $C \hookrightarrow C_n$ and $C_i \hookrightarrow C_n$ for
$i < n$  are not flat, but

\medskip


$C_n \hookrightarrow E = k[x,1/x]\,[[y]]$ is the   localization
$C_n[1/x]$ followed by

\medskip

the $(y)$-adic  completion of  $C_n[1/x]$.  Thus  $C_n
\hookrightarrow E$ is flat. These

\medskip

inclusion maps induce maps
$$
\Spec C \leftarrow \Spec C_1 \leftarrow \cdots \leftarrow \Spec C_n
\leftarrow \cdots \leftarrow \Spec E.
$$
We are interested in describing these $\Spec$ maps.



\end{slide}










\begin{slide}

\heading{GENERIC FIBER RINGS}

\medskip

Let $R \hookrightarrow S$ be an injective homomorphism of
commutative rings

\medskip

with  $R$  an integral domain. The {\bf generic fiber ring } of
the map

\medskip

$R \hookrightarrow S$ is the localization $(R \setminus
\{0\})^{-1}S$ of $S$.  With

\medskip

 $A:=k[x,y]\hookrightarrow B:=k[[y]]\,[x]\hookrightarrow
C:=k[x]\,[[y]] \hookrightarrow D := k[[x, y]], $

\medskip
the generic fiber ring of  $A \hookrightarrow R$ is one-dim.   for
$R \in \{B, C, D \}$,

\medskip


while the generic fiber ring of $R \hookrightarrow S$  is zero-dim
for $R \subseteq S$ in

\medskip


$\{B, C, D \}$.




\end{slide}




\begin{slide}

\heading{TRIVIAL GENERIC FIBER EXTENSIONS}

Let $R$ be a subring of an integral domain $S$.

{\bf  Definition.}  $R\hookrightarrow S$ is a {\bf trivial generic
fiber} extension or a

{\bf TGF} extension if

\center{$(0)\ne P\in \Spec S \implies P\cap R\ne (0).$}

\medskip

One obtains a TGF extension $S$ of $R$ by considering
$$
R\hookrightarrow T \to T/P := S,
$$ where $T$ is an
extension ring of $R$  and $P \in \Spec T$ is maximal with

respect to $P\cap R=(0)$.

Thus the generic fiber ring of $R \hookrightarrow T$ is relevant
to constructing

TGF extensions $S$ of $R$.




\end{slide}


\begin{slide}

\heading{A TGF EXTENSION}

Let $x$ and $y$ be indeterminates over a field $k$. Then

$$
R := k[[x,y]]\hookrightarrow S := k[[x]]\,[[\frac{y}{x}]] \quad
\text{ is TGF}.
$$

{\bf Proof}. It suffices to show $P \cap R \ne (0)$ for each $P \in
\Spec S$ with

\medskip



$\hgt P = 1$.
  This is clear if $x \in P$, while
if $x \not\in P$, then

\medskip



$k[[x]] \cap P = (0)$,\quad so $k[[x]] \hookrightarrow  R/(P \cap R)
\hookrightarrow  S/P$.  Now

\medskip

$S/P$ is one-dim local with residue field $k$. \quad Hence by
Cohen's

\medskip

Theorem 8,  \quad $S/P$ is finite over $k[[x]]$. Thus  $\dim R/(P
\cap R) = 1$,

\medskip


so $P \cap R \ne (0)$.


\end{slide}



\begin{slide}
\begin{center}
\textbf{Cohen's Theorem 8}
\end{center}

{\bf Theorem (Classical)} Let $I$ be an ideal of a ring $R$ and
let $M$

be an $R$-module. Assume that $R$ is complete in the $I$-adic
topology

and $\bigcap_{n=1}^\infty I^nM = (0)$. If $M/I$ is generated over
$R/I$ by elements

$\overline{w}_1, \ldots, \overline{w}_s$ and $w_i$ is a preimage
in $M$ of $\overline{w}_i$ for $1 \le i \le s$, then

$M$ is generated over $R$ by $w_1, \ldots, w_s$.

\medskip
This is useful for proving that with



 $$B:=k[[y]]\,[x]\hookrightarrow
C:=k[x]\,[[y]] \hookrightarrow D := k[[x, y]], $$


\medskip
then $R \hookrightarrow S$ is TGF for $R \subseteq S$ in $\{B, C,
D\}$.



\end{slide}

\begin{slide}
\heading{TGF EXTENSIONS}

\medskip


{\bf PROP. 1}. \quad  Let $R \hookrightarrow S$ and $S
\hookrightarrow T$ be injective maps,

\medskip

 where $R$, $S$ and $T$ are integral domains.

\begin{description}
 \item[(1)] If $R \hookrightarrow S$ and $S
\hookrightarrow T$ are TGF extensions, then so is $R \hookrightarrow
T$.

\medskip

Equivalently if $R \hookrightarrow T$ is not TGF, then at least one
of the

\medskip

extensions $R \hookrightarrow S$ or $S \hookrightarrow T$ is not
TGF.

\item[(2)] If $R \hookrightarrow T$
is TGF, then $S \hookrightarrow T$ is TGF.

\item[(3)] If the map $\Spec
T \to \Spec S$ is surjective, then $R \hookrightarrow T$

\medskip

is TGF implies $R \hookrightarrow S$ is TGF.
\end{description}


\end{slide}

\begin{slide}

\heading{A NON-TGF EXTENSION}

{\bf PROP. 2}.  $R = k[[x]][y, z] \hookrightarrow k[y, z][[x]] = S$
is not TGF.

\medskip

{\bf Proof.}  There exists  $\sigma \in k[y][[x]]$ that is
transcendental over

\medskip

$k[[x]][y]$. \quad  Let $\q = (z - \sigma x)k[ y, z][[x]]$.

\medskip

Define $\pi: k[y, z][[x]]\to  k[y, z][[x]]/\q \cong k[y][[x]]$. Thus

\medskip

$\pi(z)=\sigma x$. If $h\in \q\cap (k[[x]] [y, z])$, then $\exists
\,\, s, t \in \N$ so that

\medskip

$h=\sum_{i=0}^s \sum_{j=0}^t(\sum_{\ell\in\N}
a_{ij\ell}x^\ell)y^iz^j $, \quad
 where $a_{ij\ell}\in k$.

 \medskip

Hence  $0=\pi (h)= \sum_{i=0}^s \sum_{j=0}^t(\sum_{\ell\in\N}
a_{ij\ell}x^\ell)y^i(\sigma x)^j $.

\medskip

Since $\sigma$ is transcendental over $k[[x]][y]$, each $a_{ij\ell}
= 0$.

\medskip

Therefore  $\q\cap (k[[x]] [y, z])=(0)$, and $R \hookrightarrow S$
is not TGF.


\end{slide}





\begin{slide}

\heading{POWER SERIES RINGS}

\medskip

{\bf Lemma}.
 Let
$R[[y]]$ denote the  power series ring in the variable

\medskip

$y$ over the commutative ring $R$. Then

\medskip

\begin{description}
\item[{(1)}]
Each maximal ideal of $R[[y]]$ has the form $(\m, y)R[[y]]$,  where

\medskip

$\m$ is a maximal ideal of $R$. Thus $y$ is in every maximal ideal

\medskip


of $R[[y]]$.

\medskip

\item[{(2)}] If $R$ is Noetherian with $\dim R[[y]] = n$ and $x_1,
\ldots, x_m$ are

\medskip

indeterminates over $R[[y]]$, then $y$ is in every maximal

\medskip

ideal of  height $n+m$  of the polynomial ring

\medskip



$R[[y]]\,[x_1, \ldots,x_m]$.
\end{description}


\end{slide}



\begin{slide}

{\bf Lemma}.  Let  $R$ be an  $n$-dim.  Noetherian domain, let $y$
be an

\medskip

indeterminate over $R$,
 and let $\q$ be a prime ideal of height $n$ in

 \medskip

 the power series ring
$R[[y]]$. \quad If $y \not\in \q$, then $\q$ is contained in

\medskip

a unique maximal ideal of $R[[y]]$.

\medskip

{\bf Proof}. Let $S := R[[y]]/\q$. The assertion is clear if $\q$ is
maximal.

\medskip

Otherwise, $\dim S = 1$. Moreover,  $S$ is complete in its $yS$-adic

\medskip

topology  and every maximal ideal of $S$ is a minimal prime of the

\medskip


principal ideal $yS$. Hence $S$ is a complete semilocal ring. Since
$S$

\medskip

 is also an integral domain, it is local by [Mat., Theorem
8.15].

\medskip

Thus $\q$ is contained in a unique maximal ideal of $R[[y]]$.




\end{slide}



\begin{slide}


\begin{center}
The picture of $\Spec k[[y]]\,[x]$ is shown below: \end{center}
\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 5 pt

\centerline{
 In the diagram, $\beta$ is the cardinality of $k[[y]]$, and
$\alpha$ is the}

cardinality of the set of maximal ideals of $k[x]$; the boxed
$\beta$ means

there are cardinality $\beta$ height-one primes in that position
with

\begin{center} respect to the partial ordering. \end{center}

\end{slide}

\begin{slide}

\begin{center}
The picture of $\Spec k[x]\,[[y]]$ is shown below: \end{center}




\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%
$\boxed{\beta}$\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 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

Here $\alpha$ is the cardinality of the set of maximal ideals of
$k[x]$, and

\begin{center}
$\beta$ is the uncountable cardinal equal to the cardinality of
$k[[y]]$. \end{center}

 \end{slide}


\begin{slide}

\begin{center}
The picture of $\Spec R[[y]]$ for a one-dim Noetherian domain $R$:
\end{center}

\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

\begin{center}
Here $\kappa_i$ and $\kappa_j$ are uncountable cardinals.
\end{center}

\end{slide}

\begin{slide}

\heading{ISOMORPHIC SPECTRA}

\medskip

{\bf REMARK}.  Let $F$ be a field that is algebraic over a finite
field.

\medskip

Roger Wiegand proved that as partially ordered sets or

\medskip

topological spaces

$$ \Spec \mathbb Q[x, y] \not\cong \Spec F[x, y] \cong \Spec \mathbb  Z[y].
$$

The spectra of power series extensions in $y$ behave differently:

We have
$$\Spec \mathbb Z[[y]] \cong\Spec \mathbb Q[x]\,[[y]] \cong \Spec
F[x]\,[[y]].
$$

\end{slide}



\begin{slide}

\heading{Higher dimensional mixed power series/polynomial rings}

We display several extensions involving three variables:

\begin{eqnarray*}  \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*}


We have been able to show most of these extensions are not TGF.
\end{slide}


\begin{slide}


{\bf PROP. 3}. \quad   $k[[z]]\,[x,y] \overset\beta\hookrightarrow
k[x]\,[[z]]\,[y]$ is not TGF.

{\bf Proof}.  Fix  $\sigma \in k[x]\,[[z]]$ that is transcendental
over $k[[z]]\,[x]$.

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 trans.  over $k[[z]]\,[x]$, \quad  $x$ and
$\sigma$ are alg.  indep. over

$k((z))$. Thus each  $a_{ij\ell} = 0$.
 Therefore  $\q\cap (k[[z]] [x, y])=(0)$, and

  the
embedding $\beta$ is not TGF.

\end{slide}



\begin{slide}

{\bf QUESTION}.\quad Is $k[x,y]\,[[z]]\overset\theta\hookrightarrow
k[x,y, 1/x]\,[[z]]$ TGF?

\medskip

\medskip

\medskip


{\bf REMARK}.  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{slide}



\end{document}





\begin{slide}
\heading{Mixed Polynomial/Power Series Rings}

Consider

 $$
 A:=K[x,y]\hookrightarrow B:=K[[x]][y]\hookrightarrow
C:=K[y][[x]]
$$

$$
\hookrightarrow  D:=K[y][[x/y]]\hookrightarrow E:=K[y,1/y][[x]]
$$
\medskip

where $K$ is a field and $x$ and $y$ are indeterminates over $K$.

\medskip

We are interested in   the prime spectra of these rings and the
maps

on the spectra  determined by the inclusion maps on the rings. For

example, do there exist nonzero primes of one of the larger rings

that intersect a smaller ring in zero (i.e. non-TGF extensions).
\end{slide}

\begin{slide}

\heading{Motivation 2}

 From the introduction to a paper  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  it  also be
factored as an open immersion followed by a closed one?
\medskip

Brian Conrad observed that a counterexample can be constructed for
every triple $(R, x, p)$, where

\begin{enumerate} \item $R$ 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
 $x \in R$ is  nonzero,  the completion
of $R[1/x]$ with respect to the powers of $IR[1/x]$, denoted $S: =
R_{\{x\}}$,  is  an integral domain.


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

\medskip


\end{slide}


\begin{slide}

\heading{Motivation 2 continued}

If  $(R, x, p)$ is such a triple  with $S = R_{\{x\}}$, then the
composition $R \to S \to S/p$ determines a   map on formal spectra
$\text{Spf}(S/p)\to \text{Spf}(S) \to \text{Spf}(R)$ that is a
closed immersion followed by an open one.


For 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
like that of $R$ with respect to $x$ 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 $\Spf(S) \to \Spf(R)$.

However, the map $\Spf(S/p) \to \Spf(R)$ cannot be factored as an
open immersion followed by a closed one. For a closed immersion
into $\Spf(R)$ corresponds to a surjective map of adic rings $R
\to R/J$, where $J$ is an ideal of $R$.

\bigskip

\end{slide}

\begin{slide}
\heading{Motivation 2 continued}

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.

One of our motivations is to describe such triples $(R, x, p)$.


\end{slide}

\begin{slide}

\heading{An example of $(R, x, p)$.}

An example described in [AJL]: Let $w,x,y,z$ be indeterminates

over a field $k$ and consider
$$
R:=k[w,x,z][[y]] \hookrightarrow  S:=R_{\{x\}}= k[w,x,1/x,z][[y]].
$$

Let  $P=(w,z)R$ and $T:=R_P\subset S_{PS}:=G$.



Then  $T\subset G$ are 2-dim  regular local domains, and the
residue

field of $G$ (i.e. the fraction field of $k[x,1/x][[y]]$) is
transcendental

over that of $T$ (i.e. the fraction field of $k[x][[y]]$).


  It follows by [HR, p. 364, Theorem 1.12]
that there exist infinitely

many height-one prime ideals of $G$ in the generic fiber over $T$.
Any

one of these prime ideals of $G$ contracts in $S$ to a prime $p$
as above.


\end{slide}


\begin{slide}

\heading{Is There A Simpler Example?}
\medskip

In some correspondence to Lipman, Conrad asked:

\medskip
{\bf Question.} Is there a nonzero prime ideal of
$E:=k[x,1/x][[y]]$ that intersects $C = k[x][[y]]$ in zero?

\medskip

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.

We show there is no such prime ideal.

\end{slide}



\begin{slide}
\heading{TGF Extensions.}

Let $k$ be a field and let $x$ and $y$ be indeterminates over $k$.

\medskip


{\bf  Question.} Is  $C:=k[x][[y]]\hookrightarrow
E:=k[x,1/x][[y]]$
 a TGF extension?

We show that the answer to this  question  is ``yes''.

This is part of our analysis of  the prime spectra  of $A$, $B$,
$C$, $D$ and

$E$, and the maps among these spectra determined by the inclusion


maps on the rings.




\end{slide}

\begin{slide}
\begin{center}
\textbf{An Example.}
\end{center}

\bigskip
{\bf Example.} The extension  $R := k[[x,y]]\hookrightarrow S :=
k[[x]][[\frac{y}{x}]]$ is a

TGF extension of the type described in Hochster's question.

\medskip

 To see  this extension is TGF, it suffices to
show $P \cap R \ne (0)$

for each $P \in \Spec S$ with $\hgt P = 1$.

\medskip

  This is clear if $x \in P$, while
if $x \not\in P$, then  $k[[x]] \cap P = (0)$, so

$k[[x]] \hookrightarrow  R/(P \cap R) \hookrightarrow  S/P$ and
$S/P$ is finite over $k[[x]]$.

Therefore $\dim R/(P \cap R) = 1$, so $P \cap R \ne (0)$.

\medskip

{\bf  Remark.} The extension $k[[x, y]] \hookrightarrow k[[x,
\frac{y}{x}]]$ is, up to isomorphism,

 the same as the extension
$k[[x, xy]] \hookrightarrow k[[x, y]]$.


\end{slide}




\begin{slide}


\end{slide}

\begin{slide}


\end{slide}

\begin{slide}


\end{slide}

\end{document}





Equivalently, $(R\setminus \{0\})^{-1}S$ is the field

\medskip
of fractions of $S$.

\begin{slide}
\begin{center}
\textbf{Local Embeddings}
\end{center}
There exist local embeddings:

\medskip

$ A = K[X]_{(X)}\hookrightarrow \widehat A:=K[[X]],\qquad \widehat A
\hookrightarrow \widehat B = \widehat C = K[[X, Y]] $

\medskip

$ B = K[[X]]\,[Y]_{(X,Y)}\hookrightarrow C =
K[Y]_{(Y)}[[X]]\hookrightarrow\widehat B=\widehat C= K[[X]]\,[[Y]].
$

\medskip
Matsumura observes there exist $P\in \Spec\widehat A $ with $ \hgt P
= m-1$

and $ P \cap A = (0)$, and also that there exist $ P\in \Spec K[[X,
Y]]$

 with $\hgt P = m+n-2$ such that

 \center{$ P \cap C = (0)$ \quad or \quad
 $P \cap B = (0)$.}

However he does not address the question of whether all primes

maximal with these properties   have the same height.

\end{slide}

\begin{slide}

\heading{DEFINITION}

\medskip

Let $R$ be a subring of an integral domain $S$, we say $R
\hookrightarrow S$ is a


\medskip
{\bf trivial generic fiber} extension, or a {\bf TGF} extension, if

\medskip

$P \cap R \ne (0)$ for each nonzero prime ideal $P$ of $S$, i.e.,
the generic

\medskip
fiber of the map $\Spec S \to \Spec R$ is trivial containing  only
the

\medskip

zero prime ideal of $S$. With

\medskip

 $A:=k[x,y]\hookrightarrow B:=k[[y]]\,[x]\hookrightarrow
C:=k[x]\,[[y]] \hookrightarrow D := k[[x, y]], $

\medskip
then $A \hookrightarrow R$ is not TGF for $R \in \{B, C, D \}$,
while

\medskip
$R \hookrightarrow S$ is TGF for $R \subseteq S$ in $\{B, C, D \}$.


\end{slide}

Let $(R,\m)$ be a Noetherian local integral domain and let $\widehat
R$ denote the $\m$-adic completion of $R$.

\medskip

The {\bf generic formal fiber ring} of  $R$ is the localization

$(R \setminus (0))^{-1}\widehat R$ of $\widehat R$.

\medskip

The {\bf formal fibers} of $R$ are the fibers of the map

$\Spec \widehat R \to \Spec R$.    For $P \in \Spec R$, the  {\bf
formal fiber}  over $P$ is

$\Spec(\,(R_P/PR_P)\otimes_R\widehat R$\,).

\begin{slide}
\begin{center}
\textbf{Variations on a theme of Weierstrass}
\end{center}

The Weierstrass Preparation Theorem is our main technical tool.

Let  $P$ be a prime ideal  in the power series ring $\widehat A =
K[[X]]$, where $X = \{x_1, \ldots , x_n\}$ is a set of $n$ variables
over the field $K$.

Here $A = K[X]_{(X)}$ is the localized polynomial ring.

{\bf  Notation.} By a {\bf change of variables}, we  mean a finite
sequence of  `polynomial' change of variables  over $K$ of the form
$$
x_1 \mapsto x_1 +x_n^{e_1}, \quad \ldots \quad  x_{n-1} \mapsto
x_{n-1} + x_n^{e_{n-1}}, \quad  x_n \mapsto x_n,
$$ perhaps  followed by
$$
z_1 \mapsto z_1,\quad z_2 \mapsto z_2 + z_1^{f_2},\quad \ldots \quad
z_n \mapsto z_n+ z_1^{f_n},
 $$
where $e_i, f_i \in \N$ and $z_i$ denotes the image of $x_i$ under
the first map.

\end{slide}

\begin{slide}
\begin{center}
\textbf{The Weierstrass Preparation Theorem}
\end{center}
{\bf Theorem (Weierstrass)} Let $( R, \m)$ be a complete local ring,
let $f \in R[[x]]$ be a formal power series and let $\overline f$
denote the image of $f$ in $(R/\m)[[x]]$. Assume that $\overline f
\ne 0$ and that $\ord \overline f = s > 0$. There exists a unique
ordered pair $(u, F)$ such that $u$ is  a unit in $R[[x]]$ and  $F
\in R[x]$ is a distinguished monic polynomial of degree $s$ such
that $f = uF$.

Here $F = x^s + a_{s-1}x^{s-1} + \cdots + a_0 \in R[x]$ is {\bf
distinguished} if $a_i \in \m$ for $0 \le i \le s-1$.

\medskip

{\bf Corollary.} The ideal $fR[[x]]$ is extended from $R[x]$ and
$R[[x]]/(f)$ is a free $R$-module of rank $s$. Every $g \in R[[x]]$
is of the form $g = qf + r$, where $q \in R[[x]]$ and $r \in R[x]$
is a polynomial with $\deg r \le s-1$.

\end{slide}

\begin{slide}
\begin{center}
\textbf{Embeddings of Power Series Rings}
\end{center}
{\bf Theorem (Classical)} There exists a $K$-algebra embedding of
the formal power series ring $R :=K[[x_1, \ldots, x_n]]$ into the
formal power series ring in two variables $K[[y, z]]$.

{\bf Proof.} Let $f_1, \ldots, f_n$ be power series in $K[[y]]$ that
are algebraically independent over $K$. Define $\varphi : R \to
K[[y, z]]$ by $\varphi(x_i) = f_iz$,  for $1 \le i \le n$. Suppose
$g \in R$ is such that $\varphi(g) = 0$. Write $g =
\bigoplus_{k=0}^\infty  g_k$, where $g_k$ is a form in $K[x_1,
\ldots, x_n]$ of degree $k$. Then $\varphi(g_k) = z^kg_k(f_1,
\ldots, f_n)$ with $g_k(f_1, \ldots, f_n) \in K[[y]]$. Hence
$$
\varphi(g) = \bigoplus_{k=0}^\infty g_k(f_1, \ldots, f_n)z^k = 0
$$
implies $g_k(f_1, \ldots, f_n) = 0$ for each $k \in \N$. Since $f_1,
\ldots, f_n$ are algebraically independent over $K$, each $g_k = 0$,
so $g = 0$.


\end{slide}

{\bf Corollary.} Assume  $\varphi :(R, \m) \to (S, \n)$ is a local
homomorphism.

If $R$ is $\m$-adically  complete, $\m S$ is $\n$-primary and $S/\n$
is finite

over $R/\m$, then $S$ is a finitely generated  $R$-module.

If $\varphi : R \to K[[y, z]] = S$ is an embedding of the power
series ring in

$n > 2$ variables, then $\m S$ is not $\n$-primary.

\heading{The Generic Formal Fiber Rings of  $B$ and $C$}

\smallskip

Since  $ B := K[[X]][Y]_{(X, Y)} \hookrightarrow K[Y]_{(Y)}[[X]] :=
C $,

\smallskip

if  $P \in \Spec K[[X, Y]]$ and  $P \cap C = (0)$, then $ P \cap B =
(0)$.

\smallskip

We know  $P\in\Spec K[[X, Y]]$ maximal with respect to $P \cap B =
(0)$

\smallskip

or maximal with respect to $P \cap C = (0)\implies \hgt P=n + m -
2$.

\smallskip

Hence if $P \in \Spec K[[X, Y]]$ is maximal  with  $P \cap C = (0)$,

\smallskip

then $P$ is also maximal  with respect to $P \cap B = (0)$.

\smallskip

However, if  $n \ge 2$, the generic fiber of $B \hookrightarrow C$
is nonzero.

\smallskip

This means  $\exists \quad  P\in\Spec K[[X, Y]]$ maximal in the
generic formal

\smallskip

fiber of $B$ but $P$ not in the generic formal fiber of $C$.

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.

{\bf Lemma. } Let $\widehat A:=K[[X]]$ and let $P \in \Spec \widehat
A$ with $x_1 \not\in P$ and $\hgt P = r$, where $1 \le  r \le n-1$.
There exists  a change of variables $x_1 \mapsto z_1:=x_1$ ($x_1$ is
fixed), $x_2 \mapsto z_2,\,\hdots, \,x_n \mapsto z_n$ and a regular
sequence  $f_1,\hdots,f_r\in P$ so that, upon setting $Z_1 = \{z_1,
\ldots, z_{n-r}\}$, $Z_2 = \{z_{n-r+1}, \ldots, z_n\}$ and $Z = Z_1
\cup Z_2$,

 \begin{enumerate}
\item $(f_1, \ldots, f_r) \subset K[[Z_1]][Z_2]$ and $P$ is a
minimal prime of

 $(f_1,\hdots,f_r)\widehat A$.

\item  The $(Z_2)$-adic completion  of $K[[Z_1]]\,[Z_2]_{(Z)}$ is
identical to the

$(f_1,\hdots,f_r)$-adic completion
 and is  $\widehat A=K[[X]] = K[[Z]]$.

\item  If $P_1 := P \cap K[[Z_1]]\,[Z_2]_{(Z)}$, then $P_1\widehat
A = P$, that is, $P$ is

extended from $K[[Z_1]]\,[Z_2]_{(Z)}$.

\item The extension: $ K[[Z_1]]\hookrightarrow
K[[Z_1]]\,[Z_2]_{(Z)}/P_1\cong K[[Z]]/P$

is finite (and integral).
\end{enumerate}
