# Naive $\mathbb{A}^{1}$-connectedness of retract rational varieties

Chetan Balwe  Department of Mathematical Sciences, Indian Institute of Science
Education and Research Mohali, Knowledge City, Sector-81, Mohali 140306,
India<EMAIL_ADDRESS>and  Bandna Rani  Department of Mathematical
Sciences, Indian Institute of Science Education and Research Mohali, Knowledge
City, Sector-81, Mohali 140306, India<EMAIL_ADDRESS>
###### Abstract.

A smooth, proper, retract rational variety over a field $k$ is known to be
$\mathbb{A}^{1}$-connected. We improve on this result, in the case when $k$ is
infinite, showing that such varieties are naively $\mathbb{A}^{1}$-connected.

##  1\. Introduction

Let $k$ be a field and let $Sm_{k}$ denote site of smooth varieties over $k$,
equipped with the Nisnevich topology. To any simplicial sheaf $\mathcal{X}$ on
$Sm_{k}$, $\mathbb{A}^{1}$-homotopy theory (see [15]) associates the sheaf of
$\mathbb{A}^{1}$-connected components of $\mathcal{X}$, denoted by
$\pi_{0}^{\mathbb{A}^{1}}(\mathcal{X})$. We say that $\mathcal{X}$ is
$\mathbb{A}^{1}$-connected if $\pi_{0}^{\mathbb{A}^{1}}(\mathcal{X})$ is equal
to $\ast$, the point sheaf.

In general, the sheaf $\pi_{0}^{\mathbb{A}^{1}}(\mathcal{X})$ is difficult to
compute explicitly for general $\mathcal{X}$. We recall a related construction
— the sheaf of naively $\mathbb{A}^{1}$-connected components, which may be
viewed as a crude approximation to $\pi_{0}^{\mathbb{A}^{1}}(\mathcal{X})$
(see [3] and [5]). For any smooth scheme $U$, let $\sigma_{0},\sigma_{1}:U\to
U\times\mathbb{A}^{1}$ denote the morphisms which map $U$ isomorphically onto
the closed subschemes $U\times\\{0\\}$ and $U\times\\{1\\}$ of
$U\times\mathbb{A}^{1}$, respectively. If $\mathcal{X}$ is a sheaf of sets,
and $U$ is a smooth scheme over $k$, we say that two elements
$f,g\in\mathcal{X}(U)$ are $\mathbb{A}^{1}$-homotopic if there exists an
element $h\in\mathcal{X}(U\times\mathbb{A}^{1})$ which is mapped to $f$ and
$g$ under the two restriction maps
$\sigma_{0}^{*},\sigma_{1}^{*}:\mathcal{X}(U\times\mathbb{A}^{1})\to\mathcal{X}(U)$.
This generates an equivalence relation on $\mathcal{X}(U)$, which we denote by
$\sim_{U}$. Let $\mathcal{S}(\mathcal{X})$ be the Nisnevich sheafification of
the presheaf which maps $U$ to the set of equivalence classes
$\mathcal{X}(U)/\sim_{U}$. We say that $\mathcal{X}$ is naively
$\mathbb{A}^{1}$-connected if $\mathcal{S}(\mathcal{X})=\ast$.

The relationship between the functors $\pi_{0}^{\mathbb{A}^{1}}(-)$ and
$\mathcal{S}(-)$ is quite subtle. We have canonical morphisms of sheaves
$\mathcal{X}\to\mathcal{S}(\mathcal{X})$ and
$\mathcal{S}(\mathcal{X})\to\pi_{0}^{\mathbb{A}^{1}}(X)$, both of which are
epimorphisms. It was proved in [3] that there exists a canonical morphism

$\pi_{0}^{\mathbb{A}^{1}}(\mathcal{X})\to\mathop{\varinjlim}\limits\mathcal{S}^{n}(\mathcal{X})=:\mathcal{L}(\mathcal{X})$

which is an isomorphism if and only if $\pi_{0}^{\mathbb{A}^{1}}(X)$ is
$\mathbb{A}^{1}$-invariant. Thus, if $\mathcal{X}$ is
$\mathbb{A}^{1}$-connected, then we have
$\mathop{\varinjlim}\limits\mathcal{S}^{n}(\mathcal{X})=\ast$. A recent
example, constructed by J. Ayoub shows that
$\pi_{0}^{\mathbb{A}^{1}}(\mathcal{X})$ need not be $\mathbb{A}^{1}$-invariant
in general (see [2]). However, the map
$\pi_{0}^{\mathbb{A}^{1}}(\mathcal{X})(K)\to\mathcal{L}(\mathcal{X})(K)$ is a
bijection for any finitely generated, separable field extension $K/k$ (see
[4].

The relationship between $\pi_{0}^{\mathbb{A}^{1}}(-)$ and $\mathcal{S}$ is
more interesting when we study it in the context of proper varieties. A result
of Asok and Morel states that if $X$ is a proper variety over $k$, the map
$\mathcal{S}(X)(L)\to\pi_{0}^{\mathbb{A}^{1}}(X)(L)$ is an isomorphism for
every finitely generated, separable field extension $L/k$. A result of Morel
shows that if $\pi_{0}^{\mathbb{A}^{1}}(X)(L)=\ast$ for all finitely
generated, separable field extensions $L/k$, then $X$ is
$\mathbb{A}^{1}$-connected. Thus, we see that if $X$ is a proper variety over
$k$, then $X$ is $\mathbb{A}^{1}$-connected if and only if
$\mathcal{S}(X)(L)=\ast$ for any finitely generated, separable field extension
$L/k$. Since $X$ is proper over $k$, the latter condition is equivalent to
saying that $X$ is universally $R$-trivial.

The relationship between universal $R$-triviality and other near-rationality
properties (such as retract rationality, unirationality, etc.) of smooth,
proper varieties is not well-understood. (Note that smoothness is an important
condition in this context. For example, a cone over any variety is
$\mathbb{A}^{1}$-connected, but need not have any other near-rationality
property.) Smooth, proper, retract rational varieties over a field $k$ are
$\mathbb{A}^{1}$-connected. This was proved by Asok and Morel in the case
$char(k)=0$ (see [1, Theorem 2.3.6]) and in the general case by Kahn and
Sujatha (see [11, Theorems 8.5.1 and 8.6.2]). It is not known if the converse
is true.

We will prove the following result:

######  Theorem 1.1.

Let $k$ be an infinite field. Let $X$ be a smooth, proper, retract rational
variety over $k$. Then $\mathcal{S}(X)=\ast$.

Thus, if $X$ is a smooth, proper variety over an infinite field $k$, we have
the following implications:

$\text{retract rational}\implies\text{naively
$\mathbb{A}^{1}$-connected}\implies\text{$\mathbb{A}^{1}$-connected}$

It is not known if either of these two implications is strict. We note that if
$X$ is a proper, $\mathbb{A}^{1}$-connected variety, a result of Sawant (see
[17]) implies that $\mathcal{S}^{2}(X)=\ast$. It is easy to construct examples
of singular varieties $X$ such that $\mathcal{S}^{2}(X)=\ast$ but
$\mathcal{S}(X)\neq\ast$, but we are not aware of an example of a smooth
variety satisfying these conditions.

The above theorem says that if $X$ is a smooth, proper, retract rational
variety over an infinite field $k$, then
$\mathcal{S}(X)\to\pi_{0}^{\mathbb{A}^{1}}(X)$ is an isomorphism. This may be
contrasted with the results of [6] where we see that if $X$ is a smooth,
proper surface that is birationally ruled over a curve of genus $>0$, the
morphism $\mathcal{S}(X)\to\pi_{0}^{\mathbb{A}^{1}}(X)$ is an isomorphism if
and only if $X$ is minimal.

In Section 2, we collect some technical results regarding henselization that
are required for the proof of Theorem 1.1. At the beginning of the section, we
include a brief, informal discussion regarding the strategy of the proof. The
proof of Theorem 1.1 is presented in Section 3.

######  Notation 1.2.

  1. (1)

If $x$ is a point on a scheme $X$, the residue field at $x$ will be denoted by
$\kappa(x)$.

  2. (2)

Given a scheme $X$ (resp. an affine scheme $X={\rm Spec\,}R$), and an ideal
sheaf $\mathcal{I}$ (resp. an ideal $I\subset R$), we will denote by
$\mathcal{Z}(\mathcal{I})$ (resp. $\mathcal{Z}(I)$) the closed subscheme of
$X$ associated to the ideal sheaf $\mathcal{I}$ (resp. the ideal $I$). If
$\mathcal{L}$ is a line bundle on $X$ and $S$ is a set of sections of
$\mathcal{L}$, we may also write $\mathcal{Z}(S)$ for the closed subschemes
defined by the vanishing of the elements of $S$.

  3. (3)

Let $\mathcal{F}$ be a presheaf on $Sm_{k}$. Suppose $U$ is an essentially
smooth scheme over $k$, i.e. a filtered inverse limit of a diagram of smooth
schemes $U_{\alpha}$, in which the transition maps are étale, affine
morphisms. Then, we will define
$\mathcal{F}(U)=\mathop{\varinjlim}\limits\mathcal{F}(U_{\alpha})$. We will
primarily use this definition in the context of schemes of the form ${\rm
Spec\,}\mathcal{O}_{X,x}^{h}$ where $X$ is a smooth variety over $k$, $x$ is a
point of $X$ and $\mathcal{O}_{X,x}^{h}$ is the henselization of the local
ring $\mathcal{O}_{X,x}$. We will denote the local scheme ${\rm
Spec\,}\mathcal{O}_{X,x}^{h}$ by $X_{x}$ and the canonical morphism $X_{x}\to
X$ by $\omega_{x}$.

##  2\. Preliminaries

In this section, we gather some technical results required in our proof of
Theorem 1.1. First, we briefly discuss the role they will play in the proof.

Let $k$ be a field and let $X$ be a variety over $k$. To prove that
$\mathcal{S}(X)=\ast$, it suffices to show that if $U$ is any smooth variety
over $k$ and $u$ is any point of $U$, then $\mathcal{S}(X)(U_{u})=\ast$ (see
Notation 1.2).

If $f:U_{u}\to X$ is any morphism, and if $f(u)=x$, then $f$ factors through
the canonical morphism $\omega_{x}:X_{x}\to X$. (This follows from the
definition of henselization in subsection 2.1.) Thus, if we want to show that
$\mathcal{S}(X)=\ast$, it suffices to prove that $\mathcal{S}(X)(X_{x})=\ast$
for any point $x\in X$. In other words, we need to show that there exists a
point $x_{0}\in X(k)$ such that for any $x\in X$, there exists a chain of
$\mathbb{A}^{1}$-homotopies connecting $\omega_{x}$ to the morphism

$X_{x}\to{\rm Spec\,}k\xrightarrow{x_{0}}X\text{.}$

Thus, for any $x\in X$, we will need to construct certain
$\mathbb{A}^{1}$-homotopies of $X_{x}$ in $X$, i.e. morphisms of the form
${\rm Spec\,}R[t]\to X$ where $R=\mathcal{O}_{X,x}^{h}$. For a general variety
$X$, it may not be easy to construct such morphisms. However, we can start
with a somewhat weaker notion. A morphism ${\rm Spec\,}R[[t]]\to X$ could be
seen as the “germ” of a homotopy (or an ‘infinitesimal homotopy”, as in
subsection 2.2). If $X=\mathbb{A}^{N}_{k}$ for some $N$, there is a simple
trick to obtain a homotopy from a morphism ${\rm Spec\,}R[[t]]\to X$. Indeed,
this morphism is given by an $N$-tuple of power series with coefficients in
$R$. We can just truncate these power series at some degree to obtain an
$N$-tuple of polynomials. This gives us a morphism from ${\rm Spec\,}R[t]$
into $\mathbb{A}^{N}_{k}$, i.e. an $\mathbb{A}^{1}$-homotopy of ${\rm
Spec\,}R$ in $\mathbb{A}^{N}_{k}$. By truncating at a sufficiently high
degree, we can make sure that the $\mathbb{A}^{1}$-homotopy approximates the
infinitesimal homotopy that we started with. With a little care, this trick
can be made to work in $\mathbb{P}^{N}_{k}$.

Given a retract rational variety $X$, we have birational maps
$X\dashrightarrow\mathbb{P}^{N}_{k}$ and $\mathbb{P}^{N}_{k}\dashrightarrow
X$, the composition of which is the identity map on $X$. To prove that $X$ is
naively $\mathbb{A}^{1}$-connected, we have to construct some
$\mathbb{A}^{1}$-homotopies on $X$. In very crude terms, the idea is to first
construct an infinitesimal homotopy on $X$ and then to “lift it” to
$\mathbb{P}^{N}_{k}$, and to then use the “truncation trick” described above
to obtain an $\mathbb{A}^{1}$-homotopy.

Of course, we cannot really lift the infinitesimal homotopy to
$\mathbb{P}^{N}_{k}$, since we only have a rational map from $X$ to
$\mathbb{P}^{N}_{k}$. So one has to use a somewhat sophisticated analogue of
the “truncation trick” — we need to divide the power series in question by an
appropriately chosen Weierstrass polynomial and compute the remainders. For
this, we will need an analogue of the Weierstrass preparation theorem in the
context of henselian local rings. This is what we recall in subsection 2.1. In
subsection 2.2 we prove a lemma that will allow us construct suitable
infinitesimal homotopies on a smooth variety. This is precisely where
smoothness is crucially required. Note that Theorem 1.1 is not true for
singular varieties. (Indeed, a rational, proper variety that is not smooth is
not necessarily $\mathbb{A}^{1}$-connected.)

###  2.1. Henselization

In this subsection, we will collect some facts about henselization that we
will require in the proof of the main result.

Let $R$ be a ring and let $I$ be an ideal of $R$. We say that $(R,I)$ is a
_henselian pair_ if $I$ is contained in the Jacobson radical of $R$ and given
any étale extension $R\to R^{\prime}$ such that $R/I\to
R^{\prime}/IR^{\prime}$ is an isomorphism, there exists an $R$-algebra
homomorphism $R^{\prime}\to R$. When $R$ is a local ring with maximal ideal
$\mathfrak{m}$ such that $(R,\mathfrak{m})$ is a henselian pair, we say that
$R$ is a _henselian local ring_.

######  Fact 2.1 (See [9, Theorem 18.5.1]).

Let $U={\rm Spec\,}R$ where $R$ is a henselian local ring. Let $u$ be the
closed point of $U$. Let $f:V\to U$ be a finite morphism. Then, $V$ is the
disjoint union of the schemes ${\rm Spec\,}\mathcal{O}_{V,v}$ where $v$ ranges
over all the points of $f^{-1}(u)$.

Given any ring $R$ and an ideal $I$ of $R$, consider the category of henselian
pairs $(S,J)$ such that $S$ is an $R$-algebra and $IS\subset J$. This category
has an initial object $(R^{h},I^{h})$, called the _henselization of $R$ at
$I$_. When $R$ is a local ring with maximal ideal $\mathfrak{m}$, we will
refer to the henselization of $R$ at $\mathfrak{m}$ as just the _henselization
of $R$_.

######  Fact 2.2 (See [18, Lemma 15.12.4]).

Let $R$ be a noetherian ring and let $I$ be an ideal of $R$. Let
$(R^{h},I^{h})$ denote the henselization of $R$ at $I$. Then the canonical
homomorphism $R\to R^{h}$ induces an isomorphism of the $I$-adic completions
$\widehat{R}\to\widehat{R^{h}}$. The canonical homomorphism $R\to R^{h}$ is a
flat extension which induces an isomorphism of the $I$-adic completions
$\widehat{R}\to\widehat{R^{h}}$. Also, $R^{h}$ is a noetherian ring and so
$R^{h}\to\widehat{R}$ is a faithfully flat extension.

We will primarily be concerned with the henselizations of the local rings at
points of a variety over a field and we quote a basic result about such rings.

######  Fact 2.3 (See [9, Corollary 18.7.6]).

Let $k$ be a field. Let $X$ be a variety over $k$ and let $x$ be a point of
$X$. Then, the ring $\mathcal{O}_{X,x}^{h}$ is an excellent ring.

Let $R$ be any ring and let $t$ denote the variable. The henselization of the
ring $R[t]$ at the ideal $\langle t\rangle$ is called the ring of henselian
power series, and is denoted by $R\\{t\\}$. We will now review some results
about such rings.

First, let us take $R$ to be a noetherian ring. Then, Fact 2.2 implies that
$R\\{t\\}$ injects into $R[[t]]$ and we will identify $R\\{t\\}$ with its
image in $R[[t]]$. Since $R\\{t\\}\to R[[t]]$ is faithfully flat, it is easy
to see that an element $f\in R\\{t\\}$ is a unit if and only if it is a unit
in $R[[t]]$, i.e. if and only if its image under the quotient map $R[[t]]\to
R[[t]]/\langle t\rangle\cong R$ is a unit in $R$.

The results stated in the previous paragraph continue to hold even when $R$ is
not noetherian. Indeed, the functor $R\to R\\{t\\}$ commutes with filtered
colimits. So one can express $R$ as a filtered colimit of its finitely
generated $\mathbb{Z}$-algebras and generalize the above statements to the
non-noetherian setting. ( See [7, Subsection 2.1.2].)

If $(R,\mathfrak{m})$ is a local ring, a polynomial $f(t)\in R[t]$ is said to
be a _Weierstarss polynomial_ if it is of the form
$f(t)=t^{d}+a_{d-1}t^{d-1}+\ldots+a_{0}$ where $a_{i}\in\mathfrak{m}$ for all
$i$.

######  Fact 2.4.

Let $(R,\mathfrak{m})$ be a henselian local ring. Let $f\in
R\\{t\\}\backslash\mathfrak{m}R\\{t\\}$. Then $f$ can be uniquely factored as
$f=P\cdot u$ where $f$ is a Weierstarss polynomial and $u$ is a unit in
$R\\{t\\}$. Also, the natural homomorphism $R[t]/\langle P\rangle\to
R\\{t\\}/\langle f\rangle$ is an isomorphism.

The following lemma is an easy consequence:

######  Lemma 2.5.

Let $R$ be a henselian local ring with maximal ideal $\mathfrak{m}$. Let $I$
be a proper ideal of $R[t]$ such that the following conditions hold:

  1. (1)

The homomorphism $R\to R[t]/I$ is a finite extension.

  2. (2)

The only prime ideal of $R[t]$ containing $I$ and $\mathfrak{m}R[t]$ is
$\langle\mathfrak{m},t\rangle$.

Then, the homomorphism $R[t]/I\to R\\{t\\}/IR\\{t\\}$ is an isomorphism.

###### Proof.

Let $Z={\rm Spec\,}R[t]/I$, which we view as a closed subscheme of ${\rm
Spec\,}R[t]$. Let $x_{0}$ be the closed point of ${\rm Spec\,}R$ and let
$y_{0}$ be the point of ${\rm Spec\,}R[t]$ corresponding to the ideal
$\langle\mathfrak{m},t\rangle$.

Let $\pi:Z\to{\rm Spec\,}R$ be the morphism corresponding to the $R$-algebra
homomorphism $R\to R[t]/I$. According to condition (1), $\pi$ is a finite
morphism. Thus, if $z\in Z$ is any point, there exists a point $z_{0}$ in its
closure such that $\pi(z_{0})=x_{0}$. By (2), the closed subscheme $Z$ and the
fibre $\pi^{-1}(x_{0})$ have only the point $y_{0}$ in common. Thus, we see
that every point of $Z$ lies in the closure of $y_{0}$.

By Fact 2.1, we see that $Z$ is isomorphic to ${\rm
Spec\,}\mathcal{O}_{Z,y_{0}}$. Thus, we see that if
$S=R[t]_{\langle\mathfrak{m},t\rangle}$, then the homomorphism $R[t]/I\to
S/IS$ is an isomorphism.

The ring $R\\{t\\}$ is a local ring with maximal ideal
$\langle\mathfrak{m},t\rangle$. Thus, the canonical homomorphism $R[t]\to
R\\{t\\}$ induces a local homomorphism $S\to R\\{t\\}$. This homomorphism is
flat, and since it is a local homomorphism, it is faithfully flat. Thus, as
$IS\not\subset\mathfrak{m}S$, we see that
$IR\\{t\\}\not\subset\mathfrak{m}R\\{t\\}$. Let $f$ be an element of
$IR\\{t\\}\backslash\mathfrak{m}R\\{t\\}$. Then $f=u\cdot p$ where $u$ is a
unit in $R\\{t\\}$ and $p$ is a Weierstrass polynomial. Since $S\to R\\{t\\}$
is a faithfully flat extension, $IS=S\cap IR\\{t\\}$. Thus, $p\in IS$.

By [7, Proposition 3.1.2], the ring homomorphism $R[t]/pR[t]\to
R\\{t\\}/pR\\{t\\}$ is an isomorphism. Thus, it follows that the homomorphism
$S/pS\to R\\{t\\}/pR\\{t\\}$ is surjective. It is also injective since it is a
faithfully flat extension. Thus, it is an isomorphism.

As $pS\subset IS$ and $pR\\{t\\}\subset IR\\{t\\}$, it follows that $S/IS\to
R\\{t\\}/IR\\{t\\}$ is an isomorphism. This completes the proof. ∎

A noetherian local ring $(R,\mathfrak{m})$ is said to be an _approximation
ring_ if for any finite system of polynomial equations with coefficients in
$R$, the set of solutions in $R$ is dense, with respect to the
$\mathfrak{m}$-adic topology, in the set of solutions in the
$\mathfrak{m}$-adic completion $\widehat{R}$. We will need the following
facts:

######  Fact 2.6 (See [16, Theorem 1.3]).

Excellent henselian local rings are approximation rings.

######  Fact 2.7 (See [[16, Corollary 3.5]).

Let $(R,\mathfrak{m})$ be an approximation ring. Then $R\\{t\\}$ is an
approximation ring.

###  2.2. Infinitesimal homotopies

For any ring $R$, an _infinitesimal homotopy_ of ${\rm Spec\,}R$ in a scheme
$X$ is a morphism $h:{\rm Spec\,}R\\{t\\}\to X$. Let
$\widehat{\sigma}_{0}:{\rm Spec\,}R\to{\rm Spec\,}R\\{t\\}$ be the morphism
induced by the quotient homomorphism $R\\{t\\}\to R\\{t\\}/tR\\{t\\}\cong R$.
We say that this infinitesimal homotopy _starts_ from the morphism
$h\circ\widehat{\sigma}_{0}:{\rm Spec\,}R\to X$.

The following lemma shows that for a smooth variety $X$ and a point $x\in X$,
one can easily construct the germ of a homotopy of $X_{x}$ in $X$ starting
from the canonical morphism $\omega_{x}:X_{x}\to X$.

######  Lemma 2.8.

Let $X$ be a smooth $d$-dimensional variety over $k$. Let $x$ be a point of
$X$. Let $U$ be an open subset of $X$. Let $R=\mathcal{O}_{X,x}^{h}$ and let
$\omega_{x}:{\rm Spec\,}(R)=:X_{x}\to X$ be the canonical morphism. Then,
there exists a morphism $h:{\rm Spec\,}R\\{t\\}\to X$ starting from
$\omega_{x}$ such that

${\rm Spec\,}\kappa(x)\\{t\\}\to{\rm Spec\,}R\\{t\\}\stackrel{{\scriptstyle
h}}{{\to}}X$

maps the generic point of ${\rm Spec\,}\kappa(x)\\{t\\}$ into $U$.

###### Proof.

If $x\in U$, we may take $h$ to be the composition

${\rm Spec\,}R\\{t\\}\to{\rm
Spec\,}R\stackrel{{\scriptstyle\omega_{x}}}{{\to}}X$

where the first morphism is induced by the inclusion $R\hookrightarrow
R\\{t\\}$. Thus, we will now assume that $x\notin U$. Let $Z=X\backslash U$.

For any non-negative integer $n$, consider the functor

$U\mapsto Mor_{Sch/k}(U\times{\rm Spec\,}k[t]/\langle t^{n+1}\rangle)$

on the category of $k$-schemes. This functor is known to be representable by a
$k$-scheme of finite type (see [10]), which we denote by $J_{n}(X)$. Let
$J(X)=\mathop{\varprojlim}\limits J_{n}(X)$, where the inverse limit is
computed in the category of $k$-schemes. The quotient homomorphism $k[[t]]\to
k[t]/\langle t^{n+1}\rangle$ induces the morphism $\pi_{n}^{X}:J(X)\to
J_{n}(X)$. For $n\geq m$, let $\pi^{X}_{n,m}:J_{n}(X)\to J_{m}(X)$ denotes the
morphism induced by the quotient homomorphism $k[t]/\langle t^{n+1}\rangle\to
k[t]/\langle t^{m+1}\rangle$.

Choose a morphism $\gamma:{\rm Spec\,}\kappa(x)[[t]]\to X$ which maps the
closed point of ${\rm Spec\,}\kappa(x)[[t]]$ to $x$ and the generic point into
$U$. For any $n\geq 0$, let $\gamma_{n}$ denote the composition

${\rm Spec\,}\kappa(x)[t]/\langle t^{n+1}\rangle\to{\rm
Spec\,}\kappa(x)[[t]]\xrightarrow{\gamma}X\text{.}$

We identify $\gamma_{n}$ with a $\kappa(x)$-valued point of $J_{n}(X)$, which
we denote by $\widetilde{\gamma}_{n}$. We define $\widetilde{g}_{0}:{\rm
Spec\,}R\to J_{0}(X)=X$ to be the morphism $\omega_{x}$. For $n\geq 1$, we
will inductively construct a morphism $\widetilde{g}_{n}:{\rm Spec\,}R\to
J_{n}(X)$ such that:

  * (i)

the composition ${\rm Spec\,}\kappa(x)\to{\rm
Spec\,}R\xrightarrow{\widetilde{g}_{n}}J_{n}(X)$ is equal to
$\widetilde{\gamma}_{n}$, and

  * (ii)

the composition $\pi^{X}_{n+1,n}\circ\widetilde{g}_{n+1}$ equals
$\widetilde{g}_{n}$.

Suppose $\widetilde{g}_{n}$ has been chosen for some non-negative integer $n$.
Since $X$ is smooth, $J_{n+1}(X)\to J_{n}(X)$ is smooth. (In fact, it is an
affine bundle for all $n$ — see [14, Lemma 9.1]). So we can choose a morphism
$\widetilde{g}_{n+1}:{\rm Spec\,}R\to J_{n+1}(X)$ satisfying the conditions
(i) and (ii) (see [9, Corollary 17.16.3, (ii)]).

The collection $\\{\widetilde{g}_{n}\\}_{n\geq 0}$ defines a morphism
$\widetilde{g}:{\rm Spec\,}R\to J(X)$, which corresponds to a morphism $g:{\rm
Spec\,}R[[t]]\to X$. The restriction of $g$ to $\kappa(x)[[t]]$ is equal to
$\gamma$.

There exists an integer $n$ such that if $\gamma^{\prime}:{\rm
Spec\,}\kappa(x)[[t]]\to X$ satisfies $\pi_{n,{\rm
Spec\,}\kappa(x)}(\gamma)=\pi_{n,{\rm Spec\,}\kappa(x)}(\gamma^{\prime})$,
then $\gamma^{\prime}$ maps the generic point of ${\rm Spec\,}\kappa(x)[[t]]$
into $U$. (Indeed, if there is no such $n$, then since
$J(Z)=\mathop{\varprojlim}\limits J_{n}(X)$, it will follow that $\gamma\in
J(Z)$, which is not true.) By Facts 2.6 and 2.7, there exists a morphism
$h:{\rm Spec\,}R\\{t\\}\to X$ such that $\pi^{X}_{n}(h)=\pi^{X}_{n}(g)$. This
proves the lemma. ∎

######  Remark 2.9.

As we see in the above proof, it is very easy to construct a morphism ${\rm
Spec\,}R[[t]]\to X$. We could have used this as our notion of “infinitesimal
homotopy”, if we had an analogue of Fact 2.4 for the ring $R[[t]]$, at least
when $R$ is regular. (Such a result was proved in characteristic $0$ by Lafon
in [13].) The proof of the preparation theorem for $R\\{t\\}$ in [7] crucially
uses the fact that the functor $R\\{t\\}$ is a colimit of finite type
$R$-algebras and we do not know if it can be adapted to give an analogous
result for $R[[t]]$. So we choose to work with the ring $R\\{t\\}$ instead.

##  3\. Retract rational varieties

We will now focus on retract rational varieties. In subsection 3.1, we prove a
small lemma regarding rational curves in projective space. The proof of
Theorem 1.1 is presented in subsection 3.2.

###  3.1. Rational curves in projective space

Let us fix a base field $k$. Let $L$ be any field containing $k$. We will use
$T_{0}$ and $T_{1}$ as homogeneous coordinates on $\mathbb{P}^{1}_{L}$. In
other words, we will write $\mathbb{P}^{1}_{L}={\rm Proj}\,L[T_{0},T_{1}]$. We
will identify $\mathbb{A}^{1}_{L}={\rm Spec\,}L[t]$ with the open subscheme
$\mathbb{P}^{1}_{L}\backslash\mathcal{Z}(T_{1})$ by identifying $t$ with
$T_{0}/T_{1}$. We will denote the point $(0:1)$ of $\mathbb{P}^{1}_{L}$ by
$\mathbf{0}_{L}$ and the point $(1:0)$ by $\mathbf{\infty}_{L}$. Thus, $t$ is
a parameter at $\mathbf{0}_{L}$ and $1/t$ is a parameter at
$\mathbf{\infty}_{L}$. To avoid making the notation cumbersome, we will write
$\mathbf{0}$ and $\mathbf{\infty}$ instead of $\mathbf{0}_{L}$ and
$\mathbf{\infty}_{L}$ in the following discussion.

A morphism $\phi:\mathbb{P}^{1}_{L}\to\mathbb{P}^{N}_{k}$ can be represented
by an $(N+1)$-tuple $(P_{0},\ldots,P_{N})$ of homogeneous polynomials of a
fixed degree $d$, such that $P_{i}\neq 0$ for some $i$. (Of course, some of
the $P_{i}$’s may be equal to $0$. The zero polynomial can be assigned any
degree.) Such a representation is not unique, but if we require the
polynomials to be coprime, it is unique up to multiplication by a unit.
Dehomogenizing this $(N+1)$-tuple with respect to $T_{1}$ gives an
$(N+1)$-tuple of polynomials in $t$, which describes the restriction of $\phi$
to the open subscheme $D(T_{1})$.

Recall that given a morphism from
$\mathbb{P}^{1}_{L}\backslash\mathcal{Z}(T_{1})$ to $\mathbb{P}_{k}^{N}$, it
can be uniquely extended to a morphism
$\mathbb{P}^{1}_{L}\to\mathbb{P}^{N}_{k}$. A morphism from
$\mathbb{P}^{1}_{L}\backslash\mathcal{Z}(T_{1})$ to $\mathbb{P}^{N}_{k}$ is
given by an $(N+1)$-tuple of polynomials in $L[t]$, such that at least one of
the polynomials is non-zero. Thus, we see that a morphism
$\mathbb{P}^{1}_{k}\to\mathbb{P}^{N}_{k}$ can be represented in three ways —
using $(N+1)$-tuples of homogeneous polynomials of a same degree in
$(T_{0},T_{1})$ or by using $(N+1)$-tuples of polynomials in either $t$ or
$1/t$. (Again, note that these representations are unique up to multiplication
by a unit if we require the polynomials to be coprime.)

Given a morphism $\phi:\mathbb{P}^{1}_{k}\to\mathbb{P}^{N}_{k}$, we choose a
representation of $\phi$ by an $(N+1)$-tuple of polynomials
$(P_{0},\ldots,P_{N})$ in $k[t]$ which are coprime. Let $m\geq 0$ be any
integer. For $0\leq i\leq N$, let $P^{\prime}_{i}$ be the polynomial obtained
by truncating $P_{i}$ to degree $m$. Then, the $(N+1)$-tuple
$(P^{\prime}_{0},\ldots,P^{\prime}_{N})$ is called the $m$-jet of $\phi$ at
$\mathbf{0}$. Similarly, we can define the $m$-jet of $\phi$ at
$\mathbf{\infty}$. Note that these are well-defined up to multiplication by a
unit.

The following lemma shows that given a closed subscheme $W$ of
$\mathbb{P}_{k}^{N}$ of codimension $\geq 2$, and non-negative integers
$m_{1}$ and $m_{2}$, there exists a morphism
$\phi:\mathbb{P}^{1}_{k}\to\mathbb{P}^{N}_{k}$ such that it maps
$\mathbb{P}^{1}_{k}\backslash\\{\mathbf{0},\mathbf{\infty}\\}$ into
$\mathbb{P}^{N}_{k}\backslash W$, has a prescribed $m_{1}$-jet at $\mathbf{0}$
and a prescribed $m_{2}$-jet at $\mathbf{\infty}$.

######  Lemma 3.1.

Let $k$ be an infinite field and let $L$ be a field containing $k$. Let $N$ be
a positive integer. Let $W$ be a closed subscheme of $\mathbb{P}^{N}_{k}$ of
codimension $\geq 2$. Let $P=(P_{0},\ldots,P_{N})$ and $(Q_{0},\ldots,Q_{N})$
be $(N+1)$-tuples of polynomials in $L[t]$. Assume that $P_{i}\neq 0$ and
$Q_{j}\neq 0$ for some indices $i,j$. Let $m_{1}$ be an integer such that
$m_{1}\geq\max_{i}degP_{i}$. Assume that $P_{i}(0)\neq 0$ for some $i$. Then,
there exists an $(N+1)$-tuple $(c_{0},\ldots,c_{N})\in k^{N+1}$ such that the
following conditions hold:

  * (a)

For $0\leq i\leq N$, let
$R_{i}(t)=P_{i}(t)+t^{m_{1}+1}c_{i}+t^{m_{1}+2}Q_{i}(t)$. Then, the
polynomials $R_{0},\ldots,R_{N}$ are coprime.

  * (c)

Let $\phi:\mathbb{P}^{1}_{L}\to\mathbb{P}^{N}_{k}$ be the morphism represented
by the $(N+1)$-tuple $(R_{0},\ldots,R_{N})$. Then
$\phi(\mathbb{P}^{1}_{L}\backslash\\{\mathbf{0},\mathbf{\infty}\\})\subset\mathbb{P}^{N}_{k}\backslash
W$.

Note that if $\phi$ is as described in the lemma, the $m_{1}$-jet of $\phi$ at
$\mathbf{0}$ is $(P_{0}(t),\ldots,P_{N}(t))$ and if $m_{2}=\max_{i}\deg
Q_{j}$, then the $m_{2}$-jet of $\phi$ at $\infty$ is $(Q_{0}(1/t),\ldots
Q_{N}(1/t))$.

###### Proof.

For any point $x=(x_{0},\ldots,x_{N})$ of $\mathbb{A}^{N+1}_{k}$ and $0\leq
i\leq N$, we define

$R^{x}_{i}(t)=P_{i}(t)+x_{i}t^{m_{1}+1}+Q_{i}(t)t^{m_{1}+2}\text{.}$

Let $\phi_{x}:\mathbb{A}^{1}_{\kappa(x)}\to\mathbb{A}^{N+1}_{k}$ be the
morphism defined by

$\phi_{x}(s)=(R^{x}_{0}(s),\ldots,R^{x}_{N}(s))$

for any $s\in\mathbb{A}^{1}_{\kappa(x)}$. Let $C(W)\subset\mathbb{A}^{N+1}$ be
the cone over $W$.

We define

$B:=\\{(x,s,z)|\phi_{x}(s)=z\\}\subset\mathbb{A}^{N+1}_{k}\times(\mathbb{A}^{1}_{k}\backslash\\{0\\})\times
C(W)\text{.}$

This is a closed subset of
$\mathbb{A}^{N+1}_{k}\times(\mathbb{A}^{1}\backslash\\{0\\})\times C(W)$. Let
$pr_{1}:B\to\mathbb{A}^{N}_{k}$ be the projection map onto the first factor.
We need to show that the complement of the image of $pr_{1}$ contains some
$k$-rational point.

Let
$pr_{23}:\mathbb{A}^{N}_{k}\times(\mathbb{A}^{1}_{k}\backslash\\{0\\})\times
C(W)\to(\mathbb{A}^{1}_{k}\backslash\\{0\\})\times C(W)$ by the projection map
onto the product of the second and third factors. We would like to estimate
the dimension of the fibre $pr_{23}^{-1}(\gamma)$ where
$\gamma=(s,z)\in\mathbb{A}^{1}_{k}\backslash\\{0\\}\times C(W)$. The equation
$\phi_{x}(s)=z$ imposes $N+1$ linear conditions on $\mathbb{A}^{N+1}_{k}$.
Thus, the fibre has dimension $0$. Thus, it follows that $\dim(B)\leq
0+1+\dim(C(W))\leq N$.

It follows that the closure of $pr_{1}(B)$ is of dimension $\leq N$. The
result follows since $k$ is an infinite field. ∎

######  Remark 3.2.

This lemma is one of the main reasons for requiring the field $k$ to be
infinite in Theorem 1.1. The lemma need not hold if $k$ is a finite field
since all the $k$-rational points of $\mathbb{P}^{N}_{k}$ may be contained in
$W$.

###  3.2. Naive $\mathbb{A}^{1}$-connectedness of retract rational varieties

We will now prove the main result of this paper.

######  Theorem 3.3.

Let $k$ be an infinite field. Let $X$ be a smooth, proper, retract rational
variety over $k$. Then $\mathcal{S}(X)=\ast$.

###### Proof.

Since $X$ is retract rational, there exists a positive integer $N\geq 1$, and
rational maps $\phi:X\dashrightarrow\mathbb{P}_{k}^{N}$ and
$\psi:\mathbb{P}_{k}^{N}\dashrightarrow X$ such that $\psi\circ\phi$ is the
identity map on $X$. Since $k$ is infinite, this implies that $X(k)$ is non-
empty. Since $X$ is a smooth, proper, retract rational variety, we have
$\pi_{0}^{\mathbb{A}^{1}}(X)=\ast$, and so $\mathcal{S}(X)(k)=\ast$. Thus, to
prove that $X$ is naively $\mathbb{A}^{1}$-connected, it suffices to prove
that for any point $x$, there exists a chain of $\mathbb{A}^{1}$-homotopies of
$X_{x}$ in $X$ connecting the canonical morphism $\omega_{x}:X_{x}\to X$ to a
morphism that factors through some morphism ${\rm Spec\,}k\to X$.

Let us fix a point $x\in X$. We will denote the ring $\mathcal{O}_{X,x}^{h}$
by $R$. As in Notation 1.2, we denote ${\rm Spec\,}R$ by $X_{x}$ and the
canonical morphism $X_{x}\to X$ by $\omega_{x}$. We now set up some notation
for working with the scheme $\mathbb{P}^{1}_{R}$.

We use the notation in subsection 3.1, so that $\mathbf{0}$ and
$\mathbf{\infty}$ denote the points $(0:1)$ and $(1:0)$ of
$\mathbb{P}^{1}_{k}={\rm Proj}\,k[T_{0},T_{1}]$ respectively. Let $\sigma_{0}$
and $\sigma_{\infty}$ be the sections of the projection morphism
$\mathbb{P}^{1}_{R}\cong\mathbb{P}^{1}_{k}\times{\rm Spec\,}R\to{\rm
Spec\,}R$, mapping ${\rm Spec\,}R$ isomorphically onto the closed subschemes
$\mathcal{Z}(T_{0})=\\{\mathbf{0}\\}\times{\rm Spec\,}R$ and
$\mathcal{Z}(T_{1})=\\{\mathbf{\infty}\\}\times{\rm Spec\,}R$, respectively.
We will denote the rational function $T_{0}/T_{1}$ by $t$ and thus identify
the open subscheme $\mathbb{P}^{1}_{R}\backslash\mathcal{Z}(T_{1})$ with ${\rm
Spec\,}R[t]$.

We will construct a morphism $H:\mathbb{P}^{1}_{R}\to X$ such that
$H\circ\sigma_{0}=\omega_{x}$ and $H\circ\sigma_{\infty}$ factors through some
morphism ${\rm Spec\,}k\to X$. Clearly, this will prove the result.

There exists an ideal sheaf $\mathcal{K}$ on $\mathbb{P}^{N}_{k}$ such that if
$\pi:Y\to X$ is the blowup of $\mathbb{P}_{k}^{N}$ at $\mathcal{K}$, the map
$\chi:=\psi\circ\pi$ is a morphism from $Y$ to $X$. The sheaf $\mathcal{K}$
can be chosen so that $W:=\mathcal{Z}(\mathcal{K})$ is a variety of
codimension $\geq 2$. Let $V=\mathbb{P}_{k}^{N}\backslash W$. Let $U\subset X$
be an open subset on which $\phi$ is defined and such that $\phi(U)\subset V$.
The ideal sheaf $\mathcal{K}$ corresponds to a homogeneous ideal of
$k[X_{0},\ldots,X_{N}]$ generated by homogeneous polynomials
$p_{1},\ldots,p_{r}$. We may assume, without loss of generality that the
polynomials $p_{1},\ldots,p_{r}$ are all of the same degree. Note that $r\geq
1$.

The polynomials $p_{1},\ldots,p_{r}$ define global sections of $\mathcal{K}$,
which generate $\mathcal{K}$. Thus, we obtain a surjective morphism
$\mathcal{O}_{\mathbb{P}^{N}}^{r}\to\mathcal{K}$. Thus, we have the following
sequence of homomorphisms of sheaves of graded
$\mathcal{O}_{\mathbb{P}^{N}}$-rings

$Sym(\mathcal{O}^{r}_{\mathbb{P}^{N}})\to
Sym(\mathcal{K})\to\bigoplus_{j=0}^{\infty}\mathcal{K}^{j}\text{.}$

This gives us a closed embedding of $Y$ into
$\mathbb{P}_{k}^{N}\times\mathbb{P}_{k}^{r-1}$.

Using Lemma 2.8, we choose an infinitesimal homotopy $h:{\rm
Spec\,}R\\{t\\}\to X$ starting at $\omega_{x}$ such that $h$ maps the point
$\eta_{0}$ of ${\rm Spec\,}R\\{t\\}$, corresponding to the ideal
$\mathfrak{m}R\\{t\\}$, into $U$. This gives us a rational map $\phi\circ
h:{\rm Spec\,}R\\{t\\}\dashrightarrow\mathbb{P}^{N}$, which can be represented
by an $N$-tuple $\mathbf{f}:=(f_{0},\ldots,f_{N})$ where $f_{i}\in R\\{t\\}$.
We choose the $f_{i}$ to be coprime in the unique factorization domain
$R\\{t\\}$. Let $I$ denote the ideal $\langle f_{0},\ldots,f_{N}\rangle$. The
rational map $\phi\circ h$ is a morphism if and only if this ideal is
principal. Let $J$ denote the ideal $\langle
p_{1}(\mathbf{f}),\ldots,p_{r}(\mathbf{f})\rangle$.

Recall that we have chosen $h$ in such a way that the point $\eta_{0}$ of
${\rm Spec\,}R\\{t\\}$, corresponding to the ideal $\mathfrak{m}R\\{t\\}$, is
mapped into $U$. Thus, the rational map $f\circ h$ is well-defined on
$\eta_{0}$. Since we have chosen the $f_{i}$ to be coprime elements of the
unique factorization domain $R\\{t\\}$, it follows that at least one of the
$f_{i}$ does not vanish on $\eta_{0}$. By performing a change of coordinates
on $\mathbb{P}^{N}$, we may reduce to the situation where none of the $f_{i}$
vanishes on $\eta_{0}$. (Such a change of coordinates exists since $k$ is an
infinite field.) Thus, for each $i$, we have $f_{i}=u_{i}\widetilde{f}_{i}$
where $u_{i}$ is a unit in $R\\{t\\}$ and $\widetilde{f}_{i}$ is a Weierstrass
polynomial.

Recall that $f$ maps the point $\eta_{0}$ into $V$. The zero set of the ideal
$\langle p_{1},\ldots,p_{r}\rangle$ is contained in the complement of $V$.
Thus, at least one of the polynomials $p_{i}$ does not vanish on $\eta_{0}$.
We may assume that all the $p_{i}$’s are of the same degree. Suppose $p_{1}$
does not vanish on $\eta_{0}$. Then, for each $i\neq 1$, we can replace
$p_{i}$ by $p_{i}+\epsilon_{i}p_{1}$ where $\epsilon_{i}$ is $0$ if $p_{i}$
does not vanish at $\eta_{0}$ and is equal to $1$ otherwise. Thus, we may
assume that none of the polynomials $p_{i}$ vanishes at $\eta_{0}$.

Thus, $p_{i}(\mathbf{f})=v_{i}\cdot P_{i}$ where $v_{i}$ is a unit in
$R\\{t\\}$ and $P_{i}$ is a Weierstrass polynomial. Let
$p:=t\cdot\left(\prod_{i}\widetilde{f}_{i}\right)^{2}\cdot\left(\prod_{j}P_{j}\right)^{2}$.
Then, $p$ is an element of the ideal $IJ$ and it is a Weierstrass polynomial
with $\deg_{t}(p)\geq 1$.

For $0\leq i\leq N$, we can express $f_{i}$ in the form

$f_{i}=\alpha_{i}+p\beta_{i}$

where $\alpha_{i}\in R[t]$ with $\deg_{t}(\alpha_{i})<\deg_{t}(p)$ and
$\beta_{i}\in R\\{t\\}$.

For $0\leq i\leq N$, let $\overline{\alpha}_{i}(t)\in\kappa(x)[t]$ be the
image of $\alpha_{i}(t)$ under the quotient homomorphism $R[t]\to
R[t]/\mathfrak{m}R[t]=\kappa(x)[t]$. Let $d$ be the largest non-negative
integer such that $t^{d}$ divides $\overline{\alpha}_{i}(t)$ for all $i$. Let
$(\lambda_{0},\ldots,\lambda_{N})\in k^{N+1}$ such that
$(\lambda_{0}:\ldots:\lambda_{N})\in V(k)$. (The existence of such an
$(N+1)$-tuple $(\lambda_{0},\ldots,\lambda_{N})$ follows from the assumption
that $k$ is infinite.) We apply Lemma 3.1 to the two $(N+1)$-tuples

$(\overline{\alpha}_{0}(t)/t^{d},\overline{\alpha}_{1}(t)/t^{d},\ldots,\overline{\alpha}_{N}(t)/t^{d})\hskip
28.45274pt\text{and}\hskip 28.45274pt(\lambda_{0},\ldots,\lambda_{N})$

of polynomials in $\kappa(x)[t]$. We see that there exists an $(N+1)$-tuple
$(\mu_{0},\ldots,\mu_{N})\in k^{N+1}$ such that the $(N+1)$ polynomials
$R_{0}(t),\ldots,R_{N}(t)$ in $\kappa(x[t]$ the polynomials defined by

$R_{i}(t)=\overline{\alpha}_{i}(t)/t^{d}+(\mu_{i}+\lambda_{i}t)\cdot
t^{\deg_{t}(p)-d}\text{,}$

are coprime and define a morphism
$\phi:\mathbb{P}^{1}_{\kappa(x)}\to\mathbb{P}^{N}_{k}$ mapping
$\mathbb{P}^{1}_{\kappa(x)}\backslash\\{\mathbf{0},\mathbf{\infty}\\}$ into
$\mathbb{P}^{N}_{k}\backslash W$.

Let $\mathbf{g}=(g_{0},\ldots,g_{N})$ where
$g_{i}(t)=\alpha_{i}(t)+(\mu_{i}+\lambda_{i}t)\cdot p(t)$ for $0\leq i\leq N$.
Then, if $\overline{g}_{i}(t)\in\kappa(x)[t]$ is the image of $g_{i}(t)$ under
the quotient homomorphism $R[t]\to\kappa(x)[t]$, we see that
$\overline{g}_{i}(t)=t^{d}R_{i}(t)$ for all $i$. Thus, the following
conditions hold:

  * (A)

The polynomials $\overline{g}_{0}(t),\ldots,\overline{g}_{N}(t)$ have no
common zeros in $\mathbb{A}^{1}_{\kappa(x)}\backslash\\{0\\}$.

  * (B)

If $\overline{\mathbf{g}}$ denotes the $(N+1)$-tuple
$(\overline{g}_{0},\ldots,\overline{g}_{N})$, then the polynomials
$p_{1}(\overline{\mathbf{g}}),\ldots,p_{r}(\overline{\mathbf{g}})$ have no
common zero in $\mathbb{A}^{1}_{\kappa(x)}\backslash\\{0\\}$.

Let $\widetilde{I}$ and $\widetilde{J}$ be the ideals $\langle
g_{0},\ldots,g_{N}\rangle$ and $\langle
p_{1}(\mathbf{g}),\ldots,p_{r}(\mathbf{g})\rangle$ of $R[t]$ respectively.

For $0\leq i\leq N$,

(3.1)
$g_{i}=f_{i}+p\cdot(\mu_{i}+\lambda_{i}t-\beta_{i})=f_{i}[1+(p/f_{i})\cdot(\mu_{i}+\lambda_{i}t-\beta_{i})]\text{.}$

As $t$ divides $p/f_{i}$, it is a non-unit in $R\\{t\\}$, and so $g_{i}$ is a
unit multiple of $f_{i}$ in $R\\{t\\}$, for each $i$ . In particular, we have
$\widetilde{I}R\\{t\\}=I$.

Similarly, for $1\leq i\leq r$,

$p_{i}(\mathbf{g})-p_{i}(\mathbf{f})=\sum_{j}(g_{j}-f_{j})\cdot Q_{ij}$

where $Q_{ij}$ is some element of $R\\{t\\}$. Thus,

(3.2) $p_{i}(\mathbf{g})=p_{i}(\mathbf{f})[1+(p/p_{i}(\mathbf{f}))\cdot
Q^{\prime}_{i}]$

for some $Q^{\prime}_{i}\in R\\{t\\}$. As $t$ divides $p/p_{i}(\mathbf{f})$,
it is a non-unit in $R\\{t\\}$, and so $p_{i}(\mathbf{g})$ is a unit multiple
of $p_{i}(\mathbf{f})$ in $R\\{t\\}$. This proves that
$\widetilde{J}R\\{t\\}=J$.

We would now like to show that the $R$-algebra homomorphism
$R[t]/(\widetilde{I}\widetilde{J})\to R\\{t\\}/(IJ)$ is an isomorphism. This
statement is trivially true if $\widetilde{I}\widetilde{J}$ is the unit
ideals. Thus, let us assume for now, that $\widetilde{I}\widetilde{J}$ is not
the unit ideal. We would like to apply Lemma 2.5, and so we verify that
$\widetilde{I}\widetilde{J}$ satisfies the hypothesis of that lemma.

Condition (A), which was imposed on the $(N+1)$-tuple $\mathbf{g}$, implies
that if $\widetilde{I}$ is not the unit ideal, then the only prime ideal
containing $\widetilde{I}$ and $\mathfrak{m}R[t]$ is $\langle m,t\rangle$.
Condition (B) implies the same for $\widetilde{J}$. Since at least one of the
ideals $\widetilde{I}$ and $\widetilde{J}$ is not the unit ideal, it follows
that $\widetilde{I}\widetilde{J}$ satisfies condition (2) of Lemma 2.5.

Now, we verify that the homomorphism $R\to R[t]/(\widetilde{I}\widetilde{J})$
is a finite extension. For this, it suffices to find an element of
$\widetilde{I}\widetilde{J}$ such that its leading coefficient, (as a
polynomial in $t$) is an invertible element of $R$.

First, we note that there exists an index $i_{0}$, $0\leq i_{0}\leq N$ such
that $\lambda_{i_{0}}\neq 0$. Thus, $g_{i_{0}}$ is a polynomial in $t$ with a
leading coefficient that is a unit in $R$. Secondly, we observe that for
$1\leq j\leq r$, $p_{j}(\mathbf{g})$ is a polynomial in $t$ with degree
$\leq\deg(p_{j})\cdot(\deg_{t}(p)+1)$. The coefficient of
$t^{\deg(p_{j})\cdot(\deg_{t}(p)+1)}$ is equal to
$p_{j}(\lambda_{0},\ldots,\lambda_{N})$. Since
$(\lambda_{0}:\ldots,\lambda_{N})$ is in $V$, there exists an index $j_{0}$
such that $p_{j_{0}}(\lambda_{0},\ldots,\lambda_{N})\neq 0$. Thus,
$p_{j_{0}}(\mathbf{g})$ is a polynomial in $t$ with a leading coefficient that
is an invertible element in $R$. Thus, $g_{i_{0}}\cdot
p_{j_{0}}(\mathbf{g})\in\widetilde{I}\widetilde{J}$ is a polynomial in $t$
with a leading coefficient that is an invertible element of $R$. Thus, we see
that the homomorphism $R\to R[t]/(\widetilde{I}\widetilde{J})$ is a finite
extension.

Thus, if $\widetilde{I}\widetilde{J}$ is not the unit ideal, we may now apply
Lemma 2.5 to conclude that the morphism

$\theta:{\rm Spec\,}R\\{t\\}\to{\rm Spec\,}R[t]$

induces an isomorphism of the closed subscheme $\mathcal{Z}(IJ)\subset{\rm
Spec\,}R\\{t\\}$ with the closed subscheme
$\mathcal{Z}(\widetilde{I}\widetilde{J})\subset{\rm Spec\,}R[t]$. Of course,
as we noted above, if $\widetilde{I}\widetilde{J}$ happens to be the unit
ideal, $\mathcal{Z}(IJ)\to\mathcal{Z}(\widetilde{I}\widetilde{J})$ is
trivially an isomorphism.

Let $\tau:B\to{\rm Spec\,}R\\{t\\}$ denote the blowup of ${\rm
Spec\,}R\\{t\\}$ at the ideal $IJ$ and let
$\widetilde{\tau}:\widetilde{B}\to{\rm Spec\,}R[t]$ denote the blowup of ${\rm
Spec\,}R[t]$ at the ideal $\widetilde{I}\widetilde{J}$. Since
$\widetilde{I}\widetilde{J}R\\{t\\}=IJ$, we see that

$B\cong\widetilde{B}\times_{{\rm Spec\,}R[t]}{\rm Spec\,}R\\{t\\}\text{.}$

Let us denote the projection morphism $B\to\widetilde{B}$ by
$\theta^{\prime}$. Since $\theta$ maps $\mathcal{Z}(IJ)$ isomorphically onto
the closed subscheme $\mathcal{Z}(\widetilde{I}\widetilde{J})$ of ${\rm
Spec\,}R[t]$, it follows that $\theta^{\prime}$ maps
$\tau^{-1}(\mathcal{Z}(IJ))$ isomorphically onto
$\widetilde{\tau}^{-1}(\mathcal{Z}(\widetilde{I}\widetilde{J}))$.

Since the ideal sheaf $\tau^{-1}(I)\cdot\mathcal{O}_{B}$ is invertible, the
rational map $h^{\prime}:{\rm Spec\,}R\\{t\\}\to\mathbb{P}^{N}_{k}$ defined by
the $(N+1)$-tuple $(f_{0},\ldots,f_{N})$ lifts to a morphism
$B\to\mathbb{P}^{N}$. Since the ideal sheaf $\tau^{-1}(J)\cdot\mathcal{O}_{B}$
is also invertible, it further lifts to a morphism $h^{\prime\prime}:B\to Y$.
Thus, the diagram

$\textstyle{B\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{h^{\prime\prime}}$$\scriptstyle{\tau}$$\textstyle{Y\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\chi}$$\textstyle{{\rm
Spec\,}R\\{t\\}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{h}$$\textstyle{X}$

commutes.

Similarly, the rational map $\widetilde{h}^{\prime}:{\rm
Spec\,}R[t]\dashrightarrow\mathbb{P}^{N}_{k}$, defined by the $(N+1)$-tuple
$(g_{0},\ldots,g_{N})$ lifts to a morphism
$\widetilde{h}^{\prime\prime}:\widetilde{B}\to Y$. Thus, the diagram

$\textstyle{B\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\theta^{\prime}}$$\scriptstyle{\tau}$$\textstyle{\widetilde{B}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\widetilde{h}^{\prime\prime}}$$\scriptstyle{\widetilde{\tau}}$$\textstyle{Y\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\chi}$$\textstyle{{\rm
Spec\,}R\\{t\\}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\theta}$$\textstyle{{\rm
Spec\,}R[t]}$$\textstyle{X}$

commutes. Notice that, in the above diagram, we do not yet have a morphism
from ${\rm Spec\,}R[t]$ to $X$ making the diagram commute. We will prove that
such a morphism exists.

We have the two morphisms $h^{\prime\prime}$ and
$\widetilde{h}^{\prime\prime}\circ\theta^{\prime}$ from $B$ to $Y$. These need
not be equal. However we will show that they agree on
$\tau^{-1}(\mathcal{Z}(IJ))$. This claim is trivial if $IJ$ is the unit ideal.
Thus, we now assume that $IJ$ is not the unit ideal.

Let $z$ be any point of $\tau^{-1}(\mathcal{Z}(IJ))$. We want to prove that
that
$h^{\prime\prime}(z)=\widetilde{h}^{\prime\prime}\circ\theta^{\prime}(z)$.
Recall that we have fixed an embedding of $Y$ into
$\mathbb{P}_{k}^{N}\times\mathbb{P}_{k}^{r-1}$. Let
$pr_{1}:\mathbb{P}_{k}^{N}\times\mathbb{P}_{k}^{r-1}\to\mathbb{P}_{k}^{N}$ and
$pr_{2}:\mathbb{P}_{k}^{N}\times\mathbb{P}_{k}^{r-1}\to\mathbb{P}_{k}^{r-1}$
be the projection morphisms. It will suffice to prove that $pr_{i}\circ
h^{\prime\prime}(z)=pr_{i}\circ\widetilde{h}^{\prime\prime}\circ\theta^{\prime}(z)$
for $i=1,2$.

For any element $r\in R\\{t\\}$, we will denote its image in
$\mathcal{O}_{B,z}$ by $r$ as well. Let $\mathfrak{n}_{z}$ denote the maximal
ideal of the local ring $\mathcal{O}_{B,z}$. Since
$\tau(z)\in\mathcal{Z}(IJ)$, there exists at least one element in the set
$\\{f_{0},\ldots,f_{N},p_{1}(\mathbf{f}),\ldots,p_{r}(\mathbf{f})\\}$ which is
a non-unit in $\mathcal{O}_{B,z}$. Let us pick one such element and denote it
by $q_{z}$. Recall that we had chosen $p$ to be equal to
$t\cdot\left(\prod_{i}\widetilde{f}_{i}\right)^{2}\cdot\left(\prod_{j}P_{j}\right)^{2}$.
Thus, it follows that, in the ring $\mathcal{O}_{B,z}$, the non-unit element
$q_{z}$ divides $p/f_{i}$ for every $i$ and $p/p_{j}(\mathbf{f})$ for every
$j$. We will use this observation in the following discussion.

The restriction of $pr_{1}\circ h^{\prime\prime}$ to ${\rm
Spec\,}\mathcal{O}_{B,z}$ is given by the $(N+1)$-tuple
$(f_{0},\ldots,f_{N})$. We know that the ideal $I\cdot\mathcal{O}_{B,z}$ is
principal. Thus, there exists an index $i_{z}$ such that
$f_{i}/f_{i_{z}}\in\mathcal{O}_{B,z}$ for all $i$. Let
$f^{\prime}_{i}=f_{i}/f_{i_{z}}$ for $0\leq i\leq N$. If
$\overline{f^{\prime}_{i}}$ is the image of $f^{\prime}_{i}$ in
$\mathcal{O}_{B,z}/\mathfrak{n}_{z}=:\kappa(z)$, the composition

${\rm Spec\,}\kappa(z)\stackrel{{\scriptstyle
z}}{{\to}}B\xrightarrow{pr_{1}\circ h^{\prime\prime}}\mathbb{P}_{k}^{N}$

is given by the $(N+1)$-tuple
$(\overline{f^{\prime}_{0}},\overline{f^{\prime}_{1}},\cdots,\overline{f^{\prime}_{N}})$.
of elements in $\kappa(z)$. Note that $\overline{f^{\prime}_{i_{z}}}=1$.

Recall (see equation (3.1)) that
$g_{i}=f_{i}(1+(p/f_{i})(\mu_{i}+\lambda_{i}t-\beta_{i}))$. Let
$g^{\prime}_{i}=f^{\prime}_{i}(1+(p/f_{i})(\mu_{i}+\lambda_{i}t-\beta_{i}))$
for every $i$. Thus, we have $g_{i}=f_{i_{z}}g^{\prime}_{i}$ for every $i$.

Let $\overline{g^{\prime}_{i}}$ denote the image of $g^{\prime}_{i}$ in
$\kappa(z)$. As we observed above, $p/f_{i}$ is in $\mathfrak{n}_{z}$. Thus,
we see that $\overline{f^{\prime}_{i}}=\overline{g^{\prime}_{i}}$ for every
$i$.

The composition

${\rm Spec\,}\kappa(z)\stackrel{{\scriptstyle
z}}{{\to}}B\xrightarrow{{pr_{1}\circ\widetilde{h}^{\prime\prime}\circ\theta^{\prime}}}\mathbb{P}_{k}^{N}$

is given by the $(N+1)$-tuple
$(\overline{g^{\prime}_{0}},\overline{g^{\prime}_{1}},\cdots,\overline{g^{\prime}_{N}})$.
Since $\overline{g^{\prime}_{i}}=\overline{f^{\prime}_{i}}$ for every $i$, we
see that $pr_{1}\circ
h^{\prime\prime}(z)=pr_{1}\circ\widetilde{h}^{\prime\prime}\circ\theta^{\prime}(z)$.

Similarly, using equation (3.2), we can show that $pr_{2}\circ
h^{\prime\prime}(z)=pr_{2}\circ\widetilde{h}^{\prime\prime}\circ\theta^{\prime}(z)$
for any point $z\in\tau^{-1}(\mathcal{Z}(IJ))$. Thus, we conclude that
$h^{\prime\prime}(z)=\widetilde{h}^{\prime\prime}\circ\theta^{\prime}(z)$ for
any $z\in\tau^{-1}(\mathcal{Z}(IJ))$.

Now, suppose $z_{1}$ and $z_{2}$ are two distinct points of $\widetilde{B}$
such that $\widetilde{\tau}(z_{1})=\widetilde{\tau}(z_{2})=z$. Then, as
$\widetilde{\tau}$ is an isomorphism on the complement of
$\mathcal{Z}(\widetilde{I}\widetilde{J})$, we see that
$z\in\mathcal{Z}(\widetilde{I}\widetilde{J})$. Recall that $\theta$ maps
$\theta^{-1}(\mathcal{Z}(\widetilde{I}\widetilde{J}))=\mathcal{Z}(IJ)$
isomorphically onto $\mathcal{Z}(\widetilde{I}\widetilde{J})$ and that
$\theta^{\prime}$ maps $\tau^{-1}(\mathcal{Z}(IJ))$ maps isomorphically onto
$\widetilde{\tau}^{-1}(\mathcal{Z}(\widetilde{I}\widetilde{J}))$. Thus, there
exist unique points $y_{1}$ and $y_{2}$ in $\psi^{-1}(\mathcal{Z}(IJ))$ such
that $\theta^{\prime}(y_{i})=z_{i}$ for $i=1,2$. Also,
$\widetilde{\tau}\circ\theta^{\prime}=\theta\circ\tau$, and since $\theta$ is
an injective on $\mathcal{Z}(IJ)$, we see that $\tau(y_{1})=\tau(y_{2})$.
Since $\chi\circ h^{\prime\prime}=h\circ\tau$, we see that $\chi\circ
h^{\prime\prime}(y_{1})=\chi\circ h^{\prime\prime}(y_{2})$. Thus,

$\displaystyle\chi\circ\widetilde{h}^{\prime\prime}(z_{1})$
$\displaystyle=\chi\circ\widetilde{h}^{\prime\prime}\circ\theta^{\prime}(y_{1})$
$\displaystyle=\chi\circ h^{\prime\prime}(y_{1})$ $\displaystyle=\chi\circ
h^{\prime\prime}(y_{2})$
$\displaystyle=\chi\circ\widetilde{h}^{\prime\prime}\circ\theta^{\prime}(y_{2})=\chi\circ\widetilde{h}^{\prime\prime}(z_{2})\text{.}$

Note that $\widetilde{\tau}$ is a proper, birational morphism. Also, as ${\rm
Spec\,}R[t]$ is normal, we have
$\tau_{*}(\mathcal{O}_{\widetilde{B}})\cong\mathcal{O}_{{\rm Spec\,}R[t]}$.
Thus, we may apply [8, Lemma 8.11.1] to conclude that there exists a morphism
$\widetilde{h}:{\rm Spec\,}R[t]\to X$ such that
$\chi\circ\widetilde{h}^{\prime\prime}=\widetilde{h}\circ\widetilde{\tau}$.

Now, we will show that $\widetilde{h}$ can be extended to a morphism
$H:\mathbb{P}^{1}_{R}\to X$. (Recall that $\mathbb{P}^{1}_{R}={\rm
Proj}\,R[T_{0},T_{1}]$ contains ${\rm Spec\,}R[t]$ as the open subscheme
$\mathbb{P}^{1}_{R}\backslash({\rm Spec\,}R\times\\{\mathbf{\infty}\\})$, via
the identification $t=T_{0}/T_{1}$.)

For $0\leq i\leq N$, let $G_{i}(T_{0},T_{1})\in R[T_{0},T_{1}]$ be defined by

$G_{i}(T_{0},T_{1})=T_{1}^{\deg_{t}(p)+1}\cdot g_{i}(T_{0}/T_{1})\text{.}$

Thus, each $G_{i}$ is a homogeneous polynomial of degree $\deg_{t}(p)+1$ in
$R[T_{0},T_{1}]$. The coefficient of $T_{0}^{\deg_{t}(p)+1}$ in
$G_{i}(T_{0},T_{1})$ is $\lambda_{i}$. Recall that there exists an index
$i_{0}$ such that $\lambda_{i_{0}}$ is a non-zero element of $k$. Thus,
$G_{i_{0}}$ has no zero in the closed subscheme $\mathcal{Z}(T_{1})$ of
$\mathbb{P}^{1}_{R}$. Thus, the rational map
$H^{\prime}:\mathbb{P}^{1}_{R}\dashrightarrow\mathbb{P}^{N}_{k}$ defined by
the $(N+1)$-tuple $(G_{0},\ldots,G_{N})$ is defined on an open subscheme of
$\mathbb{P}^{1}_{R}$ containing the closed subscheme $\mathcal{Z}(T_{1})$. The
restriction of $H^{\prime}$ to the open subscheme
$\mathbb{P}^{1}_{R}\backslash\mathcal{Z}(T_{1})={\rm Spec\,}R[t]$ is given by
the $(N+1)$-tuple $(g_{0}(t),\ldots,g_{N}(t))$. Thus, it is the same as the
rational map $\widetilde{h}^{\prime}$. Thus, $H^{\prime}$ is defined on the
open subscheme $\mathbb{P}^{1}_{R}\backslash\mathcal{Z}(\widetilde{I})$. (Note
that the morphism $\mathcal{Z}(\widetilde{I})\hookrightarrow{\rm
Spec\,}R[t]\hookrightarrow\mathbb{P}^{1}_{R}$ is closed since
$\mathcal{Z}(\widetilde{I})$ is finite over ${\rm Spec\,}R$. The same is true
for the closed subschemes $\mathcal{Z}(\widetilde{J})$ and
$\mathcal{Z}(\widetilde{I}\widetilde{J})$ of ${\rm Spec\,}R[t]$. So, we view
$\mathcal{Z}(\widetilde{I})$, $\mathcal{Z}(\widetilde{J})$ and
$\mathcal{Z}(\widetilde{I}\widetilde{J})$ as closed subschemes of
$\mathbb{P}^{1}_{R}$.)

On the open subscheme ${\rm
Spec\,}R[t]\backslash\mathcal{Z}(\widetilde{I}\widetilde{J})$ of ${\rm
Spec\,}R[t]$, where $\psi\circ\widetilde{h}^{\prime}$ is well-defined, it
agrees with the restriction of the morphism $\widetilde{h}$. Thus, we see that
$\psi\circ H^{\prime}$ agrees with $\widetilde{h}$ on
$\mathbb{P}^{1}_{R}\backslash(\mathbb{Z}(\widetilde{I}\widetilde{J})\cup\mathcal{Z}(T_{1}))$.
Since $\mathbb{P}^{1}_{R}\backslash\mathcal{Z}(T_{1})$ (where $\widetilde{h}$
is defined) and
$\mathbb{P}^{1}_{R}\backslash\mathcal{Z}(\widetilde{I}\widetilde{J})$ (where
$\psi\circ H^{\prime}$ is defined) form a Zariski open cover of
$\mathbb{P}^{1}_{R}$, we see that there exists a morphism
$H:\mathbb{P}^{1}_{R}\to X$ extending $\widetilde{h}$. Now we compute the
morphisms $H\circ\sigma_{0}$ and $H\circ\sigma_{\infty}$ from ${\rm Spec\,}R$
to $X$.

The morphism $H\circ\sigma_{0}$ is the same as $\widetilde{h}\circ\sigma_{0}$.
So we will now compute $h\circ\sigma_{0}$. We will prove that it is the same
as the canonical morphism $\omega_{x}$. Note that
$h\circ\widehat{\sigma}_{0}=\omega_{x}$. (Recall from subsection 2.2 that
$\widehat{\sigma}_{0}$ is the morphism ${\rm Spec\,}R\to{\rm Spec\,}R\\{t\\}$
corresponding to the quotient homomorphism $R\\{t\\}\to
R\\{t\\}/tR\\{t\\}\cong R$.) Thus, it is enough to prove that
$\widetilde{h}\circ\sigma_{0}=h\circ\widehat{\sigma}_{0}$. For this, it will
suffice to show that if $\eta$ denotes the generic point of ${\rm Spec\,}R$,
then the two compositions

$\textstyle{{\rm
Spec\,}\kappa(\eta)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{{{\rm
Spec\,}R}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\widetilde{h}\circ\sigma_{0}}$$\scriptstyle{h\circ\widehat{\sigma}_{0}}$$\textstyle{X}$

are equal.

The rational map $h^{\prime}:{\rm
Spec\,}R\\{t\\}\dashrightarrow\mathbb{P}^{N}_{k}$ is defined on the open
subscheme ${\rm Spec\,}R\\{t\\}\backslash\mathcal{V}(\widetilde{I})$ of ${\rm
Spec\,}R\\{t\\}$ and it is represented by the $(N+1)$-tuple
$(f_{0},\ldots,f_{N})$. Since $t$ divides $p$, the image of $f_{i}$ under the
quotient homomorphism $R\\{t\\}\to R\\{t\\}/tR\\{t\\}\cong R$ is
$\alpha_{i}(0)$, i.e. the constant term in the polynomial $\alpha_{i}(t)\in
R[t]$. Since the $f_{i}$ were chosen to be coprime, we see that at least one
of the elements $\alpha_{0}(0),\ldots,\alpha_{N}(0)$ is non-zero. Thus, the
morphism $h^{\prime}\circ\widehat{\sigma}\circ\eta:{\rm
Spec\,}\kappa(\eta)\to\mathbb{P}^{N}$ is represented by the $(N+1)$-tuple
$(\alpha_{0}(0),\ldots,\alpha_{N}(0))$. Similarly, the morphism
$\widetilde{h}^{\prime}\circ\sigma_{0}\circ\eta:{\rm
Spec\,}\kappa(\eta)\to\mathbb{P}^{N}$ is also represented by the same
$(N+1)$-tuple. Composing with $\psi$, we obtain the desired conclusion that
the morphism $h\circ\widehat{\sigma}_{0}\circ\eta$ is equal to
$\widetilde{h}\circ\sigma_{0}\circ\eta$. Thus, it follows that
$\widetilde{h}\circ\sigma_{0}$ is equal to $\omega_{x}$.

The restriction of $H^{\prime}$ to the closed subscheme $\mathcal{Z}(T_{1})$
of $\mathbb{P}^{1}_{R}$ maps $\mathcal{Z}(T_{1})$ to the point
$(\lambda_{0}:\ldots:\lambda_{N})$ of $\mathbb{P}^{N}_{k}$. The morphism $H$
agrees with $H^{\prime}$ on an open subscheme of $\mathbb{P}^{1}_{R}$
containing $\mathcal{Z}(T_{1})$. Thus $H\circ\sigma_{\infty}$ maps ${\rm
Spec\,}R$ to the $k$-valued point $\psi((\lambda_{0}:\ldots:\lambda_{N}))$ of
$X$. This completes the proof. ∎

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