# On sequences arising from randomizing
subtraction games

Nicolás Capitelli  and  Francisco Somma
Universidad Nacional de Luján, Departamento de Ciencias Básicas, Argentina.

###### Abstract.

In this article, we study the behaviour of a broad family of real sequences
derived from randomized one-pile subtraction games. For any subtraction set
$S$, we allow any valid number of chips $s\in S$ to be removed at equal
probability at any given position and we study the sequences
$(a_{n}^{S})_{n\in\mathbb{N}}$ representing the probability of winning the
game from a position with $n$ chips. We characterize these sequences in terms
of linear recurrence relations and examine their behavior as
$n\rightarrow\infty$ for all finite $S$. We fully solve the cases for
subtraction sets of fewer than 3 elements and partially complete the general
case for arbitrary $S$.

###### Key words and phrases:

Sequences, Combinatorial games, Subtraction games

###### 2020 Mathematics Subject Classification:

91A46, 91A60, 11B83, 40A05

Corresponding address<EMAIL_ADDRESS>
This research was partially supported by the Department of Basic Sciences,
UNLu (Disposition CDD-CB 148/18) and University of Luján (Resolutions REC
224/19 and PHCS 084/20)

A (one-pile) subtraction game with subtraction set $S\subset\mathbb{N}$ is a
combinatorial game which is played as follows. From a pile with $n$ chips, two
players alternate moves. A move consists of removing $s$ chips from the pile
where $s\in S$. The player that has no moves left available loses. Subtraction
games are fundamental to combinatorial game theory and have a rich theoretical
background. However, analyzing them for generic subtraction sets can be quite
challenging, with many open problems still remaining (see e.g. [6]). For a
comprehensive understanding of combinatorial games, we refer the reader to the
seminal work [1] (see also [3, §1])

In this article, we examine the behaviour of a large family of real sequences
which we derive by randomizing subtraction games. Specifically, we allow one
such game to be played stochastically, with an equal probability of removing
any valid number of chips from the subtraction set $S$ at a given position. We
then let $a_{n}^{S}$ be the probability that the next player to move will win
this game from a position with a pile of $n$ chips. For each finite $S$, we
characterize the sequences $(a_{n}^{S})_{n\in\mathbb{N}}$ in terms of linear
recurrence relations and examine their behavior as $n\rightarrow\infty$. For
the case $|S|=2$, we univoquely characterize their convergence and we explicit
exhibit the limit value of the subsequences of even and odd terms when
$\lim_{n\rightarrow\infty}(a_{n}^{S})_{n\in\mathbb{N}}$ does not exist
(Theorem 2.5). As a general result, we prove convergence for all finite
subtraction sets $S$ containing at least an even number (Theorem 3.2).

##  1\. A family of sequences

Let $S\subset\mathbb{N}$ be a non-empty set. We randomize the one-pile
subtraction game of $n$ chips with subtraction set $S$ by playing it as
follows.

  1. (1)

First player to move starts from the starting position (an $n$-chips pile).

  2. (2)

Players alternate moves.

  3. (3)

At a given position (number of chips in the pile) $0\leq m\leq n$, an element
$k^{\prime}\in\\{k\in S\,|\,k\leq m\\}$ is randomly selected with equal
probability, and the next player to move removes $k^{\prime}$ chips from the
pile.

  4. (4)

The first player who is confronted with a position $m$ for which $\\{k\in
S\,|\,k\leq m\\}=\emptyset$, loses.

These randomly-played games are a particular case of those studied in [2],
which involve playing through noisy channels that can introduce errors in
transmission.

Our main concern is the sequence $a_{n}^{S}$ which represents the probability
of winning for the next player $N$ to move in a game of $n$ chips and
subtraction set $S$. Note that, for $n\geq\max\\{S\\}$, $a_{n}^{S}$ can be
computed recursively as follows.

$\displaystyle a_{n}^{S}$ $\displaystyle=$
$\displaystyle\displaystyle\sum_{i\in S}\ Prob(\text{$N$ wins from $n$ $|$ $i$
is selected})\cdot Prob(\text{$i$ is selected})$ $\displaystyle=$
$\displaystyle\displaystyle\sum_{i\in S}\ (1-Prob(\text{$P$ wins from
$n-i$}))\cdot\frac{1}{|S|}$ $\displaystyle=$
$\displaystyle\displaystyle\sum_{i\in S}(1-a_{n-i}^{S})\cdot\frac{1}{|S|}$
$\displaystyle=$ $\displaystyle 1-\frac{1}{|S|}\displaystyle\sum_{i\in
S}a_{n-i}^{S}$

Here $P$ denotes the previous player, $Prob(X)$ the probability of event $X$
and $Prob(X|Y)$ the conditional probability of $X$ given $Y$. Once more, this
line of reasoning mirrors that found in [2].

We see that the sequence $a_{n}^{S}$ satisfies a non-homogeneous linear
recurrence of order $\max\\{S\\}$. To determine the initial values of
$a_{i}^{S}$ for $i<\max\\{S\\}$, we proceed as follows. Let
$S=\\{k_{1},\ldots,k_{t}\\}$ with $k_{1}<\cdots<k_{t}$. Thus, for $i<k_{t}$,
$a_{i}^{S}$ represents the probability of the next player winning from
position $i$ in a game with subtraction set
$S-\\{k_{t}\\}=\\{k_{1},\ldots,k_{t-1}\\}$. Therefore, the sequence
$a_{n}^{S}$ can be determined inductively based on the size of $S$. For $t=1$,
there is no randomness, and $a_{n}^{S}$ coincides with the Sprague-Grundy
function of the subtraction game with subtraction set $S={k_{1}}$ (see (6)
below for an explicit formula for this function). For $t>1$ we let:

$a_{n}^{S}=\begin{cases}a_{n}^{S-\\{k_{t}\\}}&0\leq n<k_{t}\\\
1-\frac{1}{t}\sum_{i=1}^{t}a_{n-k_{i}}^{S}&n\geq k_{t}\end{cases}$ (2)

Our main focus in this article is to analyze the behavior of $a_{n}^{S}$ for
large $n$, which serves as an indicator of the game’s fairness. It’s
straighfoward to verify that convergence of $a_{n}^{S}$ implies convergence to
$\frac{1}{2}$. However, this will also emerge as a consequence of our analysis
of the recursive relation (2). A solution for this non-homogenous linear
recurrence is of the form

$a_{n}^{S}=\alpha_{1}z_{1}^{n}+\alpha_{2}z_{2}^{n}+\cdots+\alpha_{l}z_{k_{t}}^{n}+\frac{1}{2}$
(3)

where $\\{z_{i}\\}_{1\leq i\leq k_{t}}$ are the complex roots of the
characteristic polynomial

$\chi_{S}(x)=tx^{k_{t}}+\sum_{i=1}^{t}x^{k_{t}-k_{i}}$ (4)

and $\alpha_{i}\in\mathbb{C}$ are the solutions to the system

$\begin{cases}\alpha_{1}+\alpha_{2}+\cdots+\alpha_{k_{t}}+\frac{1}{2}=a_{0}^{S}\\\
\alpha_{1}z_{1}+\alpha_{2}z_{2}+\cdots+\alpha_{k_{t}}z_{k_{t}}+\frac{1}{2}=a_{1}^{S}\\\
\alpha_{1}z_{2}^{1}+\alpha_{2}z_{2}^{2}+\cdots+\alpha_{k_{t}}z_{k_{t}}^{2}+\frac{1}{2}=a_{2}^{S}\\\
\hskip 86.72377pt\vdots\\\
\alpha_{1}z_{1}^{{k_{t}}-1}+\alpha_{2}z_{2}^{{k_{t}}-1}+\cdots+\alpha_{k_{t}}z_{k_{t}}^{{k_{t}}-1}+\frac{1}{2}=a_{{k_{t}}-1}^{S}\end{cases}$
(5)

(see, e.g., [5, §1] for a glimpse of recursive sequences theory). In the
following sections we’ll study the behaviour of
$(a_{n}^{S})_{n\in\mathbb{N}_{0}}$, completely solving the case $t=2$ and
partially the general case for finite $S$.

######  Remark 1.1.

Note that in order to study the limit of $(a_{n}^{S})_{n\in\mathbb{N}_{0}}$,
we may restrict our attention to the case where
$\gcd\\{k_{1},\ldots,k_{t}\\}=1$. Indeed, let
$m=\gcd\\{k_{1},\ldots,k_{t}\\}$. Consider the subsequence defined as
$b_{n}=a^{S}_{mn}$. Then

$b_{n}=a^{S}_{mn}=1-\frac{1}{t}\sum_{i=1}^{t}a_{mn-k_{i}}^{S}$

Since $mn-k_{i}=mn-m(k_{i}/m)=m(n-k_{i}/m)$ then $a_{mn-
k_{i}}^{S}=b_{n-k_{i}/m}$. Hence,

$b_{n}=1-\frac{1}{t}\sum_{i=1}^{t}b_{n-k_{i}/m},$

which is the sequence associated to the one-pile subtraction game of $n$ chips
and subtraction set $S^{\prime}=\\{\frac{k_{1}}{m},\ldots,\frac{k_{t}}{m}\\}$.
Since $\lim_{n\rightarrow\infty}a_{n}=\lim_{n\rightarrow\infty}b_{n}$ the
claim is proven.

As it was pointed out earlier, for $S=\\{k\\}$, the game is devoid of
randomness, with either no moves or just one available. In this case, the
sequence adopts the following straightforward closed form:

$a_{n}^{\\{k\\}}=\dfrac{1+(-1)^{\lfloor\frac{n}{k}\rfloor+1}}{2}$ (6)

That is, $a_{n}^{\\{k\\}}$ consists of alternating blocks of zeros and ones,
each block being of size $k$.

##  2\. The case $S=\\{k,l\\}$

In this section with shall completely solve the case of a subtraction set with
two elements. Namely, we will univoquely determine for which subtraction sets
$S$ the sequence $a_{n}^{S}$ converges and for which it does not. Moreover, in
the latter scenario, the subsequences of even and odd terms converge, and we
explicitly exhibit these limits for each $S$. Our approach to solving this
case mirrors our strategy for addressing the general case.

Throughout this section we let $S=\\{k,l\\}\subset\mathbb{N}$ with $k<l$ and
$\gcd\\{k,l\\}=1$ (see Remark 1.1). We also shall write $a_{n}$ for
$a_{n}^{S}=a_{n}^{\\{k,l\\}}$ for simplicity. In this case, Equations (2) and
(6) give us the following recursive sequence.

$a_{n}=\begin{cases}\frac{1+(-1)^{\lfloor\frac{n}{k}\rfloor+1}}{2}&0\leq
n<l\\\ 1-\frac{1}{2}(a_{n-k}+a_{n-l})&n\geq l\end{cases}$ (7)

The following properties of the characteristic polynomial
$\chi(z)=2z^{l}+z^{l-k}+1$ of this recurrence will help us characterize the
convergence of $(a_{n})_{n\in\mathbb{N}_{0}}$ for each election of $k$ and
$l$. In what follows we let $G_{i}\subset\mathbb{C}$ denote the group of $i$th
roots of unity.

######  Proposition 2.1.

All the complex roots of $\chi(x)$ have norm less than 1, with the exception
of the root $z=-1$ which happens only for the case $k,l\equiv 1\,(2)$.
Furthermore, in this last case, all roots are simple.

###### Proof.

Let $z$ be a root of $\chi(x)$. By Lagrange’s bound (see e.g. [8, IV]) we have
that

$|z|\leq\max\left\\{1,\sum_{i=0}^{d-1}\left|\frac{c_{i}}{c_{l}}\right|\right\\}=1,$
(8)

where $c_{0},\ldots,c_{d_{t}}$ denote the coefficients of $\chi(x)$. Suppose
$|z|=1$. Write $-1=2z^{l}+z^{l-k}=z^{l-k}(2z^{k}+1)$. Since $|z|=1$ then
$|2z^{k}+1|=1$. Therefore $2z^{k}+1$ lies in the intersection of the circle of
center $(1,0)$ and radius $2$ and the circle of center $(0,0)$ and radius 1.
Hence $2z^{k}+1=-1$. We conclude that $z^{l-k}=1$ and that $z^{l}=-1$.
Particularly, $z$ is a root of unity of order $2l$. Therefore $z\in
G_{l-k}\cap G_{2l}=G_{\gcd\\{l-k,2l\\}}$. Since $\gcd\\{l-k,l\\}=1$ then

$G_{\gcd\\{l-k,2l\\}}=\begin{cases}G_{1}&l-k\equiv 1\,(2)\\\ G_{2}&l-k\equiv
0\,(2)\end{cases}$

Now, $\chi(1)\neq 0$ so if $l-k\equiv 1\,(2)$ the $\chi$ has no roots of norm
1 and if $l-k\equiv 0\,(2)$ the only possible root of norm 1 is $z=-1$. Since
$\chi(-1)=0$ for this last case (which implies $l$ odd), the result is
settled.

It remains to show the simplicity of the all of the roots for the case $l$ and
$k$ odd. In this case, $z=-1$ is a simple root since $\chi^{\prime}(-1)\neq
0$. Let $z\neq-1$ and suppose $z$ is a root of $\chi^{\prime}(x)$. Then
$z^{l-k-1}(2lz^{k}+l-k)=0$, which implies that $z^{k}\in\mathbb{R}$. If $z$ is
also a root of $\chi(x)$ then $0=2z^{l}+z^{l-k}+1=z^{l-k}(2z^{k}+1)+1$ which
implies $z^{l-k}\in\mathbb{R}$ and $z^{l}\in\mathbb{R}$. This means that
$\arg(z)=\frac{r\pi}{l-k}$ for some $0\leq r<2(l-k)$ and
$\arg(z)=\frac{r^{\prime}\pi}{l}$ for some $0\leq r^{\prime}<2l$
simultaneously. Hence $lr=(l-k)r^{\prime}$. Since $\gcd\\{l-k,l\\}=1$ then
$l-k$ divides $r$. Since $0\leq r<2(l-k)$ then either $r=0$ o $r=l-k$, from
which $\arg(z)=0$ or $\arg(z)=\pi$. So $z\in\mathbb{R}$. In the first case,
$z>0$ but $\chi(x)$ doesn’t have real positive roots. In the second case,
since $z\neq-1$ and $-1$ is a root for this case, then it must cancel the
polynomial $\chi(x)/(x+1)$. An easy verification shows that

$\frac{\chi(x)}{x+1}=\sum_{i=0}^{l-k-1}(-1)^{i}x^{i}+2\sum_{i=l-k}^{l-1}(-1)^{i}x^{i}$
(9)

But since $z<0$ then $(-1)^{i}z^{i}=(-z)^{i}>0$ for all $i$. Therefore, $z$
does not anulate this quotient and cannot be a multiple root of $\chi(x)$.∎

######  Corollary 2.2.

The sequence $(a_{n})_{n\in\mathbb{N}_{0}}$ defined in (7) converges to
$\frac{1}{2}$ if $l\not\equiv k\,(2)$.

###### Proof.

If $l\not\equiv k(2)$ then Proposition 2.1 implies that all roots of
$\chi(z)=2z^{l}+z^{l-k}+1$ have norm less than $1$ (since if $l$ is odd then
$l-k$ is also odd). Hence, $\lim_{n\rightarrow\infty}a_{n}=\frac{1}{2}$ by
equation (3). ∎

The case $l,k\equiv 1\,(2)$ remains to be addressed. We tackle this in the
following steps. Under these conditions, Proposition 2.1 gives $z_{1}=-1$ as
the only root of $\chi(x)$ of norm 1, while all remaining roots possess norm
less than 1. From Equation (3) we conclude that

$\lim_{n\rightarrow\infty}a_{n}=\lim_{n\rightarrow\infty}\alpha_{1}(-1)^{n}+\frac{1}{2}.$
(10)

We shall explicity compute $\alpha_{1}$ next. Note that the linear system (5)
becomes:

$\begin{pmatrix}1&1&1&\cdots&1\\\ -1&z_{2}&z_{3}&\cdots&z_{l}\\\
1&z_{2}^{2}&z_{3}^{2}&\cdots&z_{l}^{2}\\\ \vdots&\vdots&\vdots&&\vdots\\\
-1&z_{2}^{l-2}&z_{3}^{l-2}&\cdots&z_{l}^{l-2}\\\
1&z_{2}^{l-1}&z_{3}^{l-1}&\cdots&z_{l}^{l-1}\end{pmatrix}\begin{pmatrix}\alpha_{1}\\\
\alpha_{2}\\\ \alpha_{3}\\\ \vdots\\\ \alpha_{l-1}\\\
\alpha_{l}\end{pmatrix}=\begin{pmatrix}a_{0}-\frac{1}{2}\\\
a_{1}-\frac{1}{2}\\\ a_{2}-\frac{1}{2}\\\ \vdots\\\ a_{l-2}-\frac{1}{2}\\\
a_{l-1}-\frac{1}{2}\end{pmatrix}$ (11)

Since all roots of $\chi(x)$ are simple (again by Proposition 2.1) then this
Vandermonde matrix is invertible. The inverse is given by

$\begin{pmatrix}L_{00}&\cdots&L_{l-10}\\\ \vdots&&\vdots\\\
L_{0l-1}&\cdots&L_{l-1l-1}\end{pmatrix}$

where
$L_{j}(x)=\sum_{i=0}^{l-1}L_{ij}x^{i}=\frac{f(x)}{(x-z_{j+1})f^{\prime}(z_{j+1})}$
and $f(x)=\prod_{i=1}^{l}(x-z_{i})$ (see e.g. [4, §0.9] for information on
Vandermonde matrices). We conclude that:

$\begin{pmatrix}\alpha_{1}\\\ \vdots\\\
\alpha_{l}\end{pmatrix}=\begin{pmatrix}L_{00}&\cdots&L_{l-10}\\\
\vdots&&\vdots\\\
L_{0l-1}&\cdots&L_{l-1l-1}\end{pmatrix}\begin{pmatrix}a_{0}-\frac{1}{2}\\\
\vdots\\\ a_{l-1}-\frac{1}{2}\end{pmatrix}$

Since we are solely interested in $\alpha_{1}$ we only need consider

$L_{0}(x)=\dfrac{f(x)}{(x-z_{1})f^{\prime}(z_{1})}=\dfrac{f(x)}{(x+1)f^{\prime}(-1)}.$
(12)

Since $\chi(x)=2f(x)$ then, by equation (9), we get

$\dfrac{f(x)}{(x+1)}=\frac{1}{2}\sum_{i=0}^{l-k-1}(-1)^{i}x^{i}+\sum_{i=l-k}^{l-1}(-1)^{i}x^{i}$

Also, $f^{\prime}(z_{1})=f^{\prime}(-1)=\frac{1}{2}(l+k)=\frac{l+k}{2}$. Hence

$L_{0}(x)=\frac{2}{l+k}\left(\frac{1}{2}\sum_{i=0}^{l-k-1}(-1)^{i}x^{i}+\sum_{i=l-k}^{l-1}(-1)^{i}x^{i}\right)$

We conclude that the coefficients of $L_{0}(x)$ are then given by

$L_{i0}=\begin{cases}\dfrac{(-1)^{i}}{l+k}&0\leq i<l-k\\\ &\\\
\dfrac{2(-1)^{i}}{l+k}&l-k\leq i<l\\\ \end{cases}$

Hence

$\displaystyle\alpha_{1}$ $\displaystyle=$
$\displaystyle\sum_{i=0}^{l-k-1}\dfrac{(-1)^{i}}{l+k}\left(a_{i}-\frac{1}{2}\right)+\sum_{i=l-k}^{l-1}\dfrac{2(-1)^{i}}{l+k}\left(a_{i}-\frac{1}{2}\right)$
$\displaystyle=$
$\displaystyle\dfrac{1}{l+k}\sum_{i=0}^{l-k-1}(-1)^{i}\left(a_{i}-\frac{1}{2}\right)+\dfrac{2}{l+k}\sum_{i=l-k}^{l-1}(-1)^{i}\left(a_{i}-\frac{1}{2}\right)$
$\displaystyle=$
$\displaystyle\dfrac{1}{l+k}\left(\sum_{i=0}^{l-k-1}(-1)^{i}\left(a_{i}-\frac{1}{2}\right)+2\sum_{i=l-k}^{l-1}(-1)^{i}\left(a_{i}-\frac{1}{2}\right)\right)$
$\displaystyle=$
$\displaystyle\dfrac{1}{l+k}\left(\sum_{i=0}^{l-k-1}(-1)^{i}a_{i}-\sum_{i=0}^{l-k-1}(-1)^{i}\frac{1}{2}+2\sum_{i=l-k}^{l-1}(-1)^{i}a_{i}-2\sum_{i=l-k}^{l-1}(-1)^{i}\frac{1}{2}\right)$
$\displaystyle=$
$\displaystyle\dfrac{1}{l+k}\left(\sum_{i=0}^{l-k-1}(-1)^{i}a_{i}+2\sum_{i=l-k}^{l-1}(-1)^{i}a_{i}-\sum_{i=0}^{l-k-1}(-1)^{i}\frac{1}{2}-2\sum_{i=l-k}^{l-1}(-1)^{i}\frac{1}{2}\right)$

Now, since $l-k-1$ is odd then the sum $\sum_{i=0}^{l-k-1}(-1)^{i}\frac{1}{2}$
is zero and since $l-k$ and $l-1$ are even then the sum
$\sum_{i=l-k}^{l-1}(-1)^{i}\frac{1}{2}$ is $\frac{1}{2}$. We conclude

$\alpha_{1}=\dfrac{1}{l+k}\left(\sum_{i=0}^{l-k-1}(-1)^{i}a_{i}+2\sum_{i=l-k}^{l-1}(-1)^{i}a_{i}-1\right)=\dfrac{1}{l+k}\left(\sum_{i=0}^{l-1}(-1)^{i}a_{i}+\sum_{i=l-k}^{l-1}(-1)^{i}a_{i}-1\right)$

We compute these sums in the following

######  Lemma 2.3.

Let $a_{j}=\dfrac{1+(-1)^{\lfloor\frac{j}{k}\rfloor+1}}{2}$ for $0\leq j\leq
l-1$. Then,

  1. (1)

$\displaystyle\sum_{i=0}^{l-1}(-1)^{i}a_{i}=\begin{cases}-\frac{1}{2}\left\lfloor\frac{l}{k}\right\rfloor&\left\lfloor\frac{l}{k}\right\rfloor\equiv
0\,(2)\\\
-\frac{1}{2}\left(\left\lfloor\frac{l}{k}\right\rfloor-1\right)&\left\lfloor\frac{l}{k}\right\rfloor\equiv
1\,(2)\end{cases}$

  2. (2)

$\displaystyle\sum_{i=l-k}^{l-1}(-1)^{i}a_{i}=\begin{cases}-k\left(\left\lfloor\frac{l}{k}\right\rfloor+1\right)+l&\left\lfloor\frac{l}{k}\right\rfloor\equiv
0\,(2)\\\
-l+k\left\lfloor\frac{l}{k}\right\rfloor&\left\lfloor\frac{l}{k}\right\rfloor\equiv
1\,(2)\end{cases}$

###### Proof.

Let $s$ be such that $sk\leq i\leq(s+1)k-1$ and note that

$(-1)^{i}a_{i}=\begin{cases}0&s\equiv 0\,(2)\\\ (-1)^{i}&s\equiv
1\,(2)\end{cases}$

So, we have blocks of zeros and $\pm 1$ alternating. All the blocks of $\pm 1$
begin at $i=sk$ with odd $s$, so the leading term of each of these blocks is
$-1$. Since there are $(s+1)k-1-sk+1=k$ integers in each interval and $k$ is
odd then the sum of each block of $\pm 1$ is $-1$. Let us write
$l=k\lfloor\frac{l}{k}\rfloor+r$, where $r$ is the remainder of the division
of $l$ by $k$. Then, we have $\lfloor\frac{l}{k}\rfloor$ alternating blocks of
zeros and $\pm 1$ and a truncated block of size $r$. More concretely,

  * •

If $\lfloor\frac{l}{k}\rfloor\equiv 0\,(2)$ then the truncated block is of
zeros and we have $\frac{1}{2}\lfloor\frac{l}{k}\rfloor$ complete blocks of
$\pm 1$. In this case,
$\sum_{i=0}^{l-1}(-1)^{i}a_{i}=-\frac{1}{2}\lfloor\frac{l}{k}\rfloor$.

  * •

If $\lfloor\frac{l}{k}\rfloor\equiv 1\,(2)$ then the truncated block is of
$\pm 1$ and we have $\frac{1}{2}(\lfloor\frac{l}{k}\rfloor-1)$ complete blocks
of $\pm 1$. The truncated block has $r=l-k\lfloor\frac{l}{k}\rfloor$ $\pm 1$.
Since $l$, $k$ and $\lfloor\frac{l}{k}\rfloor$ are odd then $r$ is even. So
the sum of the truncated block of $r$ $\pm 1$ cancels and
$\sum_{i=0}^{l-1}(-1)^{i}a_{i}=-\frac{1}{2}(\lfloor\frac{l}{k}\rfloor-1)$.

Now, for the sum $\sum_{i=l-k}^{l-1}(-1)^{i}a_{i}$ note that there are
$l-1-(l-k)+1=k$ sumands. Since if $\lfloor\frac{l}{k}\rfloor\equiv 0\,(2)$ the
truncated block is of zeros then we must have $r$ zeros and $k-r$ $\pm 1$.
Since $r=l-k\lfloor\frac{l}{k}\rfloor$ then we have
$k-r=k-(l-k\lfloor\frac{l}{k}\rfloor)=k(\lfloor\frac{l}{k}\rfloor+1)-l$ $\pm
1$. On the other hand, if $\lfloor\frac{l}{k}\rfloor\equiv 1\,(2)$, then the
truncated block is of $\pm 1$ and we have $r=l-k\lfloor\frac{l}{k}\rfloor$
$\pm 1$. This concludes de proof. ∎

######  Corollary 2.4.

For odd $k,l$

$\alpha_{1}=\begin{cases}\frac{1}{l+k}\left(-\frac{1}{2}\left\lfloor\frac{l}{k}\right\rfloor-k\left(\left\lfloor\frac{l}{k}\right\rfloor+1\right)+l-1\right)&\left\lfloor\frac{l}{k}\right\rfloor\equiv
0\,(2)\\\ &\\\
\frac{1}{l+k}\left(-\frac{1}{2}\left(\left\lfloor\frac{l}{k}\right\rfloor-1\right)-l+k\left\lfloor\frac{l}{k}\right\rfloor-1\right)&\left\lfloor\frac{l}{k}\right\rfloor\equiv
1\,(2)\\\ \end{cases}$ (13)

In particular, $\alpha_{1}\neq 0$ for any choice of coprime odd $k$ and $l$.

###### Proof.

Note that $\alpha_{1}<0$ since
$-\frac{1}{2}\left\lfloor\frac{l}{k}\right\rfloor<0$,
$k\left(\left\lfloor\frac{l}{k}\right\rfloor+1\right)+l\leq 0$ and $-1<0$.
Also, $-\frac{1}{2}\left(\left\lfloor\frac{l}{k}\right\rfloor-1\right)\leq 0$,
$-l+k\left\lfloor\frac{l}{k}\right\rfloor<0$ and $-1<0$. ∎

Combining Remark 1.1, Equation (10), and Corollaries 2.2 and 2.4 we obtain

######  Theorem 2.5.

Let $S=\\{k,l\\}\subset\mathbb{N}$ and let $\alpha_{1}$ be as in Equation
(13). The sequence (7) conveges (to $\frac{1}{2}$) only when $kl\equiv
0\,(2)$. When $k,l\equiv 1\,(2)$ the subsequences $b_{n}=a_{2n}$ and
$c_{n}=a_{2n-1}$ converge to $\frac{1}{2}+\alpha_{1}$ and
$\frac{1}{2}-\alpha_{1}$ respectively.

##  3\. The General Case

We now study the problem for a generic subtraction set
$S=\\{k_{1},\ldots,k_{t}\\}\subset\mathbb{N}$ with $k_{i}<k_{j}$ if $i<j$ and
$\gcd\\{k_{1},\ldots,k_{t}\\}=1$. In this situation, we shall prove that
$(a_{n}^{S})_{n\in\mathbb{N}_{0}}$ converges for every $S$ that has at least
one even element (see Theorem 3.2). However, we do not show that an all-odd
subtraction set necesarily do not converge (although we conjeture it) nor we
provide the exact limits of the even and odd subsequences for the
(conjectured) non-convergent case.

Consider the general equation (2). We shall apply the same reasoning as in the
previous section to identify when the sequences converge in terms of the
properties of the roots of the characteristic polynomial $\chi_{S}(x)$ given
in (4). To simplify formulae, in the following we let

$d_{i}:=\begin{cases}k_{t}-k_{i}&1\leq i\leq t-1\\\ k_{t}&i=t\end{cases}.$

Note that $\gcd\\{k_{1},\ldots,k_{t}\\}=1$ if and only if
$\gcd\\{d_{1},\ldots,d_{t}\\}=1$.

We have the following (partial) generalization of Proposition 2.1.

######  Proposition 3.1.

All the complex roots of $\chi_{S}(x)=tx^{d_{t}}+\sum_{i=1}^{t-1}x^{d_{i}}+1$
have norm less than 1, with the exception of the simple root $z=-1$ which
happens only for the case $d_{t}\equiv 1\,(2)$ and $d_{i}\equiv 0\,(2)$ for
all $1\leq i\leq t-1$.

###### Proof.

Let $z$ be a root of $\chi_{S}(x)$. Again by Lagrange’s bound we have

$|z|\leq\max\left\\{1,\sum_{i=0}^{d_{t}-1}\left|\frac{c_{i}}{c_{d_{t}}}\right|\right\\}=1,$
(14)

where $c_{0},\ldots,c_{d_{t}}$ denote the coefficients of $\chi_{S}(x)$.
Assume $|z|=1$ and write

$tz^{d_{t}}+\sum_{i=1}^{t-1}z^{d_{i}}+1=0.$

Now $tz^{d_{t}}$ lies in the circle of center $(0,0)$ and radius $t$ and
$\sum_{i=1}^{t-1}z^{d_{i}}+1$ lies inside the circle of center $(1,0)$ and
radius $t-1$. The only way that $\sum_{i=1}^{t-1}z^{d_{i}}+1$ can bring an
element from distance $t$ to $0$ is if $z^{d_{i}}=1$ for all $i=1,\ldots,t-1$
and, consequently, $tz^{d_{t}}=-t$. We conclude that $z\in G_{d_{i}}$ for all
$i=1,\ldots,t-1$ and $z\in G_{2d_{t}}$. That is,

$z\in\left(\bigcap_{i=1}^{t-1}G_{d_{i}}\right)\cap
G_{2d_{t}}=G_{\gcd\\{d_{1},\ldots,d_{t-1},2d_{t}\\}}=\begin{cases}G_{1}&\exists\,d_{i}\equiv
1\,(2)\\\ G_{2}&d_{i}\equiv 0\,(2)\,\forall\,i=1,\dots t-1\end{cases}$

Now, $1$ is not a root of $\chi_{S}(x)$ so we conclude that $d_{i}\equiv
0\,(2)\,\forall\,i=1,\dots t-1$. Note that this implies that $d_{t}=k_{t}$ is
odd, since if $k_{t}$ were even and also all the $d_{i}=k_{t}-k_{i}$ then all
$k_{i}$ would be even, contradicting that $\gcd\\{k_{1},\ldots,k_{t}\\}=1$. We
conclude that $k_{t}$ must be odd and, hence, all $k_{i}$ with
$i=1,\ldots,t-1$ must also be odd (for $d_{i}$ to be all even). We conclude
that the only root of $\chi_{S}(x)$ of norm 1 is $z=-1$ for the case
$k_{i}\equiv 1\,(2)$ for all $i=1,\dots,t$. Furthermore, since
$\chi^{\prime}(-1)\neq 0$ then $-1$ is a simple root. ∎

######  Theorem 3.2.

The sequence $a_{n}^{S}$ given in (2) converges (to $\frac{1}{2}$) if
$k_{i}\equiv 0\,(2)$ for some $i=1,\ldots,t$.

###### Proof.

For the mentioned case, Proposition 3.1 asserts that all roots of
$\chi_{S}(z)$ have norm less than $1$. Hence,
$\lim_{n\rightarrow\infty}a_{n}=\frac{1}{2}$ by equation (3). ∎

We now provide some remarks for the case not covered by Theorem 3.2. On one
hand, in order to determine if $(a_{n}^{S})_{n\in\mathbb{N}_{0}}$ does or not
converge in this setting it must be shown whether the coefficient $\alpha_{1}$
of $(-1)^{n}$ in Equation (3) is or is not zero (respectively). We did this in
the case $t=2$ by explicitly finding $\alpha_{1}$ and showing it is always
negative. However, finding and explicity expression for this coefficient is
rather difficult in this more general setting. We could proceed with the same
arguments as in the case $t=2$ but under the assumption that all roots in this
particular case are simple, which we believe is the case.

######  Conjecture 3.3.

The roots of $\chi_{S}(x)=tx^{d_{t}}+\sum_{i=1}^{t-1}x^{d_{i}}+1$ are all
simple whenever $d_{t}\equiv 1\,(2)$ and $d_{i}\equiv 0\,(2)$ for all $1\leq
i\leq t-1$.

Indeed, we again solve the system (11), where $z_{1}=-1$. This system is
invertible by Conjecture 3.3, so the same arguments go through to compute the
coefficients $L_{0}(x)$ as in Equation (12). In this case, an easy computation
shows that

$\frac{\chi_{S}(z)}{x+1}=\sum_{j=1}^{t}\sum_{i=d_{t-j+1}}^{d_{t-j}-1}j(-1)^{i}x^{i}.$

Also, since $d_{t}$ is odd and $d_{i}$ are even for all $i=1,\ldots,t-1$, then

$\chi^{\prime}(-1)=td_{t}(-1)^{d_{t}-1}+\sum_{i=1}^{t-1}(-1)^{d_{i}-1}d_{i}=td_{t}-\sum_{i=1}^{t-1}d_{i}=\sum_{i=1}^{t}d_{i}=\sum_{i=1}^{t}k_{i}.$

Hence, since $\chi(z)=tf(x)$ we conclude that

$L_{0}(x)=\dfrac{t}{k_{1}+\cdots+k_{t}}\left(\sum_{j=1}^{t}\sum_{i=d_{t-j+1}}^{d_{t-j}-1}\dfrac{j}{t}(-1)^{i}x^{i}\right).$

So, the coefficients for $L_{0}(x)$ are:

$L_{i0}=\begin{cases}\dfrac{j(-1)^{i}}{k_{1}+\cdots+k_{t}}&d_{t-j+1}\leq
i<d_{t-j}\,(j=1,\ldots,t)\end{cases}$

We conclude that

$\alpha_{1}=\sum_{j=1}^{t}\sum_{i=d_{t-j+1}}^{d_{t-j}-1}\dfrac{j(-1)^{i}}{k_{1}+\cdots+k_{t}}\left(a_{i}^{S}-\frac{1}{2}\right).$

This sum is challenging to address because the initial coefficients
$a_{i}^{S}$ align with the first $k_{t}-1$ values of the sequence
$a_{n}^{S-\\{k_{t}\\}}$, which no longer consists of blocks of zeros and ones.
However, other non-necesarily explicit arguments could be found to detemine
whether coefficient $\alpha_{1}$ in Equation (3) is or is not null.

######  Question 3.4.

Is $\alpha_{1}\neq 0$ for every $S=\\{k_{1},\ldots,k_{t}\\}$ with
$k_{1}<\cdots<k_{t}$ and $\gcd\\{k_{1},\ldots,k_{t}\\}=1$?

##  4\. Closing remarks

In this section we outline some open problems, potential avenues for further
exploration, and conjectures that may suggest future research paths.

###  4.1. The problem for all finite $S$.

In this paper we have addressed the study of the behaviour of the sequences
$(a_{n}^{S})_{n\in\mathbb{N}_{0}}$ for all finite $S$, completely determining
it for $|S|<3$ and partially for a generic $S$. A possible path to complete
this last case would be to go through Conjecture 3.3 and Question 3.4 above
(although probably no computing explicity the coefficient $\alpha_{1}$
corresponding to the root $-1$ of $\chi_{S}$). Additionally, for some
particular choices of $S$ and for the complete case $|S|=3$, it may be
possible even to find the limit of the even and odd subsequences of
$(a_{n}^{S})_{n\in\mathbb{N}_{0}}$.

###  4.2. Dynamic Subtraction Sets

A more general class of subtraction games is obtained by allowing the
subtraction set depend on the oponent’s previous move (see e.g. [7]). For
example, at a given position, the player whose turn it is to move can remove 1
chip or the entire pile; or a legal move is to remove any number of chips but
not exceeding the amount removed by his opponent on the previous turn.
Sequences derived from this problems would also satisfy a recurrence relation
and can be attacked with some of the techniques presented in this article.

For example, the case with subtraction set $S=\\{1,n\\}$, which allows as a
valid move to either remove 1 chip or the entire pile at each given position
$n$, gives rise to the recurrence

$a_{n}^{S}=\begin{cases}0&n=0\\\ 1-\frac{1}{2}a_{n-1}^{S}&n\geq 1\end{cases}$

This is a linear recursion of order 1, so the solution can be computed
straightforwardly as $\alpha_{1}z_{1}^{n}+\beta$ where $z_{1}$ is the root of
the characteristic polynomial $\chi(x)=x-\frac{1}{2}$ and $\beta=\frac{2}{3}$
is a particular solution of the recursive formula. We find that
$a_{n}^{S}=\alpha_{1}(\frac{1}{2})^{n}+\frac{2}{3}$, which converges to
$\frac{2}{3}$. This evidences that with dynamic substraction sets we no longer
have converges to $\frac{1}{2}$.

On the other hand, the (trivial) case for $S=\\{1,\ldots,n\\}$ is
straightforward. Indeed, in this situation:

$a_{n}=\begin{cases}a_{0}=0,\,a_{1}=1&\\\
1-\dfrac{1}{n}(a_{0}+a_{1}+\cdots+a_{n-1})&n\geq 2\end{cases}$

It’s easy to see by induction that $a_{n}=\frac{1}{2}$ for every $n\geq 2$. So
the game is balanced for each $n\geq 2$.

###  4.3. Several piles

A next step can also be studying substraction games of more than one pile. In
this case, and additional level of randomness must be considered as to which
pile is selected to perform the move. Also, we may require different
subtraction set for different piles. In the more general case, assume there
are $r$ piles and we let
$\mathcal{S}=\\{S_{j}\\}_{j=1,\dots,r}\subset\mathbb{N}$ be a collection of
non-empty sets, set $S_{j}$ being the subtraction set for pile $j$. We
represent a position of this game $(n_{1},\ldots,n_{r})$, pile $j$ containing
$n_{j}$ chips. A similar argument as in Equation (1) can be performed to
obtain multi-indexed sequences
$(a_{n_{1},\ldots,n_{r}}^{\mathcal{S}})_{n_{i}\in\mathbb{N}_{0}}$ representing
the probability of winning for the next player to move from the position
$(n_{1},\ldots,n_{r})$. Let $Prob(X,n_{1},\ldots,n_{r},j,i)$ stand for the
conditional probability that player $X$ wins from position
$(n_{1},\ldots,n_{r})$ given that move $i$ in pile $j$ is selected. Then, for
$n_{i}\geq\max\\{S_{i}\\}$, we have:

$\displaystyle a_{n_{1},\ldots,n_{r}}^{\mathcal{S}}$ $\displaystyle=$
$\displaystyle\displaystyle\sum_{j=1}^{r}\sum_{i\in
S_{j}}Prob(N,n_{1},\ldots,n_{j-i},\ldots,n_{r},j,i)\cdot Prob(\text{$i$ is
selected})$ $\displaystyle=$
$\displaystyle\displaystyle\sum_{j=1}^{r}\sum_{i\in S_{j}}(1-Prob(\text{$P$
wins from
$(n_{1},\ldots,n_{j-i},\ldots,n_{r})$}))\cdot\frac{1}{r}\cdot\frac{1}{|S_{j}|}$
$\displaystyle=$ $\displaystyle\displaystyle\sum_{j=1}^{r}\sum_{i\in
S_{j}}(1-a_{n_{1},\ldots,n_{j-i},\ldots,n_{r}}^{\mathcal{S}})\cdot\frac{1}{r}\cdot\frac{1}{|S_{j}|}$
$\displaystyle=$
$\displaystyle\frac{1}{r}\displaystyle\sum_{j=1}^{r}\frac{1}{|S_{j}|}\sum_{i\in
S_{j}}(1-a_{n_{1},\ldots,n_{j-i},\ldots,n_{r}}^{\mathcal{S}})$
$\displaystyle=$
$\displaystyle\frac{1}{r}\displaystyle\sum_{j=1}^{r}\left(1-\frac{1}{|S_{j}|}\sum_{i\in
S_{j}}a_{n_{1},\ldots,n_{j-i},\ldots,n_{r}}^{\mathcal{S}}\right)$
$\displaystyle=$ $\displaystyle
1-\frac{1}{r}\displaystyle\sum_{j=1}^{r}\frac{1}{|S_{j}|}\sum_{i\in
S_{j}}a_{n_{1},\ldots,n_{j-i},\ldots,n_{r}}^{\mathcal{S}}$

It would be interesting to look into the stardard $r$=pile NIM, since in this
case $S_{j}=\\{1,\ldots,n\\}$ for all $j=1,\ldots,r$ and the recursive
relation becomes:

$a_{n_{1},\ldots,n_{r}}^{\mathcal{S}}=1-\frac{1}{rn}\displaystyle\sum_{j=1}^{r}\sum_{i=1}^{n}a_{n_{1},\ldots,n_{j-i},\ldots,n_{r}}^{\mathcal{S}}$

### Acknowledgements

The authors would like to thank Melina Privitelli for her useful contributions
to the computational aspects of this paper.

## References

  * [1] E. R. Berlekamp, J. H. Conway and R. K. Guy. Winning Ways for Your Mathematical Plays. AK Peters (2001).
  * [2] N. A. Capitelli and M. L. Privitelli. Playing through a noisy channel (and knowing it). arXiv:2012.06768.
  * [3] T. S. Ferguson. Course in Game Theory. World Scientific (2020).
  * [4] R.A. Horn and C.R. Johnson Matrix analysis. Cambridge university press. (2012).
  * [5] R. L. Graham, D. E. Knuth and O. Patashnik. Concrete Mathematics: A foundation for computer science. Reading, MA, USA: Addison-Wesley Professional, 2nd Edition (1994).
  * [6] R. J. Nowakowski. Unsolved problems in combinatorial games. Games of no chance, 5, 125-168 (2019).
  * [7] A. J. Schwenk. Take-away games. Fibonacci Quart. 8, 225-234 (1970).
  * [8] C. K. Yap. Fundamental problems of algorithmic algebra. Oxford University Press, Vol. 49 (2000).


# On the integrality of étale extensions of polynomial rings

Lázaro Orlando Rodríguez Díaz  Instituto de Matemática, Universidade Federal
do Rio de Janeiro, RJ, Brazil<EMAIL_ADDRESS>
###### Abstract.

We use a valuation theorem for Noetherian rings, recently obtained by
Rangachev, to prove that for any non-integral étale extension $A\subset B$ of
polynomial rings over an algebraically closed field of characteristic zero,
the integral closure $\overline{A}$ of $A$ in $B$ has infinite cyclic divisor
class group, where the generator is a prime divisor equal to the complement of
$\textrm{Spec}(B)$ in $\textrm{Spec}(\overline{A})$. Moreover, this prime
divisor coincides with the ramification divisor of the finite extension
$A\subset\overline{A}$. We apply this result to carry out Wright’s geometric
approach for two-dimensional non-integral étale extensions. It follows from
our main theorem and the work of Miyanishi that $\textrm{Spec}(\overline{A})$
is a smooth affine surface. We show that $\textrm{Spec}(\overline{A})$ is an
$\mathbb{A}^{1}$-bundle over $\mathbb{P}^{1}$, more precisely a Danilov-
Gizatullin surface of index three. Based on Wright’s analysis of which of
these affine surfaces can factorize an étale morphism of the complex affine
plane and his description of its affine coordinate rings we prove that any
two-dimensional complex étale extension is integral.

##  1\. Introduction

The problem we want to address is the following: when an étale extension of
polynomial rings $A=k[{y}_{1},\,\ldots,\,{y}_{n}]\subset
B=k[{x}_{1},\,\ldots,\,{x}_{n}]$ over an algebraically closed field $k$ of
characteristic zero is integral? It is well-known that, if $B$ is integral
over $A$ then $B=A$, [16, Theorem 47], [17, Theorem 3.3]. Hence, the problem
becomes: how far is the integral closure $\overline{A}$ of $A$ in
$\mathrm{Frac}(B)$ from $B$? On the other hand, it is known that the integral
closure of $A$ in $\mathrm{Frac(B)}$ is the intersection
$\overline{A}=\cap\mathcal{V}$ of the family of valuations rings $\mathcal{V}$
of $\mathrm{Frac(B)}$ that contains $A$, [5, Theorem 19.8]. Therefore, we can
write $\overline{A}=B\cap\left(\cap\mathcal{V}\right)$, where the intersection
is taken over all valuations rings $\mathcal{V}$ of $\mathrm{Frac}(B)$ that
contain $A$ and does not contain $B$. The question can be reformulated as: how
many valuations rings are there?, can we describe them?

Recently, Rangachev [14] proved that if $A\subset B$ are integral domains, $A$
is Noetherian and $B$ is a finitely generated $A$-algebra then there are a
finite number of uniquely determined discrete valuation rings
$\mathcal{V}_{i}$ each of which is a localization of $\overline{A}$ at a
height one prime ideal, that is,
$\overline{A}=B\cap\left(\cap_{i=1}^{r}\mathcal{V}_{i}\right)$,
$\mathcal{V}_{i}=\overline{A}_{\mathfrak{p}_{i}}$,
$\mathrm{ht}(\mathfrak{p}_{i})=1$. Rangachev’s result is stated with such
generality that $\overline{A}$ may fail to be Noetherian. However, if we
assume $\overline{A}$ is Noetherian, we derive from Rangachev’s theorem and
his description of $\mathrm{Supp}_{\overline{A}}\left(B/\overline{A}\right)$
as a Zariski closed set that there is only one discrete valuation ring in the
above decomposition. More precisely, we prove the following lemma:

######  Lemma.

Let $A\subset B$ be integral domains such that $A$ is Noetherian and $B$ is a
finitely generated $A$-algebra. Denote by $\overline{A}$ the integral closure
of $A$ in $B$. Suppose $\overline{A}$ is a Noetherian ring, $\overline{A}\neq
B$ and $\mathrm{Frac}(\overline{A})=\mathrm{Frac}(B)$. Then
$\overline{A}=B\cap\overline{A}_{\mathfrak{p}}$,
$\mathrm{Ass}_{\overline{A}}\left(B/\overline{A}\right)=\\{\mathfrak{p}\\}$,
$\mathrm{ht}(\mathfrak{p})=1$, and
$\mathrm{Supp}_{\overline{A}}\left(B/\overline{A}\right)=V\left(\mathfrak{p}\right)$.

Going back to our original problem, we use this lemma to prove our main
result:

######  Theorem.

Let $k$ be an algebraically closed field of characteristic zero. Let
$A=k[{y}_{1},\,\ldots,\,{y}_{n}]\subset B=k[{x}_{1},\,\ldots,\,{x}_{n}]$ be an
étale extension of polynomial rings (each in $n$ indeterminates). Denote by
$\overline{A}$ the integral closure of $A$ in $B$. Suppose that
$\overline{A}\neq B$. Then
$\operatorname{Spec}(\overline{A})\setminus\operatorname{Spec}(B)=V\left(\mathfrak{p}\right)$,
where $\mathfrak{p}$ is a prime ideal of height one in $\overline{A}$, and the
divisor class group
$\mathrm{Cl}(\operatorname{Spec}(\overline{A}))\cong\mathbb{Z}\cong\left\langle
V\left(\mathfrak{p}\right)\right\rangle$.

In Section 3 we apply this theorem to carry out Wright’s geometric approach
[18] for two-dimensional non-integral étale extensions. Wright proved that,
given an étale extension $A=\mathbb{C}[p,q]\subset B=\mathbb{C}[x,y]$ (that
satisfies some assumptions, see Remark 3.4) such that $\overline{A}\neq B$,
then we can construct a normal affine variety $V$, containing
$\textrm{Spec}(B)$ as an open subvariety, such that $V$ admits a map to
$\mathbb{P}^{1}$ making it an $\mathbb{A}^{1}$-bundle except over one point of
$\mathbb{P}^{1}$ whose fiber is $V\setminus\textrm{Spec}(B)$ set-
theoretically. Our main theorem overcomes the obstacles encountered by Wright
to prove that $V$ is an $\mathbb{A}^{1}$-bundle over $\mathbb{P}^{1}$. The
difficulties are: $\operatorname{Spec}(\overline{A})$ could have
singularities, and even if it did not have any singularities, the fiber at
infinity may be non-reduced.

Due to Miyanishi’s work, our main result implies that
$\operatorname{Spec}(\overline{A})$ is a smooth surface, see Proposition 3.3.
In Proposition 3.5 we proved that given an étale extension
$A=\mathbb{C}[p,q]\subset B=\mathbb{C}[x,y]$ such that $\overline{A}\neq B$,
then the smooth affine surface $\operatorname{Spec}(\overline{A})$ admits the
structure of an $\mathbb{A}^{1}$-bundle over $\mathbb{P}^{1}$,
$\pi:\operatorname{Spec}(\overline{A})\to\mathbb{P}^{1}$ such that the open
embedding of $\operatorname{Spec}(B)\cong\mathbb{A}^{2}$ in
$\operatorname{Spec}(\overline{A})$ coincides with the complement of the fiber
of $\pi$ over the infinity and
$\pi|_{\operatorname{Spec}(B)}:\mathbb{A}^{2}\to\mathbb{P}^{1}\setminus\\{\infty\\}=\mathbb{A}^{1}$
is one of the standard projections. In particular,
$\operatorname{Spec}(\overline{A})$ is a Danilov-Gizatullin surface. On the
other hand, we deduce from our main theorem that the Picard group of
$\operatorname{Spec}(\overline{A})$ is generated by its canonical class, see
Corollary 2.6. Among the Danilov-Gizatullin surfaces the only one whose Picard
group is generated by the canonical class are those of index three. Then
$\operatorname{Spec}(\overline{A})$ is in fact a Danilov-Gizatullin surface of
index 3, see Proposition 3.6.

We determine the explicit form of the integral closure using Wright’s explicit
description of the coordinates ring of an $\mathbb{A}^{1}$-bundle over
$\mathbb{P}^{1}$ and Wright’s characterization of when such bundles can
factorize an étale morphism of the affine plane. More precisely, we prove that
if $A=\mathbb{C}[p,q]\subset B=\mathbb{C}[x,y]$ is an étale extension such
that $\overline{A}\neq B$, then
$\overline{A}=\mathbb{C}[y,xy,x^{2}y,x^{3}y+\alpha x]$, where
$\alpha\in\mathbb{C}$, $\alpha\neq 0$; in particular
$p,q\in\mathbb{C}[y,xy,x^{2}y,x^{3}y+\alpha x]$, see Proposition 3.10. Wright
noted that $\mathbb{C}[y,xy,x^{2}y,x^{3}y+\alpha x]$ is a graded algebra with
$\textrm{deg}\;x=-1$ and $\textrm{deg}\;y=2$. Using this observation we show
that $\mathbb{C}[y,xy,x^{2}y,x^{3}y+\alpha x]$ does not contain polynomials
that are regular in both variables, see Lemma 3.11. Nevertheless, we can
always choose a linear automorphism that makes, for example, the polynomial
$p$ regular in both variables, from there we arrive at the following result:

######  Theorem.

Every étale extension of polynomial rings
$\mathbb{C}[p,q]\subset\mathbb{C}[x,y]$ is integral.

##  2\. Main theorem

Recently, Rangachev proved a valuation theorem for Noetherian rings under
quite general hypothesis.

######  Theorem 2.1.

[14, Theorem 1.1] Let $A\subset B$ be integral domains, suppose $A$ is
Noetherian and $B$ is a finitely generated $A$-algebra. Denote by
$\overline{A}$ the integral closure of $A$ in $B$. Then one of the following
holds:

  * (i)

$\overline{A}=B$;

  * (ii)

$\mathrm{Ass}_{\overline{A}}\left(B/\overline{A}\right)=\\{(0)\\}$;

  * (iii)

$\mathrm{Ass}_{\overline{A}}\left(B/\overline{A}\right)=\\{\mathfrak{p}_{1},\ldots,\mathfrak{p}_{r}\\}$,
$\mathrm{ht}(\mathfrak{p}_{i})=1$ for all $i$, and
$\overline{A}=B\cap\left(\cap_{i=1}^{r}\mathcal{V}_{i}\right)$, where
$\mathcal{V}_{i}=\overline{A}_{\mathfrak{p}_{i}}$.

Also, under the same hypothesis, he provides a description of the support of
$B/\overline{A}$ as a closed set.

######  Proposition 2.2.

[14, Proposition 2.5] Assume we are in the situation of Theorem 2.1. Denote by
$\mathcal{I}_{B/\overline{A}}$ the intersection of all elements in
$\mathrm{Ass}_{\overline{A}}\left(B/\overline{A}\right)$. Then
$\mathrm{Supp}_{\overline{A}}\left(B/\overline{A}\right)=V\left(\mathcal{I}_{B/\overline{A}}\right)$.

The Rangachev’s theorem is stated with such generality that $\overline{A}$ may
fail to be Noetherian. However, if we assume $\overline{A}$ is Noetherian, we
derive from Rangachev’s results the following lemma:

######  Lemma 2.3.

Let $A\subset B$ be integral domains such that $A$ is Noetherian and $B$ is a
finitely generated $A$-algebra. Denote by $\overline{A}$ the integral closure
of $A$ in $B$. Suppose $\overline{A}$ is a Noetherian ring, $\overline{A}\neq
B$ and $\mathrm{Frac}(\overline{A})=\mathrm{Frac}(B)$. Then
$\mathrm{Ass}_{\overline{A}}\left(B/\overline{A}\right)=\\{\mathfrak{p}\\}$,
$\mathrm{ht}(\mathfrak{p})=1$, $\overline{A}=B\cap\overline{A}_{\mathfrak{p}}$
and
$\mathrm{Supp}_{\overline{A}}\left(B/\overline{A}\right)=V\left(\mathfrak{p}\right)$.

###### Proof.

As $\overline{A}\neq B$, this rule out the case (i) above. We always have that
$\mathrm{Ass}_{\overline{A}}\left(B/\overline{A}\right)\subset\mathrm{Supp}_{\overline{A}}\left(B/\overline{A}\right)$,
if $\mathrm{Ass}_{\overline{A}}\left(B/\overline{A}\right)=\\{(0)\\}$, then
$(0)\in\mathrm{Supp}_{\overline{A}}\left(B/\overline{A}\right)$; however, this
contradicts that $\mathrm{Frac}(\overline{A})=\mathrm{Frac(B)}$. Then by
Rangachev’s Theorem 2.1 we are left with the third possibility, that is,
$\mathrm{Ass}_{\overline{A}}\left(B/\overline{A}\right)=\\{\mathfrak{p}_{1},\ldots,\mathfrak{p}_{r}\\}$,
$\mathrm{ht}(\mathfrak{p}_{i})=1$, $\forall i$. By Rangachev’s Proposition
2.2,
$\mathrm{Supp}_{\overline{A}}\left(B/\overline{A}\right)=V\left(\mathcal{I}_{B/\overline{A}}\right)$,
where
$\mathcal{I}_{B/\overline{A}}=\mathfrak{p}_{1}\cap\ldots\cap\mathfrak{p}_{r}$.
As $\overline{A}$ is a Noetherian ring, any minimal element of
$\mathrm{Supp}_{\overline{A}}\left(B/\overline{A}\right)$ is in
$\mathrm{Ass}_{\overline{A}}\left(B/\overline{A}\right)$ by Matsumura [11,
Theorem 9], then
$\mathcal{I}_{B/\overline{A}}\in\mathrm{Ass}_{\overline{A}}\left(B/\overline{A}\right)$,
and consequently $r=1$. ∎

We are going to use this lemma to prove our main theorem:

######  Theorem 2.4.

Let $k$ be an algebraically closed field of characteristic zero. Let
$A=k[{y}_{1},\,\ldots,\,{y}_{n}]\subset B=k[{x}_{1},\,\ldots,\,{x}_{n}]$ be an
étale extension of polynomial rings (each in $n$ indeterminates). Denote by
$\overline{A}$ the integral closure of $A$ in $B$. Suppose that
$\overline{A}\neq B$. Then
$\operatorname{Spec}(\overline{A})\setminus\operatorname{Spec}(B)=V\left(\mathfrak{p}\right)$,
where $\mathfrak{p}$ is a prime ideal of height one in $\overline{A}$, and the
divisor class group
$\mathrm{Cl}(\operatorname{Spec}(\overline{A}))\cong\mathbb{Z}\cong\left\langle
V\left(\mathfrak{p}\right)\right\rangle$.

###### Proof.

The extension $A\subset B$ is étale, so it is quasi-finite and by Zariski’s
Main Theorem [15, Corollaire 2, p. 42], the morphism
$\operatorname{Spec}(B)\to\operatorname{Spec}(\overline{A})$ is an open
immersion. Also, the étaleness of the extension implies that
$\mathrm{Frac(A)}\subset\mathrm{Frac(B)}$ is a finite separable algebraic
extension of fields, then $\overline{A}$ is a finite module over $A$, in
particular $\overline{A}$ is Noetherian. It is clear that
$\mathrm{Frac}(\overline{A})=\mathrm{Frac}(B)$.

Then the étale extension $A\subset B$ satisfies the hypothesis of our Lemma
2.3, therefore, there exist a prime ideal $\mathfrak{p}$ of height one in
$\overline{A}$ such that $\overline{A}=B\cap\overline{A}_{\mathfrak{p}}$,
$\mathrm{Ass}_{\overline{A}}\left(B/\overline{A}\right)=\\{\mathfrak{p}\\}$,
and
$\mathrm{Supp}_{\overline{A}}\left(B/\overline{A}\right)=V\left(\mathfrak{p}\right)$.
From this we can derive the following consequences:

  * -

$\operatorname{Spec}(\overline{A})\setminus\operatorname{Spec}(B)=V\left(\mathfrak{p}\right)$,
where we identify $\operatorname{Spec}(B)$ with its image in
$\operatorname{Spec}(\overline{A})$. In fact, denote by
$g:\operatorname{Spec}(B)\hookrightarrow\operatorname{Spec}(\overline{A})$ the
open immersion. We are going to show that
$\operatorname{Im}g=\operatorname{Spec}(\overline{A})\setminus\mathrm{Supp}_{\overline{A}}\left(B/\overline{A}\right)$.
Suppose that
$\mathfrak{p}\in\operatorname{Im}g\subset\operatorname{Spec}(\overline{A})$,
that is, there exist $\mathfrak{q}\in\operatorname{Spec}(B)$ such that
$\mathfrak{p}=\mathfrak{q}\cap\overline{A}$. As $g$ is an open immersion we
have that $\overline{A}_{\mathfrak{p}}=B_{\mathfrak{q}}=B_{\mathfrak{p}}$, see
for example [4, Exercise 2.3.9], therefore
$\mathfrak{p}\notin\mathrm{Supp}_{\overline{A}}\left(B/\overline{A}\right)$.
Conversely, suppose that $\mathfrak{p}\in\operatorname{Spec}(\overline{A})$
and
$\mathfrak{p}\notin\mathrm{Supp}_{\overline{A}}\left(B/\overline{A}\right)$,
that is, $\overline{A}_{\mathfrak{p}}=B_{\mathfrak{p}}$. Then $B\subset
B_{\mathfrak{p}}=\overline{A}_{\mathfrak{p}}$, define the prime ideal
$\mathfrak{r}:=\mathfrak{p}\overline{A}_{\mathfrak{p}}\cap B$, then
$\mathfrak{r}\in\operatorname{Spec}(B)$ and
$\mathfrak{r}\cap\overline{A}=\mathfrak{p}$, therefore
$\mathfrak{p}\in\operatorname{Im}g$. The assertion now follows because we
already know that
$\mathrm{Supp}_{\overline{A}}\left(B/\overline{A}\right)=V\left(\mathfrak{p}\right)$.

  * -

$\mathrm{Cl}(\operatorname{Spec}(\overline{A}))\cong\mathbb{Z}\cong\left\langle
V\left(\mathfrak{p}\right)\right\rangle$. In fact,
$\operatorname{Spec}(\overline{A})$ is a noetherian normal affine scheme,
$V(\mathfrak{p})$ is an irreducible closed subset of codimension one, and
$\operatorname{Spec}(\overline{A})\setminus\operatorname{Spec}(B)=V\left(\mathfrak{p}\right)$,
then by Hartshorne [8, Chap. II, Prop.6.5] there is an exact sequence
$\mathbb{Z}\to\mathrm{Cl}(\operatorname{Spec}(\overline{A}))\to\mathrm{Cl}(\operatorname{Spec}(B))\to
0$, where the first map is defined by $1\to 1\cdot V(\mathfrak{p})$. Note that
$\mathrm{Cl}(\operatorname{Spec}(B))=0$ because $B$ is a unique factorization
domain, then $\mathrm{Cl}(\operatorname{Spec}(\overline{A}))$ is generated by
$V\left(\mathfrak{p}\right)$. On the other hand, $V\left(\mathfrak{p}\right)$
cannot be of finite order in $\mathrm{Cl}(\operatorname{Spec}(\overline{A}))$,
because it is known [1, Proposition 6] that if
$\mathrm{Cl}(\operatorname{Spec}(\overline{A}))$ were a torsion group (i.e.,
$\overline{A}$ were almost factorial) then $A=B$, but we are assuming that
$\overline{A}\neq B$. We have proved the assertion.

∎

Suppose we are in the situation of Theorem 2.4. Then the étale morphism
$\mathbb{A}^{n}=\operatorname{Spec}(B)\xrightarrow{f}\mathbb{A}^{n}=\operatorname{Spec}(A)$
induced by the étale extension $A\subset B$ factorizes as $f=hg$,
$\mathbb{A}^{n}=\operatorname{Spec}(B)\xhookrightarrow{g}\operatorname{Spec}(\overline{A})\xrightarrow{h}\mathbb{A}^{n}=\operatorname{Spec}(A)$,
where $g$ is an open immersion and $h$ is finite. We denote by $R_{h}$ the
ramification divisor of the morphism $h$, since $f$ is étale, we have that
$R_{h}\subset\operatorname{Spec}(\overline{A})\setminus\operatorname{Spec}(B)=V\left(\mathfrak{p}\right)$.
By the purity of the branch locus it follows that $R_{h}=V(\mathfrak{p})$.
Then we have proved:

######  Corollary 2.5.

Assume we are in the situation of Theorem 2.4. Then the ramification divisor
$R_{h}\subset\operatorname{Spec}(\overline{A})$ of the finite morphism
$\operatorname{Spec}(\overline{A})\xrightarrow{h}\mathbb{A}^{n}=\operatorname{Spec}(A)$
coincides with the prime divisor
$V\left(\mathfrak{p}\right)=\operatorname{Spec}(\overline{A})\setminus\operatorname{Spec}(B)$.

######  Corollary 2.6.

Assume we are in the situation of Theorem 2.4. Suppose
$\operatorname{Spec}(\overline{A})$ is smooth and denote by
$K_{\operatorname{Spec}(\overline{A})}$ its canonical divisor. Then
$K_{\operatorname{Spec}(\overline{A})}=V\left(\mathfrak{p}\right)$, that is,
the Picard group of $\operatorname{Spec}(\overline{A})$ is generated by the
canonical class.

###### Proof.

We have that
$K_{\operatorname{Spec}(\overline{A})}=h^{*}(K_{\mathbb{A}^{2}})+R_{h}$ by the
ramification formula [9, Theorem 5.5] applied to the dominant morphism
$\operatorname{Spec}(\overline{A})\xrightarrow{h}\mathbb{A}^{2}=\operatorname{Spec}(A)$.
Then $K_{\operatorname{Spec}(\overline{A})}=R_{h}$. The conclusion now follows
from Corollary 2.5. ∎

##  3\. Two-dimensional étale extensions

In this section $k=\mathbb{C}$, the field of complex numbers. We are going to
apply Theorem 2.4 to carry out Wright’s geometric approach for two-dimensional
non-integral étale extensions [18]. Wright proved that if we start with an
étale extension $A=\mathbb{C}[p,q]\subset B=\mathbb{C}[x,y]$ that is not
integral, it is possible to construct a normal affine variety that admits the
structure of an $\mathbb{A}^{1}$-fibration over $\mathbb{P}^{1}$, more
precisely:

######  Theorem 3.1.

[18, Theorem 4.3] Let $A=\mathbb{C}[p,q]\subset B=\mathbb{C}[x,y]$ be an étale
extension such that $\overline{A}\neq B$, then there exist a normal affine
variety $V$ containing $U=\mathbb{A}^{2}$ as an open subvariety having the
following properties:

  * i)

$F=V\setminus U$ is a rational curve whose normalization is $\mathbb{A}^{1}$
and each singular point of $F$ has one-point desingularization.

  * ii)

there is a map $\pi:V\to\mathbb{P}^{1}$ such that $F$ is the set-theoretic
fiber of a point $z\in\mathbb{P}^{1}$, and the restriction map
$\pi|_{U}:U\to\mathbb{P}^{1}\setminus\\{z\\}=\mathbb{A}^{1}$ is the projection
onto a coordinate line.

  * iii)

there is a map $f:V\to\mathbb{A}^{2}$ such that $f|_{U}$ is étale.

Moreover, if we assume that $\overline{A}$ is smooth, then $V$ can be chosen
to be smooth and $F\cong\mathbb{A}^{1}$. Even more, [18, Remark p. 605] if $F$
has multiplicity one in the fiber, that is, $\pi^{-1}(z)=\mathbb{A}^{1}$
scheme-theoretically, then $V$ is an $\mathbb{A}^{1}$-bundle over
$\mathbb{P}^{1}$ via the map $\pi:V\to\mathbb{P}^{1}$.

In order to get an $\mathbb{A}^{1}$-bundle over $\mathbb{P}^{1}$ we first need
to prove that the integral closure is smooth. We are going to show that this
is always the case. To achieve this we apply the following result of Gurjar
and Miyanishi. In fact, the specific case we need from this result, that is,
when $X=\mathbb{A}^{2}$, was proved by Miyanishi in [13, Chap. I, Sect. 6].

######  Theorem 3.2.

[7, Theorem 1] Let $\varphi:X_{u}\to X_{l}$ be an étale morphism, where
$X_{u}=X$ and $X_{l}=X$ are two copies of an irreducible normal affine surface
$X$ defined over $\mathbb{C}$. Let $\widetilde{X}$ be the normalization of
$X_{l}$ in the function field of $X_{u}$. Then $\widetilde{X}\setminus X_{u}$
is a disjoint union of irreducible curves $C_{i}$, each of which is isomorphic
to $\mathbb{A}^{1}$. Further, each singularity of $\widetilde{X}$ which is not
contained in $X_{u}$ is a cyclic quotient singularity. Any irreducible
component of $\widetilde{X}\setminus X_{u}$ contains at most one such singular
point.

Consider now an étale extension $A=\mathbb{C}[p,q]\subset B=\mathbb{C}[x,y]$,
or equivalently, the corresponding étale morphism
$f=(p,q):\mathbb{A}^{2}=\operatorname{Spec}(B)\to\mathbb{A}^{2}=\operatorname{Spec}(A)$.
Suppose that $\overline{A}\neq B$, by our main Theorem 2.4 we have that
$\operatorname{Spec}(\overline{A})\setminus\operatorname{Spec}(B)=V\left(\mathfrak{p}\right)$
is reduced and irreducible. Applying the Gurjar-Miyanishi’s Theorem 3.2 to the
morphism
$f=(p,q):\mathbb{A}^{2}=\operatorname{Spec}(B)\to\mathbb{A}^{2}=\operatorname{Spec}(A)$
implies that
$\operatorname{Spec}(\overline{A})\setminus\operatorname{Spec}(B)\cong\mathbb{A}^{1}$
scheme-theoretically, and consequently $\operatorname{Spec}(\overline{A})$ has
no singularities at all. We have proved the following:

######  Proposition 3.3.

Let $A=\mathbb{C}[p,q]\subset B=\mathbb{C}[x,y]$ be an étale extension of
polynomial rings. Denote by $\overline{A}$ the integral closure of $A$ in $B$.
Then $\overline{A}$ is a smooth $\mathbb{C}$-algebra.

It was previously unknown whether the integral closure $\overline{A}$ was
smooth. Even in the case of smooth integral closure, it was unknown whether
the special fiber would be reduced, see [18, Remark pag. 605]. This is another
contribution of our main Theorem 2.4,
$\operatorname{Spec}(\overline{A})\setminus\operatorname{Spec}(B)$ is reduced
and irreducible. It follows now from Wright’s construction that the variety
$V$ in Wright’s Theorem 3.1 is exactly $\operatorname{Spec}(\overline{A})$,
and it admits a map to $\mathbb{P}^{1}$ making it an $\mathbb{A}^{1}$-bundle
except over one point of $\mathbb{P}^{1}$ whose fiber is exactly
$F=\operatorname{Spec}(\overline{A})\setminus\operatorname{Spec}(B)\cong\mathbb{A}^{1}$,
now scheme-theoretically. Therefore, we have proved that
$\operatorname{Spec}(\overline{A})$ has the structure of an
$\mathbb{A}^{1}$-bundle over $\mathbb{P}^{1}$, such that the open embedding of
$\operatorname{Spec}(B)\cong\mathbb{A}^{2}$ in
$\operatorname{Spec}(\overline{A})$ coincides with the complement of the fiber
of $\pi$ over the infinity and
$\pi|_{\operatorname{Spec}(B)}:\mathbb{A}^{2}\to\mathbb{P}^{1}\setminus\\{\infty\\}=\mathbb{A}^{1}$
is one of the standard projections. Nevertheless, we must make the following
observation:

######  Remark 3.4.

In his Theorem 3.1, Wright assumed that the curves $p=0$ and $q=0$ each have
two points at infinity in $\mathbb{P}^{2}$ that coincide with the points at
infinity on the lines $x=0$ and $y=0$. He need this in his proof to
accommodate the variety $V$ within the blowups performed in the resolution of
the birational map associated to the étale extension. It is always possible to
make such an assumption at the cost of modifying the original étale extension.
Its means that if we start with an arbitrary non-integral étale extension
$A\subset B$, then we probably need to change to a different non-integral
étale extension $A^{\prime}\subset B^{\prime}$ by means of an automorphism, if
we want to construct $V$ from $\operatorname{Spec}(\overline{A^{\prime}})$ as
Wright does.

Let’s prove, using our main Theorem 2.4, that we don’t need to make such
assumptions. Proving it this way will be useful later to explicitly describe
the integral closure and to prove Theorem 3.12.

######  Proposition 3.5.

Let $A=\mathbb{C}[p,q]\subset B=\mathbb{C}[x,y]$ be an étale extension such
that $\overline{A}\neq B$, then $\operatorname{Spec}(\overline{A})$ is a
smooth affine surface that admits the structure of an $\mathbb{A}^{1}$-bundle
over $\mathbb{P}^{1}$,
$\pi:\operatorname{Spec}(\overline{A})\to\mathbb{P}^{1}$ such that the open
embedding of $\operatorname{Spec}(B)\cong\mathbb{A}^{2}$ in
$\operatorname{Spec}(\overline{A})$ coincides with the complement of the fiber
of $\pi$ over the infinity and
$\pi|_{\operatorname{Spec}(B)}:\mathbb{A}^{2}\to\mathbb{P}^{1}\setminus\\{\infty\\}=\mathbb{A}^{1}$
is one of the standard projections.

###### Proof.

By Theorem 2.4 and Proposition 3.3 we have that
$\operatorname{Spec}(\overline{A})$ is a smooth affine surface such that
$\operatorname{Spec}(\overline{A})\setminus\operatorname{Spec}(B)\cong\mathbb{A}^{1}$
scheme-theoretically, lets denote this affine line by
$F:=\operatorname{Spec}(\overline{A})\setminus\operatorname{Spec}(B)$. On the
other hand, $\operatorname{Spec}(B)\cong\mathbb{A}^{2}$ admits two
$\mathbb{A}^{1}$-fibrations over $\mathbb{A}^{1}$ given by the standard
projections $\pi_{i}:\operatorname{Spec}(B)\to\mathbb{A}^{1}$, $i=1,2$,
corresponding to the containments $\mathbb{C}[x]\subset\mathbb{C}[x,y]$ and
$\mathbb{C}[y]\subset\mathbb{C}[x,y]$ respectively. It is known [13, Chap. I,
Lemma 4.3 ] that each $\pi_{i}$ extends to an $\mathbb{A}^{1}$-fibration
$\widetilde{\pi}_{i}:\operatorname{Spec}(\overline{A})\to C$ where
$C\cong\mathbb{A}^{1}$ or $C\cong\mathbb{P}^{1}$.

We are going to argue as in [6, Lemma 5.2] to show that at least one of the
$\pi_{i}$’s extends to an $\mathbb{A}^{1}$-fibration over $\mathbb{P}^{1}$.
Suppose that both fibrations $\pi_{1}$ and $\pi_{2}$ extends to
$\mathbb{A}^{1}$-fibrations over $\mathbb{A}^{1}$, that is,
$\widetilde{\pi}_{1}:\operatorname{Spec}(\overline{A})\to\operatorname{Spec}(\mathbb{C}[x])$
and
$\widetilde{\pi}_{2}:\operatorname{Spec}(\overline{A})\to\operatorname{Spec}(\mathbb{C}[y])$.
Then $F$ is contained in some fiber of $\widetilde{\pi}_{1}$ and in some fiber
of $\widetilde{\pi}_{2}$. This means that $x$ and $y$ are constant along $F$.
Therefore, any function
$f\in\Gamma(\operatorname{Spec}(\overline{A}),\mathcal{O}_{\operatorname{Spec}(\overline{A})})\subset\Gamma(\operatorname{Spec}(B),\mathcal{O}_{\operatorname{Spec}(B)})=\mathbb{C}[x,y]$
is also constant along $F$. This leads to a contradiction, because given any
two points of $F$ there is a regular function of
$\operatorname{Spec}(\overline{A})$ that separates them.

Then at least one of the $\pi_{i}$’s extends to an $\mathbb{A}^{1}$-fibration
over $\mathbb{P}^{1}$. Without loss of generality, suppose that it is
$\pi_{1}$ that extends to an $\mathbb{A}^{1}$-fibration
$\widetilde{\pi}_{1}:\operatorname{Spec}(\overline{A})\to\mathbb{P}^{1}$. Then
$\widetilde{\pi}_{1}^{-1}(\infty)=\operatorname{Spec}(\overline{A})\setminus\operatorname{Spec}(B)\cong\mathbb{A}^{1}$,
scheme-theoretically. Therefore, $\widetilde{\pi}_{1}$ is an
$\mathbb{A}^{1}$-bundle over $\mathbb{P}^{1}$ that satisfies the desired
properties. ∎

It is known that every $\mathbb{A}^{1}$-bundle $\pi:V\to\mathbb{P}^{1}$, where
$V$ is a affine nonsingular surface, can be realized as the complement
$V=\mathbb{F}_{n}\setminus S$ of an ample section $S$, where
$\mathbb{F}_{n}=\mathbb{P}\left(\mathcal{O}_{\mathbb{P}^{1}}\oplus\mathcal{O}_{\mathbb{P}^{1}}(-n)\right)$,
$n\geq 0$, is a Hirzebruch surface, and the canonical projection
$\tilde{\pi}_{n}:\mathbb{F}_{n}\to\mathbb{P}^{1}$ extends $\pi$. In this case,
$S=C_{0}+\frac{1}{2}(S^{2}+n)F$, where $C_{0}$ is a section of
$\tilde{\pi}_{n}$ with self-intersection $C_{0}^{2}=-n$, $F$ is a fiber of
$\tilde{\pi}_{n}$, and $S^{2}\geq n+2$. Moreover, $n$ and $S^{2}$ are uniquely
determined by $V$ and $\pi$, see [3], [12, Lemma 5.5.1], [18, Theorem 2.3].

Danilov and Gizatullin proved that the isomorphism class of the complement
$V=\mathbb{F}_{n}\setminus S$ of an ample section $S$ in a Hirzebruch surface
$\mathbb{F}_{n}$ depends only on the self-intersection number $S^{2}$ and
neither on $n$ nor on the choice of the section $S$, see [3, Theorem 5.8.1],
[2, Corollary 4.8]. These affine surfaces $V$ are called Danilov-Gizatullin
surfaces of index $S^{2}$.

The Picard group of the Hirzebruch surface $\mathbb{F}_{n}$ is
$\operatorname{Pic}(\mathbb{F}_{n})\cong\mathbb{Z}^{2}$, freely generated by
the class of $C_{0}$ and the class of a fiber $F$ of $\tilde{\pi}_{n}$, and
$C_{0}^{2}=-n$, $F^{2}=0$ and $C_{0}F=1$. The canonical divisor is given by
$K_{\mathbb{F}_{n}}=-2C_{0}-(n+2)F=-2S+(S^{2}-2)F$. Therefore, the Picard
group of the Danilov-Gizatullin surface $\mathbb{F}_{n}\setminus S$ is freely
generated by the class of $F|_{\mathbb{F}_{n}\setminus S}$,
$\operatorname{Pic}(\mathbb{F}_{n}\setminus S)\cong\mathbb{Z}$, and the
canonical divisor is given by $K_{\mathbb{F}_{n}\setminus
S}=(S^{2}-2)F|_{\mathbb{F}_{n}\setminus S}$.

######  Proposition 3.6.

Let $A=\mathbb{C}[p,q]\subset B=\mathbb{C}[x,y]$ be an étale extension such
that $\overline{A}\neq B$, then $\operatorname{Spec}(\overline{A})$ is a
Danilov-Gizatullin surface of index 3.

###### Proof.

It follows from Proposition 3.5 that $\operatorname{Spec}(\overline{A})$ is a
Danilov-Gizatullin surface
$\operatorname{Spec}(\overline{A})\cong\mathbb{F}_{n}\setminus S$. On the
other hand, we know by Corollary 2.6 that the Picard group of
$\operatorname{Spec}(\overline{A})$ is generated by the canonical divisor
$K_{\operatorname{Spec}(\overline{A})}$. It follows that
$(S^{2}-2)F|_{\mathbb{F}_{n}\setminus S}=F|_{\mathbb{F}_{n}\setminus S}$, this
forces $S^{2}=3$. ∎

Now, the problem is whether a Danilov-Gizatullin surface (equivalently, an
$\mathbb{A}^{1}$-bundle over $\mathbb{P}^{1}$) can factorizes an étale
morphism of the complex affine plane. This is precisely the situation Wright
conjectured should not happen. _Geometric Formulation of Wright’s conjecture_
: let $V$ be an affine variety which is an $\mathbb{A}^{1}$-bundle over
$\mathbb{P}^{1}$, $U=V\setminus F$, where $F$ is a fiber in $V$. There does
not exist $f:V\to\mathbb{A}^{2}$ such that $f|_{U}$ is étale, [18, Conjecture
3.2].

To address his conjecture, Wright provides an explicit description of the
coordinate ring of an affine $\mathbb{A}^{1}$-bundle over $\mathbb{P}^{1}$,
that is, he describes $\Gamma(V)$ as a subring of $\mathbb{C}[x,y]$,
corresponding to the containment of $U\cong\mathbb{A}^{2}$ in $V$. Let’s
recall his construction.

######  Theorem 3.7.

[18, Theorem 3.1] Let $V$ be an affine variety which is an
$\mathbb{A}^{1}$-bundle over $\mathbb{P}^{1}$ with structure map
$\pi:V\to\mathbb{P}^{1}$. Pick $x$ such that the function field of
$\mathbb{P}^{1}$ is $\mathbb{C}(x)$, and let
$U_{0}=\pi^{-1}\left(\operatorname{Spec}\mathbb{C}[x]\right)$,
$U_{1}=\pi^{-1}\left(\operatorname{Spec}\mathbb{C}[x^{-1}]\right)$. Then
$V=U_{0}\cup U_{1}$ and there exist $y\in\Gamma(V)$ such that
$U_{0}=\operatorname{Spec}\mathbb{C}[x,y]$,
$U_{1}=\operatorname{Spec}\mathbb{C}[x^{\prime},y^{\prime}]$ where

$\displaystyle x^{\prime}=x^{-1},\quad
y^{\prime}=x^{m}y+\alpha_{1}x^{m-1}+\alpha_{2}x^{m-2}+\cdots+\alpha_{m-1}x,$

where $m\geq 2$ and $\alpha_{1},\ldots,\alpha_{m-1}\in\mathbb{C}$ not all
zero. Moreover

$\displaystyle\Gamma(V)=\mathbb{C}[t_{0},t_{1},\ldots,t_{m}],$

where

$\displaystyle t_{0}$ $\displaystyle=y,$ (3.1) $\displaystyle t_{1}$
$\displaystyle=xy,$ $\displaystyle t_{2}$ $\displaystyle=x^{2}y+\alpha_{1}x,$
$\displaystyle t_{3}$ $\displaystyle=x^{3}y+\alpha_{1}x^{2}+\alpha_{2}x,$
$\displaystyle\ldots$ $\displaystyle t_{m}$
$\displaystyle=x^{m}y+\alpha_{1}x^{m-1}+\alpha_{2}x^{m-2}+\cdots+\alpha_{m-1}x\;(=y^{\prime}).$

Conversely, every ring of the form $\mathbb{C}[t_{0},t_{1},\ldots,t_{m}]$
above can be made into the coordinate ring of an $\mathbb{A}^{1}$-bundle over
$\mathbb{P}^{1}$.

In view of this description, Wright’s conjecture can be rephrased as follows.
_Algebraic formulation of Wright’s conjecture_ : there does not exist a pair
of polynomials
$p,q\in\mathbb{C}[t_{0},t_{1},\ldots,t_{m}]\subset\mathbb{C}[x,y]$, where
$t_{0},t_{1},\ldots,t_{m}$ are as in (3.1)
($\alpha_{1},\ldots,\alpha_{m-1}\in\mathbb{C}$ not all zero), with
$\frac{\partial(p,q)}{\partial(x,y)}$ non-vanishing (i.e., constant) on
$\mathbb{A}^{2}$, [18, Conjecture 3.2].

######  Remark 3.8.

The number of generators of the algebra $\Gamma(V)$, that is, the integer
$m+1$, is completely determined by the $\mathbb{A}^{1}$-bundle
$\pi:V\to\mathbb{P}^{1}$. In fact, let $V=\mathbb{F}_{n}\setminus S$ be the
corresponding Danilov-Gizatullin surface, then $m=S^{2}$, see [18, p. 599,
above the Conjecture 3.2] and [2, Proposition 3.5].

Wright succeeded in proving the following case of his conjecture.

######  Theorem 3.9.

[18, Theorem 3.3] Wright’s conjecture is true in the case where the
coefficient $\alpha_{1}$ is non-zero.

Putting together the previous results we obtain the following explicit
description of the integral closure of a non-integral étale extension:

######  Proposition 3.10.

Let $A=\mathbb{C}[p,q]\subset B=\mathbb{C}[x,y]$ be an étale extension such
that $\overline{A}\neq B$, then
$\overline{A}=\mathbb{C}[y,xy,x^{2}y,x^{3}y+\alpha x]$, where
$\alpha\in\mathbb{C}$, $\alpha\neq 0$. In particular
$p,q\in\mathbb{C}[y,xy,x^{2}y,x^{3}y+\alpha x]$.

###### Proof.

By Proposition 3.5 we have that $V=\operatorname{Spec}(\overline{A})$ admits
the structure of an $\mathbb{A}^{1}$-bundle over $\mathbb{P}^{1}$ such that
the open embedding of $\operatorname{Spec}(B)\cong\mathbb{A}^{2}$ in
$\operatorname{Spec}(\overline{A})$ coincides with the complement of the fiber
over the infinity and
$\pi|_{\operatorname{Spec}(B)}:\mathbb{A}^{2}\to\mathbb{P}^{1}\setminus\\{\infty\\}=\mathbb{A}^{1}$
is one of the standard projections. Moreover, in Proposition 3.6 we proved
that $V$ is a Danilov-Gizatullin surface of index 3, therefore, $m=3$ by
Remark 3.8. It now follows from Wright’s Theorem 3.7 that
$p,q\in\Gamma(V)=\mathbb{C}[y,xy,x^{2}y+\alpha_{1}x,x^{3}y+\alpha_{1}x^{2}+\alpha_{2}x]$,
where $\alpha_{1},\alpha_{2}\in\mathbb{C}$ cannot be both zero. By Wright’s
Theorem 3.9 we have that $\alpha_{1}=0$. ∎

At this point it would only remain to prove the following case of Wright’s
conjecture: there does not exist a pair of polynomials
$p,q\in\mathbb{C}[y,xy,x^{2}y,x^{3}y+\alpha x]$, $\alpha\in\mathbb{C}$,
$\alpha\neq 0$ such that $\frac{\partial(p,q)}{\partial(x,y)}$ is constant.
This was precisely the simplest unknown case of Wright’s conjecture [18, pag.
601]. We are going to use Proposition 3.10 and Wright’s observation that
$\mathbb{C}[y,xy,x^{2}y,x^{3}y+\alpha x]$ is a graded algebra with
$\textrm{deg}\;x=-1$ and $\textrm{deg}\;y=2$ to prove that in the end this
case is also true.

With respect to this grading the generators of
$\mathbb{C}[y,xy,x^{2}y,x^{3}y+\alpha x]$ have degrees: $\textrm{deg}\;y=2$,
$\textrm{deg}\;xy=1,\textrm{deg}\;x^{2}y=0$ and $\textrm{deg}\;(x^{3}y+\alpha
x)=-1$. Let $f(x,y)\in\mathbb{C}[y,xy,x^{2}y,x^{3}y+\alpha x]$ be a
homogeneous polynomial of weighted degree $-m<0$. We can write:

$\displaystyle f(x,y)$
$\displaystyle=\sum_{2l_{1}+l_{2}-l_{4}=-m}f_{l_{1}l_{2}l_{3}l_{4}}y^{l_{1}}(xy)^{l_{2}}(x^{2}y)^{l_{3}}(x^{3}y+\alpha
x)^{l_{4}},\quad\quad l_{i}\in\mathbb{Z}_{\geq 0},\quad
f_{l_{1}l_{2}l_{3}l_{4}}\in\mathbb{C},$ $\displaystyle f(x,y)$
$\displaystyle=\sum_{2l_{1}+l_{2}-l_{4}=-m}f_{l_{1}l_{2}l_{3}l_{4}}y^{l_{1}}(xy)^{l_{2}}(x^{2}y)^{l_{3}}(x^{3}y+\alpha
x)^{2l_{1}+l_{2}+m},$ $\displaystyle f(x,y)$ $\displaystyle=(x^{3}y+\alpha
x)^{m}\sum_{2l_{1}+l_{2}-l_{4}=-m}f_{l_{1}l_{2}l_{3}l_{4}}(x^{6}y^{3}+2\alpha
x^{4}y^{2}+\alpha^{2}x^{2}y)^{l_{1}}(x^{4}y^{2}+\alpha
x^{2}y)^{l_{2}}(x^{2}y)^{l_{3}},$

then $f(x,y)=(x^{3}y+\alpha x)^{m}g(z)$, where $g(z)\in\mathbb{C}[z]$ and
$z=x^{2}y$.

For the next definition, let’s consider the usual grading, that is,
$\textrm{deg}\;x=1$ and $\textrm{deg}\;y=1$. We say a polynomial
$p(x,y)\in\mathbb{C}[x,y]$ of total degree $n$ is regular in $x$ if it
contains a term in $x^{n}$. It is well known that for any polynomial $p(x,y)$
it is always possible to find an invertible linear transformation $x=av+bw$,
$y=cv+dw$, ($a,b,c,d\in\mathbb{C}$, $ad-cd\neq 0$) such that
$r(v,w)=p(av+bw,cv+dw)$ is regular in $v$ and $w$, see [10, (2.7), p. 5].

######  Lemma 3.11.

The algebra $\mathbb{C}[y,xy,x^{2}y,x^{3}y+\alpha x]$, $\alpha\in\mathbb{C}$,
$\alpha\neq 0$ does not contain polynomials that are regular in both
variables.

###### Proof.

Suppose $p(x,y)\in\mathbb{C}[y,xy,x^{2}y,x^{3}y+\alpha x]$ is a polynomial of
total (usual) degree $n$ that is regular in both variables. Then in particular
$p(x,y)$ contains a term in $x^{n}$ and consequently $p(x,y)$ has a non-zero
homogeneous component of weighted degree $-n<0$. On the other hand, we already
observe that all homogeneous polynomials of weighted degree $-n<0$ in
$\mathbb{C}[y,xy,x^{2}y,x^{3}y+\alpha x]$ are of the form
$f(x,y)=(x^{3}y+\alpha x)^{n}g(z)$, where $g(z)\in\mathbb{C}[z]$ and
$z=x^{2}y$. This is a contradiction because $p(x,y)$ is a polynomial of total
(usual) degree $n$. ∎

######  Theorem 3.12.

Every étale extension of polynomial rings
$\mathbb{C}[p,q]\subset\mathbb{C}[x,y]$ is integral.

###### Proof.

Suppose that there exist an étale extension $A=\mathbb{C}[p,q]\subset
B=\mathbb{C}[x,y]$ that is not integral, that is, $\overline{A}\neq B$
(equivalently the corresponding étale morphism
$(p,q):\mathbb{A}^{2}=\operatorname{Spec}(B)\to\mathbb{A}^{2}=\operatorname{Spec}(A)$
is not injective, [16, Theorem 46]). Choose an invertible linear substitution
$x=av+bw$, $y=cv+dw$, ($a,b,c,d\in\mathbb{C}$, $ad-cd\neq 0$) such that
$r(v,w)=p(av+bw,cv+dw)$ is regular in both variables $v$ and $w$. After
composing the morphism $(p,q)$ with this linear automorphism we obtain another
étale extension $\mathbb{C}[r,s]\subset\mathbb{C}[v,w]$ that is also non-
integral (equivalently an étale morphism
$(r,s):\mathbb{A}^{2}=\operatorname{Spec}(\mathbb{C}[v,w])\to\mathbb{A}^{2}=\operatorname{Spec}(\mathbb{C}[r,s])$
that is not injective), where $r(v,w)=p(av+bw,cv+dw)$ and
$s(v,w)=q(av+bw,cv+dw)$. Applying Proposition 3.10 to the non-integral étale
extension $\mathbb{C}[r,s]\subset\mathbb{C}[v,w]$ we have that
$r,s\in\mathbb{C}[w,vw,v^{2}w,v^{3}w+\beta v]$, where $\beta\in\mathbb{C}$,
$\beta\neq 0$. By construction the polynomial $r$ is regular in $v$ and $w$,
however, by Lemma 3.11 the algebra $\mathbb{C}[w,vw,v^{2}w,v^{3}w+\beta v]$
does not contain polynomials regular in both variables, a contradiction. Hence
$\overline{A}=B$. ∎

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