# Subresultants and the Shape Lemma

David A. Cox  Department of Mathematics & Statistics, 31 Quadrangle Drive,
Amherst College, Amherst, MA 01002, USA<EMAIL_ADDRESS>https://dacox.people.amherst.edu/ and  Carlos D’Andrea  Departament de
Matemàtiques i Informàtica, Universitat de Barcelona. Gran Via 585, 08007
Barcelona, Spain & Centre de Recerca Matemàtica, Edifici C, Campus Bellaterra,
08193 Bellaterra, Spain<EMAIL_ADDRESS>http://www.ub.edu/arcades/cdandrea.html

###### Abstract.

In nice cases, a zero-dimensional complete intersection ideal over a field has
a Shape Lemma. There are also cases where the ideal is generated by the
resultant and first subresultant polynomials of the generators. This paper
explores the relation between these representations and studies when the
resultant generates the elimination ideal. We also prove a Poisson formula for
resultants arising from the hidden variable method.

###### Key words and phrases:

Shape Lemma, resultant, subresultant, Poisson formula

###### 2010 Mathematics Subject Classification:

Primary 13P10; Secondary 13P15

††This paper is dedicated to Teresa Krick on the occasion of the TeresaFest
2021 conference in her honor.

##  1\. Introduction

This paper will study the relation between the Shape Lemma and subresultants.
We begin with an example that introduces some of the key players:

######  Example 1.1.

In the polynomial ring $\mathbb{C}[x_{1},x_{2}]$, let

$f_{1}=x_{1}^{2}-x_{2}-1,\ f_{2}=x_{1}^{2}+x_{1}x_{2}-2.$

Computing a lex Gröbner basis of $I=\langle f_{1},f_{2}\rangle$ with
$x_{2}\prec x_{1}$ gives

$\displaystyle I=\langle r(x_{2}),\,$ $\displaystyle
x_{1}-g_{1}(x_{2})\rangle,\quad r(x_{2})=x_{2}^{3}+2x_{2}-1,\
g_{1}(x_{2})=x_{2}^{2}+1.$

The structure of this basis makes it easy to find the solutions and is
encapsulated by saying that $I$ _has a Shape Lemma_. Notice also that
$I\cap\mathbb{C}[x_{2}]=\langle r(x_{2})\rangle$.

A more classical approach to finding nice generators of $I$ uses resultants
and subresultant polynomials. Regarding $f_{1},f_{2}$ as polynomials in
$x_{1}$ with coefficients in $A=\mathbb{C}[x_{2}]$, the zeroth subresultant
polynomial is the resultant, and the first subresultant polynomial is linear
in $x_{1}$. These polynomials will be denoted $R_{\text{\it f}\hskip
0.9pt}(x_{2})$ and $p_{1}(x_{1},x_{2})$ respectively in what follows. By
standard determinantal formulas,

$\displaystyle R_{\text{\it f}\hskip 0.9pt}(x_{2})$
$\displaystyle=-x_{2}^{3}-2x_{2}+1$ $\displaystyle p_{1}(x_{1},x_{2})$
$\displaystyle=x_{2}\cdot x_{1}+x_{2}-1.$

(One could also use the SubresultantPolynomials command in _Mathematica_
[Math21].) These polynomials always lie in $I$, and in this case, they
actually generate. To see why, note that $r(x_{2})=-R_{\text{\it f}\hskip
0.9pt}(x_{2})$, and since $x_{2}$ is relatively prime to $R_{\text{\it
f}\hskip 0.9pt}(x_{2})=-x_{2}^{3}-2x_{2}+1$, we have a Bézout identity

$A\cdot R_{\text{\it f}\hskip 0.9pt}+B\cdot x_{2}=1,\quad A=1,\
B=x_{2}^{2}+2.$

One computes without difficulty that

$x_{1}+B\cdot(x_{2}-1)=Ax_{1}\cdot R_{\text{\it f}\hskip 0.9pt}+B\cdot
p_{1}\in\langle R_{\text{\it f}\hskip 0.9pt},p_{1}\rangle,$

and also that

$x_{1}-g_{1}(x_{2})=x_{1}-x_{2}^{2}-1=x_{1}+B\cdot(x_{2}-1)+R_{\text{\it
f}\hskip 0.9pt}\in\langle R_{\text{\it f}\hskip 0.9pt},p_{1}\rangle.$

Since we know $\langle R_{\text{\it f}\hskip 0.9pt},p_{1}\rangle\subseteq
I=\langle r(x_{2}),x_{1}-g(x_{2})\rangle$, equality follows.

This example has two features that lead to interesting questions:

  * •

$I$ has a Shape Lemma representation $I=\langle
r(x_{2}),x_{1}-g(x_{2})\rangle$ and a subresultant representation $I=\langle
R_{\text{\it f}\hskip 0.9pt}(x_{2}),p_{1}(x_{1},x_{2})\rangle$. How often does
this happen?

  * •

$r(x_{2})$ generates the elimination ideal $I\cap\mathbb{C}[x_{2}]$, which
implies that $R_{\text{\it f}\hskip 0.9pt}(x_{2})$ also generates
$I\cap\mathbb{C}[x_{2}]$. How often does this happen?

Our goal is to study these questions when $I=\langle
f_{1},\dots,f_{n}\rangle\subseteq\mathbb{K}[x_{1},\dots,x_{n}]$ is a zero-
dimensional complete intersection and $\mathbb{K}$ is algebraically closed. We
will always assume that $n\geq 2$.

The paper is structured as follows. Sections 2 and 3 provide background
material on the Shape Lemma and resultants that will be used in Sections 4 and
5, where the main theorems of the paper are proved. In the remainder of this
introduction, we will describe the contents of Sections 2–5 in more detail and
discuss how our results relate to previous work.

Section 2 studies zero-dimensional ideals
$I\subseteq\mathbb{K}[x_{1},\dots,x_{n}]$ of the form

$I=\langle r(x_{n}),x_{1}-g_{1}(x_{n}),\dots,x_{n-1}-g_{n-1}(x_{n})\rangle$

for polynomials
$r(x_{n}),g_{1}(x_{n}),\dots,g_{n-1}(x_{n})\in\mathbb{K}[x_{n}]$. We say that
$I$ _has a Shape Lemma with respect to $x_{n}$_ when this happens. Since our
results and examples will always be with respect to $x_{n}$, we will simply
say “$I$ has a Shape Lemma” for the rest of the paper. Section 2 will
characterize when $I$ has a Shape Lemma, following [BMMT94] for Lemmas 2.1 and
2.2, and adding Lemma 2.3 as suggested by one of the reviewers.

Resultants take center stage in Section 3. We will use the classical
multivariable resultant

$\mathrm{Res}_{d_{1},\ldots,d_{n}}(g_{1},\ldots,g_{n})\in A,$

where $g_{1},\ldots,g_{n}$ are homogeneous polynomials in
$A[x_{0},\ldots,x_{n-1}]$ of respective degrees $d_{1},\dots,d_{n}$, and $A$
is an integral domain. We assume $d_{i}\geq 1$ for all $i$.

Given an ideal $I=\langle
f_{1},\ldots,f_{n}\rangle\subseteq\mathbb{K}[x_{1},\ldots,x_{n}]$ as above, we
want to think of the last variable $x_{n}$ as a constant, similar to what we
did in Example 1.1. So regard $f_{1},\dots,f_{n}$ as lying in
$A[x_{1},\ldots,x_{n-1}]$ with $A=\mathbb{K}[x_{n}]$, and let
$f_{1}^{h},\ldots,f_{n}^{h}\in A[x_{0},x_{1},\ldots,x_{n-1}]$ be their
homogenizations with a new variable $x_{0}$ up to degrees $d_{1},\ldots,d_{n}$
respectively, where $d_{i}=\deg_{x_{1},\ldots,x_{n-1}}(f_{i})$ for
$i=1,\dots,n$. Define

$R_{\text{\it f}\hskip
0.9pt}(x_{n}):=\mathrm{Res}_{d_{1},\ldots,d_{n}}(f_{1}^{h},\ldots,f_{n}^{h})\in\mathbb{K}[x_{n}].$

This resultant will appear in the theorems proved in Sections 3, 4 and 5.
Since the variable $x_{n}$ is “hidden” in the coefficients, the resultant
$R_{\text{\it f}\hskip 0.9pt}(x_{n})$ is an instance of the so-called “hidden
variable method” (see for example [CLO05, Chapter 3, §5]). Extra care must be
taken because in some cases (like in [CLO05]), the degrees
$d_{1},\ldots,d_{n}$ used to compute a hidden variable resultant are the total
degrees of $f_{1},\ldots,f_{n}$. Note that in our case, we use a smaller
degree sequence that takes into account only the first $n-1$ variables.

The main purpose of Section 3 is to give a Poisson-style formula for
$R_{\text{\it f}\hskip 0.9pt}(x_{n}).$ Since
$f_{1}^{h},\ldots,f_{n}^{h}\in\mathbb{K}[x_{0},\ldots,x_{n-1},x_{n}]$ are
homogeneous with respect to $x_{0},\dots,x_{n-1}$, they define a variety

$\mathbf{V}(f_{1}^{h},\ldots,f_{n}^{h})\subseteq\mathbb{P}^{n-1}_{\mathbb{K}}\times_{\mathbb{K}}\mathbb{A}_{\mathbb{K}}^{1}.$

Here is the main result of Section 3, which is of independent interest:

######  Theorem 1.1.

If
$\mathbf{V}(f_{1}^{h},\dots,f_{n}^{h})\subseteq\mathbb{P}^{n-1}_{\mathbb{K}}\times_{\mathbb{K}}\mathbb{A}_{\mathbb{K}}^{1}$
is finite, then there is a nonzero constant $c\in\mathbb{K}$ such that

(1.1) $R_{\text{\it f}\hskip
0.9pt}(x_{n})=c\\!\\!\prod_{\xi\in\mathbf{V}(f_{1}^{h},\dots,f_{n}^{h})}\\!\\!(x_{n}-\xi_{n})^{m_{\xi}},$

where $\xi=([\xi_{0}\hskip 0.85pt{:}\hskip 0.85pt\ldots\hskip 0.85pt{:}\hskip
0.85pt\xi_{n-1}],\xi_{n})\in\mathbb{P}^{n-1}_{\mathbb{K}}\times_{\mathbb{K}}\mathbb{A}_{\mathbb{K}}^{1}$
and $m_{\xi}$ is the Hilbert-Samuel multiplicity of $\xi$.

Properties of resultants guarantee that $R_{\text{\it f}\hskip
0.9pt}(x_{n})\in\langle
f_{1},\dots,f_{n}\rangle\cap\mathbb{K}[x_{n}]=I\cap\mathbb{K}[x_{n}]$. Section
4 will study when $R_{\text{\it f}\hskip 0.9pt}(x_{n})$ generates
$I\cap\mathbb{K}[x_{n}]$. The ideal $I\subseteq\mathbb{K}[x_{1},\dots,x_{n}]$
gives
$\mathbf{V}(I)=\mathbf{V}(f_{1},\dots,f_{n})\subseteq\mathbb{A}_{\mathbb{K}}^{n}$,
where the affine space $\mathbb{A}_{\mathbb{K}}^{n}$ has coordinates
$x_{1},\dots,x_{n}$. This lies in
$\mathbb{P}^{n-1}_{\mathbb{K}}\times_{\mathbb{K}}\mathbb{A}_{\mathbb{K}}^{1}$,
with complement defined by $x_{0}=0$, the “points at $\infty$” in
$\mathbb{P}^{n-1}_{\mathbb{K}}\times_{\mathbb{K}}\mathbb{A}_{\mathbb{K}}^{1}$.
From this point of view, elements of $\mathbf{V}(f_{1}^{h},\dots,f_{n}^{h})$
with $x_{0}=0$ will be regarded as “solutions at $\infty$” of
$f_{1}=\cdots=f_{n}=0$.

The main result of Section 4 describes how solutions at $\infty$ and the Shape
Lemma interact with $R_{\text{\it f}\hskip 0.9pt}(x_{n})$ and the elimination
ideal $I\cap\mathbb{K}[x_{n}]$:

######  Theorem 1.2.

Let $I=\langle f_{1},\dots,f_{n}\rangle$ be a zero-dimensional ideal such that
the map $\mathbf{V}(I)\to\mathbb{A}_{\mathbb{K}}^{1}$ given by projection onto
the $n$th coordinate is injective as a map of sets. Then any two of the
following three conditions imply the third:

  1. (1)

$I$ has a Shape Lemma.

  2. (2)

$f_{1},\dots,f_{n}$ have no solutions at $\infty$.

  3. (3)

$I\cap\mathbb{K}[x_{n}]=\langle R_{\text{\it f}\hskip 0.9pt}(x_{n})\rangle$.

In Section 5, subresultants enter the picture. The _critical degree_ of the
system $f_{1}=\dots=f_{n}=0$ is $\rho=d_{1}+\cdots+d_{n}-n$, where
$d_{i}=\deg_{x_{1},\ldots,x_{n-1}}(f_{i})$ as above. For every monomial
$x^{\alpha}$ in $x_{0},\dots,x_{n-1}$ of degree $\rho$, there is a _scalar
subresultant_ $s_{\alpha}(x_{n})\in\mathbb{K}[x_{n}]$ (it is scalar with
respect to $x_{0},\dots,x_{n-1}$). Then, if $\rho\geq 1$, define

$s_{i}(x_{n}):=s_{\alpha(i)}(x_{n}),\text{ where
}x^{\alpha(i)}=x_{0}^{\rho-1}x_{i}\text{ for }i=0,\dots,n-1,$

which leads to the _first subresultant polynomials_

$p_{i}(x_{i},x_{n}):=s_{0}(x_{n})x_{i}-s_{i}(x_{n})\in\mathbb{K}[x_{i},x_{n}]$

for $i=1,\dots,n-1$. Note that the coefficient of $x_{i}$ in
$p_{i}(x_{i},x_{n})$ is $s_{0}(x_{n})$, independent of $i$. This will be
important in what follows. When $n=2$, $p_{1}(x_{1},x_{2})$ agrees with the
subresultant polynomial that appeared in Example 1.1.

The first main result of Section 5 describes the optimal interaction between
the ideal $I$, the resultant $R_{\text{\it f}\hskip 0.9pt}(x_{n})$, and first
subresultant polynomials $p_{i}(x_{i},x_{n})$:

######  Theorem 1.3.

Assume that $I=\langle f_{1},\dots,f_{n}\rangle$ is zero-dimensional with
$\rho\geq 1$. Then the following are equivalent:

  1. (1)

$I$ has a Shape Lemma and no solutions at $\infty$.

  2. (2)

$I\cap\mathbb{K}[x_{n}]=\langle R_{\text{\it f}\hskip 0.9pt}(x_{n})\rangle$
and $\gcd(R_{\text{\it f}\hskip 0.9pt}(x_{n}),s_{0}(x_{n}))=1$.

  3. (3)

$I=\langle R_{\text{\it f}\hskip
0.9pt}(x_{n}),p_{1}(x_{1},x_{n}),\ldots,p_{n-1}(x_{n-1},x_{n})\rangle$ and
$I\cap\mathbb{K}[x_{n}]=\langle R_{\text{\it f}\hskip 0.9pt}(x_{n})\rangle$.

Furthermore, when these conditions are all true, we have

$I\cap\mathbb{K}[x_{i_{1}},\dots,x_{i_{\ell}},x_{n}]=\langle R_{\text{\it
f}\hskip
0.9pt}(x_{n}),p_{i_{0}}(x_{i_{0}},x_{n}),\dots,p_{i_{\ell}}(x_{i_{\ell}},x_{n})\rangle$

whenever $1\leq i_{1}<\cdots<i_{\ell}<n$.

Notice that (3) is the nicest case: the ideal is generated by the resultant
and first subresultant polynomials, and the elimination ideal is generated by
the resultant. The miracle is that when this holds, explicit generators can be
given for _all_ elimination ideals that do not eliminate $x_{n}$. Also, as we
did in Example 1.1, in this situation one can recover the lexicographic
Gröbner basis of all of these ideals by computing
$x_{i_{j}}-s_{i_{j}}(x_{n})\cdot(s_{0}(x_{n})^{-1}\\!\\!\\!\mod R_{\text{\it
f}\hskip 0.9pt}(x_{n})),\,j=1,\ldots,\ell.$

The second main theorem of Section 5 assumes only that $I$ is generated by the
resultant $R_{\text{\it f}\hskip 0.9pt}(x_{n})$ and first subresultant
polynomials $p_{i}(x_{i},x_{n})=s_{0}(x_{n})x_{i}-s_{i}(x_{n})$:

######  Theorem 1.4.

Let $I=\langle f_{1},\dots,f_{n}\rangle$ be zero-dimensional with $\rho\geq 1$
and assume that $I=\langle R_{\text{\it f}\hskip
0.9pt}(x_{n}),p_{1}(x_{1},x_{n}),\dots,p_{n-1}(x_{n-1},x_{n})\rangle$. Then:

  1. (1)

$\gcd(R_{\text{\it f}\hskip
0.9pt}(x_{n}),s_{0}(x_{n}),\dots,s_{n-1}(x_{n}))=1$.

  2. (2)

$I$ has a Shape Lemma.

Furthermore, the following conditions are equivalent:

  1. (3)

$I\cap\mathbb{K}[x_{n}]=\langle R_{\text{\it f}\hskip 0.9pt}(x_{n})\rangle$.

  2. (4)

$\gcd(R_{\text{\it f}\hskip 0.9pt}(x_{n}),s_{0}(x_{n}))=1$.

  3. (5)

$f_{1},\dots,f_{n}$ have no solutions at $\infty$.

The case $n=2$ has one special feature. If
$f_{1},f_{2}\in\mathbb{K}[x_{1},x_{2}]$, then their leading coefficients with
respect to $x_{1}$ are polynomials in $x_{2}$. In Section 4, we will show that
$f_{1},f_{2}$ have no solutions at $\infty$ if and only if these leading
coefficients are relatively prime in $\mathbb{K}[x_{2}]$. Thus solutions at
$\infty$ are easy to detect when $n=2$.

Let us revisit Example 1.1 in light of what we now know:

######  Example 1.2.

For $f_{1}=x_{1}^{2}-x_{2}-1,\
f_{2}=x_{1}^{2}+x_{1}x_{2}-2\in\mathbb{C}[x_{1},x_{2}]$, we began Example 1.1
by computing that $I=\langle f_{1},f_{2}\rangle=\langle
x_{1}^{2}-x_{2}-1,x_{1}^{2}+x_{1}x_{2}-2\rangle$. Thus $I$ has a Shape Lemma.
Also, the leading coefficients of $f_{1},f_{2}$ with respect to $x_{1}$ are
both equal to $1$, which implies that there are no solutions at $\infty$. By
Theorem 1.3, we immediately conclude that $I=\langle R_{\text{\it f}\hskip
0.9pt}(x_{2}),p_{1}(x_{1},x_{2})\rangle$ and $I\cap\mathbb{C}[x_{2}]=\langle
R_{\text{\it f}\hskip 0.9pt}(x_{2})\rangle$.

In closing, we mention that the results of this paper can be modified to apply
when $\mathbb{K}$ is an arbitrary field. For example, in Theorems 1.2, 1.3 and
1.4, injectivity means that projection induces an injection on points over
$\bar{\mathbb{K}}$, and Theorem 1.1 needs to be formulated in terms of the
irreducible polynomials that define the image of
$\mathbf{V}(f_{1}^{h},\dots,f_{n}^{h})\to\mathbb{A}_{\mathbb{K}}^{1}$. We
prefer to assume that $\mathbb{K}$ is algebraically closed since this makes
the relation between the algebra and the geometry easier to see.

### Previous Work

The representation of algebraic varieties by polynomials having a “Shape
Lemma” has a long history in Computational Algebra. In 1826, Abel solved
$\chi(x,y)=\theta(x,y)=0$ by eliminating $y$ via a Poisson formula and then
expressing $y$ as a rational function of $x$ (see [Abel81, p. 148]). A more
general version of this idea is due to Kronecker (see the introduction of
[GLS01] for a reference and more history). In [Can88], $u$-resultants are used
to compute this representation, while [GLS01] introduces geometric resolutions
of varieties to simplify the computations. In this representation, the focus
is on equations for the variety $\mathbf{V}(I)$ and not on the ideal $I$
itself, so these results are limited to radical ideals with points in some
kind of general position (see for instance [BU99, Theorem 7.4]). In [Rou99], a
“Rational Univariate Representation,” which also takes into account the
multiplicities of the points, is introduced and studied from a computational
point of view. A generalization to a sparse RUR can be found in [MST17].

Our current conception of the Shape Lemma began with [GM89] for a radical
zero-dimensional ideal, though the name “Shape Lemma” came later. The history
of the Shape Lemma is discussed in [BMMT94], which also characterizes ideals
having a Shape Lemma in terms of the geometry of the points in
$\mathbf{V}(I)$. We will review and use some of their results in Section 2.
Also, comparing the monic generator of $I\cap\mathbb{K}[x_{n}]$ with
$R_{\text{\it f}\hskip 0.9pt}(x_{n})$ is a classical exercise in basic
Computational Algebra, see for instance Exercise 3 of Chapter 3, §6 in the
first three editions of [CLO07]. It is clear that both multiplicities and
roots at infinity play a decisive role here. In [MRZ16], directional
multiplicities are used to explain the differences between the degrees of
these two polynomials. In [GRZ13], the connection between the elimination
ideal and univariate resultants of two generators is explored.

The use of subresultants for the Shape Lemma has been done already by Habicht
in [Hab48]. In [GV91], this method is explained and used to produce another
Gröbner-free/resultant-friendly computation of the Shape Lemma for a radical
ideal with points in general position. In a different context (overdetermined
$n$ homogeneous polynomials in $n$ variables), [Sza08] uses multivariable
subresultants to describe the roots of a polynomial system.

As already noted, Poisson-type formulæ for resultants are important tools for
solving polynomial systems. For the classical homogeneous case, this goes back
to Poisson in 1802. See [Jou91, Proposition 2.7] for the presentation of this
formula for generic polynomials. Whether or not one can apply Poisson to a
given polynomial system depends on where the solutions are. For the classical
resultant, such a formula is valid if there are no solutions at infinity. In
[DS15, Theorem 1.1], Poisson has been extended to sparse resultants and its
validity has been shown for systems having all of their roots in
$(\mathbb{K}^{\times})^{n}$ (no solutions at any infinity of the associated
toric variety).

### Acknowledgements

Our calculations were done with the aid of Mathematica [Math21]. We are
grateful to the reviewers whose suggestions led to improvements in Sections 2,
3 and 5.

C. D’Andrea was supported by the Spanish MICINN research project
PID2019-104047GB-I00, the Spanish State Research Agency, through the Severo
Ochoa and María de Maeztu Program for Centers and Units of Excellence in R&D
(CEX2020-001084-M), and the European H2020-MSCA-ITN-2019 research project
GRAPES.

##  2\. The Shape Lemma

As in Section 1, a zero-dimensional ideal
$I\subseteq\mathbb{K}[x_{1},\dots,x_{n}]$ has a Shape Lemma if it is of the
form $I=\langle
r(x_{n}),x_{1}-g_{1}(x_{n}),\dots,x_{n-1}-g_{n-1}(x_{n})\rangle$. We write the
generators in this order because to find the solutions, one first solves
$r(x_{n})=0$ and then uses the roots $\xi_{n}$ to find the coordinates
$\xi_{1}=g_{1}(\xi_{n}),\dots,\xi_{n-1}=g_{n-1}(\xi_{n})$ of the solution
$\xi=(\xi_{1},\dots,\xi_{n})$.

Here we recall some basic facts about the Shape Lemma, following [BMMT94].
(Although [BMMT94] assumes characteristic zero, this assumption is not used in
the results we cite from their paper.). We say that a point $\xi$ of a zero-
dimensional scheme over $\mathbb{K}$ is _curvilinear_ if it is either smooth
or has a one-dimensional Zariski tangent space. See [Cox05] for more on
curvilinear singularities.

We begin with a lemma that combines several results from [BMMT94]:

######  Lemma 2.1.

Let $I=\langle
r(x_{n}),x_{1}-g_{1}(x_{n}),\dots,x_{n-1}-g_{n-1}(x_{n})\rangle\subseteq\mathbb{K}[x_{1},\dots,x_{n}]$
have a Shape Lemma. Then the map

$\mathbf{V}(I)\subseteq\mathbb{A}_{\mathbb{K}}^{n}\longrightarrow\mathbb{A}_{\mathbb{K}}^{1}$

given by projection onto the $x_{n}$-axis is injective as a map of sets.
Furthermore, for every point $\xi=(\xi_{1},\dots,\xi_{n})\in\mathbf{V}(I)$, we
have:

  1. (1)

$\xi$ is curvilinear.

  2. (2)

The Hilbert-Samuel multiplicity of $\xi\in\mathbf{V}(I)$ equals the length of
the local ring $\mathcal{O}_{\mathbf{V}(I),\xi}$.

  3. (3)

The Hilbert-Samuel multiplicity of $\xi\in\mathbf{V}(I)$ equals the
multiplicity of $\xi_{n}$ as a root of $r(x_{n})$.

  4. (4)

The projection $\mathbf{V}(I)\to\mathbb{A}_{\mathbb{K}}^{1}$ induces an
isomorphism of Zariski tangent spaces

$T_{\xi}(\mathbf{V}(I))\xrightarrow{\,\sim\,}T_{\xi_{n}}(\mathbf{V}(r)).$

###### Proof.

For the projection $\mathbf{V}(I)\to\mathbb{A}_{\mathbb{K}}^{1}$, note that
$x_{i}-g_{i}(x_{n})\in I$ implies that for $i=1,\dots,n-1$, the $i$th
coordinate of a point in $\mathbf{V}(I)$ is determined by its $n$th
coordinate. Injectivity follows immediately. Also note that (2) is true since
$I$ is a complete intersection. It remains to prove (1), (3) and (4).

Since $I\cap\mathbb{K}[x_{n}]=\langle r(x_{n})\rangle$, we have an injection

(2.1) $\mathbb{K}[x_{n}]/\langle
r(x_{n})\rangle\longhookrightarrow\mathbb{K}[x_{1},\dots,x_{n}]/I,$

which is onto since $[x_{i}]=[g_{i}(x_{n})]$ in
$\mathbb{K}[x_{1},\dots,x_{n}]/I$ for $i=1,\dots,n-1$. Thus (2.1) is an
isomorphism. Write $r(x_{n})=\prod_{j=1}^{N}(x_{n}-\xi_{n}^{(j)})^{e_{j}}$
with distinct $\xi_{n}^{(j)}$ (we may assume that $r(x_{n})$ is monic). Then
the points of $\mathbf{V}(I)$ are given by
$\xi^{(j)}=(\xi_{1}^{(j)},\xi_{2}^{(j)},\dots,\xi_{n}^{(j)})$ for
$j=1,\dots,N$, where $\xi_{i}^{(j)}=g_{i}(\xi_{n}^{(j)})$ for $i=1,\dots,n-1$.
Using the isomorphism (2.1), we obtain

$\mathbb{K}[x_{1},\dots,x_{n}]/I\simeq\mathbb{K}[x_{n}]/\langle
r(x_{n})\rangle\simeq\prod_{j=1}^{N}\mathbb{K}[x_{n}]/\langle(x_{n}-\xi_{n}^{(j)})^{e_{j}}\rangle.$

This expresses $\mathbb{K}[x_{1},\dots,x_{n}]/I$ as a product of local rings,
and in particular, the local ring
$\mathcal{O}_{\mathbf{V}(I),\xi^{(j)}}\simeq(\mathbb{K}[x_{1},\dots,x_{n}]/I)_{\xi^{(j)}}$
is isomorphic to
$\mathbb{K}[x_{n}]/\langle(x_{n}-\xi_{n}^{(j)})^{e_{j}}\rangle$ via projection
onto the $x_{n}$-axis. This implies that $\xi^{(j)}$ is curvilinear and its
Hilbert-Samuel multiplicity, which equals the length of
$\mathcal{O}_{\mathbf{V}(I),\xi}$ by (2), is simply the multiplicity of
$\xi_{n}^{(j)}$ as a root of $r(x_{n})$. This proves assertion (3).

Finally, since
$(\mathbb{K}[x_{1}\dots,x_{n}]/I)_{\xi^{(j)}}\simeq\mathbb{K}[x_{n}]/\langle(x_{n}-\xi_{n}^{(j)})^{e_{j}}\rangle$
is induced by the projection, we get an induced isomorphism on Zariski tangent
spaces since the Zariski tangent space of a local ring with maximal ideal
$\mathfrak{m}$ is $\mathfrak{m}/\mathfrak{m}^{2}$. This proves assertions (1)
and (4). ∎

The converse of Lemma 2.1 is true for a zero-dimensional ideal $I$, namely if
the projection onto the $x_{n}$-axis is injective as a map of sets and
conditions (1)–(4) of the lemma are satisfied for
$I\cap\mathbb{K}[x_{n}]=\langle r(x_{n})\rangle$, then $I$ has a Shape Lemma.
In fact, we have a slightly stronger result as follows:

######  Lemma 2.2.

Suppose that $I\subseteq\mathbb{K}[x_{1},\dots,x_{n}]$ is a zero-dimensional
ideal and let $I\cap\mathbb{K}[x_{n}]=\langle r(x_{n})\rangle$. If the
projection $\mathbf{V}(I)\to\mathbb{A}^{1}_{\mathbb{K}}$ onto the $x_{n}$-axis
is injective as a map of sets, then the following conditions are equivalent:

  1. (1)

For every $\xi=(\xi_{1},\dots,\xi_{n})\in\mathbf{V}(I)$, the Hilbert-Samuel
multiplicity of $\xi$ equals the multiplicity of $\xi_{n}$ as a root of
$r(x_{n})$.

  2. (2)

For every $\xi=(\xi_{1},\dots,\xi_{n})\in\mathbf{V}(I)$, the length of
$\mathcal{O}_{\mathbf{V}(I),\xi}$ equals the multiplicity of $\xi_{n}$ as a
root of $r(x_{n})$.

  3. (3)

For every $\xi=(\xi_{1},\dots,\xi_{n})\in\mathbf{V}(I)$, the induced map on
Zariski tangent spaces $T_{\xi}(\mathbf{V}(I))\to T_{\xi_{n}}(\mathbf{V}(r))$
is an isomorphism.

  4. (4)

$I$ has a Shape Lemma.

###### Proof.

By Lemma 2.1, (4) implies (1), (2) and (3). It remains to prove that (1), (2)
and (3) each imply (4). Since $I\cap\mathbb{K}[x_{n}]=\langle
r(x_{n})\rangle$, we get the injection (2.1). Thus, to prove (4), it suffices
to show that (2.1) is an isomorphism since the elements of
$\mathbb{K}[x_{1},\dots,x_{n}]/I$ represented by $x_{1},\dots,x_{n-1}$ would
be in the image of (2.1). This would give the desired
$g_{1}(x_{n}),\dots,g_{n-1}(x_{n})$.

Let $\mathbf{V}(I)=\\{\xi^{(1)},\dots,\xi^{(N)}\\}$ with
$\xi^{(j)}=(\xi_{1}^{(j)},\dots,\xi_{n}^{(j)})$, and write

(2.2)
$\mathbb{K}[x_{1},\dots,x_{n}]/I=\prod_{j=1}^{N}\mathcal{O}_{\mathbf{V}(I),\xi^{(j)}}.$

Let $m_{j}$ be the Hilbert-Samuel multiplicity of $\xi^{(j)}$ and $k_{j}$ be
the length $\dim_{\mathbb{K}}\mathcal{O}_{\mathbf{V}(I),\xi^{(j)}}$ of
$\mathcal{O}_{\mathbf{V}(I),\xi^{(j)}}$. Then
$\dim_{\mathbb{K}}\mathbb{K}[x_{1},\dots,x_{n}]/I=\sum_{j=1}^{N}k_{j}$, and
since $k_{j}\leq m_{j}$ for all $j$, the injection (2.1) implies

(2.3) $\begin{array}[]{c}\begin{aligned}
\deg(r(x_{n}))&=\dim_{\mathbb{K}}\mathbb{K}[x_{n}]/\langle r(x_{n})\rangle\\\
&\leq\dim_{\mathbb{K}}\mathbb{K}[x_{1},\dots,x_{n}]/I=\sum_{j=1}^{N}k_{j}\leq\sum_{j=1}^{N}m_{j}.\end{aligned}\end{array}$

Now assume that (1) is true. Since $r(x_{n})\in I$ and
$\xi^{(j)}=(\xi_{1}^{(j)},\dots,\xi_{n}^{(j)})$, we have $r(\xi_{n}^{(j)})=0$,
and then assumption (1) implies that $\xi_{n}^{(j)}$ is a root of multiplicity
$m_{j}$ of $r(x_{n})$. Thus $(x_{n}-\xi_{n}^{(j)})^{m_{j}}$ divides
$r(x_{n})$. But the $\xi_{n}^{(j)}$ are distinct by our injectivity
assumption, so that $\prod_{j=1}^{N}(x_{n}-\xi_{n}^{(j)})^{m_{j}}$ divides
$r(x_{n})$. Thus

$\deg(r(x_{n}))\geq\sum_{j=1}^{N}m_{j},$

which is an equality by virtue of (2.3). Thus (2.1) is an injection where both
source and target have the same dimension, proving that it is an isomorphism.

Similarly, if (2) is true, then $\xi_{n}^{(j)}$ is a root of multiplicity
$k_{j}$ of $r(x_{n})$, so that $\deg(r(x_{n}))\geq\sum_{j=1}^{N}k_{j}$. Using
(2.3) as in the previous paragraph, we again see that (2.1) is an isomorphism.

Finally, assume that (3) is true. Then Proposition 5 of [BMMT94] implies that
$I$ has a Shape Lemma. For completeness, we sketch the proof. Write $I$ as the
intersection of primary ideals $Q_{j}$ whose radicals are the maximal ideals
of the $\xi^{(j)}$. The isomorphism on Zariski tangent spaces implies that
$\xi^{(j)}$ is curvilinear. Also, with $\ell_{1}=x_{n}-\xi_{n}^{(j)}$ and
$\ell_{i}=x_{i}$ for $i=1,\dots,{n-1}$, [BMMT94, Proposition 4(10)] implies
that

$Q_{j}=\langle(x_{n}-\xi_{n}^{(j)})^{m_{j}},x_{i}-g_{ij}(x_{n}),\,i=1,\dots,n-1\rangle,\text{
where }g_{ij}(x_{n})\in\mathbb{K}[x_{n}].$

It follows that
$Q_{j}\cap\mathbb{K}[x_{n}]=\langle(x_{n}-\xi_{n}^{(j)})^{m_{j}}\rangle$.
Since $r(x_{n})\in I\cap\mathbb{K}[x_{n}]\subseteq
Q_{j}\cap\mathbb{K}[x_{n}]$, we see that $(x_{n}-\xi_{n}^{(j)})^{m_{j}}$
divides $r(x_{n})$. From here, the proof of (1) $\Rightarrow$ (3) shows that
(2.1) is an isomorphism. ∎

Our final lemma uses the scheme-theoretic fibers of the projection morphism
$\mathbf{V}(I)\to\mathbb{A}^{1}_{\mathbb{K}}$ to characterize when $I$ has a
Shape Lemma. Since $\mathbb{K}$ is algebraically closed, a closed point
$\mathfrak{p}\in\mathrm{Spec}(A)=\mathrm{Spec}(\mathbb{K}[x_{n}])=\mathbb{A}_{\mathbb{K}}^{1}$
can be written as $\mathfrak{p}=\langle x_{n}-\lambda\rangle$ for some
$\lambda\in\mathbb{K}$. For simplicity, set $V=\mathbf{V}(I)$. Then the
scheme-theoretic fiber $V_{\mathfrak{p}}$ is the subscheme of
$V=\mathrm{Spec}(\mathbb{K}[x_{1},\dots,x_{n}]/I)$ defined as follows. When
$V$ is finite, the decomposition (2.2) can be written

$\mathbb{K}[x_{1},\dots,x_{n}]/I\simeq\prod_{\xi\in V}\mathcal{O}_{V,\xi}.$

Take $\xi=(\xi_{1},\dots,\xi_{n})\in V$. If $\xi_{n}\neq\lambda$, then
$x_{n}-\lambda$ is nonvanishing at $\xi$ and hence invertible in
$\mathcal{O}_{V,\xi}$. Thus
$\mathfrak{p}\mathcal{O}_{V,\xi}=\mathcal{O}_{V,\xi}$ when
$\xi_{n}\neq\lambda$. It follows that the fiber above $\mathfrak{p}$ is given
by

(2.4)
$V_{\mathfrak{p}}=\mathrm{Spec}(A/\mathfrak{p}\otimes_{A}\mathbb{K}[x_{1},\dots,x_{n}]/I)=\mathrm{Spec}\bigg{(}\\!\prod_{\xi\in
V,\,\xi_{n}=\lambda}\\!\\!\mathcal{O}_{V,\xi}/\mathfrak{p}\mathcal{O}_{V,\xi}\bigg{)}.$

######  Lemma 2.3.

The zero-dimensional ideal $I$ has a Shape Lemma if an only if for all closed
points $\mathfrak{p}\in\mathbb{A}_{\mathbb{K}}^{1}$, the fiber
$V_{\mathfrak{p}}$ is either empty or consists of a single reduced point.

###### Proof.

First suppose that $I$ has a Shape Lemma and take $\mathfrak{p}=\langle
x_{n}-\lambda\rangle\subseteq\mathbb{A}_{\mathbb{K}}^{1}$ as above. If
$\mathfrak{p}$ is not in the image of $V\to\mathbb{A}^{1}_{\mathbb{K}}$, then
$\lambda\neq\xi_{n}$ for all $\xi\in V$, so that the fiber $V_{\mathfrak{p}}$
is empty by (2.4). On the other hand, if $\mathfrak{p}$ is in the image of
$V\to\mathbb{A}^{1}_{\mathbb{K}}$, there is a unique point $\xi\in\mathbf{V}$
with $\xi_{n}=\lambda$ since $V\to\mathbb{A}^{1}_{\mathbb{K}}$ is injective as
a map of sets by Lemma 2.1. Then the fiber (2.4) becomes
$V_{\mathfrak{p}}=\mathrm{Spec}(\\!\mathcal{O}_{V,\xi}/\mathfrak{p}\mathcal{O}_{V,\xi})$.
The proof of Lemma 2.1 implies that

$\mathcal{O}_{V,\xi}\simeq\mathbb{K}[x_{n}]/\langle(x_{n}-\lambda)^{m_{\xi}}\rangle.$

The maximal ideal of
$\mathbb{K}[x_{n}]/\langle(x_{n}-\lambda)^{m_{\xi}}\rangle$ is generated by
the image of $x_{n}-\lambda$, so the same is true for $\mathcal{O}_{V,\xi}$,
so that $\mathfrak{p}\mathcal{O}_{V,\xi}$ is the maximal ideal of
$\mathcal{O}_{V,\xi}$. Thus
$\\!\mathcal{O}_{V,\xi}/\mathfrak{p}\mathcal{O}_{V,\xi}\simeq\mathbb{K}$,
which proves that $V_{\mathfrak{p}}$ consists of a single reduced point.

Conversely, suppose that every fiber $V_{\mathfrak{p}}$ is either empty or
consists of a single reduced point. If $V_{\mathfrak{p}}$ is nonempty, there
is $\xi\in V$ with $x_{n}=\lambda$. Since $x_{n}-\lambda$ vanishes at $\xi$,
it gives an element of the maximal ideal
$\mathfrak{m}_{\xi}\subseteq\mathcal{O}_{V,\xi}$. Hence
$\mathfrak{p}\mathcal{O}_{V,\xi}\subseteq\mathfrak{m}_{\xi}$, which proves
that $\mathcal{O}_{V,\xi}/\mathfrak{p}\mathcal{O}_{V,\xi}$ is a nonzero ring.
Since $V_{\mathfrak{p}}$ consists of a single reduced point, it follows that
there is a unique point $\xi\in V$ that maps to $\mathfrak{p}$. We conclude
that $V\to\mathbb{A}_{\mathbb{K}}^{1}$ is injective as map of sets.

Let us examine further the case when $V_{\mathfrak{p}}$ consists of a single
reduced point. This means
$\mathcal{O}_{V,\xi}/\mathfrak{p}\mathcal{O}_{V,\xi}\simeq\mathbb{K}$, so that
$\mathfrak{p}\mathcal{O}_{V,\xi}$ equals the maximal ideal
$\mathfrak{m}_{\xi}\subseteq\mathcal{O}_{V,\xi}$. Let
$I\cap\mathbb{K}[x_{n}]=\langle r(x_{n})\rangle$ with $r(x_{n})$ monic, and
write $r(x_{n})=\prod_{\xi\in V}(x_{n}-\xi_{n})^{e_{\xi}}$. Then for $\xi\in
V$ with $\xi_{n}=\lambda$ the injection (2.1) gives an injection of local
rings

$\varphi:\mathbb{K}[x_{n}]/\langle(x_{n}-\lambda)^{e_{\xi}}\rangle\longhookrightarrow\mathcal{O}_{V,\xi}.$

Once we prove that $\varphi$ is an isomorphism for all $\xi\in V$, Lemma 2.2
will imply that $I$ has a Shape Lemma. It suffices so show that $\varphi$ is
surjective. We will regard $\varphi$ as an inclusion

$\mathcal{R}_{\xi}:=\mathbb{K}[x_{n}]/\langle(x_{n}-\lambda)^{e_{\xi}}\rangle\subseteq\mathcal{O}_{V,\xi}.$

Then $M=\mathcal{O}_{V,\xi}/\mathcal{R}_{\xi}$ is a finitely generated
$\mathcal{R}$-module, and we have $M/\mathfrak{p}M=0$ since
$\mathcal{R}_{\xi}/\mathfrak{p}\mathcal{R}_{\xi}\simeq\mathcal{O}_{V,\xi}/\mathfrak{p}\mathcal{O}_{V,\xi}\simeq\mathbb{K}$.
Thus $M=0$ by Nakayama’s Lemma. We conclude that
$\mathcal{R}_{\xi}=\mathcal{O}_{V,\xi}$, which proves that $\varphi$ is
surjective. ∎

##  3\. A Resultant Formula

The goal of this section is to prove Theorem 1.1. The resultant $R_{\text{\it
f}\hskip 0.9pt}(x_{n})$ featured in this theorem uses the classical
multivariable resultant. Recall that if $A$ is an integral domain and
$g_{1},\ldots,g_{n}\in A[x_{0},\ldots,x_{n-1}]$ are homogeneous of respective
degrees $d_{1},\ldots,d_{n}$, their resultant
$\mathrm{Res}_{d_{1},\ldots,d_{n}}(g_{1},\ldots,g_{n})\in A$ vanishes if and
only if the system $g_{1}=\ldots=g_{n}=0$ has a solution in
$\mathbb{P}_{\overline{k(A})}^{n-1}$. where $\overline{k(A)}$ is the algebraic
closure of the field of fractions of $A$. The general theory of resultants is
developed in [Jou91]. See also [CLO05, Chapter 3] when $A=\mathbb{C}$.

As in Section 1, we regard
$f_{1},\ldots,f_{n}\in\mathbb{K}[x_{1},\ldots,x_{n}]$ as lying in
$A[x_{1},\ldots,x_{n-1}]$ with $A=\mathbb{K}[x_{n}]$ and let
$f_{1}^{h},\ldots,f_{n}^{h}\in A[x_{0},x_{1},\ldots,x_{n-1}]$ be their
homogenizations with respect to the new variable $x_{0}$ up to degrees
$d_{1},\ldots,d_{n}$ respectively, with
$d_{i}=\deg_{x_{1},\ldots,x_{n-1}}(f_{i})$ for $i=1,\dots,n$. Then

$R_{\text{\it f}\hskip
0.9pt}(x_{n}):=\mathrm{Res}_{d_{1},\ldots,d_{n}}(f_{1}^{h},\ldots,f_{n}^{h})\in\mathbb{K}[x_{n}].$

Being homogeneous with respect to $x_{0},\dots,x_{n-1}$, the polynomials
$f_{1}^{h},\ldots,f_{n}^{h}$ define

$\mathbf{V}(f_{1}^{h},\ldots,f_{n}^{h})\subseteq\mathbb{P}^{n-1}_{\mathbb{K}}\times_{\mathbb{K}}\mathbb{A}_{\mathbb{K}}^{1}.$

Theorem 1.1 states that when $\mathbf{V}(f_{1}^{h},\dots,f_{n}^{h})$ is
finite,

(3.1) $R_{\text{\it f}\hskip
0.9pt}(x_{n})=c\\!\\!\prod_{\xi\in\mathbf{V}(f_{1}^{h},\dots,f_{n}^{h})}\\!\\!(x_{n}-\xi_{n})^{m_{\xi}}$

for some $c\in\mathbb{K}^{\times}$. Here, $\xi=([\xi_{0}\hskip 0.85pt{:}\hskip
0.85pt\ldots\hskip 0.85pt{:}\hskip
0.85pt\xi_{n-1}],\xi_{n})\in\mathbb{P}^{n-1}_{\mathbb{K}}\times_{\mathbb{K}}\mathbb{A}_{\mathbb{K}}^{1}$,
and $m_{\xi}$ is the Hilbert-Samuel multiplicity of $\xi$.

Our proof of (3.1) will use some results from [SS96]. We will need the
following algebraic characterization of when
$\mathbf{V}(f_{1}^{h},\dots,f_{n}^{h})$ is finite:

######  Proposition 3.1.

$V=\mathbf{V}(f_{1}^{h},\dots,f_{n}^{h})\subseteq\mathbb{P}^{n-1}_{\mathbb{K}}\times_{\mathbb{K}}\mathbb{A}^{1}_{\mathbb{K}}$
is finite if and only if $f_{1}^{h},\dots,f_{n}^{h}$ form a regular sequence
in $A[x_{0},x_{1},\dots,x_{n-1}]$.

###### Proof.

If $f_{1}^{h},\dots,f_{n}^{h}$ form a regular sequence, then for each
$i=0,\dots,n-1$, their dehomogenizations in
$\mathbb{K}[x_{0},\dots,\hat{x}_{i},\dots,x_{n-1},x_{n}]$ form a regular
sequence and hence define a zero-dimensional subscheme of
$U_{i}\times_{\mathbb{K}}\mathbb{A}_{\mathbb{K}}^{1}$, where
$U_{i}\subseteq\mathbb{P}^{n-1}_{\mathbb{K}}$ is the affine open defined by
$x_{i}\neq 0$. The $U_{i}$ cover $\mathbb{P}^{n-1}_{\mathbb{K}}$, which proves
that $V$ is finite.

Conversely, assume that $V$ is finite. Let $A_{\mathfrak{m}}$ be the
localization of $A=\mathbb{K}[x_{n}]$ at a maximal ideal
$\mathfrak{m}\subseteq A$. It suffices to prove that the homogeneous elements
$f_{1}^{h},\dots,f_{n}^{h}$ form a regular sequence in
$A_{\mathfrak{m}}[x_{0},\dots,x_{n-1}]$. To prove this, let
$I_{\mathfrak{m}}\subseteq A_{\mathfrak{m}}[x_{0},\dots,x_{n-1}]$ be generated
by $f_{1}^{h},\dots,f_{n}^{h}$. Then
$\mathbf{V}(I_{\mathfrak{m}})\subseteq\mathbb{P}^{n-1}_{\mathbb{K}}\times_{\mathbb{K}}\mathrm{Spec}(A_{\mathfrak{m}})$.
Since
$V=\mathbf{V}(f_{1}^{h},\dots,f_{n}^{h})\subseteq\mathbb{P}^{n-1}_{\mathbb{K}}\times_{\mathbb{K}}\mathbb{A}^{1}_{\mathbb{K}}$
is finite, $\mathbf{V}(I_{\mathfrak{m}})$ is either empty, or nonempty and
finite. We consider each case separately.

First suppose that $\mathbf{V}(I_{\mathfrak{m}})=\emptyset$. We claim that
$A_{\mathfrak{m}}[x_{0},\dots,x_{n-1}]/I_{\mathfrak{m}}$ is finite over
$A_{m}$. This is easy, since
$\mathbf{V}(I_{\mathfrak{m}})\subseteq\mathrm{Proj}(A_{\mathfrak{m}}[x_{0},\dots,x_{n-1}])$
satisfies

$\mathbf{V}(I_{\mathfrak{m}})=\emptyset\iff\langle
x_{0},\dots,x_{n-1}\rangle\subseteq\sqrt{I_{\mathfrak{m}}}\iff\langle
x_{0},\dots,x_{n-1}\rangle^{N}\subseteq I_{\mathfrak{m}}\ \text{for some}\ N$

by standard properties of $\mathrm{Proj}$. If follows that
$A_{\mathfrak{m}}[x_{0},\dots,x_{n-1}]/I_{\mathfrak{m}}$ is finite over
$A_{m}$ when $\mathbf{V}(I_{\mathfrak{m}})=\emptyset$. By (3) $\Rightarrow$
(1) of Theorem 7.3 of [SS01], $I_{\mathfrak{m}}$ is a complete intersection,
so that $f_{1}^{h},\dots,f_{n}^{h}$ form a regular sequence in
$A_{\mathfrak{m}}[x_{0},\dots,x_{n-1}]$.

Next suppose that $\mathbf{V}(I_{\mathfrak{m}})$ is nonempty and finite. Here,
we will use the theory of ∗local rings from [BH93, Section 1.5]. A proper
homogeneous ideal is a graded ring $R$ is _∗ maximal _if the only strictly
larger homogenous ideal is $R$ itself, and $R$ is _∗ local_ if it has a unique
∗maximal ideal. The basic idea is that ∗local rings are the graded
counterparts of local rings.

In our situation, $R=A_{\mathfrak{m}}[x_{0},\dots,x_{n-1}]$. Since
$A_{\mathfrak{m}}$ is a local ring with maximal ideal
$\mathfrak{m}A_{\mathfrak{m}}$, it is easy to see that $R$ is ∗local with
∗maximal ideal $\mathfrak{M}=\mathfrak{m}A_{\mathfrak{m}}+\langle
x_{0},\dots,x_{n-1}\rangle$. The fact that $\mathfrak{M}$ is maximal in the
usual sense simplifies some features of the theory. For example, the _∗
dimension_ of $R$ is defined to be the height of the ∗maximal ideal
$\mathfrak{M}$. Since $\mathfrak{M}$ is maximal, this is just the dimension of
$R$, so that ${}^{*}\\!\dim R=\dim R=n+1$.

The quotient $R/I_{\mathfrak{m}}$ is also ∗local, and its ∗maximal ideal is
again maximal in the usual sense. Thus ${}^{*}\\!\dim R/I_{\mathfrak{m}}=\dim
R/I_{\mathfrak{m}}=1$ since $V(I_{\mathfrak{m}})\subseteq\mathrm{Proj}(R)$ is
finite and nonempty. Hence

(3.2) ${}^{*}\\!\dim R/I_{\mathfrak{m}}=1=(n+1)-n={}^{*}\\!\dim R-n.$

However, we also know that $R$ is Cohen-Macaulay and that $I_{\mathfrak{m}}$
is generated by $n$ elements. If we replace $R$ with a Cohen-Macaulay local
ring and ∗$\dim$ by $\dim$, then it is well known that (3.2) implies that the
$n$ generators form a regular sequence (for example, this follows from Theorem
2.12 and Proposition A.4 of [BH93]). Since the same result holds in the ∗local
setting, we conclude that $f_{1}^{h},\dots,f_{n}^{h}$ form a regular sequence
in $R$, as desired. ∎

###### Proof of Theorem 1.1.

We need to prove (3.1). The finiteness of
$V=\mathbf{V}(f_{1}^{h},\dots,f_{n}^{h})$ implies that the resultant
$R_{\text{\it f}\hskip 0.9pt}(x_{n})$ is nonzero and hence can be written

$R_{\text{\it f}\hskip
0.9pt}(x_{n})=c\\!\\!\\!\\!\prod_{\lambda\in\mathbf{V}(R_{\text{\it f}\hskip
0.9pt}(x_{n}))}\\!\\!\\!\\!(x_{n}-\lambda)^{e_{\lambda}}$

for integers $e_{\lambda}\geq 1$ and a nonzero constant $c\in\mathbb{K}$.

We know that $R_{\text{\it f}\hskip 0.9pt}(x_{n})$ vanishes at every point of
$V$, and since $\mathbb{K}$ is algebraically closed, the universal property of
the resultant implies that any root $\lambda$ of $R_{\text{\it f}\hskip
0.9pt}(x_{n})$ comes from a solution $\xi=([\xi_{0}\hskip 0.85pt{:}\hskip
0.85pt\ldots\hskip 0.85pt{:}\hskip 0.85pt\xi_{n-1}],\xi_{n})\in V$ with
$\xi_{n}=\lambda$. Thus (3.1) will follow once we prove that

(3.3) $e_{\lambda}=\\!\\!\sum_{\xi\in V,\,\xi_{n}=\lambda}\\!\\!m_{\xi},$

where $m_{\xi}$ is the Hilbert-Samuel multiplicity of $\xi$. The sum on the
right is finite since $V$ is finite by assumption.

Since $V$ is finite, Proposition 3.1 implies that $f_{1}^{h},\dots,f_{n}^{h}$
form a regular sequence in $A[x_{0},\dots,x_{n-1}]$. This allows us to use the
results of [SS96]. The _resultant ideal_ $\mathfrak{R}\subseteq A$ is
generated by $R_{\text{\it f}\hskip 0.9pt}(x_{n})$. We also have the
$A$-algebra of global sections

$B=\Gamma(V,\mathcal{O}_{V})=\prod_{\xi\in V}\mathcal{O}_{V,\xi}.$

Note that $V=\mathrm{Spec}(B)$.

A root $\lambda\in\mathbb{K}$ of $R_{\text{\it f}\hskip 0.9pt}(x_{n})$ gives a
point $\mathfrak{p}\in\mathrm{Spec}(A)$ for $\mathfrak{p}=\langle
x_{n}-\lambda\rangle\subseteq A$. Localizing the $A$-module $B$ at
$\mathfrak{p}$ gives

$B_{\mathfrak{p}}=\\!\\!\prod_{\xi\in
V,\,\xi_{n}=\lambda}\\!\\!\mathcal{O}_{V,\xi}.$

In the local ring $A_{\mathfrak{p}}$, the resultant ideal
$\mathfrak{R}=\langle R_{\text{\it f}\hskip 0.9pt}(x_{n})\rangle$ localizes to
$\mathfrak{R}_{\mathfrak{p}}=\langle(x_{n}-\lambda)^{e_{\lambda}}\rangle$.
Thus

$A_{\mathfrak{p}}/\mathfrak{R}_{\mathfrak{p}}=A_{\mathfrak{p}}/\langle(x_{n}-\lambda)^{e_{\lambda}}\rangle.$

Since $\mathfrak{p}$ has codimension 1, [SS96, Theorem 2.1] implies that
$\mathrm{length}(B_{\mathfrak{p}})=\mathrm{length}(A_{\mathfrak{p}}/\mathfrak{R}_{\mathfrak{p}})$.
Combining this with the previous paragraph, we obtain

$\displaystyle\\!\\!\sum_{\xi\in
V,\,\xi_{n}=\lambda}\\!\\!\mathrm{length}(\mathcal{O}_{V,\xi})$
$\displaystyle=\mathrm{length}(B_{\mathfrak{p}})$
$\displaystyle=\mathrm{length}(A_{\mathfrak{p}}/\mathfrak{R}_{\mathfrak{p}})=\mathrm{length}(A_{\mathfrak{p}}/\langle(x_{n}-\lambda)^{e_{\lambda}}\rangle)=e_{\lambda}.$

Since $V=\mathbf{V}(f_{1}^{h},\dots,f_{n}^{h})$ is a local complete
intersection, the length of $\mathcal{O}_{V,\xi}$ is the Hilbert-Samuel
multiplicity $m_{\xi}$, and (3.3) follows immediately. This completes the
proof of Theorem 1.1. ∎

######  Remark 3.1.

The proof just given uses the localization $B_{\mathfrak{p}}$ to express the
exponent $e_{\lambda}$ as a sum of Hilbert-Samuel multiplicities. It is
natural to ask how this relates to the fiber $V_{\mathfrak{p}}$ of the
projection
$V\hookrightarrow\mathbb{P}^{n-1}_{\mathbb{K}}\times_{\mathbb{K}}\mathbb{A}^{1}_{\mathbb{K}}\to\mathbb{A}^{1}_{\mathbb{K}}$.
Since $V=\mathrm{Spec}(B)$, the fiber above the maximal ideal
$\mathfrak{p}=\langle
x_{n}-\lambda\rangle\in\mathrm{Spec}(A)=\mathbb{A}^{1}_{\mathbb{K}}$ is the
affine scheme

$V_{\mathfrak{p}}=\mathrm{Spec}(A/\mathfrak{p}\otimes_{A}B)=\mathrm{Spec}(B/\mathfrak{p}B)=\mathrm{Spec}\bigg{(}\\!\prod_{\xi\in
V,\,\xi_{n}=\lambda}\\!\\!\mathcal{O}_{V,\xi}/\mathfrak{p}\mathcal{O}_{V,\xi}\bigg{)}$

since $\mathfrak{p}\mathcal{O}_{V,\xi}=\mathcal{O}_{V,\xi}$ when
$\xi_{n}\neq\lambda$. Thus

$\deg V_{\mathfrak{p}}=\mathrm{length}(B/\mathfrak{p}B)=\\!\\!\sum_{\xi\in
V,\,\xi_{n}=\lambda}\\!\\!\mathrm{length}(\mathcal{O}_{V,\xi}/\mathfrak{p}\mathcal{O}_{V,\xi})\leq\\!\\!\sum_{\xi\in
V,\,\xi_{n}=\lambda}\\!\\!\mathrm{length}(\mathcal{O}_{V,\xi}).$

Hence the degree of the fiber is bounded above by the sum of the lengths of
the local rings, and it is easy to see that equality holds if and only if
$\mathfrak{p}B_{\mathfrak{p}}=0$ (see [SS96, Theorem 2.6]).

For an example where $\mathfrak{p}B_{\mathfrak{p}}\neq 0$, consider
$f_{1}=x_{1}^{2}-x_{2}^{2}$ and $f_{2}=x_{1}^{2}+x_{2}^{2}$ in
$\mathbb{C}[x_{1},x_{2}]$. These homogenize to
$f_{1}^{h}=x_{1}^{2}-x_{0}^{2}x_{2}^{2}$ and
$f_{2}^{h}=x_{1}^{2}+x_{0}^{2}x_{2}^{2}$ in $A[x_{0},x_{1}]$,
$A=\mathbb{C}[x_{2}]$, with resultant $R_{\text{\it f}\hskip
0.9pt}(x_{2})=4x_{2}^{4}$. To analyze the scheme
$V=\mathbf{V}(f_{1}^{h},f_{2}^{h})$, note that the equations
$x_{1}^{2}-x_{0}^{2}x_{2}^{2}=x_{1}^{2}+x_{0}^{2}x_{2}^{2}=0$ imply that
$x_{1}=0$, so that $x_{0}\neq 0$ since $x_{0},x_{1}$ are homogeneous
coordinates for $\mathbb{P}^{1}_{\mathbb{K}}$. Thus, there are no solutions at
$\infty$, and since $\langle f_{1},f_{2}\rangle=\langle
x_{1}^{2},x_{2}^{2}\rangle$, we obtain

$V=\mathbf{V}(f_{1},f_{2})=\mathrm{Spec}(B),\quad
B=\mathbb{C}[x_{1},x_{2}]/\langle x_{1}^{2},x_{2}^{2}\rangle.$

Thus $V$ consists of a single point of multiplicity $4$. However, for
$\mathfrak{p}=\langle x_{2}\rangle\in\mathrm{Spec}(\mathbb{C}[x_{2}])$, we
have $B_{\mathfrak{p}}=B$, and the fiber is

$V_{\mathfrak{p}}=\mathrm{Spec}(B/\mathfrak{p}B),\quad
B/\mathfrak{p}B=B/x_{2}B\simeq\mathbb{C}[x_{1},x_{2}]/\langle
x_{1}^{2},x_{2}\rangle,$

which consists of a single point of multiplicity $2\neq 4$. The discrepancy
arises because $\mathfrak{p}B_{\mathfrak{p}}\neq 0$.

##  4\. The Elimination Ideal and Resultants

As in Section 1, $f_{1},\dots,f_{n}\in\mathbb{K}[x_{1},\dots,x_{n}]$ give
$R_{\text{\it f}\hskip
0.9pt}(x_{n})=\mathrm{Res}_{d_{1},\dots,d_{n}}(f_{1}^{h},\dots,f_{n}^{h})$. It
is well known that $R_{\text{\it f}\hskip 0.9pt}(x_{n})$ belongs to the
elimination ideal $I\cap\mathbb{K}[x_{n}]$ for $I=\langle
f_{1},\dots,f_{n}\rangle$. Easy examples show that $R_{\text{\it f}\hskip
0.9pt}(x_{n})$ does not always generate $I\cap\mathbb{K}[x_{n}]$. Even when
the Shape Lemma holds, $R_{\text{\it f}\hskip 0.9pt}(x_{n})$ may fail to
generate $I\cap\mathbb{K}[x_{n}]$, though in this case, Theorem 1.2 to be
proved below explains what goes wrong.

For the moment, assume only that $I=\langle f_{1},\dots,f_{n}\rangle$ is zero-
dimensional, so that
$\mathbf{V}(f_{1},\dots,f_{n})\subseteq\mathbb{A}_{\mathbb{K}}^{n}$ is finite.
Let $I\cap\mathbb{K}[x_{n}]=\langle r(x_{n})\rangle$. Then $r(x_{n})\mid
R_{\text{\it f}\hskip 0.9pt}(x_{n})$, which implies that every root of
$r(x_{n})$ is also a root of $R_{\text{\it f}\hskip 0.9pt}(x_{n})$. It is
natural to inquire about the converse, i.e., whether every root of
$R_{\text{\it f}\hskip 0.9pt}(x_{n})$ is also a root of $r(x_{n})$.

So suppose that $\xi_{n}$ is a root of $R_{\text{\it f}\hskip 0.9pt}(x_{n})$.
Then

(4.1) $0=R_{\text{\it f}\hskip
0.9pt}(\xi_{n})=\mathrm{Res}_{d_{1},\dots,d_{n}}(f_{1}^{h}(x_{0},\dots,x_{n-1},\xi_{n}),\dots,f_{n}^{h}(x_{0},\dots,x_{n-1},\xi_{n})).$

Recall that $d_{i}=\deg_{x_{1},\dots,x_{n-1}}(f_{i})$ and
$f_{i}^{h}(x_{0},\dots,x_{n})$ is homogeneous of degree $d_{i}$ in
$x_{0},\dots,x_{n-1}$. By the universal property of the multivariable
resultant, (4.1) implies that the equations

$f_{1}^{h}(x_{0},\dots,x_{n-1},\xi_{n})=\cdots=f_{n}^{h}(x_{0},\dots,x_{n-1},\xi_{n})=0$

have a nontrivial solutions
$(\xi_{0},\dots,\xi_{n-1})\in\mathbb{P}^{n-1}_{\mathbb{K}}$. If $\xi_{0}\neq
0$, then we can assume that $\xi_{0}=1$, and it follows easily that
$(\xi_{1},\dots,\xi_{n})\in\mathbf{V}(f_{1},\dots,f_{n})$. Thus $r(\xi_{n})=0$
since $r\in I$. On the other hand, if $\xi_{0}=0$, then this solution may
cause $R_{\text{\it f}\hskip 0.9pt}(x_{n})$ to relate poorly to $r(x_{n})$.
Here are two simple examples with $n=2$:

######  Example 4.1.

Let $f_{1}=1+2x_{1}+x_{2}+2x_{1}x_{2}$ and $f_{2}=3+x_{1}+x_{2}+x_{1}x_{2}$ in
$\mathbb{C}[x_{1},x_{2}]$. One computes that $I=\langle
f_{1},f_{2}\rangle=\langle x_{2}+5,x_{1}+\tfrac{1}{2}\rangle$, which has a
Shape Lemma. Thus $I\cap\mathbb{K}[x_{2}]=\langle x_{2}+5\rangle$. However,

$R_{\text{\it f}\hskip
0.9pt}(x_{2})=\mathrm{Res}_{1,1}(f_{1}^{h},f_{2}^{h})=(x_{2}+1)(x_{2}+5),$

so that $R_{\text{\it f}\hskip 0.9pt}(x_{2})$ does not generate
$I\cap\mathbb{C}[x_{2}]$ because of the extraneous factor of $x_{2}+1$. The
reason for this factor is easy to see. Since
$f_{1}^{h}(x_{0},x_{1},x_{2})=x_{0}+2x_{1}+x_{0}x_{2}+2x_{1}x_{2}$ and
$f_{2}^{h}(x_{0},x_{1},x_{2})=3x_{0}+x_{1}+x_{0}x_{2}+x_{1}x_{2}$, we have

$\displaystyle f_{1}^{h}(0,x_{1},x_{2})=f_{2}^{h}(0,x_{1},x_{2})=0$
$\displaystyle\Longleftrightarrow 2x_{1}+2x_{1}x_{2}=x_{1}+x_{1}x_{2}=0$
$\displaystyle\Longleftrightarrow 2x_{1}(1+x_{2})=x_{1}(1+x_{2})=0.$

But $x_{0},x_{1}$ are homogeneous coordinates, so $x_{0}=0$ implies $x_{1}\neq
0$, and then we can assume $x_{1}=1$. The solution
$(x_{0},x_{1},x_{2})=(0,1,-1)$ “at $\infty$” accounts for the extraneous
factor of $x_{2}+1$ in the resultant.

######  Example 4.2.

Let $f_{1}=x_{1}^{2}x_{2}+x_{1}+x_{2}+1$ and
$f_{2}=x_{1}^{3}x_{2}^{2}+x_{1}-x_{2}+1$ in $\mathbb{C}[x_{1},x_{2}]$. One
computes that

$I=\langle f_{1},f_{2}\rangle=\langle
x_{2}(x_{2}^{4}-3x_{2}^{3}-x_{2}^{2}-5x_{2}+6),4x_{1}+x_{2}^{4}+3x_{2}^{3}+x_{2}^{2}-x_{2}+4\rangle.$

Thus $(-1,0)$ is a solution of $f_{1}=f_{2}=0$, and since
$x_{2}^{4}-3x_{2}^{3}-x_{2}^{2}-5x_{2}+6$ is irreducible over $\mathbb{Q}$,
its roots are distinct and give four more solutions. Since $I$ has a Shape
Lemma, all solutions have multiplicity one by Lemma 2.1. However,

$R_{\text{\it f}\hskip
0.9pt}(x_{3})=\mathrm{Res}_{2,3}(f_{1}^{h},f_{2}^{h})=x_{2}^{2}(x_{2}^{4}-3x_{2}^{3}-x_{2}^{2}-5x_{2}+6),$

does not generate the elimination ideal $I\cap\mathbb{C}[x_{2}]=\langle
x_{2}(x_{2}^{4}-3x_{2}^{3}-x_{2}^{2}-5x_{2}+6)\rangle$. One can check that
$(x_{0},x_{1},x_{2})=(0,1,0)$ is a solution “at $\infty$.” This explains why
the exponent of $x_{2}$ in the resultant is strictly bigger than the exponent
that appears in the generator of the elimination ideal.

In the setting of this paper, here is precisely what we mean by “at $\infty$”:

######  Definition 4.1.

Polynomials $f_{1},\dots,f_{n}\in\mathbb{K}[x_{1},\dots,x_{n}]$ have a
solution at $\infty$ with respect to $x_{1},\dots,x_{n-1}$ if there are
$\xi_{1},\dots,\xi_{n}\in\mathbb{K}$ with
$(\xi_{1},\dots,\xi_{n-1})\neq(0,\dots,0)$ such that the homogenizations
$f_{1}^{h},\dots,f_{n}^{h}\in\mathbb{K}[x_{0},\dots,x_{n}]$ satisfy

$f_{1}^{h}(0,\xi_{1},\dots,\xi_{n-1},\xi_{n})=\cdots=f_{n}^{h}(0,\xi_{1},\dots,\xi_{n-1},\xi_{n})=0.$

We will write “solution at $\infty$” when the context is clear. Solutions at
$\infty$ are easy to understand from a geometric point of view. Being
homogeneous with respect to $x_{0},x_{1},\dots,x_{n-1}$ means that
$f_{1}^{h},\dots,f_{n}^{h}$ define
$\mathbf{V}(f_{1}^{h},\dots,f_{n}^{h})\subseteq\mathbb{P}^{n-1}_{\mathbb{K}}\times_{\mathbb{K}}\mathbb{A}_{\mathbb{K}}^{1}$,
and decomposing $\mathbb{P}^{n-1}_{\mathbb{K}}$ into the affine space
$\mathbb{A}_{\mathbb{K}}^{n-1}$ (where $x_{0}\neq 0$) and the hyperplane at
$\infty$ (where $x_{0}=0$) gives the disjoint union

(4.2) $\displaystyle\mathbf{V}(f_{1}^{h},\dots,f_{n}^{h})$
$\displaystyle=\mathbf{V}(f_{1},\dots,f_{n})\cup\\{\text{solutions with
}x_{0}=0\\}$
$\displaystyle=\mathbf{V}(f_{1},\dots,f_{n})\cup\\{\text{solutions at
}\infty\\}.$

Recall that Theorem 1.2 states that if $I=\langle f_{1},\dots,f_{n}\rangle$ is
a zero-dimensional ideal such that projection
$\mathbf{V}(I)\to\mathbb{A}_{\mathbb{K}}^{1}$ onto the $n$th coordinate is
injective as a map of sets, then any two of the following three conditions
imply the third:

  1. (1)

$I$ has a Shape Lemma.

  2. (2)

$f_{1},\dots,f_{n}$ have no solutions at $\infty$.

  3. (3)

$I\cap\mathbb{K}[x_{n}]=\langle R_{\text{\it f}\hskip 0.9pt}(x_{n})\rangle$.

In other words, if any one of the above conditions holds, then the other two
are equivalent.

###### Proof of Theorem 1.2.

First note that when (1) holds, $I$ can be written in the form $\langle
r(x_{n}),x_{1}-g_{1}(x_{n}),\dots,x_{n-1}-g_{n-1}(x_{n})\rangle$, which
implies $I\cap\mathbb{K}[x_{n}]=\langle r(x_{n})\rangle$. By our injectivity
hypothesis, the roots $\xi_{n}$ of $r(x_{n})$ are indexed by solutions
$\xi=(\xi_{1},\dots,\xi_{n})\in\mathbf{V}(I)=\mathbf{V}(f_{1},\dots,f_{n})$,
and by Lemma 2.1, the multiplicity of $\xi_{n}$ as a root of $r(x_{n})$ is the
multiplicity $m_{\xi}$ of $\xi$ as a solution. Since $r(x_{n})$ can be assumed
to be monic, we see that (1) allows us to write

(4.3) $r(x_{n})=\prod_{\xi\in\mathbf{V}(f_{1},\dots,f_{n})}\\!\\!\
(x_{n}-\xi_{n})^{m_{\xi}}.$

(1) & (2) $\Rightarrow$ (3): By (1), $r(x_{n})$ is given by (4.3). Note that
$\mathbf{V}(I)$ is finite since $I$ is zero-dimensional. Combining this with
(2) and (4.2), we see that $\mathbf{V}(f_{1}^{h},\dots,f_{n}^{h})$ is also
finite. Then the resultant formula in Theorem 1.1 implies that for some
nonzero constant $c$,

$R_{\text{\it f}\hskip
0.9pt}(x_{n})=c\\!\\!\prod_{\xi\in\mathbf{V}(f_{1}^{h},\dots,f_{n}^{h})}\\!\\!\
(x_{n}-\xi_{n})^{m_{\xi}}=c\\!\\!\prod_{\xi\in\mathbf{V}(f_{1},\dots,f_{n})}\\!\\!(x_{n}-\xi_{n})^{m_{\xi}}$

where the second equality follows since there are no solutions at $\infty$. By
(4.3), $R_{\text{\it f}\hskip 0.9pt}(x_{n})=c\hskip 1.0ptr(x_{n})$, and (3)
follows.

(2) & (3) $\Rightarrow$ (1): As in the previous paragraph, our hypothesis and
(2) imply that $R_{\text{\it f}\hskip
0.9pt}(x_{n})=c\prod_{\xi\in\mathbf{V}(f_{1},\dots,f_{n})}(x_{n}-\xi_{n})^{m_{\xi}}$.
By (3), this is $r(x_{n})$ up to a constant, so that by our injectivity
hypothesis, for every $\xi=(\xi_{1},\dots,\xi_{n})\in\mathbf{V}(I)$, the
multiplicity of $\xi\in\mathbf{V}(I)$ equals the multiplicity of $\xi_{n}$ as
a root of $r(x_{n})$. Then (1) follows from Lemma 2.2.

(1) & (3) $\Rightarrow$ (2): There are two cases where this implication is
easy. First, if there are only finitely many solutions at $\infty$, then the
product formula of Theorem 1.1 shows that solutions at infinity contribute
factors of positive degree to the resultant, which makes it easy to see that
$\deg(R_{\text{\it f}\hskip 0.9pt}(x_{n}))>\deg(r(x_{n}))$ when there are
solutions at $\infty$. A second easy case is when there is a solution at
$\infty$ that does not lie above the roots of $r(x_{n})$. Here, the universal
property of the resultant implies that $R_{\text{\it f}\hskip 0.9pt}(x_{n})$
has more distinct roots than $r(x_{n})$, so again the two cannot be equal.

However, when there are infinitely many points at $\infty$, all of which lie
over roots of $r(x_{n})$ (Example 4.3 shows that this can happen), the two
previous cases do not apply. Hence we need to take a different approach. We
will show that if (1) holds and there is a solution at $\infty$, then
$\deg(R_{\text{\it f}\hskip 0.9pt}(x_{n}))>\deg(r(x_{n}))$, which implies that
$I\cap\mathbb{K}[x_{n}]\neq\langle R_{\text{\it f}\hskip
0.9pt}(x_{n})\rangle$. Our proof will use a deformation argument to reduce to
the case when $\mathbf{V}(f_{1}^{h},\dots,f_{n}^{h})$ is finite

Suppose that there is a point $\xi^{*}=([0\hskip 0.85pt{:}\hskip
0.85pt\xi^{*}_{1}\hskip 0.85pt{:}\hskip 0.85pt\ldots\hskip 0.85pt{:}\hskip
0.85pt\xi^{*}_{n-1}],\xi^{*}_{n})\in\mathbf{V}(f_{1}^{h},\ldots,f_{n}^{h})$.
Without loss of generality we may assume $\xi^{*}_{1}\neq 0$. Then define

$\displaystyle g_{1}$ $\displaystyle=x_{0}^{d_{1}}$ $\displaystyle g_{2}$
$\displaystyle=(\xi^{*}_{2}x_{1}-\xi^{*}_{1}x_{2})^{d_{2}}$ $\displaystyle\
\,\vdots$ $\displaystyle g_{n-1}$
$\displaystyle=(\xi^{*}_{n-1}x_{1}-\xi^{*}_{1}x_{n-1})^{d_{n-1}}$
$\displaystyle g_{n}$ $\displaystyle=x_{1}^{d_{n}}(x_{n}-\xi^{*}_{n}).$

Let $\varepsilon$ be a new parameter, and consider the deformed system

(4.4) $f^{h}_{i,\varepsilon}:=f^{h}_{i}+\varepsilon\cdot g_{i},\quad
i=1,\dots,n,$

We will consider solutions of the deformed system over the algebraically
closed field $\widehat{\mathbb{K}}$ whose elements $\hat{h}$ consist of formal
series

$\hat{h}=\sum_{e\in S}c_{e}\varepsilon^{e}$

where $S\subset\mathbb{Q}$ is a well-ordered subset depending on $\hat{h}$,
$c_{e}\in\mathbb{K}$ for all $e\in S$, and $S$ has the property that for some
positive integer $m$,

$mS\subseteq\begin{cases}\mathbb{Z}&\mathrm{ch}(\mathbb{K})=0\\\
\bigcup_{k=0}^{\infty}\tfrac{1}{p^{k}}\mathbb{Z}&\mathrm{ch}(\mathbb{K})=p>0.\end{cases}$

When $\mathrm{ch}(\mathbb{K})=0$, $\widehat{\mathbb{K}}$ is the field of
Puiseux series $\bigcup_{m=1}^{\infty}\mathbb{K}((\varepsilon^{1/m}))$, which
is known to be algebraically closed. When $\mathrm{ch}(\mathbb{K})=p>0$,
$\widehat{\mathbb{K}}$ is algebraically closed by [Ray68]. The paper [Ked01]
describes a smaller algebraically closed field, but we prefer
$\widehat{\mathbb{K}}$ because it is easier to describe. (References to other
proofs that $\widehat{\mathbb{K}}$ is algebraically closed can be found in
[Ked01].)

The solutions of the deformed system give
$\mathbf{V}(f_{1,\varepsilon}^{h},\ldots,f_{n,\varepsilon}^{h})\subseteq\mathbb{P}^{n-1}_{\widehat{\mathbb{K}}}\times_{\widehat{\mathbb{K}}}\mathbb{A}_{\widehat{\mathbb{K}}}^{1}$.
Note that $\xi^{*}=([0\hskip 0.85pt{:}\hskip 0.85pt\xi^{*}_{1}\hskip
0.85pt{:}\hskip 0.85pt\ldots\hskip 0.85pt{:}\hskip
0.85pt\xi^{*}_{n-1}],\xi^{*}_{n})\in\mathbf{V}(f_{1,\varepsilon}^{h},\ldots,f_{n,\varepsilon}^{h})$
by the construction of the $f_{i,\varepsilon}^{h}$.

An element $\hat{h}=\sum_{s\in S}c_{s}\varepsilon^{s}\in\widehat{\mathbb{K}}$
can be regarded as a generalized Puiseux series. We often need to take the
limit as $\varepsilon\to 0$. To explain what this means, suppose that
$\hat{h}\neq 0$, so that $\hat{h}=c_{s}\hskip 1.0pt\varepsilon^{s}+\cdots$,
where $s$ is the minimal element of $S$ such that $c_{s}\neq 0$ (remember that
$S$ is well-ordered). Thus all other nonzero terms of the expansion are of the
form $c_{s^{\prime}}\,\varepsilon^{s^{\prime}}$, with
$c_{s^{\prime}}\in\mathbb{K}$ and $s^{\prime}>s$ in $S$. We define
$\mathrm{val}(\hat{h})=s$. Note that $\mathrm{val}$ is a discrete valuation on
any subfield of $\widehat{\mathbb{K}}$ that is finitely generated over
$\mathbb{K}$. The limit $\lim_{\varepsilon\to 0}\hat{h}$ means setting
$\varepsilon=0$ in $\hat{h}$. If we write $\hat{h}=c_{s}\hskip
1.0pt\varepsilon^{s}+\cdots$ as above, then

$\lim_{\varepsilon\to 0}\hat{h}=\begin{cases}\infty&s<0\\\ c_{0}&s=0\\\
0&s>0.\end{cases}$

In particular, $\lim_{\varepsilon\to 0}\hat{h}\in\mathbb{K}^{\times}$ if and
only if $\mathrm{val}(\hat{h})=0$.

We claim that
$\mathbf{V}(f_{1,\varepsilon}^{h},\ldots,f_{n,\varepsilon}^{h})\subseteq\mathbb{P}^{n-1}_{\widehat{\mathbb{K}}}\times_{\widehat{\mathbb{K}}}\mathbb{A}_{\widehat{\mathbb{K}}}^{1}$
is finite. To prove this, first note that $\lim_{\varepsilon\to 0}R_{\text{\it
f}_{\varepsilon}\hskip 0.8pt}(x_{n})=R_{\text{\it f}\hskip 0.9pt}(x_{n})\neq
0$. Thus $R_{\text{\it f}_{\varepsilon}\hskip 0.8pt}(x_{n})\neq 0$, so that
only finitely many last coordinates occur in
$\mathbf{V}(f_{1,\varepsilon}^{h},\ldots,f_{n,\varepsilon}^{h})$. If
$\mathbf{V}(f_{1,\varepsilon}^{h},\ldots,f_{n,\varepsilon}^{h})$ is infinite,
there must thus be some $\hat{\xi}_{n}\in\widehat{\mathbb{K}}$ such that

$\mathbf{V}(f_{1,\varepsilon}^{h}|_{x_{n}=\hat{\xi}_{n}},\ldots,f_{n,\varepsilon}^{h}|_{x_{n}=\hat{\xi}_{n}})\subseteq\mathbb{P}^{n-1}_{\widehat{\mathbb{K}}}$

has positive dimension. In particular, this variety must meet the hyperplane
$x_{1}=0$, so that ignoring $f_{n,\varepsilon}$ for the moment, the $n-1$
homogeneous equations

$f_{1,\varepsilon}^{h}|_{x_{1}=0,x_{n}=\hat{\xi}_{n}}=\cdots=f_{n-1,\varepsilon}^{h}|_{x_{1}=0,x_{n}=\hat{\xi}_{n}}=0$

in $n-1$ variables $x_{0},x_{2},\dots,x_{n-1}$ have a nontrivial solution in
$\mathbb{P}^{n-2}_{\widehat{\mathbb{K}}}$. Hence

(4.5)
$\mathrm{Res}_{d_{1},\ldots,d_{n-1}}(f_{1,\varepsilon}^{h}|_{x_{1}=0,x_{n}=\hat{\xi}_{n}},\ldots,f_{n-1,\varepsilon}^{h}|_{x_{1}=0,x_{n}=\hat{\xi}_{n}})=0.$

However, it is easy to see that

$\mathrm{Res}_{d_{1},\ldots,d_{n-1}}(f_{1,\varepsilon}^{h}|_{x_{1}=0,x_{n}=\hat{\xi}_{n}},\ldots,f_{n-1,\varepsilon}^{h}|_{x_{1}=0,x_{n}=\hat{\xi}_{n}})=(-\xi^{*}_{1})^{N}\varepsilon^{M}\\!+\mbox{lower
terms in}\,\varepsilon,$

with $N,M\in\mathbb{N}$. This is clearly nonzero, contradicting (4.5). So
$\mathbf{V}(f_{1,\varepsilon}^{h},\dots,f_{n,\varepsilon}^{h})$ must be
finite.

By Theorem 1.1, it follows that there is a nonzero
$c(\varepsilon)\in\widehat{\mathbb{K}}$ such that

$R_{\text{\it f}_{\varepsilon}\hskip
0.8pt}(x_{n})=c(\varepsilon)\\!\\!\prod_{\hat{\xi}\in\mathbf{V}(f_{1,\varepsilon}^{h},\ldots,f^{h}_{n,\varepsilon})}\\!\\!(x_{n}-\hat{\xi}_{n})^{m_{\hat{\xi}}}$

The limit $\lim_{\varepsilon\to 0}\hat{\xi}=\lim_{\varepsilon\to
0}([\hat{\xi}_{0}\hskip 0.85pt{:}\hskip 0.85pt\dots\hskip 0.85pt{:}\hskip
0.85pt\hat{\xi}_{n-1}],\hat{\xi}_{n})$ is a point $\xi$ in the complete
variety $\mathbb{P}^{n-1}_{\mathbb{K}}\times\mathbb{P}^{1}_{\mathbb{K}}$. Let
homogeneous coordinates be $x_{0},\dots,x_{n-1}$ for
$\mathbb{P}^{n-1}_{\mathbb{K}}$ and $x_{n},x_{n+1}$ for
$\mathbb{P}^{1}_{\mathbb{K}}$, where $x_{n+1}$ is a new variable. Recall that
$\mathbf{V}(f_{1}^{h},\dots,f_{n}^{h})$ is a subvariety of
$\mathbb{P}^{n-1}_{\mathbb{K}}\times_{\mathbb{K}}\mathbb{A}_{\mathbb{K}}^{1}$.
Thus, if
$\hat{\xi}\in\mathbf{V}(f_{1,\varepsilon}^{h},\ldots,f^{h}_{n,\varepsilon})$,
then $\lim_{\varepsilon\to 0}\hat{\xi}=\xi$, where

$\xi\in\mathbf{V}(f_{1}^{h},\dots,f_{n}^{h})\cup\\{x_{n+1}=0\\}=\underbrace{\mathbf{V}(f_{1},\dots,f_{n})}_{A}\cup\underbrace{\\{x_{0}=0\\}\cup\\{x_{n+1}=0\\}}_{B}.$

In what follows, we often write $\hat{\xi}\to\xi$ when
$\hat{\xi}\in\mathbf{V}(f_{1,\varepsilon}^{h},\ldots,f^{h}_{n,\varepsilon})$
and $\lim_{\varepsilon\to 0}\hat{\xi}=\xi$.

Using this notation, we can write $R_{\text{\it f}_{\varepsilon}\hskip
0.8pt}(x_{n})$ as the product $R_{\text{\it f}_{\varepsilon}\hskip
0.8pt}(x_{n})=RS$, where

$R=\\!\\!\prod_{\hat{\xi}\to\xi\in
A}\\!\\!(x_{n}-\hat{\xi}_{n})^{m_{\hat{\xi}}}\ \text{ and }\
S=c(\varepsilon)\\!\\!\prod_{\hat{\xi}\to\xi\in
B}\\!\\!(x_{n}-\hat{\xi}_{n})^{m_{\hat{\xi}}}.$

Note that $x_{n}-\xi_{n}^{*}$ is a factor of $S$ since $\xi^{*}=([0\hskip
0.85pt{:}\hskip 0.85pt\xi_{1}^{*}\hskip 0.85pt{:}\hskip 0.85pt\dots\hskip
0.85pt{:}\hskip
0.85pt\xi_{n-1}^{*}],\xi_{n}^{*})\in\mathbf{V}(f_{1,\varepsilon}^{h},\ldots,f^{h}_{n,\varepsilon})$.

The next step is to apply $\lim_{\varepsilon\to 0}$ to the equation
$R_{\text{\it f}_{\varepsilon}\hskip 0.8pt}(x_{n})=RS$. On the left-hand side
of this equation, we already observed that $\lim_{\varepsilon\to
0}R_{\text{\it f}_{\varepsilon}\hskip 0.8pt}(x_{n})=R_{\text{\it f}\hskip
0.9pt}(x_{n})$, which is a nonzero polynomial in $x_{n}$. To take the limit on
the right-hand side, we first focus on $\lim_{\varepsilon\to 0}R$.

All solutions $\xi\in A$ arise as limits $\hat{\xi}\to\xi$ for some
$\hat{\xi}\in\mathbf{V}(f_{1,\varepsilon}^{h},\dots,f_{n,\varepsilon}^{h})$ in
a way that is compatible with multiplicities. Note also that
$\xi=(\xi_{1},\dots,\xi_{n})=([1\hskip 0.85pt{:}\hskip 0.85pt\xi_{1}\hskip
0.85pt{:}\hskip 0.85pt\dots\hskip 0.85pt{:}\hskip
0.85pt\xi_{n-1}],\xi_{n})\in\mathbb{P}^{n-1}_{\mathbb{K}}\times_{\mathbb{K}}\mathbb{A}_{\mathbb{K}}^{1}$,
and then $\hat{\xi}=([\hat{\xi}_{0}\hskip 0.85pt{:}\hskip 0.85pt\dots\hskip
0.85pt{:}\hskip 0.85pt\hat{\xi}_{n-1}],\hat{\xi}_{n})\to\xi$ implies that
$\lim_{\varepsilon\to 0}\hat{\xi}_{n}=\xi_{n}$.

Now fix $\xi\in\mathbf{V}(f_{1}^{h},\dots,f_{n}^{h})$ and consider all
$\hat{\xi}\in\mathbf{V}(\hat{f}_{1},\dots,\hat{f}_{n})$ such that
$\hat{\xi}\to\xi$. The previous paragraph implies that

(4.6) $\lim_{\varepsilon\to 0}\prod_{\hat{\xi}\to\xi\in
A}(x_{n}-\hat{\xi}_{n})^{m_{\hat{\xi}}}=(x_{n}-\xi_{n})^{m_{\xi}},$

and since $A=\mathbf{V}(f_{1},\dots,f_{n})$, it follows immediately that

$\lim_{\varepsilon\to 0}R=\lim_{\varepsilon\to 0}\prod_{\hat{\xi}\to\xi\in
A}(x_{n}-\hat{\xi}_{n})^{m_{\hat{\xi}}}\
=\prod_{\xi\in\mathbf{V}(f_{1},\dots,f_{n})}\\!\\!\
(x_{n}-\xi_{n})^{m_{\xi}}=r(x_{n}),$

where the last equality follows from (4.3). This is a nonzero polynomial, so
that

$R_{\text{\it f}\hskip 0.9pt}(x_{n})=\lim_{\varepsilon\to 0}R_{\text{\it
f}_{\varepsilon}\hskip 0.8pt}(x_{n})=\lim_{\varepsilon\to
0}RS=r(x_{n})\lim_{\varepsilon\to 0}S.$

Since $S$ is a polynomial in $x_{n}$, it follows that $\lim_{\varepsilon\to
0}S$ is also a polynomial in $x_{n}$. We noted above that $x_{n}-\xi_{n}^{*}$
is a factor of $S$. Being independent of $\varepsilon$, it becomes a factor of
$\lim_{\varepsilon\to 0}S$. This shows that $\deg(R_{\text{\it f}\hskip
0.9pt}(x_{n}))>\deg(r(x_{n}))$, concluding the proof of (1) & (3)
$\Rightarrow$ (2). ∎

######  Example 4.3.

Let $f_{1}=1+x_{1}+x_{1}x_{2}x_{3}$, $f_{2}=x_{2}+x_{1}^{2}x_{3}$ and
$f_{3}=1+x_{1}+2x_{2}+x_{2}^{2}x_{3}$ in $\mathbb{C}[x_{1},x_{2},x_{3}]$. Then
one computes that

$I=\langle f_{1},f_{2},f_{3}\rangle=\langle
x_{3}^{2}+6x_{3},x_{1}+\tfrac{2}{9}x_{3}+1,x_{2}+\tfrac{1}{9}x_{3}\rangle.$

Thus $I$ has a Shape Lemma with solutions $(-1,0,0)$ and
$(\tfrac{1}{3},\tfrac{2}{3},-6)$, both of multiplicity one. Note also that
$I\cap\mathbb{C}[x_{3}]=\langle r(x_{3})\rangle$ for
$r(x_{3})=x_{3}(x_{3}+6)$.

However, when we homogenize, we get

$f_{1}^{h}=x_{0}^{2}+x_{0}x_{1}+x_{1}x_{2}x_{3},\
f_{2}^{h}=x_{0}x_{2}+x_{1}^{2}x_{3},\
f_{3}^{h}=x_{0}^{2}+x_{0}x_{1}+2x_{0}x_{2}+x_{2}^{2}x_{3}$

in $A[x_{0},x_{1},x_{2}]$, $A=\mathbb{C}[x_{3}]$. Using the classical formula
for the resultant of three ternary quadrics (see, for example [CLO05, (2.8)]),
one obtains

$R_{\text{\it f}\hskip
0.9pt}(x_{3})=\mathrm{Res}_{2,2,2}(f_{1}^{h},f_{2}^{h},f_{3}^{h})=x_{3}^{7}(x_{3}+6).$

By Theorem 1.2, there must be at least one solution at $\infty$. In fact,
there are a lot, since $f_{i}^{h}(0,x_{1},x_{2},0)=0$, giving a projective
line of solutions at $\infty$, and it is easy to see that these are the only
solutions at $\infty$.

In the elimination ideal $I\cap\mathbb{C}[x_{3}]=\langle
x_{3}^{2}+6x_{3}\rangle=\langle x_{3}(x_{3}+6)\rangle$, the affine solution
$(-1,0,0)$ contributes the factor of $x_{3}$. But in the resultant, the
solutions at $\infty$ (all of which have $x_{3}=0$) cause the exponent of
$x_{3}$ to increase from $1$ to $7$. This is mysterious. It would be nice to
have a theoretical explanation of the exponent.

We conclude this section with the special case $n=2$, where solutions at
$\infty$ are easy to understand. Given
$f_{1},f_{2}\in\mathbb{K}[x_{1},x_{2}]$, write them as

$f_{i}(x_{1},x_{2})=\sum_{j=0}^{d_{i}}a_{ij}(x_{2})x_{1}^{j},\quad
d_{i}=\deg_{x_{1}}(f_{i}),$

and let $\mathrm{lc}_{x_{1}}(f_{i})=a_{id_{i}}(x_{2})$ be the leading
coefficient of $f_{i}$ with respect to $x_{1}$. Note that
$\mathrm{lc}_{x_{1}}(f_{i})\in\mathbb{K}[x_{2}]$ is nonzero. Then

$f_{i}^{h}(x_{0},x_{1},x_{2})=\sum_{j=0}^{d_{i}}a_{ij}(x_{2})x_{0}^{d_{i}-j}x_{1}^{j},$

and $f_{i}^{h}(0,x_{1},x_{2})=\mathrm{lc}(f_{i},x_{1})x_{1}^{d_{i}}$. The
solutions at $\infty$ lie in
$\mathbf{V}(f_{1}^{h},f_{2}^{h})\subseteq\mathbb{P}^{1}_{\mathbb{K}}\times_{\mathbb{K}}\mathbb{A}_{\mathbb{K}}^{1}$,
so that any solution with $x_{0}=0$ must have $x_{1}\neq 0$. Hence we can
assume $x_{1}=1$, and then solutions at $\infty$ are $([0\hskip
0.85pt{:}\hskip 0.85pt1],\xi)$, where

(4.7) $\mathrm{lc}_{x_{1}}(f_{1})(\xi)=\mathrm{lc}_{x_{1}}(f_{2})(\xi)=0.$

This leads to the following corollary of Theorem 1.2 when $n=2$:

######  Corollary 4.1.

Let $I=\langle f_{1},f_{2}\rangle\subseteq\mathbb{K}[x_{1},x_{2}]$ be a zero-
dimensional ideal such that the map
$\mathbf{V}(I)\to\mathbb{A}_{\mathbb{K}}^{1}$ given by projection onto the
second coordinate is injective as a map of sets. Then any two of the following
three conditions imply the third:

  1. (1)

$I$ has a Shape Lemma

  2. (2)

$\gcd(\mathrm{lc}_{x_{1}}(f_{1}),\mathrm{lc}_{x_{1}}(f_{2}))=1$.

  3. (3)

$I\cap\mathbb{K}[x_{2}]=\langle R_{\text{\it f}\hskip 0.9pt}(x_{2})\rangle$

###### Proof.

This follows immediately from Theorem 1.2 since the analysis of solutions at
$\infty$ given in (4.7) shows that
$\gcd(\mathrm{lc}_{x_{1}}(f_{1}),\mathrm{lc}_{x_{1}}(f_{2}))=1$ if and only if
there are no solutions at $\infty$. ∎

##  5\. The Shape Lemma and Subresultants

In this final section of the paper, we highlight the role of subresultants.
The classical theory of subresultants of two univariate polynomials goes back
to the work of Jacobi [Jac36] and Sylvester [Syl40]. Modern accounts can be
found in [AJ06, vzGL03], both of which contain references to many other papers
on subresultants. In the multivariable case, we will follow the definition and
presentation given in [Cha95], where general statements and results are
presented. For our purposes it will be enough to focus on the multivariable
version of the “first subresultant polynomial”.

Let $A$ be an integral domain with field of fractions $k(A)$. Let
$g_{1},\ldots,g_{n}\in A[x_{0},\ldots,x_{n-1}]$ be homogeneous polynomials of
respective degrees $d_{1},\ldots,d_{n}$, and define
$\rho:=d_{1}+\cdots+d_{n}-n$ to be the _critical degree_ of the system. If the
multivariable resultant of these polynomials is nonzero, the degree $\rho$
piece of the graded ring $k(A)[x_{0},\ldots,x_{n-1}]/\langle
g_{1},\ldots,g_{n}\rangle$ has dimension one. While these conditions are not
equivalent (there are systems with nontrivial solutions that also satisfy this
property), in general one expects this dimension to be equal to one.

For any monomial $x^{\alpha}=x_{0}^{\alpha_{0}}\ldots x_{n-1}^{\alpha_{n-1}}$
of degree $\rho$, there exists $s_{\alpha}\in A$ that is a polynomial in the
coefficients of $g_{1},\ldots,g_{n}$ and vanishes if and only if the class of
$x^{\alpha}$ fails to be a basis of $(k(A)[x_{0},\ldots,x_{n-1}]/\langle
g_{1},\ldots,g_{n}\rangle)_{\rho}$. We call $s_{\alpha}$ the _scalar
subresultant_ associated to $x^{\alpha}$.

######  Proposition 5.1 (Theorems $1$ and $2$ in [Cha95]).

With notation as above:

  1. (1)

$s_{\alpha}x^{\beta}-s_{\beta}x^{\alpha}\in\langle
g_{1},\ldots,g_{n}\rangle\subseteq A[x_{0},\ldots,x_{n-1}]$ for all
$\alpha,\beta$ of degree $\rho$.

  2. (2)

$s_{\alpha}=0$ for all $\alpha$ of degree $\rho$ if and only if

$\dim\hskip 0.5pt(k(A)[x_{0},\ldots,x_{n-1}]/\langle
g_{1},\ldots,g_{n}\rangle)_{\rho}>1.$

Note that $\rho=0$ if and only if $d_{1}=\cdots=d_{n}=1$, in which case the
monomial $1$ is the only one of critical degree. Its subresultant is then
defined as $s_{(0,\ldots,0)}=1$, which fulfills the conditions of Proposition
5.1. In what follows, we will always assume that $\rho\geq 1$, For readers
interested in the computational aspects of multivariable subresultants, we
recommend the treatment given in [GV91, Cha94, Cha95].

In this paper, we are dealing with polynomials
$f_{1},\ldots,f_{n}\in\mathbb{K}[x_{1},\ldots,x_{n}]$. Their homogenizations
with respect to $x_{1},\dots,x_{n-1}$ are $f_{1}^{h},\ldots,f_{n}^{h}\in
A[x_{0},\ldots,x_{n-1}]$ for $A=\mathbb{K}[x_{n}]$. Thus
$s_{\alpha}=s_{\alpha}(x_{n})\in\mathbb{K}[x_{n}]$ for all $\alpha$ of degree
$\rho$. In this situation, Proposition 5.1 gives the following useful result:

######  Proposition 5.2.

Assume that $I=\langle f_{1},\ldots,f_{n}\rangle$ is zero-dimensional with
$\rho\geq 1$. Given $\xi_{n}\in\mathbb{K}$, the following are equivalent:

  1. (1)

$R_{\text{\it f}\hskip 0.9pt}(\xi_{n})=0$ and $s_{\alpha}(\xi_{n})\neq 0$

  2. (2)

The fiber of
$\mathbf{V}(f_{1}^{h},\dots,f_{n}^{h})\to\mathbb{A}_{\mathbb{K}}^{1}$ over
$\xi_{n}$ consists of a single reduced point given by $\xi=([\xi_{0}\hskip
0.85pt{:}\hskip 0.85pt\dots\hskip 0.85pt{:}\hskip
0.85pt\xi_{n-1}],\xi_{n})\in\mathbf{V}(f_{1}^{h},\dots,f_{n}^{h})$ and
$\xi^{\alpha}\neq 0$, where
$\xi^{\alpha}=\xi_{0}^{\alpha_{0}}\dots\xi_{n-1}^{\alpha_{n-1}}$.

###### Proof.

(1) $\Rightarrow$ (2): By the universal property of resultants, $R_{\text{\it
f}\hskip 0.9pt}(\xi_{n})=0$ implies that the fiber has at least one point
$\xi$, and since $s_{\alpha}(\xi_{n})\neq 0$, (2) of Proposition 5.1 implies
that the Hilbert function of the fiber above the specialized system in the
critical degree coincides with the one of a complete intersection of a single
point. It follows that the fiber consists of a single reduced point. To show
that $\xi^{\alpha}\neq 0$, pick $x^{\beta}$ of degree $\rho$ such that
$\xi^{\beta}\neq 0$. If $\beta=\alpha$ we are done. Otherwise, by (1) of
Proposition 5.1, we have
$s_{\alpha}(\xi_{n})\xi^{\beta}=s_{\beta}(\xi_{n})\xi^{\alpha}$. Since
$s_{\alpha}(\xi_{n})\neq 0\neq\xi^{\beta},$ the same is true for
$\xi^{\alpha}$.

(2) $\Rightarrow$ (1): If the fiber over $\xi_{n}$ is nonempty, then
$R_{\text{\it f}\hskip 0.9pt}(\xi_{n})$ must be zero. In addition, if it is a
single reduced point, by [Cha94, Corollaire 2], the dimension of the $\rho$-th
degree part of $\mathbb{K}[x_{0},\ldots,x_{n-1}]/\langle
f_{1}^{h}(x,\xi_{n}),\dots,f_{n}^{h}(x,\xi_{n})\rangle$ must be equal to one.
As $\xi^{\alpha}\neq 0,$ the monomial $x_{0}^{\alpha_{0}}\ldots
x_{n-1}^{\alpha_{n-1}}$ is a basis of this $\mathbb{K}$-vector space, which
implies that $s_{\alpha}(\xi_{n})\neq 0$ because of the definition of
$s_{\alpha}$ given above. ∎

For our purposes, certain scalar subresultants are especially useful. Suppose
that $\rho\geq 1$ and consider the monomials
$x^{\alpha(i)}=x_{0}^{\rho-1}x_{i}$, $i=0,\dots,n-1$, of degree $\rho$. For
simplicity, the scalar subresultant $s_{\alpha(i)}(x_{n})$ will be denoted
$s_{i}(x_{n})$ in what follows. Proposition 5.1 implies that the polynomials
$s_{0}(x_{n})x_{0}^{\rho-1}x_{i}-s_{i}(x_{n})x_{0}^{\rho}$ belong to the ideal
$\langle
f_{1}^{h},\ldots,f_{n}^{h}\rangle\subset\mathbb{K}[x_{0},\ldots,x_{n-1}]$ for
$i=1,\ldots,n-1$. Setting $x_{0}=1$ gives the following polynomials:

(5.1) $p_{i}(x_{i},x_{n}):=s_{0}(x_{n})\,x_{i}-s_{i}(x_{n})\in\langle
f_{1},\ldots,f_{n}\rangle\cap\mathbb{K}[x_{i},x_{n}],\quad i=1,\dots,n-1.$

These are the _first subresultant polynomials_ of $f_{1},\dots,f_{n}$.

The main theorems of this section involve the ideal

(5.2) $\langle R_{\text{\it f}\hskip
0.9pt}(x_{n}),p_{1}(x_{1},x_{n}),\ldots,p_{n-1}(x_{n-1},x_{n})\rangle\subseteq
I$

generated by the resultant and first subresultant polynomials. But before
giving the proofs, we need the following general lemma about the inclusion
(5.2):

######  Lemma 5.1.

Given $d(x_{n}),d_{0}(x_{n}),\dots,d_{n-1}(x_{n})\in\mathbb{K}[x_{n}]$ with
$d(x_{n})\neq 0$, set

(5.3) $J:=\langle
d(x_{n}),d_{0}(x_{n})x_{1}-d_{1}(x_{n}),\dots,d_{0}(x_{n})x_{n-1}-d_{n-1}(x_{n})\rangle.$

Then $J$ is zero-dimensional if and only if
$\gcd(d(x_{n}),d_{0}(x_{n}),\dots,d_{n-1}(x_{n}))=1$. Furthermore, if these
conditions are satisfied and $J\subseteq
I\subseteq\mathbb{K}[x_{1},\dots,x_{n}]$, then:

  1. (1)

$I$ has a Shape Lemma.

  2. (2)

If the generator of $I\cap\mathbb{K}[x_{n}]$ lies in $J$, then $I=J$.

###### Proof.

First assume that $J$ is zero-dimensional. Any nontrivial common divisor
$h(x_{n})$ of $d(x_{n}),d_{0}(x_{n}),\dots,d_{n-1}(x_{n})$ divides
$p_{i}(x_{i},x_{n})$ for $i=1,\dots,n-1$, so that $J\subseteq\langle
h(x_{n})\rangle$, which is impossible when $J$ is zero-dimensional.
Conversely, assume that the gcd condition is satisfied. It suffices to prove
that $\mathbf{V}(J)$ is finite. Take
$\xi=(\xi_{1},\dots,\xi_{n})\in\mathbf{V}(J)$. Then $d(\xi_{n})=0$, and since
$d(x_{n})$ is nonzero, there are only finitely many choices of $\xi_{n}$. If
$d_{0}(\xi_{n})=0$, then for $i=1,\dots,n-1$, we have

$0=d_{0}(\xi_{n})\xi_{i}-d_{i}(\xi_{n})=0\cdot\xi_{i}-d_{i}(\xi_{n})=-d_{i}(\xi_{n}),$

which shows that $x_{n}-\xi_{n}$ is a common divisor of
$d(x_{n}),d_{0}(x_{n}),\dots,d_{n-1}(x_{n})$, a contradiction. Thus
$d_{0}(\xi_{n})\neq 0$, and then $\xi_{i}=d_{i}(\xi_{n})/d_{0}(\xi_{n})$ for
$i=1,\dots,n-1$ shows that $\xi$ is uniquely determined by $\xi_{n}$. Hence
$\mathbf{V}(J)$ is finite.

For (1), consider all representations (5.3) of $J$ that satisfy the gcd
condition $\gcd(d(x_{n}),d_{0}(x_{n}),\dots,d_{n-1}(x_{n}))=1$, and pick one
where $\deg(d(x_{n}))$ is minimal. Let
$e_{0}(x_{n}):=\gcd(d(x_{n}),d_{0}(x_{n}))$. Then we can write
$d(x_{n})=e_{0}(x_{n})e(x_{n})$ and $d_{0}(x_{n})=e_{0}(x_{n})E(x_{n})$, and
we have a Bézout identity

(5.4) $A(x_{n})\,e(x_{n})+B(x_{n})\,E(x_{n})=1.$

Also define $e_{i}(x_{n})=B(x_{n})d_{i}(x_{n})$ for $i=1,\ldots,n-1$. We claim
that

(5.5) $\gcd(e(x_{n}),e_{0}(x_{n}),\dots,e_{n-1}(x_{n}))=1.$

If not, there is $\xi_{n}\in\mathbb{K}$ that makes them all vanish. Then
$e(\xi_{n})=0$ and (5.4) imply that $B(\xi_{n})\neq 0$. But
$e_{0}(x_{n})=\gcd(d(x_{n}),d_{0}(x_{n}))$ implies

(5.6)
$\gcd(e_{0}(x_{n}),d_{1}(x_{n}),\dots,d_{n-1}(x_{n}))=\gcd(d(x_{n}),d_{0}(x_{n}),\dots,d_{n-1}(x_{n}))=1.$

Since $e_{0}(\xi_{n})=0$, we must have $d_{i}(\xi_{n})\neq 0$ for some $1\leq
i\leq n-1$, and then $e_{i}(\xi_{n})=B(\xi_{n})d_{i}(\xi_{n})\neq 0$, a
contradiction. This proves (5.5).

We next claim that

(5.7) $J=\langle
e(x_{n}),e_{0}(x_{n})x_{1}-e_{1}(x_{n}),\dots,e_{0}(x_{n})x_{n-1}-e_{n-1}(x_{n})\rangle.$

In what follows, we will omit “$(x_{n})$” for simplicity. To prove (5.7), let
$J^{\prime}$ denote the ideal on the right. For the inclusion
$J^{\prime}\subseteq J$, first note that (5.4) implies

$e_{0}\hskip 1.0ptx_{i}-e_{i}=e_{0}(A\,e+B\,E)x_{i}-Bd_{i}=Ax_{i}\cdot
d+B\cdot(d_{0}x_{i}-d_{i})\in J$

since $d=e_{0}e$, $d_{0}=e_{0}E$, and $e_{i}=Bd_{i}$. Showing that $e\in J$
will take more work. First observe that

$dx_{i}-e(e_{0}x_{i}-e_{i})=ee_{i},\ i=1,\dots,n-1.$

The left-hand side lies in $J$ by what we just proved, so $ee_{i}=eBd_{i}\in
J$. Then

(5.8)
$d_{0}x_{i}-d_{i}-E(e_{0}x_{i}-e_{i})=-d_{i}+EBd_{i}=(-1+EB)d_{i}=-Aed_{i}$

proves that $Aed_{i}\in J$. We showed above that $Bed_{i}\in J$, and since $A$
and $B$ are relatively prime by (5.4), we get $ed_{i}\in J$ for
$i=1,\dots,n-1$. However, we also have $ee_{0}\in J$. Since
$e_{0},d_{1},\dots,d_{n-1}$ are relatively prime by (5.6), it follows that
$e\in J$, completing the proof of $J^{\prime}\subseteq J$.

For $J\subseteq J^{\prime}$, note that $d=e_{0}e\in J^{\prime}$. Also, for
$i=1,\ldots,n-1$, $J^{\prime}$ contains $e$ and $e_{0}x_{i}-e_{i}$, so that
$d_{0}x_{i}-d_{i}\in J^{\prime}$ by (5.8). Thus $J\subseteq J^{\prime}$, and
(5.7) is proved.

When we combine (5.6) and (5.7) with the minimality of $\deg(d(x_{n}))$ in
(5.3), we see that $\deg(e(x_{n}))\geq\deg(d(x_{n}))$. Since
$d(x_{n})=e_{0}(x_{n})e(x_{n})$, it follows that
$e_{0}(x_{n})=\gcd(d(x_{n}),d_{0}(x_{n}))$ is constant, i.e.,
$e_{0}(x_{n})=1$. Then (5.7) becomes

(5.9) $J=\langle
e(x_{n}),x_{1}-e_{1}(x_{n}),\dots,x_{n-1}-e_{n-1}(x_{n})\rangle.$

so that $x_{i}-e_{i}(x_{n})\in J$ for $i=1,\dots,n-1$.

Now let $I$ be an ideal containing $J$ as in the statement of the lemma. Then
$I$ is also zero-dimensional, so that $I\cap\mathbb{K}[x_{n}]=\langle
r(x_{n})\rangle$ for some nonzero $r(x_{n})\in\mathbb{K}[x_{n}]$. It is
straighforward to show that
$G=\\{r(x_{n}),\,x_{1}-e_{1}(x_{n}),\ldots,x_{n-1}-e_{n-1}(x_{n})\\}$ is a
Gröbner basis of $I$ for lex order with $x_{n}\prec x_{n-1}\prec\ldots\prec
x_{1}$. Since a Gröbner basis is a basis, it follows that $I$ has a Shape
Lemma, proving (1).

For (2), let $I\cap\mathbb{K}[x_{n}]=\langle r(x_{n})\rangle$ with $r(x_{n})$
monic. The hypothesis of (2) implies $r(x_{n})\in J$, so that the Gröbner
basis of $I$ constructed in the previous paragraph lies in $J$. The equality
$I=J$ follows immediately. ∎

######  Remark 5.1.

Implicit in the proof of Lemma 5.1 is an algorithm that constructs a Shape
Lemma basis of the ideal (5.3) when
$\gcd(d(x_{n}),d_{0}(x_{n}),\dots,d_{n-1}(x_{n}))=1$. Note also that the
algorithm takes only one step when $\gcd(d(x_{n}),d_{0}(x_{n}))=1$.

Lemma 5.1 gives a nice consequence of (5.2):

######  Corollary 5.1.

$I$ has a Shape Lemma when $\gcd(R_{\text{\it f}\hskip
0.9pt}(x_{n}),s_{0}(x_{n}),\dots,s_{n-1}(x_{n}))=1$.

###### Proof.

Since $p_{i}(x_{i},x_{n})=s_{0}(x_{n})x_{i}-s_{i}(x_{n})$, (5.2) can be
written

$\langle R_{\text{\it f}\hskip
0.9pt}(x_{n}),s_{0}(x_{n})x_{1}-s_{1}(x_{n}),\ldots,s_{0}(x_{n})x_{n-1}-s_{n-1}(x_{n})\rangle\subseteq
I.$

Then our gcd hypothesis and Lemma 5.1 imply that $I$ has a Shape Lemma. ∎

The first main result of Section 5 is Theorem 1.3, which says that if
$I=\langle f_{1},\dots,f_{n}\rangle$ is zero-dimensional and $\rho\geq 1$,
then the conditions

  1. (1)

$I$ has a Shape Lemma and $f_{1},\dots,f_{n}$ have no solutions at $\infty$.

  2. (2)

$I\cap\mathbb{K}[x_{n}]=\langle R_{\text{\it f}\hskip 0.9pt}(x_{n})\rangle$
and $\gcd(R_{\text{\it f}\hskip 0.9pt}(x_{n}),s_{0}(x_{n}))=1$.

  3. (3)

$I=\langle R_{\text{\it f}\hskip
0.9pt}(x_{n}),p_{1}(x_{1},x_{n}),\ldots,p_{n-1}(x_{n-1},x_{n})\rangle$ and
$I\cap\mathbb{K}[x_{n}]=\langle R_{\text{\it f}\hskip 0.9pt}(x_{n})\rangle$.

are equivalent, and when these conditions are all true, we also have

$I\cap\mathbb{K}[x_{i_{1}},\dots,x_{i_{\ell}},x_{n}]=\langle R_{\text{\it
f}\hskip
0.9pt}(x_{n}),p_{i_{0}}(x_{i_{0}},x_{n}),\dots,p_{i_{\ell}}(x_{i_{\ell}},x_{n})\rangle$

whenever $1\leq i_{1}<\cdots<i_{\ell}<n$.

###### Proof of Theorem 1.3.

(1) $\Rightarrow$ (2): Assume that $I$ has a Shape Lemma with no solutions at
$\infty$. The implication (1) & (2) $\Rightarrow$ (3) of Theorem 1.2 implies
that $I\cap\mathbb{K}[x_{n}]=\langle R_{\text{\it f}\hskip
0.9pt}(x_{n})\rangle$ (the injectivity hypothesis of Theorem 1.2 is satisfied
since $I$ has a Shape Lemma).

Now suppose that $\xi_{n}$ is a root of $R_{\text{\it f}\hskip 0.9pt}(x_{n})$.
Since there are no solutions at $\infty$, it follows that there is a solution
$\xi=([1\hskip 0.85pt{:}\hskip 0.85pt\xi_{1}\hskip 0.85pt{:}\hskip
0.85pt\dots\hskip 0.85pt{:}\hskip 0.85pt\xi_{n-1}],\xi_{n})$. Since $I$ has a
Shape Lemma, the fiber over $\xi_{n}$ consists of a single smooth point by
Lemma 2.3. For $x^{\alpha(0)}=x_{0}^{\rho}$, we clearly have
$\xi^{\alpha(0)}\neq 0$, so that $s_{\alpha(0)}(\xi_{n})\neq 0$ by Proposition
5.2. But by definiton, $s_{0}(\xi_{n})=s_{\alpha(0)}(x_{n})$. Thus
$s_{0}(\xi_{n})\neq 0$ whenever $R_{\text{\it f}\hskip 0.9pt}(\xi_{n})=0$,
which proves that $\gcd(R_{\text{\it f}\hskip 0.9pt}(x_{n}),s_{0}(x_{n}))=1$.

(2) $\Rightarrow$ (3): Assume $I\cap\mathbb{K}[x_{n}]=\langle R_{\text{\it
f}\hskip 0.9pt}(x_{n})\rangle$ and $\gcd(R_{\text{\it f}\hskip
0.9pt}(x_{n}),s_{0}(x_{n}))=1$. Then $J=\langle R_{\text{\it f}\hskip
0.9pt}(x_{n}),p_{1}(x_{1},x_{n}),\dots,p_{n-1}(x_{n-1},x_{n})\rangle$
satisfies the the gcd condition of Lemma 5.1. Since $J\subset I$ and the
generator of $I\cap\mathbb{K}[x_{n}]$ lies in $J$, the lemma implies $I=J$.

(3) $\Rightarrow$ (1): Here, we assume that $I=\langle R_{\text{\it f}\hskip
0.9pt}(x_{n}),p_{1}(x_{1},x_{n}),\ldots,p_{n-1}(x_{n-1},x_{n})\rangle$ and
$I\cap\mathbb{K}[x_{n}]=\langle R_{\text{\it f}\hskip 0.9pt}(x_{n})\rangle$.
Since $I$ is zero-dimensional, Lemma 5.1 implies that $I$ has a Shape Lemma.

Since $I$ has a Shape Lemma, Lemma 2.1 implies that the injectivity hypothesis
of Theorem 1.2 is satisfied. Then this theorem and
$I\cap\mathbb{K}[x_{n}]=\langle R_{\text{\it f}\hskip 0.9pt}(x_{n})\rangle$
imply that there are no solutions at $\infty$, completing the proof that (3)
$\Rightarrow$ (1).

Finally, suppose that the conditions of Theorem 1.3 are all true. Then
$R_{\text{\it f}\hskip 0.9pt}(x_{n})$ generates $I\cap\mathbb{K}[x_{n}]$ and
is relatively prime with $s_{0}(x_{n})$. Fix $1\leq i_{1}<\cdots<i_{\ell}<n$
and observe that $\langle R_{\text{\it f}\hskip
0.9pt}(x_{n}),p_{i_{0}}(x_{i_{0}},x_{n}),\dots,p_{i_{\ell}}(x_{i_{\ell}},x_{n})\rangle$
can be written as

(5.10) $\langle R_{\text{\it f}\hskip
0.9pt}(x_{n}),s_{0}(x_{n})x_{i_{0}}-s_{i_{0}}(x_{n}),\dots,s_{o}(x_{n})x_{i_{\ell}}-s_{i_{\ell}}(x_{n})\rangle$

This lies in $I\cap\mathbb{K}[x_{i_{1}},\dots,x_{i_{\ell}},x_{n}]$. Since
$\gcd(R_{\text{\it f}\hskip 0.9pt}(x_{n}),s_{0}(x_{n}))=1$, we have

$\gcd(R_{\text{\it f}\hskip
0.9pt}(x_{n}),s_{0}(x_{n}),s_{i_{1}}(x_{n}),\dots,s_{i_{\ell}}(x_{n}))=1,$

and the generator of
$(I\cap\mathbb{K}[x_{i_{1}},\dots,x_{i_{\ell}},x_{n}])\cap\mathbb{K}[x_{n}]=I\cap\mathbb{K}[x_{n}]=\langle
R_{\text{\it f}\hskip 0.9pt}(x_{n})\rangle$ is in (5.10). Then (2) of Lemma
5.1 gives the desired equality

$I\cap\mathbb{K}[x_{i_{1}},\dots,x_{i_{\ell}},x_{n}]=\langle R_{\text{\it
f}\hskip
0.9pt}(x_{n}),p_{i_{0}}(x_{i_{0}},x_{n}),\dots,p_{i_{\ell}}(x_{i_{\ell}},x_{n})\rangle.\qed$

######  Example 5.1.

Let $f_{1}=1+x_{1}x_{3}$, $f_{2}=1+x_{2}x_{3}$, and
$f_{3}=1+x_{1}^{2}x_{3}+x_{2}^{2}x_{3}+x_{3}^{2}$ in
$\mathbb{C}[x_{1},x_{2},x_{3}]$. One checks that
$f_{i}^{h}(0,x_{1},x_{2},0)=0$, so there are solutions at $\infty$ (in fact,
infinitely many). We compute a Gröbner basis for $I$ to get a Shape Lemma

$I=\langle f_{1},f_{2},f_{3}\rangle=\langle
2+x_{3}+x_{3}^{3},\,x_{1}-\tfrac{1}{2}-\tfrac{1}{2}x_{3}^{2},\,x_{2}-\tfrac{1}{2}-\tfrac{1}{2}x_{3}^{2}\rangle.$

On the other hand, $\rho=1$, and computing resultants and subresultants gives

$\displaystyle R_{\text{\it f}\hskip 0.9pt}(x_{3})$
$\displaystyle=x_{3}^{3}(2+x_{3}+x_{3}^{3})$ $\displaystyle s_{0}(x_{3})$
$\displaystyle=s_{(1,0,0)}(x_{3})=x_{3}^{2}$ $\displaystyle s_{1}(x_{3})$
$\displaystyle=s_{(0,1,0)}(x_{3})=-x_{3}$ $\displaystyle s_{2}(x_{3})$
$\displaystyle=s_{(0,0,1)}(x_{3})=-x_{3},$

so that $p_{1}(x_{1},x_{3})=x_{3}f_{1}$ and $p_{2}(x_{2},x_{3})=x_{3}f_{2}.$
In this case, $x_{3}$ is a factor of all the scalar subresultants. It is fun
to see how all three conditions of Theorem 1.3 fail in this case.

The second main result of this section is Theorem 1.4, which assumes only that
the ideal is generated by the resultant and first subresultant polynomials.
More precisely, for a zero-dimensional ideal $I=\langle
f_{1},\dots,f_{n}\rangle$ with $\rho\geq 1$, Theorem 1.4 asserts that if
$I=\langle R_{\text{\it f}\hskip
0.9pt}(x_{n}),p_{1}(x_{1},x_{n}),\dots,p_{n-1}(x_{n-1},x_{n})\rangle$, then

  1. (1)

$\gcd(R_{\text{\it f}\hskip
0.9pt}(x_{n}),s_{0}(x_{n}),\dots,s_{n-1}(x_{n}))=1$.

  2. (2)

$I$ has a Shape Lemma.

In addition, the following are equivalent:

  1. (3)

$I\cap\mathbb{K}[x_{n}]=\langle R_{\text{\it f}\hskip 0.9pt}(x_{n})\rangle$

  2. (4)

$\gcd(R_{\text{\it f}\hskip 0.9pt}(x_{n}),s_{0}(x_{n}))=1$, where
$p_{i}(x_{i},x_{n})=s_{0}(x_{n})x_{i}-s_{i}(x_{n})$.

  3. (5)

$f_{1},\dots,f_{n}$ have no solutions at $\infty$.

###### Proof of Theorem 1.4.

As in the proof of Corollary 5.1, $I$ can be written

(5.11) $I=\langle R_{\text{\it f}\hskip
0.9pt}(x_{n}),s_{0}(x_{n})x_{1}-s_{1}(x_{n}),\ldots,s_{0}(x_{n})x_{n-1}-s_{n-1}(x_{n})\rangle.$

Then Lemma 5.1 implies that $\gcd(R_{\text{\it f}\hskip
0.9pt}(x_{n}),s_{0}(x_{n}),\dots,s_{n-1}(x_{n}))=1$ since $I$ has dimension
zero. This proves (1), and then (2) follows from Corollary 5.1.

It remains to show that our hypothesis on $I$ implies the equivalence of (3),
(4), and (5). Given Theorem 1.3, this is easy:

  * •

(3) $\Rightarrow$ (4) follows from (3) $\Rightarrow$ (2) of Theorem 1.3.

  * •

(4) $\Rightarrow$ (3) follows since the Bézout identity for $\gcd(R_{\text{\it
f}\hskip 0.9pt}(x_{n}),s_{0}(x_{n}))=1$ and (5.11) imply that $I=\langle
R_{\text{\it f}\hskip
0.9pt}(x_{n}),x_{1}-g_{1}(x_{n}),\ldots,x_{n-1}-g_{n-1}(x_{n})\rangle$ for
some $g_{1}(x_{n}),\dots,g_{n-1}(x_{n})\in\mathbb{K}[x_{n}]$. This easily
implies (3).

  * •

(3) $\Rightarrow$ (5) follows from (3) $\Rightarrow$ (1) of Theorem 1.3.

  * •

(5) $\Rightarrow$ (3) follows from (1) $\Rightarrow$ (2) of Theorem 1.3 since
$I$ has a Shape Lemma.∎

We conclude this section with a discussion of the case $n=2$. In Section 4, we
noted that $f_{1},f_{2}$ have no solution at $\infty$ if and only if
$\gcd(\mathrm{lc}_{x_{1}}(f_{1}),\mathrm{lc}_{x_{1}}(f_{2}))=1$. In
particular, this allows us to replace the condition “no solution at $\infty$”
with the easier-to-check condition
“$\gcd(\mathrm{lc}_{x_{1}}(f_{1}),\mathrm{lc}_{x_{1}}(f_{2}))=1$” in Theorems
1.3 and 1.4.

######  Example 5.2.

Let $f_{1}=x_{2}x_{1}^{2}+x_{1}+x_{2}^{2}+x_{2}$ and $f_{2}=x_{2}x_{1}+1$ in
$\mathbb{C}[x_{1},x_{2}]$. Since
$\mathrm{lc}_{x_{1}}(f_{1})=\mathrm{lc}_{x_{1}}(f_{2})=x_{2}$, there is a
solution at $\infty$. Hence (1) in Theorem 1.3 is false, so that (2) and (3)
are also false.

It is instructive to see exactly how (1), (2) and (3) fail in this case. A
Gröbner basis calculation shows that $I=\langle f_{1},f_{2}\rangle=\langle
x_{2}+1,x_{1}-1\rangle$ has a Shape Lemma, so one part of (1) is true while
the other part is false. For (2), we have

$R_{\text{\it f}\hskip 0.9pt}(x_{2})=x_{2}^{3}(x_{2}+1)\ \text{ and }\
p_{1}(x_{1},x_{2})=x_{2}x_{1}+1,$

so that $s_{0}(x_{2})=x_{2}$. Hence $\gcd(R_{\text{\it f}\hskip
0.9pt}(x_{2}),s_{0}(x_{2}))=x_{2}\neq 1$, and

$I\cap\mathbb{C}[x_{2}]=\langle
x_{2}+1,x_{1}-1\rangle\cap\mathbb{C}[x_{2}]=\langle x_{2}+1\rangle\neq\langle
R_{\text{\it f}\hskip 0.9pt}(x_{2})\rangle$

because $R_{\text{\it f}\hskip 0.9pt}(x_{2})=x_{2}^{3}(x_{2}+1)$ has an
extraneous factor of $x_{2}^{3}$. It follows that both parts of (2) are false.
As for (3), a Gröbner basis calculation reveals that

$\langle R_{\text{\it f}\hskip 0.9pt}(x_{2}),p_{1}(x_{1},x_{2})\rangle=\langle
x_{2}^{3}(x_{2}+1),x_{2}x_{1}+1\rangle=\langle x_{2}+1,x_{1}-1\rangle.$

Hence $I=\langle R_{\text{\it f}\hskip
0.9pt}(x_{2}),p_{1}(x_{1},x_{2})\rangle$ is true in this case. Thus one part
of (3) is true and the other part is false. Theorem 1.4 explains what is going
on:

  * •

$I=\langle R_{\text{\it f}\hskip 0.9pt}(x_{2}),p_{1}(x_{1},x_{2})\rangle$
implies that $I$ has a Shape Lemma. This accounts for the parts above that are
true.

  * •

$I=\langle R_{\text{\it f}\hskip 0.9pt}(x_{2}),p_{1}(x_{1},x_{2})\rangle$
implies that (3), (4), and (5) of Theorem 1.4 are equivalent. So
$I\cap\mathbb{C}[x_{2}]\neq\langle R_{\text{\it f}\hskip 0.9pt}(x_{2})\rangle$
guarantees that there are solutions at $\infty$ and that $\gcd(R_{\text{\it
f}\hskip 0.9pt}(x_{2}),s_{0}(x_{2}))\neq 1$. This explains the parts above
that are false.

Our final result shows that when $n=2$, sometimes just knowing the resultant
is enough to guarantee that the conditions of Theorem 1.3 are all true:

######  Proposition 5.3.

If $f_{1},f_{2}\in\mathbb{K}[x_{1},x_{2}]$ satisfy $\rho\geq 1$ and the
resultant $R_{\text{\it f}\hskip 0.9pt}(x_{2})$ has degree
$\deg(f_{1})\cdot\deg(f_{2})$ with distinct roots, then the conditions of
Theorem 1.3 are all true. In particular, $I=\langle f_{1},f_{2}\rangle$ has a
Shape Lemma, $I=\langle R_{\text{\it f}\hskip
0.9pt}(x_{2}),p_{1}(x_{1},x_{2})\rangle$, and $I\cap\mathbb{K}[x_{2}]=\langle
R_{\text{\it f}\hskip 0.9pt}(x_{2})\rangle$.

###### Proof.

We first show that $f_{1},f_{2}$ are relatively prime. The nonvanishing of
$R_{\text{\it f}\hskip 0.9pt}(x_{2})$ implies that $f_{1},f_{2}$ cannot have a
common factor that involves $x_{1}$. But a common factor involving only
$x_{2}$ would appear with exponent $d_{1}+d_{2}=\rho+2\geq 3$ in the
resultant, which is impossible since $R_{\text{\it f}\hskip 0.9pt}(x_{2})$ has
distinct roots. It follows that $I=\langle f_{1},f_{2}\rangle$ is zero-
dimensional.

Let $M_{i}=\deg(f_{i})$ and recall that $d_{i}=\deg_{x_{1}}(f_{i})$. In this
case, Bézout [Bez64] proved in 1764 that

$\deg(R_{\text{\it f}\hskip 0.9pt}(x_{2}))\leq
M_{1}d_{2}+M_{2}d_{1}-d_{1}d_{2}=M_{1}M_{2}-(M_{1}-d_{1})(M_{2}-d_{2}).$

Since the degree in the left is $M_{1}M_{2}$ by hypothesis, we must have
either $M_{1}=d_{1}$ or $M_{2}=d_{2}$. If $M_{1}=d_{1}$, then
$\mathrm{lc}(f_{1},x_{1})$ is a nonzero constant, which implies that
$\gcd(\mathrm{lc}(f_{1},x_{1}),\mathrm{lc}(f_{2},x_{1}))=1$. Similarly,
$\gcd(\mathrm{lc}(f_{1},x_{1}),\mathrm{lc}(f_{2},x_{1}))=1$ when
$M_{2}=d_{2}$. It follows that there are no solutions at $\infty$.

The universal property of the resultant implies that the $M_{1}M_{2}$ distinct
roots of $R_{\text{\it f}\hskip 0.9pt}(x_{2})$ extend to solutions in
$\mathbb{P}^{1}\times_{\mathbb{K}}\mathbb{A}_{\mathbb{K}}^{1}$. There are no
solutions at $\infty$ since
$\gcd(\mathrm{lc}(f_{1},x_{1}),\mathrm{lc}(f_{2},x_{1}))=1$, so that the
$M_{1}M_{2}$ distinct roots of $R_{\text{\it f}\hskip 0.9pt}(x_{2})$ extend to
solutions of $f_{1}=f_{2}=0$. Thus we have $M_{1}M_{2}$ elements of
$\mathbf{V}(f_{1},f_{2})$ with distinct $x_{2}$-coordinates. If we homogenize
$f_{i}(x_{1},x_{2})$ to $F_{i}(x_{1},x_{2},x_{3})$, Bézout’s theorem for
$\mathbb{P}^{2}_{\mathbb{K}}$ implies that $F_{1}=F_{2}=0$ has $M_{1}M_{2}$
solutions in $\mathbb{P}^{2}$, counting multiplicity. Yet we just constructed
$M_{1}M_{2}$ affine solutions in $\mathbb{K}^{2}$ with distinct
$x_{2}$-coordinates. Thus:

  * •

All solutions of $F_{1}=F_{2}=0$ lie in $\mathbb{K}^{2}$ and have multiplicity
one.

  * •

The solutions in $\mathbf{V}(f_{1},f_{2})$ have distinct $x_{2}$-coordinates.

The second bullet shows that the injectivity hypothesis of Lemma 2.2 is
satisfied. Furthermore, if we let $I\cap\mathbb{K}[x_{2}]=\langle
r(x_{2})\rangle$, then $r(x_{2})$ divides $R_{\text{\it f}\hskip
0.9pt}(x_{2})$. Since the latter has distinct roots, the same is true for
$r(x_{2})$, so its roots all have multiplicity one. All solutions of
$f_{1}=f_{2}=0$ have multiplicity one by the first bullet, so that $I$ has a
Shape Lemma by Lemma 2.2. We showed above that there are no solutions at
$\infty$, so condition (1) of Theorem 1.3 is satisfied, and we are done. ∎

######  Example 5.3.

Let $f_{1}=x_{1}^{2}+x_{2}^{3}$ and $f_{2}=1+x_{2}+x_{1}^{3}$ in
$\mathbb{C}[x_{1},x_{2}]$. Then one computes that

$R_{\text{\it f}\hskip 0.9pt}(x_{2})=x_{2}^{9}+x_{2}^{2}+2x_{2}+1.$

This polynomial has degree $9=\deg(f_{1})\cdot\deg(f_{2})$ and discriminant
$384126317\neq 0$. By Proposition 5.3, $I=\langle f_{1},f_{2}\rangle$ has a
Shape Lemma, $I=\langle R_{\text{\it f}\hskip
0.9pt}(x_{2}),p_{1}(x_{1},x_{2})\rangle$, and $I\cap\mathbb{C}[x_{2}]=\langle
R_{\text{\it f}\hskip 0.9pt}(x_{2})\rangle$.

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