# Upper Bounds on Resolvent Degree and Its Growth Rate

Alexander J. Sutherland111This work was supported in part by the National
Science Foundation under Grant No. DMS-1944862.

###### Abstract

For each $n$, let $\text{RD}(n)$ denote the minimum $d$ for which there exists
a formula for the general polynomial of degree $n$ in algebraic functions of
at most $d$ variables. In 1945, Segre called for a better understanding of the
large $n$ behavior of $\text{RD}(n)$. In this paper, we provide improved
thresholds for upper bounds on $\text{RD}(n)$. Our techniques build upon
classical algebraic geometry to provide new upper bounds for small $n$ and, in
doing so, fix gaps in the proofs of A. Wiman and G.N. Chebotarev in [Wim1927]
and [Che1954].

###### Contents

  1. 1 Introduction
  2. 2 Polar Cones
    1. 2.1 An Introduction to Polar Cones
    2. 2.2 Resolvent Degree
    3. 2.3 Iterated Polar Cones and $k$-Polar Points
  3. 3 New Upper Bounds on $\text{RD}(n)$
    1. 3.1 Tschirnhaus Transformations
    2. 3.2 New Bounds From Iterated Polar Cones
    3. 3.3 New Bounds from Moduli Spaces
    4. 3.4 Upper Bounds on the Bounding Function $G(m)$
  4. 4 Comparison with Prior Bounds
  5. 5 Appendices
    1. 5.1 Explicit Bounds on $\text{RD}(n)$
    2. 5.2 Explicit Approximations of $F(m)/G(m)$
    3. 5.3 Proof of Technical Lemma

##  1 Introduction

Consider the following classical problem:

######  Problem 1.1.

Given a general polynomial $z^{n}+a_{1}z^{n-1}+\cdots+a_{n-1}z+a_{n}$,
determine a root of the polynomial in terms of $a_{1},\dotsc,a_{n}$ in the
simplest manner possible.

Recent work on this problem has been cast within the framework of resolvent
degree (as in [FW2019]). Informally, $\text{RD}(n)$ is the minimal $d$ for
which there exists a formula for the roots of the generic degree $n$
polynomial using algebraic functions of at most $d$ variables (see Definitions
2.13 and 2.14). While there are currently no non-trivial lower bounds on
$\text{RD}(n)$ and it is possible in theory that $\text{RD}(n)=1$ for all $n$,
there is a history of determining upper bounds on $\text{RD}(n)$. This
includes the work of Bring [Bri1786], Hamilton [Ham1836], Sylvester [Syl887,
SH1887, SH1888], Klein [Kle1884, Kle1887, Kle1905], and Hilbert
([Hil1927]).222Chen, He, and McKay provide a modern English translation of
[Bri1786] (originally in Latin) in [CHM2017] and the author provides a modern
English translation of [Kle1905] (originally in German) in [Sut2019]. Indeed,
Hamilton proved that $\text{RD}(6)\leq 2$, $\text{RD}(7)\leq 3$, and
$\text{RD}(8)\leq 4$. Hilbert’s sextic conjecture, Hilbert’s 13th problem, and
Hilbert’s octic conjecture, respectively, predict that these upper bounds are
sharp. In Section 3 of [Seg1945], Segre indicated the following problem:

######  Problem 1.2.

(Segre, 1945)
Determine the large $n$ behavior of $\text{RD}(n)$.

The current upper bounds on $\text{RD}(n)$ are found in [Wol2021], where
Wolfson introduces a function $F(m)$ such that $\text{RD}(n)\leq n-m$ for all
$n\geq F(m)$ (Definition 5.4 and Theorem 5.6, [Wol2021]). In this paper, we
construct a similar function $G(m)$. For $21\leq n\leq 3,632,428,800$, we use
methods which build upon the work of Segre. By using a theorem of Debarre and
Manivel (see Theorem 3.20, or Theorem 2.1 of [DM1998]), we are able to
streamline Wolfson’s method to obtain better thresholds on upper bounds on
$\text{RD}(n)$ for $n\geq 348,489,068,134$. For $3,632,428,801\leq n\leq
348,489,068,133$, we obtain the same bounds as Wolfson’s method. More
specifically, we prove the following results:

######  Theorem 1.3.

(Key Properties of $G(m)$)
The function $G(m)$ of Definition 3.26 has the following properties:

  1. 1.

For each $m\geq 1$ and $n\geq G(m)$, $\text{RD}(n)\leq n-m$.

  2. 2.

For each $d\geq 4$, $G(2d^{2}+7d+6)\leq\frac{(2d^{2}+7d+5)!}{d!}$. In
particular, for $d\geq 4$ and $n\geq\frac{(2d^{2}+7d+5)!}{d!}$,

$\text{RD}(n)\leq n-2d^{2}-7d-6.$

  3. 3.

For each $m\geq 1$, $G(m)\leq F(m)$ with equality only when
$m\in\left\\{1,2,3,4,5,15,16\right\\}$ and

$\lim\limits_{m\rightarrow\infty}\frac{F(m)}{G(m)}=\infty.$

The first statement of Theorem 1.3 is Theorem 3.27 and, in light of this
statement, we refer to $F(m)$ and $G(m)$ as “bounding functions” for
$\text{RD}(n)$. Hamilton was the first to provide a bounding function for
$\text{RD}(n)$ (see [OEIS2021]). Brauer greatly improved Hamilton’s result
when he showed that $\text{RD}(n)\leq n-m$ for $n\geq(m-1)!+1$ in [Bra1975].
Wolfson improved upon these bounds with his function $F(m)$, but did not
provide an upper bound on $F(m)$ in terms of elementary functions. The
construction of $G(m)$ is simpler than $F(m)$, but is not as explicit as
Brauer’s condition. The second statement of Theorem 1.3 gives an upper bound
on $\text{RD}(n)$ and its growth rate using elementary functions; it is
Theorem 3.28.

The third statement of Theorem 1.3 shows that $G(m)$ provides better
asymptotic bounds than $F(m)$; it is Theorem 4.1. Note that the inequalities
when $1\leq m\leq 4$ (which imply $\text{RD}(n)=1$ for $2\leq n\leq 5$) are
due to the classical solutions of general polynomials of low degree. Recall
that Hilbert’s Octic Conjecture predicts that $\text{RD}(8)=4$ and, if true,
would imply that the bounds $G(5)=F(5)=9$ cannot be improved. We expect that
the general thresholds obtained by $G(m)$ are not sharp (including the cases
of $G(15)=F(15)$ and $G(16)=F(16)$).

#### Historical Remarks

Many of the approaches above rely on the theory of Tschirnhaus
transformations. In [Wol2021], Wolfson consolidates the history of Tschirnhaus
transformations and establishes upper bounds on $\text{RD}(n)$ by determining
points of Tschirnhaus complete intersections over field extensions of bounded
resolvent degree. In [Wim1927], Wiman used these methods to give an argument
that implied $\text{RD}(n)\leq n-5$ for $n\geq 9$ and Chebotarev333Note that
this is G.N. Chebotarev, son of Nikolai Chebotarev. extended Wiman’s idea to
argue that $\text{RD}(n)\leq n-6$ for $n\geq 21$ in [Che1954]. Dixmier noted
in [Dix1993], however, that Wiman’s argument has a gap and provided an
algebraic argument to conclude that $\text{RD}(n)\leq n-5$ for $n\geq 9$.
Moreover, the gap found in Wiman’s argument is also present in the argument of
Chebotarev.

Building on the work of Segre, we introduce (iterated) polar cones for
hypersurfaces and intersections of hypersurfaces, which we use to determine
linear subvarieties on intersections of hypersurfaces over extensions of
bounded resolvent degree. In doing so, we fix the gaps in the arguments of
Chebotarev and provide a geometric fix for the arguments of Wiman.

#### Outline of the Paper

In Section 2, we use the language of polar cones (respectively, iterated polar
cones) and highlight their connection to lines (respectively, $k$-planes) on
intersections of hypersurfaces. In Section 3, we use techniques involving
iterated polar cones and moduli spaces to construct a function $G(m)$ which
yields the upper bounds on $\text{RD}(n)$. In Section 4, we compare these new
bounds to those of Wolfson. In Appendix 5.1, we give explicit values of $G(m)$
for small values of $m$. In Appendix 5.2, we provide additional data on the
ratio $F(m)/G(m)$. In Appendix 5.3, we provide the proof of a technical lemma
from Section 2.

#### Conventions

  1. 1.

We restrict to fields $K$ which are finitely generated $\mathbb{C}$-algebras.
The interested reader could instead fix an arbitrary algebraically closed
field $F$ of characteristic zero (in lieu of $\mathbb{C}$) and the statements
(relative to $F$) would hold.

  2. 2.

For varieties, we follow the conventions of [Har2010]. Namely, we define a
projective (respectively, affine) variety to be a closed algebraic set in
$\mathbb{P}_{K}^{n}$ (respectively, $\mathbb{A}_{K}^{n}$). When we simply say
variety, we mean a quasi-projective variety. In particular, we do not assume
that varieties are irreducible.

  3. 3.

For a collection of homogeneous polynomials $f_{1},\dotsc,f_{s}\in
K[x_{0},\dotsc,x_{r}]$, we write $\mathbb{V}(f_{1},\dotsc,f_{s})$ for the
subvariety of $\mathbb{P}_{K}^{r}$ determined by the conditions
$f_{1}=\cdots=f_{s}=0$.

  4. 4.

Given a subvariety $V\subseteq\mathbb{P}_{K}^{r}$, we use the notation $V(K)$
to refer to the set of $K$-rational points of $V$.

  5. 5.

We write $K_{n}$ to mean $\mathbb{C}(a_{1},\dotsc,a_{n})$, a purely
transcendental extension of $\mathbb{C}$ with transcendence basis
$a_{1},\dotsc,a_{n}$.

  6. 6.

We write $\text{Gr}(k,r)$ to denote the Grassmannian of $k$-dimensional
subspaces of $\mathbb{C}^{r}$ and $\mathcal{G}\text{r}(k,r)$ for the space of
$k$-planes in $\mathbb{P}_{\mathbb{C}}^{r}$. In particular,
$\mathcal{G}\text{r}(k,r)\cong\text{Gr}(k+1,r+1)$.

  7. 7.

Given a polynomial ring $K[x_{0},\dotsc,x_{r}]$ over a field $K$, we write

     * •

$K[x_{0},\dotsc,x_{r}]_{(d)}$ for the vector space of degree $d$ polynomials,

     * •

$K[x_{0},\dotsc,x_{r}]_{(d)}^{\vee}$ for its dual space,

     * •

$S^{*}\left(K[x_{0},\dotsc,x_{r}]_{(d)}^{\vee}\right)$ for the corresponding
free commutative $K$-algebra, and

     * •

$S^{*}\left(K[x_{0},\dotsc,x_{r}]_{(d)}^{\vee}\right)^{\text{GL}(K,r+1)}$ for
the associated graded ring of $\text{GL}(K,r+1)$-invariants.

  8. 8.

We write $\log$ to mean the base $e$ logarithm.

With respect to convention 2, note that for generic choices of
$f_{1},\dotsc,f_{s}$, $\mathbb{V}(f_{1},\dotsc,f_{s})$ is a complete
intersection. However, there are examples of such choices which are not
complete intersections, such as the twisted cubic curve. Our methods apply to
all such choices and to avoid confusion, we generally use the terminology
“intersection of hypersurfaces” instead of “complete intersection” to refer to
a variety of the form $\mathbb{V}(f_{1},\dotsc,f_{s})$. Additionally, we note
that conventions 5 and 6 are primarily for Definition 3.12 and Remark 3.14,
respectively.

#### Acknowledgements

I thank Jesse Wolfson for his generous support. Next, I thank Joshua Jordan
for several key conversations and general support. I thank Benson Farb, Hannah
Knight, Curt McMullen, and Zinovy Reichstein for helpful comments on a draft.
Finally, I thank Kenneth Ascher, Claudio Gómez-Gonzáles, and Roman Vershynin
for helpful conversations.

##  2 Polar Cones

###  2.1 An Introduction to Polar Cones

In [Seg1945], Segre refers to a “well-known fact” which is the main ingredient
for many of his proofs. This fact may no longer be as well-known as it once
was. Definitions 2.1, 2.2, and 2.4 are necessary to state the fact, which we
include as Lemma 2.5 and refer to as Bertini’s Lemma (for Hypersurfaces),
since the reference Segre gives for this fact is [Ber1923].

######  Definition 2.1.

(Polars of a Polynomial)
Let $f\in K[x_{0},\dotsc,x_{r}]$ be a homogeneous polynomial of degree $d$ and
$P\in\mathbb{P}^{r}(K)$. For every $n\in\mathbb{Z}_{\geq 1}$, let

$\displaystyle[n]=\left\\{0,1,\dotsc,n\right\\},$
$\displaystyle[n]^{*}=\left\\{1,\dotsc,n\right\\}.$

For each $0\leq k\leq d-1$, we set

$\displaystyle I_{k}:=\left\\{(i_{0},\dotsc,i_{r})\in\mathbb{Z}_{\geq
0}^{r+1}\ |\ i_{0}+\cdots+i_{r}=k\right\\}.$

We have $r+1$ first partial derivatives $\frac{\partial}{\partial
x_{0}},\dotsc,\frac{\partial}{\partial x_{r}}$. To obtain an (ordered)
$k^{th}$ partial derivative of $f$, we make $k$ successive choices of first
partial derivatives. Hence, the set of all (ordered) $k^{th}$ partial
derivatives is indexed by
$I_{k}^{*}=\text{Hom}_{{\textbf{Set}}}([k]^{*},[r])$. For any $\iota\in
I_{k}^{*}$ and $0\leq j\leq r$, we set

$\displaystyle|\iota|(j):=\left|\iota^{-1}(j)\right|$

and note that $(|\iota|(0),\dotsc,|\iota|(r))\in I_{k}$. Additionally, we use
the shorthand

$\displaystyle\partial_{0}^{j_{0}}\cdots\partial_{l}^{j_{l}}f:=\frac{\partial^{j_{0}+\cdots+j_{l}}f}{\partial
x_{0}^{j_{0}}\cdots\partial x_{l}^{j_{l}}}.$

With this notation set, the $k^{th}$ polar of $f$ at $P$ is the homogeneous
polynomial of degree $d-k$

$\displaystyle t(k,f,P)(y_{0},\dotsc,y_{r}):=\sum\limits_{\iota\in
I_{d-k}^{*}}\left(\partial_{0}^{|\iota|(0)}\cdots\partial_{r}^{|\iota|(r)}f\right)\biggr{\rvert}_{P}\
y_{0}^{|\iota|(0)}\cdots y_{r}^{|\iota(r)|}.$

######  Definition 2.2.

(Polars of a Hypersurface)
Let $V=\mathbb{V}(f)$ be a hypersurface in $\mathbb{P}_{K}^{r}$ and $P\in
V(K)$. The $k^{th}$ polar of $V$ at $P$ is

$\displaystyle
T(k,f,P):=\mathbb{V}\left(t(k,f,P)\right)\subseteq\mathbb{P}_{K}^{r}.$

######  Example 2.3.

Using the conventions of Definition 2.2, we observe that

$\displaystyle t(0,f,P)(x_{0},\dotsc,x_{r})$
$\displaystyle=f(x_{0},\dotsc,x_{r}),$ $\displaystyle
t(d,f,P)(x_{0},\dotsc,x_{r})$ $\displaystyle=f(P),$

so the $0^{th}$ polar of $\mathbb{V}(f)$ at $P$ is $T(0,f,P)=V$ for all $P$
and the $d^{th}$ polar of $\mathbb{V}(f)$ at $P$ is

$\displaystyle T(d,f,P)=\begin{cases}\mathbb{P}_{K}^{r},&\text{if
}P\in\mathbb{V}(f),\\\ \emptyset,&\text{if }P\not\in\mathbb{V}(f).\end{cases}$

Moreover, if $V$ is smooth at $P$, then $T(d-1,f,P)$ is the tangent hyperplane
of $V$ at $P$; this motivates the use of $T(k,f,P)$ for polar hypersurfaces
and hence $t(k,f,P)$ for their defining polynomials.

######  Definition 2.4.

(Polar Cone of a Hypersurface)
Let $V=\mathbb{V}(f)$ be a degree $d$ hypersurface in $\mathbb{P}_{K}^{r}$ and
$P\in V(K)$. The (first) polar cone of $V$ at $P$ is

$\displaystyle\mathcal{C}(V;P):=\bigcap\limits_{k=0}^{d-1}T(k,f,P).$

We can now state Lemma 2.5, which motivates the terminology of the previous
definition.

######  Lemma 2.5.

(Bertini’s Lemma for Hypersurfaces)444See Chapter 8 of [Ber1923] for the
reference given in [Seg1945].
Let $V=\mathbb{V}(f)$ be a hypersurface in $\mathbb{P}_{K}^{r}$ and $P\in
V(K)$. Then, $\mathcal{C}(V;P)$ is a cone with vertex $P$ which is contained
in $V$.

######  Example 2.6.

(Lines on Cubic Surfaces)
Let $V\subseteq\mathbb{P}_{K}^{3}$ be a smooth cubic surface. If
$K=\overline{K}$, then $V$ contains exactly 27 lines $L_{1},\dotsc,L_{27}$. If
$P\in V(K)$ lies on exactly one line $L$ (respectively, exactly two lines
$L_{1},L_{2}$), then $\mathcal{C}(V;P)=L$ (respectively,
$\mathcal{C}(V;P)=L_{1}\cup L_{2}$). Most points $Q\in V(K)$, however, do not
lie on any line and in such a case $\mathcal{C}(V;Q)$ is the point $Q$ with
multiplicity 6. We will generally consider fields $K$ which are not
algebraically closed, in which case we may need to pass to an extension of $K$
to obtain a line.

######  Remark 2.7.

For an alternative modern perspective on Definitions 2.1 and 2.2, Remark 2.3,
and Lemma 2.8, see p.5-6 of [Dol2012] \- in particular, equations 1.8 through
1.11. Note that Lemma 2.5 follows directly from the following technical lemma.
The proof of Lemma 2.8 can be found in Appendix 5.3; it is not exceedingly
complicated, but is notationally cumbersome.

######  Lemma 2.8.

(Technical Lemma)
Let $P,Q\in\mathbb{P}^{r}(K)$ and $f\in K[x_{0},\dotsc,x_{r}]$ be a
homogeneous polynomial of degree $d$. Applying a projective change of
coordinates as necessary, we assume that

$\displaystyle P$ $\displaystyle=[1:p_{1}:\cdots:p_{r}],$ $\displaystyle Q$
$\displaystyle=[1:q_{1}:\cdots:q_{r}],$

so that the line determined by $P$ and $Q$ is

$\displaystyle L(P,Q)(K)=\left\\{[1:\lambda p_{1}+\mu q_{1}:\cdots:\lambda
p_{r}+\mu q_{r}]\ |\ [\lambda:\mu]\in\mathbb{P}^{1}(K)\right\\}.$

For any point $R_{\lambda:\mu}=[1:\lambda p_{1}+\mu q_{1}:\cdots:\lambda
p_{r}+\mu q_{r}]\in L(P,Q)(K)$,

$f\left(R_{\lambda:\mu}\right)=f(\lambda P)+f(\mu
Q)+\sum\limits_{k=1}^{d-1}\frac{1}{k!}t(d-k,f,\lambda P)(\mu Q).$ (1)

######  Remark 2.9.

Lemma 2.5 implies that for every point
$Q\in\mathcal{C}(V;P)(K)\setminus\\{P\\}$, the line $L(P,Q)$ determined by $P$
and $Q$ lies in $V$. Furthermore, a line $L$ lies on an intersection of
hypersurfaces $\mathbb{V}(f_{1},\dotsc,f_{n})$ exactly when it lies on each
$\mathbb{V}(f_{i})$. This motivates Definition 2.10 and implies Lemma 2.11.

######  Definition 2.10.

(Polar Cone of an Intersection of Hypersurfaces)
Let $V=\mathbb{V}(f_{1},\dotsc,f_{n})\subseteq\mathbb{P}_{K}^{r}$ be an
intersection of hypersurfaces and $P\in V(K)$. The (first) polar cone of $V$
at $P$ is

$\displaystyle\mathcal{C}(V;P):=\bigcap\limits_{i=1}^{n}\mathcal{C}(\mathbb{V}(f_{i});P).$

######  Lemma 2.11.

(Bertini’s Lemma for Intersections of Hypersurfaces)
Let $V=\mathbb{V}(f_{1},\dotsc,f_{n})\subseteq\mathbb{P}_{K}^{r}$ be an
intersection of hypersurfaces and $P\in V(K)$. Then, $\mathcal{C}(V;P)$ is a
cone with vertex $P$ which is contained in $V$.

######  Remark 2.12.

(Passing to Extensions)
We continue using the notation of Lemma 2.11. It is easy to see that when $K$
is algebraically closed and $\dim\left(\mathcal{C}(V;P)\right)\geq 1$, the set
$\mathcal{C}(V;P)(K)\setminus\\{P\\}$ is not empty. Thus, any choice of
$Q\in\mathcal{C}(V;P)\setminus\\{P\\}$ determines a line on $V$ containing
$P$. When $K$ is not algebraically closed, however, it may be that
$\mathcal{C}(V;P)(K)=\\{P\\}$. In what follows, we will often extend scalars
to an extension $L/K$ over which we can determine an $L$-rational point.

###  2.2 Resolvent Degree

Resolvent degree is an invariant introduced independently by Brauer in
[Bra1975] and Arnol’d-Shimura in [AS1976]. Farb and Wolfson summarize the
history of resolvent degree and significantly broaden the resolvent degree
framework in [FW2019], which we refer the reader to for background on
resolvent degree. We now recall the definition of resolvent degree for finite
field extensions and for generically finite, dominant, rational maps of
$\mathbb{C}$-varieties. We refer the reader to Section 4 of [Wol2021] for how
to extend this definition to arbitrary dominant, rational maps. We also remind
the reader that in this paper, we work only with fields that are finitely
generated $\mathbb{C}$-algebras.

######  Definition 2.13.

(Resolvent Degree of Field Extensions)
Let $L/K$ be a finite extension of $\mathbb{C}$-fields. The resolvent degree
of $L/K$, denoted $\text{RD}(L/K)$, is the minimum $d$ for which there is a
tower of finite field extensions

$\displaystyle K=E_{0}\hookrightarrow
E_{1}\hookrightarrow\cdots\hookrightarrow E_{\ell}$

such that $L$ embeds into $E_{\ell}$ over $K$ and for each $E_{i+1}/E_{i}$,
there is a finite extension of $\widetilde{F}_{i}/F_{i}$ with
$\text{tr.deg}_{\mathbb{C}}(F_{i})\leq d$ such that

$\displaystyle E_{i+1}\cong E_{i}\otimes_{F_{i}}\widetilde{F}_{i}.$

######  Definition 2.14.

(Resolvent Degree of Generically Finite, Dominant Maps)
Let $Y\dashrightarrow X$ be a generically finite, dominant, rational map of
$\mathbb{C}$-varieties. The resolvent degree of $Y\dashrightarrow X$, denoted
$\text{RD}(Y\dashrightarrow X)$, is the minimum $d$ for which there is a tower
of generically finite, dominant, rational maps

$\displaystyle E_{\ell}\dashrightarrow\cdots\dashrightarrow
E_{1}\dashrightarrow E_{0}\subseteq X$

such that $E_{0}\subseteq X$ is a dense Zariski open; $E_{\ell}\dashrightarrow
E_{0}$ factors through $Y\dashrightarrow X$; and for each
$\pi_{i}:E_{i+1}\dashrightarrow E_{i}$, there exists a surjective morphism
$\widetilde{Z}_{i}\rightarrow Z_{i}$ with $\dim\left(Z_{i}\right)\leq d$, a
Zariski open $E_{i}^{\circ}\subseteq E_{i}$, and a morphism
$E_{i}^{\circ}\rightarrow Z_{i}$ such that

$\displaystyle\pi_{i}^{-1}\left(E_{i}^{\circ}\right)\cong
E_{i}^{\circ}\times_{Z_{i}}\widetilde{Z}_{i}.$

######  Remark 2.15.

(Compatibility of Resolvent Degree Definitions)
We note that Definitions 2.13 and 2.14 are equivalent. For an irreducible,
affine $\mathbb{C}$-variety $X$, the equivalence of definitions is derived
from the classical equivalence which sends $X$ to its function field
$\mathbb{C}(X)$; the general case follows from invariance of resolvent degree
under birational equivalence. We refer the reader to [FW2019] for details.

######  Remark 2.16.

($\text{RD}(n)$ Notation)
We write $\text{RD}(n)$ for the resolvent degree of the general degree $n$
polynomial. Recall that $K_{n}=\mathbb{C}(a_{1},\dotsc,a_{n})$. If
$f(z)=z^{n}+a_{1}z^{n-1}+\cdots+a_{n-1}z+a_{n}\in K_{n}[z]$, and
$K_{n}^{\prime}=K_{n}[z]/(f(z))$, then

$\displaystyle\text{RD}(n)=\text{RD}(K_{n}^{\prime}/K_{n})=\text{RD}\left(\text{Spec}\left(K_{n}^{\prime}\right)\rightarrow\text{Spec}\left(K_{n}\right)\right).$

######  Remark 2.17.

(RD of a Composition)
We will frequently use the following field-theoretic version of Lemma 2.7 of
[FW2019] without explicit reference: given a tower of field extensions

$\displaystyle E_{0}\hookrightarrow E_{1}\hookrightarrow\cdots E_{k},$

the resolvent degree of the composition is the maximum of the resolvent degree
of the components, e.g.

$\displaystyle\text{RD}(E_{k}/E_{0})=\max\left\\{\text{RD}(E_{j}/E_{j-1})\ |\
1\leq j\leq k\right\\}.$

######  Lemma 2.18.

(Upper Bound on $\text{RD}(L/K)$)
Let $L/K$ be a degree $\ell$ field extension. Then,
$\text{RD}\left(L/K\right)\leq\text{RD}(\ell)$.

###### Proof.

As $L$ and $K$ are $\mathbb{C}$-fields, they have characteristic zero. Hence,
the Primitive Element Theorem yields that we need only solve a degree $\ell$
polynomial to determine a primitive element $\zeta$ of $L/K$. The isomorphism
$L\cong K(\zeta)$ then establishes the claim. ∎

We will frequently make use of the previous lemma without explicit reference
after giving an upper bound on the degree of an extension.

######  Proposition 2.19.

(Determining Rational Points over Extensions)
Let $V\subseteq\mathbb{P}_{K}^{r}$ be a degree $d$ subvariety. Then, there is
an extension $L/K$ with $\text{RD}(L/K)\leq\text{RD}(d)$ over which we can
determine a rational point of $V$ .

###### Proof.

Set $\ell=\dim(V)$. For a generic $(r-\ell)$-plane $\Lambda$, the intersection
$V\cap\Lambda$ has dimension 0 and thus has $d$ $\overline{K}$-points (with
multiplicity) in any algebraic closure $\overline{K}$ of $K$; we denote these
points by $Q_{1},\dotsc,Q_{d}$. Observe that the polynomial

$\displaystyle
f(z)=\left(z-Q_{1}\right)\left(z-Q_{2}\right)\cdots\left(z-Q_{d}\right)$

has coefficients defined over $K$. Let $m(z)$ be an irreducible factor of
$f(z)$ over $K$. We set $L:=K[z]/(m(z))$ and observe that we can determine an
$L$-point of $V$ by construction. Lemma 2.18 yields that
$\text{RD}(L/K)\leq\text{RD}(d)$. ∎

######  Remark 2.20.

(Extensions Given By Solving Polynomials)
Given Proposition 2.19, we henceforth say that we can determine a point of $V$
by solving a degree $d$ polynomial.

###  2.3 Iterated Polar Cones and $k$-Polar Points

As we have established, for an intersection of hypersurfaces $V$ and $P\in
V(K)$, points of $\mathcal{C}(V;P)\setminus\\{P\\}$ determine lines on $V$.
Next, we will iterate the polar cone construction in a method to determine
$k$-planes on intersections of hypersurfaces (as in Lemma 2.24). To do so, we
must first introduce additional definitions and notation.

######  Remark 2.21.

(Linear Subvarieties)

  * •

For any points $P_{0},\dotsc,P_{k}\in\mathbb{P}^{r}(K)$, we denote the linear
subsvariety they determine by $L(P_{0},\dotsc,P_{k})$. Note that
$L(P_{0},\dotsc,P_{k})$ has dimension at most $k$.

  * •

Henceforth, when we say $k$-plane, we mean a linear subvariety of
$\mathbb{P}^{r}$ of dimension $k$. When $k=1$, we say line instead of 1-plane
and when $k=2$, we say plane instead of 2-plane.

######  Definition 2.22.

(Iterated Polar Cones & $k$-Polar Points)
Let $V\subseteq\mathbb{P}_{K}^{r}$ be an intersection of hypersurfaces and
$P_{0}\in V(K)$; we set $\mathcal{C}^{1}(V;P_{0}):=\mathcal{C}(V;P_{0})$.
Given additional points $P_{1},\dotsc,P_{k-1}\in V(K)$ such that

$P_{\ell-1}\in\mathcal{C}^{\ell-1}(V;P_{0},\dotsc,P_{\ell-2})\setminus
L(P_{0},\dotsc,P_{\ell-2})$ (2)

for $0\leq\ell\leq k-1$, the $k^{th}$ polar cone of $V$ at
$P_{0},\dotsc,P_{k-1}$ is

$\displaystyle\mathcal{C}^{k}(V;P_{0},\dotsc,P_{k-1}):=\mathcal{C}\left(\mathcal{C}^{k-1}(V;P_{0},\dotsc,P_{k-2}),P_{k-1}\right).$

We refer to an ordered collection of points $\left(P_{0},\dotsc,P_{k}\right)$
that satisfy (2) for each $0\leq\ell\leq k-1$ as a $k$-polar point of $V$.

When the points in question have already been specified, we simply refer to
_the_ $k^{th}$ polar cone of $V$. When such points exist but have not been
specified, we refer to _a_ $k^{th}$ polar cone of $V$. It is occasionally
useful to refer to $V$ itself as a zeroth polar cone of $V$ (at any rational
point of $V$).

######  Remark 2.23.

(Iterated Polar Cones Are Nested)
Note that if $\left(P_{0},\dotsc,P_{k}\right)$ is a $k$-polar point of an
intersection of hypersurfaces $V\subseteq\mathbb{P}_{K}^{r}$, then

$\displaystyle\mathcal{C}^{k}(V;P_{0},\dotsc,P_{k-1})\subseteq\mathcal{C}^{k-1}(V;P_{0},\dotsc,P_{k-2})\subseteq\cdots\subseteq\mathcal{C}^{2}(V;P_{0},P_{1})\subseteq\mathcal{C}^{1}(V;P_{0})=\mathcal{C}(V;P_{0}).$

######  Lemma 2.24.

(Polar Point Lemma)
Let $V\subseteq\mathbb{P}_{K}^{r}$ be an intersection of hypersurfaces and let
$\left(P_{0},\dotsc,P_{k}\right)$ be a $k$-polar point of $V$. Then,
$L(P_{0},\dotsc,P_{k})\subseteq V$ is a $k$-plane.

###### Proof.

We prove the claim by induction on $k$ and observe that the case of $k=1$
follows immediately from Lemma 2.11. Now, consider arbitrary $k>1$ and let
$\left(P_{0},\dotsc,P_{k}\right)$ be a $k$-polar point of $V$. Then,
$\left(P_{1},\dotsc,P_{k}\right)$ is a $(k-1)$-polar point of
$\mathcal{C}(V;P_{0})$. Recall that $\mathcal{C}(V;P_{0})$ is a cone and
$L(P_{1},\dotsc,P_{k})\subseteq\mathcal{C}(V;P_{0})$ is a $(k-1)$-plane which
does not contain the vertex $P_{0}$, hence $L(P_{0},\dotsc,P_{k})\subseteq V$
is a $k$-plane. ∎

######  Definition 2.25.

(Type of an Intersection of Hypersurfaces)
Given an intersection of hypersurfaces $V\subseteq\mathbb{P}_{K}^{r}$, we say
that $V$ is of type $\left[\begin{matrix}d&d-1&\cdots&2&1\\\
\ell_{d}&\ell_{d-1}&\cdots&\ell_{2}&\ell_{1}\end{matrix}\right]$ if $V$ has
multi-degree

$\displaystyle(\underbrace{d,\dotsc,d}_{\ell_{d}\text{
many}},\underbrace{d-1,\dotsc,d-1}_{\ell_{d-1}\text{
many}},\cdots,\underbrace{2,\dotsc,2}_{\ell_{2}\text{
many}},\underbrace{1,\dotsc,1}_{\ell_{1}\text{ many}}).$

If $\ell_{j}=0$ for any $1\leq j\leq d-1$, we may omit it in the presentation;
e.g. an intersection of four quadrics is of type $\left[\begin{matrix}2\\\
4\end{matrix}\right]$. When $d\geq 2$ and each $\ell_{j}=1$, we abbreviate the
notation and say $V$ is of type $(1,\dotsc,d)$. When $d\geq 2$, $\ell_{1}=0$,
and the other $\ell_{j}=1$, we say $V$ is of type $(2,\dotsc,d)$.

Consider an intersection of hypersurfaces $V=V_{d}\cap\cdots\cap V_{1}$ of
type $(1,\dotsc,d)$ in $\mathbb{P}_{K}^{r}$ with $\deg(V_{j})=j$. Note that
$V_{1}\cong\mathbb{P}_{K}^{r-1}$ and thus we can also consider $V$ as an
intersection of hypersurfaces of type $(2,\dotsc,d)$ inside
$V_{1}\cong\mathbb{P}_{K}^{r-1}$.

######  Proposition 2.26.

(Type of a $k^{th}$ Polar Cone of an Intersection of Type $(1,\dotsc,d)$)
Let $V\subseteq\mathbb{P}_{K}^{r}$ be an intersection of hypersurfaces of type
$(1,\dotsc,d)$ and take $k\geq 1$. A $k^{th}$ polar cone
$\mathcal{C}^{k}(V;P_{0},\dotsc,P_{k-1})$ is of type

$\displaystyle\left[\begin{matrix}d&d-1&d-2&\cdots&3&2&1\\\
1&k+1&\binom{k+2}{2}&\cdots&\binom{k+d-3}{d-3}&\binom{k+d-2}{d-2}&\binom{k+d-1}{d-1}\end{matrix}\right]$

for $r\geq\binom{k+d}{d-1}$.

###### Proof.

We proceed by induction on $k$ and note that the $k=1$ case follows from the
definition of a polar cone of an intersection of hypersurfaces, along with the
observation that $\binom{1+j}{j}=1+j$ for each $2\leq j\leq n-1$.

Now, suppose the claim is true for an arbitrary $k$. The number of
hypersurfaces of degree $j$ in a $(k+1)^{st}$ polar cone is exactly the number
of hypersurfaces of degree at least $j$ in a $k^{th}$ polar cone, e.g.
$\sum\limits_{i=0}^{j}\binom{k+i}{i}$. However,

$\displaystyle\sum\limits_{i=0}^{j}\binom{k+i}{i}=\binom{k+j+1}{j}=\binom{(k+1)+j}{j},$

by induction on $j$, using that $\binom{a}{b}=\binom{a-1}{b}+\binom{a-1}{b-1}$
for positive integers $a>b$. This combinatorial argument also yields that
$\sum\limits_{i=0}^{d-1}\binom{k+i}{i}=\binom{k+d}{d-1}$. ∎

In Subsection 3.2, we will use iterated polar cones to establish new upper
bounds on $\text{RD}(n)$. We first use iterated polar cones to determine
$k$-planes on intersections of quadrics.

######  Proposition 2.27.

($k$-Polar Points Intersections of Quadrics)
Let $V\subseteq\mathbb{P}_{K}^{r}$ be an intersection of hypersurfaces of type
$\left[\begin{matrix}2\\\ \ell\end{matrix}\right]$ and take $k\geq 1$. For
$r\geq(k+1)\ell+k$, we can determine a $k$-polar point
$\left(P_{0},\dotsc,P_{k}\right)$ over an extension $L/K$ with
$\text{RD}(L/K)\leq\text{RD}(2^{\ell})$. Moreover, for any point $P\in V(K)$,
we can determine a $k$-plane containing $P$ over such an extension $L$.

###### Proof.

When $r>(k+1)\ell+k$, we can restrict to an arbitrary $(k+1)\ell+k$-plane in
$\mathbb{P}_{K}^{r}$ and hence it suffices to consider the case where
$r=(k+1)\ell+k$. Note that for a quadric hypersurface, a $k^{th}$ polar cone
consists of the original quadric and $k$ hyperplanes $\left(\text{e.g. is of
type }\left[\begin{matrix}2&1\\\ 1&k\end{matrix}\right]\right)$. Hence, a
$k^{th}$ polar cone of $V$ has type $\left[\begin{matrix}2&1\\\
\ell&k\ell\end{matrix}\right]$. We proceed by induction on $k$.

When $k=1$ and $r=2\ell+1$, $\dim(V)\geq\ell+1>0$ and Proposition 2.19 allows
us to determine a point $P_{0}\in V(L_{1})$ over an extension $L_{1}/K$ of
degree $2^{\ell}$. The polar cone $\mathcal{C}(V;P_{0})$ is of type
$\left[\begin{matrix}2&1\\\ \ell&\ell\end{matrix}\right]$, hence

$\displaystyle\dim\left(\mathcal{C}(V;P_{0})\right)\geq(2\ell+1)-(2\ell)=1>0.$

We use Proposition 2.19 to determine a point
$P_{1}\in\mathcal{C}(V;P_{0})(L)\setminus\\{P_{0}\\}$ over an extension
$L/L_{1}$ of resolvent degree at most $\text{RD}\left(2^{\ell}\right)$. Note
that $\left(P_{0},P_{1}\right)$ is a 1-polar point of $V$ by construction.

We now consider the case of an arbitrary $k>1$. By induction, we pass to an
extension $L_{1}/K$ of degree at most $2^{\ell}$ to determine a $(k-1)$-polar
point $\left(P_{0},\dotsc,P_{k-1}\right)$ of $V$. Observe that

$\displaystyle\dim\left(\mathcal{C}^{k}(V;P_{0},\dotsc,P_{k-1})\right)\geq(k+1)\ell+k-(k+1)\ell\geq
k.$

By Proposition 2.19, we can pass to an extension $L/L_{1}$ of degree at most
$2^{\ell}$ to determine an $L$-rational point $P_{k}$ of
$\mathcal{C}^{k}(V;P_{0},\dotsc,P_{k-1})\setminus L(P_{0},\dotsc,P_{k-1})$. By
construction, $\left(P_{0},\dotsc,P_{k}\right)$ is a $k$-polar point of $V$.

For the final claim, note that we can replace $P_{0}$ with $P$ in the proof
and then Lemma 2.24 yields $L(P,P_{1},\dotsc,P_{k})$ is a suitable $k$-plane.
∎

##  3 New Upper Bounds on $\text{RD}(n)$

###  3.1 Tschirnhaus Transformations

We recall pertinent information about Tschirnhaus transformations here and
refer the reader to [Wol2021] for a more complete treatment. Recall that
$K_{n}=\mathbb{C}(a_{1},\dotsc,a_{n})$, a purely transcendental extension of
$\mathbb{C}$ with transcendence basis $a_{1},\dotsc,a_{n}$.

######  Definition 3.1.

(General Polynomials)
The general polynomial of degree $n$ is the polynomial

$\displaystyle\phi_{n}(z)=z^{n}+a_{1}z^{n-1}+\cdots+a_{n-1}z+a_{n}\in
K_{n}[z].$

######  Definition 3.2.

(Tschirnhaus Transformations)
A Tschirnhaus transformation of the general degree $n$ polynomial is an
isomorphism of $K_{n}$-fields

$\displaystyle K_{n}[z]/(\phi_{n}(z))\cong K_{n}[z]/(\psi(z)),$

where

$\displaystyle\psi(z)=z^{n}+b_{1}z^{n-1}+\cdots+b_{n-1}z+b_{n}.$

It has type $(j_{1},\dotsc,j_{k})$ if $b_{j_{1}}=\cdots=b_{j_{k}}=0$.

######  Remark 3.3.

(Description of Tschirnhaus Transformations)
Given primitive elements $\zeta_{\phi}$ of $K_{n}[z]/(\phi_{n}(z))$ and
$\zeta_{\psi}$ of $K_{n}[z]/(\psi(z))$, every $K_{n}$-algebra isomorphism
$\Upsilon:K_{n}[z]/(\phi_{n}(z))\rightarrow K_{n}[z]/(\psi(z))$ is determined
by $\Upsilon(\zeta_{\phi})$, which is a $K_{n}$-linear combination of powers
of $\zeta_{\psi}$:

$\Upsilon(\zeta_{\phi})=w_{0}+w_{1}\zeta_{\psi}+\cdots+w_{n-1}\zeta_{\psi}^{n-1}.$

Note that $\Upsilon$ is an isomorphism exactly when there is some $j\geq 1$
such that $w_{j}\not=0$. Let $\mathbb{A}_{K_{n}}^{n}$ be the affine space with
coordinates $w_{0},\dotsc,w_{n-1}$ and denote the $w_{0}$-axis of
$\mathbb{A}_{K_{n}}^{n}$ by $\mathbb{A}_{K_{n},0}^{1}$. Corollary 3.3 of
[Wol2021] shows that the space of Tschirnhaus transformations is exactly

$\widetilde{\mathcal{T}}_{K_{n}}^{n}=\mathbb{A}_{K_{n}}^{n}\setminus\mathbb{A}_{K_{n},0}^{1}.$

However, we need only work with Tschirnhaus transformations up to re-scaling.
Let $\mathbb{P}_{K_{n}}^{n-1}$ denote the projective space with coordinates
$w_{0},\dotsc,w_{n-1}$ and observe that the space of Tschirnhaus
transformations up to re-scaling is

$\mathcal{T}_{K_{n}}^{n}=\mathbb{P}_{K_{n}}^{n-1}\setminus\left\\{[1:0:\cdots:0]\right\\}.$

Moreover, each $b_{m}$ in Definition 3.2 is a homogenous polynomial of degree
$m$ in the $w_{0},\dotsc,w_{n-1}$ with coefficients in $K_{n}$ (e.g. an
element an of $K_{n}[w_{0},\dotsc,w_{n-1}]_{(m)}$). In the following
definition, we build upon the language used in [Wol2021].

######  Definition 3.4.

(Tschirnhaus Complete Intersections)
Fix $n\in\mathbb{Z}_{\geq 1}$. For any $m\in\left\\{1,\dotsc,n\right\\}$, the
$m^{th}$ extended Tschirnhaus hypersurface is

$\tau_{m}:=\mathbb{V}(b_{m})\subseteq\mathbb{P}_{K_{n}}^{n},$

and the $m^{th}$ extended Tschirnhaus complete intersection is

$\tau_{1,\dotsc,m}:=\tau_{1}\cap\cdots\cap\tau_{m}\subseteq\mathbb{P}_{K_{n}}^{n}.$

Similarly, the $m^{th}$ Tschirnhaus hypersurface is

$\tau_{m}^{\circ}:=\tau_{m}\cap\mathcal{T}_{K_{n}}^{n}=\tau_{m}\setminus\left\\{[1:0:\cdots:0]\right\\},$

and the $m^{th}$ Tschirnhaus complete intersection is

$\tau_{1,\dotsc,m}^{\circ}:=\tau_{1,\dotsc,m}\cap\mathcal{T}_{K_{n}}^{n}\setminus\left\\{[1:0:\cdots:0]\right\\}.$

######  Remark 3.5.

(RD Bounds from Tschirnhaus Transformations)
Given a Tschirnhaus transformation $\Upsilon$ (up to re-scaling) of type
$(1,\dotsc,m-1)$ of $\phi_{n}(z)$ (i.e. a point of
$\tau_{1,\dotsc,m-1}^{\circ}$), we need only consider the normal form

$z^{n}+b_{m}z^{n-m}+\cdots+b_{n-1}z+b_{n},$ (3)

as $\Upsilon^{-1}$ takes the roots of (3) to the roots of $\phi_{n}$. We can
then re-scale the roots over $\sqrt[n]{b_{n}}$ (over a cyclic extension) and
arrive at the normal form

$z^{n}+c_{m}z^{n-m}+\cdots+c_{n-1}z+1.$ (4)

Note that (4) is an algebraic function of $n-m$ variables and so if we can
determine an $L$-rational point of $\tau_{1,\dotsc,m-1}^{\circ}$ over an
extension $L/K_{n}$ of sufficiently small resolvent degree, we can conclude
that $\text{RD}(n)\leq n-m$.

###  3.2 New Bounds From Iterated Polar Cones

######  Definition 3.6.

(Sub-Intersections of Given Level)
Let $V\subseteq\mathbb{P}_{K}^{r}$ be an intersection of hypersurfaces of type
$\left[\begin{matrix}d&d-1&\cdots&2&1\\\
\ell_{d}&\ell_{d-1}&\cdots&\ell_{2}&\ell_{1}\end{matrix}\right]$ defined by

$\displaystyle
V=\bigcap\limits_{i=1}^{d}\bigcap\limits_{j=1}^{\ell_{i}}\mathbb{V}(f_{i,j}),$

with $\deg(f_{i,j})=i$ for all $1\leq j\leq\ell_{i}$. For each $1\leq
d^{\prime}\leq d$, the sub-intersection of $V$ of level $d^{\prime}$ is the
intersection of all defining hypersurfaces of $V$ of degree $d^{\prime}$ and
is denoted $V_{d^{\prime}}$, i.e.

$\displaystyle
V_{d^{\prime}}:=\mathbb{V}(f_{d^{\prime},1})\cap\cdots\cap\mathbb{V}(f_{d^{\prime},\ell_{d^{\prime}}}).$

######  Theorem 3.7.

(The $n-6$ Bound)
For $n\geq 21$, $\text{RD}(n)\leq n-6$.

###### Proof.

First, suppose that we can determine a plane
$\Lambda\subseteq\tau_{1,2,3}^{\circ}$ over an extension of $K_{n}$ of low
resolvent degree. Observe that
$\deg\left(\Lambda\cap\tau_{1,\dotsc,5}\right)=20$ and, by Proposition 2.19,
we can solve a polynomial of degree at most 20 to determine a point $Q$ of
$\Lambda\cap\tau_{1,\dotsc,5}$.

From Lemma 2.24, it suffices to determine a 2-polar point
$(P_{0},P_{1},P_{2})$ on $\tau_{1,2,3}\subseteq\mathcal{T}^{n}$ such that
$L(P_{0},P_{1},P_{2})\subseteq\tau_{1,2,3}^{\circ}$ for $n=20$ over a suitable
extension. Indeed, we show the stronger claim that we can determine such a
2-polar point of $\tau_{1,2,3}$ when $n=19$ over an extension of resolvent
degree at most $\text{RD}(12)$, which implies the result for larger $n$.

Recall that when $n=19$, we work in $\mathbb{P}_{K_{n}}^{18}$ (as in Remark
3.3). To ensure that the plane $\Lambda$ associated to the 2-polar point we
determine lies in $\tau_{1,2,3}^{\circ}$, we pass to a hyperplane $H$ which
does not contain $[1:0:\cdots:0]$. We can determine a point
$P_{0}\in(\tau_{1,2,3}\cap H)(L_{1})$, where $L_{1}/K_{n}$ is an extension of
degree 6, by Proposition 2.19. The polar cone $\mathcal{C}(\tau_{1,2,3}\cap
H;P_{0})$ has type $\left[\begin{matrix}3&2&1\\\ 1&2&4\end{matrix}\right]$ and
so

$\displaystyle\dim\left(\mathcal{C}(\tau_{1,2,3};P_{0})\right)\geq 18-7=11.$

Hence, we can determine a point $P_{1}\in\mathcal{C}(\tau_{1,2,3}\cap
H;P_{0})(L_{2})\setminus\\{P_{0}\\}$ over a degree 12 extension $L_{2}/L_{1}$.

The second polar cone $\mathcal{C}^{2}(\tau_{1,2,3}\cap H;P_{0},P_{1})$ has
type $\left[\begin{matrix}3&2&1\\\ 1&3&7\end{matrix}\right]$. As
$\mathcal{C}^{2}(\tau_{1,2,3}\cap H;P_{0},P_{1})_{1}$ is the intersection of 7
hyperplanes, $\mathcal{C}^{2}(\tau_{1,2,3}\cap
H;P_{0}P_{1})_{1}\cong\mathbb{P}_{L_{2}}^{11}$. Denote one of the quadrics
defining $\mathcal{C}^{2}(\tau_{1,2,3}\cap H;P_{0},P_{1})$ by $U$. Applying
Proposition 2.27 to $U\cap\mathcal{C}^{2}(\tau_{1,2,3}\cap H;P_{0},P_{1})_{1}$
(with $\ell=1$ and $k=5$) yields that we can determine a 5-plane
$\Lambda^{\prime}\subseteq U\cap\mathcal{C}^{2}(\tau_{1,2,3}\cap
H;P_{0},P_{1})_{1}$ over a quadratic extension $L_{3}/L_{2}$. Note that
$\mathcal{C}^{2}(\tau_{1,2,3}\cap H;P_{0},P_{1})\cap\Lambda^{\prime}$ has type
$\left[\begin{matrix}3&2\\\ 1&2\end{matrix}\right]$ inside $\Lambda^{\prime}$
and

$\displaystyle\dim\left(\mathcal{C}^{2}(\tau_{1,2,3}\cap
H;P_{0},P_{1})\cap\Lambda^{\prime}\right)\geq 5-3\geq 2.$

Thus, we can determine a point $P_{2}\in\left(\mathcal{C}^{2}(\tau_{1,2,3}\cap
H;P_{0},P_{1})\cap\Lambda^{\prime}\right)\setminus L(P_{0},P_{1})$ over an
extension $L_{4}/L_{3}$ of degree

$\displaystyle\deg\left(\mathcal{C}^{2}(\tau_{1,2,3}\cap
H;P_{0},P_{1})\cap\Lambda^{\prime}\right)=3\cdot 2^{2}=12,$

and $\left(P_{0},P_{1},P_{2}\right)$ is a 2-polar point of $\tau_{1,2,3}\cap
H$ by construction. ∎

######  Remark 3.8.

(Fixing Gaps in [Che1954])
As noted in Section 1, Chebotarev gave an argument that $\text{RD}(n)\leq n-6$
for $n\geq 21$ in [Che1954]. However, his argument had gaps (similar to those
of Wiman in [Wim1927] and which are also in the translation [Sut2021A]). There
are two specific issues. First, Chebotarev uses a geometric argument of Wiman
which assumed without proof that certain intersections of hypersurfaces in
affine space were generic (see p.190-191 of [Che1954] or p.3 of the
translation [Sut2021B]). Additionally, Lemma 1 and Lemma 2 of [Che1954] (Lemma
2.1 and Lemma 2.2 of the translation [Sut2021B]) also assume, but do not
prove, that certain intersections of affine hypersurfaces are generic. The
above proof of Theorem 3.7 fixes the issues associated to the use of Lemma 1
and Lemma 2 in the proof of the un-numbered theorem of [Che1954] (Theorem 2.3
of [Sut2021B]). Additionally, this argument can be adapted to give a geometric
proof of Wiman’s claims, although Dixmier has already given an algebraic proof
in the appendix of [Dix1993].

######  Remark 3.9.

(Notation for Theorem 3.10)
We move from finding a plane on $\tau_{1,2,3}^{\circ}$ to finding $k$-planes
on $\tau_{1,2,3,4}^{\circ}$ to improve the bound on $n$ for which
$\text{RD}(n)\leq n-m$ for $m$ between $7$ and $14$. To simplify the statement
of Theorem 3.10, we introduce two functions
$\rho:\\{1,2,3,4,5,6,7,8,9\\}\rightarrow\mathbb{Z}\text{ and
}\eta:\\{1,2,3,4,5,6,7,8,9\\}\rightarrow\mathbb{Z}$. For each $k$, $\rho(k)$
is the ambient dimension needed in our method to determine a $k$-plane on
$\tau_{1,2,3,4}$ and $\eta(k)$ is the degree of the largest extension needed.
The functions $\rho$ and $\eta$ are defined by the following values:

$k$ | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9
---|---|---|---|---|---|---|---|---|---
$\rho(k)$ | 25 | 60 | 264 | 806 | 1773 | 8905 | 34546 | 77040 | 612581
$\eta(k)$ | 36 | 108 | 324 | 972 | 2916 | 8748 | 26244 | 78732 | 236196

######  Theorem 3.10.

(The $n-7,\dotsc,n-14$ Bounds)

  1. 1.

For $n\geq 109$, $\text{RD}(n)\leq n-7$.

  2. 2.

For $n\geq 325$, $\text{RD}(n)\leq n-8$.

  3. 3.

For each $9\leq\ell\leq 14$ and $n>\frac{(\ell-1)!}{24}$, $\text{RD}(n)\leq
n-\ell$.

###### Proof.

We claim that we can determine a $k$-plane
$\Lambda\subseteq\tau_{1,2,3,4}^{\circ}$ over an extension of $K_{n}$ of
resolvent degree at most $\text{RD}(\eta(k))$ when $n\geq\rho(k)$. Given this
claim, $\Lambda\cap\tau_{1,\dotsc,k+4}$ has degree $\frac{(k+4)!}{24}$. By
Proposition 2.19, we can solve a polynomial of degree at most
$\frac{(k+4)!}{24}$ to determine a point $Q$ of
$\Lambda\cap\tau_{1,\dotsc,k+4}^{\circ}$. Then, the new bounds on resolvent
degree follow immediately from Remark 3.5. Note that for the $n-7$ bound
(respectively, $n-8$ bound), $k=2$ (respectively, $k=3$) and
$\eta(k)\geq\frac{(k+4)!}{24}$.

It remains to determine the $k$-planes on $\tau_{1,2,3,4}^{\circ}$ and, by
Lemma 2.24, it suffices to construct a $k$-polar point $(P_{0},\dotsc,P_{k})$
of $\tau_{1,2,3,4}$ which does not contain $[1:0:\cdots:0]$. For each $k$, it
suffices to prove the claim when $n=\rho(k)$, as we can always restrict to a
$\mathbb{P}_{K_{n}}^{\rho(k)}$ when $n>\rho(k)$. In each case, we immediately
pass to a hypersurface which does not contain $[1:0:\cdots:0]$ and, to
simplify notation, the computations that follow all start within this
hypersurface. Recall that the space of Tschirnhaus transformations (up to re-
scaling) is $\mathbb{P}_{K_{n}}^{n-1}$ and we are passing to a hypersurface,
so for each $k$ we work in a $\mathbb{P}_{K_{n}}^{\rho(k)-2}$. In each case,
the extensions will be enumerated as $L_{j}$ and we always begin with
$L_{1}/K_{n}$. Similarly, the $k$-planes we determine will be enumerated as
$\Lambda_{\ell}$ and we always begin with $\Lambda_{1}$. Additionally,
Proposition 2.26 gives the type of the relevant polar cone.

#### Case: $k=1$.

Recall that $\rho(1)=25$ and thus we work in $\mathbb{P}_{K_{n}}^{23}$. By
Proposition 2.19, we can determine a point $P_{0}$ of $\tau_{1,2,3,4}$ over an
extension $L_{1}/K_{n}$ of degree at most 24. The polar cone
$\mathcal{C}(\tau_{1,2,3,4};P_{0})$ has type $\left[\begin{matrix}4&3&2&1\\\
1&2&3&4\end{matrix}\right]$. Note that $\mathcal{C}(\tau_{1,2,3,4};P_{0})_{1}$
is an intersection of 4 hyperplanes and thus
$\mathcal{C}(\tau_{1,2,3,4};P_{0})_{1}\cong\mathbb{P}_{L_{1}}^{19}$. Since
$\mathcal{C}(\tau_{1,2,3,4};P_{0})_{2}\cap\mathcal{C}(\tau_{1,2,3,4};P_{0})_{1}$
has type $\left[\begin{matrix}2\\\ 3\\\ \end{matrix}\right]$ inside
$\mathcal{C}(\tau_{1,2,3,4};P_{0})_{1}$ and

$19=(4+1)(3)+4,$

Proposition 2.27 (with $k=4$ and $\ell=3$) allows us to determine a 4-plane
$\Lambda_{1}\subseteq\mathcal{C}(\tau_{1,2,3,4};P_{0})_{2}\cap\mathcal{C}(\tau_{1,2,3,4};P_{0})_{1}$
over an extension $L_{2}/L_{1}$ of degree at most 8. Observe that
$\mathcal{C}(\tau_{1,2,3,4};P_{0})\cap\Lambda_{1}$ has type
$\left[\begin{matrix}4&1\\\ 1&2\end{matrix}\right]$, hence

$\dim\left(\mathcal{C}(\tau_{1,2,3,4};P_{0})\cap\Lambda_{1}\right)\geq 4-3\geq
1.$

Consequently, we can determine a rational point $P_{1}$ of
$\mathcal{C}(\tau_{1,2,3,4};P_{0})\setminus\\{P_{0}\\}$ over an extension
$L_{3}/L_{2}$ of degree at most 36. By construction, $(P_{0},P_{1})$ is a
1-polar point of $\tau_{1,2,3,4}$.

#### Case: $k=2$.

Note that $\rho(2)=60$, so we work in $\mathbb{P}_{K_{n}}^{58}$. From the
$k=1$ case, we pass to an extension $L_{1}/K_{n}$ of resolvent degree at most
$\text{RD}(36)$ and assume that we have a 1-polar point $(P_{0},P_{1})$. The
second polar cone $\mathcal{C}^{2}(\tau_{1,2,3,4};P_{0},P_{1})$ has type
$\left[\begin{matrix}4&3&2&1\\\ 1&3&6&10\end{matrix}\right]$. Observe that
$\mathcal{C}^{2}(\tau_{1,2,3,4};P_{0},P_{1})_{1}$ is an intersection of 10
hyperplanes, hence
$\mathcal{C}^{2}(\tau_{1,2,3,4};P_{0},P_{1})_{1}\cong\mathbb{P}_{L_{1}}^{48}$.
Moreover,
$\mathcal{C}^{2}(\tau_{1,2,3,4};P_{0},P_{1})_{2}\cap\mathcal{C}^{2}(\tau_{1,2,3,4};P_{0},P_{1})_{1}$
has type $\left[\begin{matrix}2\\\ 6\end{matrix}\right]$ in
$\mathcal{C}^{2}(\tau_{1,2,3,4};P_{0},P_{1})_{1}$ and

$48=(5+1)(6)+5.$

Thus, applying Proposition 2.27 (with $k=5$ and $\ell=6$) allows us to
determine a 5-plane
$\Lambda_{1}\subseteq\mathcal{C}^{2}(\tau_{1,2,3,4};P_{0},P_{1})_{2}\cap\mathcal{C}^{2}(\tau_{1,2,3,4};P_{0},P_{1})_{1}$
over an extension $L_{2}/L_{1}$ of degree at most 64. Note that
$\mathcal{C}^{2}(\tau_{1,2,3,4};P_{0},P_{1})\cap\Lambda_{1}$ has type
$\left[\begin{matrix}4&3\\\ 1&3\end{matrix}\right]$ inside $\Lambda_{1}$,
hence

$\dim\left(\mathcal{C}^{2}(\tau_{1,2,3,4};P_{0},P_{1})\cap\Lambda_{1}\right)\geq
6-4\geq 2.$

As a result, we can determine a rational point $P_{2}$ of
$\mathcal{C}^{2}(\tau_{1,2,3,4};P_{0},P_{1})\setminus L(P_{0},P_{1})$ over an
extension $L_{3}/L_{2}$ of degree at most 108 and our method ensures that
$(P_{0},P_{1},P_{2})$ is a 2-polar point of $\tau_{1,2,3,4}$.

#### Case: $k=3$.

Recall that $\rho(3)=264$ and so we work in $\mathbb{P}_{K_{n}}^{262}$. Using
the $k=2$ case, we pass to an extension $L_{1}/K_{n}$ of resolvent degree at
most $\text{RD}(108)$ and assume that we have a 2-polar point
$(P_{0},P_{1},P_{2})$. Observe that the third polar cone
$\mathcal{C}^{3}(\tau_{1,2,3,4};P_{0},P_{1},P_{2})$ has type
$\left[\begin{matrix}4&3&2&1\\\ 1&4&10&20\end{matrix}\right]$. Thus,
$\mathcal{C}^{3}(\tau_{1,2,3,4};P_{0},P_{1},P_{2})_{1}$ is an intersection of
35 hyperplanes and
$\mathcal{C}^{3}(\tau_{1,2,3,4};P_{0},P_{1},P_{2})_{1}\cong\mathbb{P}_{L_{1}}^{242}$.
Inside $\mathcal{C}^{3}(\tau_{1,2,3,4};P_{0},P_{1},P_{2})_{1}$,
$\mathcal{C}^{3}(\tau_{1,2,3,4};P_{0},P_{1},P_{2})_{2}\cap\mathcal{C}^{3}(\tau_{1,2,3,4};P_{0},P_{1},P_{2})_{1}$
has type $\left[\begin{matrix}2\\\ 10\end{matrix}\right]$. However,
$2^{8}=256$ is the largest power of 2 less than $4\cdot 3^{4}=324$. As a
result, we split $\mathcal{C}^{3}(\tau_{1,2,3,4};P_{0},P_{1},P_{2})_{2}$ into
two intersections of quadrics:

$\mathcal{C}^{3}(\tau_{1,2,3,4};P_{0},P_{1},P_{2})_{2}=W_{1}\cap W_{2},$

where $W_{1}$ has type $\left[\begin{matrix}2\\\ 2\end{matrix}\right]$ and
$W_{2}$ has type $\left[\begin{matrix}2\\\ 8\end{matrix}\right]$. Next,
observe that

$242=(80+1)(2)+80,$

and thus applying Proposition 2.27 (with $k=75$ and $\ell=2$) allows us to
determine an 80-plane $\Lambda_{1}\subseteq
W_{1}\cap\mathcal{C}^{3}(\tau_{1,2,3,4};P_{0},P_{1},P_{2})_{1}$ over an
extension $L_{2}/L_{1}$ of degree at most 4. Also, $W_{2}\cap\Lambda_{1}$ has
type $\left[\begin{matrix}2\\\ 8\end{matrix}\right]$ in $\Lambda_{1}$ and

$80=(8+1)(8)+8.$

We then apply Proposition 2.27 (with $k=8$ and $\ell=8$) to determine an
8-plane $\Lambda_{2}\subseteq W_{2}\cap\Lambda_{1}$ over an extension
$L_{3}/L_{2}$ of degree at most 256. Now,
$\mathcal{C}^{3}(\tau_{1,2,3,4};P_{0},P_{1},P_{2})\cap\Lambda_{2}$ has type
$\left[\begin{matrix}4&3\\\ 1&4\end{matrix}\right]$ in $\Lambda_{2}$.
Consequently,

$\dim\left(\mathcal{C}^{3}(\tau_{1,2,3,4};P_{0},P_{1},P_{2})\cap\Lambda_{2}\right)\geq
8-5\geq 3$

and we can determine a rational point $P_{3}$ of
$\mathcal{C}^{3}(\tau_{1,2,3,4};P_{0},P_{1},P_{2})\setminus
L(P_{0},P_{1},P_{2})$ over an extension $L_{4}/L_{3}$ of degree at most 324.
By design, $(P_{0},P_{1},P_{2},P_{3})$ is a 3-polar point of $\tau_{1,2,3,4}$.

#### Case: $k=4$.

Observe that $\rho(4)=806$, hence we work in $\mathbb{P}_{K_{n}}^{804}$. By
the $k=3$ case, we pass to an extension $L_{1}/K_{n}$ of resolvent degree at
most $\text{RD}(324)$ and assume that we have a 3-polar point
$(P_{0},P_{1},P_{2},P_{3})$. Observe that the fourth polar cone
$\mathcal{C}^{4}(\tau_{1,2,3,4};P_{0},\dotsc,P_{3})$ has type
$\left[\begin{matrix}4&3&2&1\\\ 1&5&15&35\end{matrix}\right]$. Thus,
$\mathcal{C}^{4}(\tau_{1,2,3,4};P_{0},\dotsc,P_{3})_{1}$ is an intersection of
35 hyperplanes and
$\mathcal{C}^{4}(\tau_{1,2,3,4};P_{0},\dotsc,P_{3})_{1}\cong\mathbb{P}_{L_{1}}^{769}$.
Note that
$\mathcal{C}^{4}(\tau_{1,2,3,4};P_{0},\dotsc,P_{3})_{2}\cap\mathcal{C}^{4}(\tau_{1,2,3,4};P_{0},\dotsc,P_{3})_{1}$
has type $\left[\begin{matrix}2\\\ 15\end{matrix}\right]$ in
$\mathcal{C}^{4}(\tau_{1,2,3,4};P_{0},\dotsc,P_{3})_{1}$ and $2^{9}=512$ is
the largest power of 2 less than $4\cdot 3^{5}=972$. Consequently, we split
$\mathcal{C}^{4}(\tau_{1,2,3,4};P_{0},\dotsc,P_{3})_{2}$ into two
intersections of quadrics:

$\mathcal{C}^{4}(\tau_{1,2,3,4};P_{0},\dotsc,P_{3})_{2}=W_{1}\cap W_{2},$

where $W_{1}$ has type $\left[\begin{matrix}2\\\ 6\end{matrix}\right]$ and
$W_{2}$ has type $\left[\begin{matrix}2\\\ 9\end{matrix}\right]$. We have that

$769=(109+1)(6)+109,$

and so by applying Proposition 2.27 (with $k=109$ and $\ell=6$), we can
determine a 109-plane $\Lambda_{1}\subseteq
W_{1}\cap\mathcal{C}^{4}(\tau_{1,2,3,4};P_{0},\dotsc,P_{3})_{1}$ over an
extension $L_{2}/L_{1}$ of degree at most 64. Similarly,
$W_{2}\cap\Lambda_{1}$ has type $\left[\begin{matrix}2\\\
9\end{matrix}\right]$ in $\Lambda_{1}$ and

$109=(10+1)(9)+10.$

By applying Proposition 2.27 (with $k=10$ and $\ell=9$), we determine an
8-plane $\Lambda_{2}\subseteq W_{2}\cap\Lambda_{1}$ over an extension
$L_{3}/L_{2}$ of degree at most 512. It follows that
$\mathcal{C}^{4}(\tau_{1,2,3,4};P_{0},\dotsc,P_{3})\cap\Lambda_{2}$ has type
$\left[\begin{matrix}4&3\\\ 1&5\end{matrix}\right]$ in $\Lambda_{2}$.
Consequently,

$\dim\left(\mathcal{C}^{4}(\tau_{1,2,3,4};P_{0},\dotsc,P_{3})\cap\Lambda_{2}\right)\geq
10-6\geq 4$

and we can determine a rational point $P_{4}$ of
$\mathcal{C}^{4}(\tau_{1,2,3,4};P_{0},\dotsc,P_{3})\setminus
L(P_{0},P_{1},P_{2},P_{3})$ over an extension $L_{4}/L_{3}$ of degree at most
972. Indeed, $(P_{0},\dotsc,P_{4})$ is a 4-polar point of $\tau_{1,2,3,4}$.

#### Case: $k=5$.

Observe that $\rho(5)=1773$, hence we work in $\mathbb{P}_{K_{n}}^{1771}$.
From the $k=4$ case, we pass to an extension $L_{1}/K_{n}$ of resolvent degree
at most $\text{RD}(972)$ and assume that we have a 4-polar point
$(P_{0},\dotsc,P_{4})$. The fifth polar cone
$\mathcal{C}^{5}(\tau_{1,2,3,4};P_{0},\dotsc,P_{4})$ has type
$\left[\begin{matrix}4&3&2&1\\\ 1&6&21&56\end{matrix}\right]$. Hence,
$\mathcal{C}^{5}(\tau_{1,2,3,4};P_{0},\dotsc,P_{4})_{1}$ is an intersection of
56 hyperplanes and
$\mathcal{C}^{5}(\tau_{1,2,3,4};P_{0},\dotsc,P_{4})_{1}\cong\mathbb{P}_{L_{1}}^{1715}$.
Note that
$\mathcal{C}^{5}(\tau_{1,2,3,4};P_{0},\dotsc,P_{4})_{2}\cap\mathcal{C}^{5}(\tau_{1,2,3,4};P_{0},\dotsc,P_{4})_{1}$
has type $\left[\begin{matrix}2\\\ 21\end{matrix}\right]$ in
$\mathcal{C}^{5}(\tau_{1,2,3,4};P_{0},\dotsc,P_{4})_{1}$. Observe that
$2^{11}=2048$ is the largest power of 2 less than $4\cdot 3^{6}=2916$ and so
we split $\mathcal{C}^{5}(\tau_{1,2,3,4};P_{0},\dotsc,P_{4})_{2}$ into two
intersections of quadrics:

$\mathcal{C}^{5}(\tau_{1,2,3,4};P_{0},\dotsc,P_{4})_{2}=W_{1}\cap W_{2},$

where $W_{1}$ has type $\left[\begin{matrix}2\\\ 10\end{matrix}\right]$ and
$W_{2}$ has type $\left[\begin{matrix}2\\\ 11\end{matrix}\right]$. Note that

$1715=(155+1)(10)+155,$

and, by applying Proposition 2.27 (with $k=155$ and $\ell=10$), we can
determine a 155-plane $\Lambda_{1}\subseteq
W_{1}\cap\mathcal{C}^{5}(\tau_{1,2,3,4};P_{0},\dotsc,P_{4})_{1}$ over an
extension $L_{2}/L_{1}$ of degree at most 1024. Also, $W_{2}\cap\Lambda_{1}$
has type $\left[\begin{matrix}2\\\ 11\end{matrix}\right]$ in $\Lambda_{1}$ and

$155=(12+1)(11)+12,$

so we can apply Proposition 2.27 (with $k=12$ and $\ell=11$) to determine a
10-plane $\Lambda_{2}\subseteq W_{2}\cap\Lambda_{1}$ over an extension
$L_{3}/L_{2}$ of degree at most 2048. It follows that
$\mathcal{C}^{5}(\tau_{1,2,3,4};P_{0},\dotsc,P_{4})\cap\Lambda_{2}$ has type
$\left[\begin{matrix}4&3\\\ 1&6\end{matrix}\right]$ in $\Lambda_{2}$.
Consequently,

$\dim\left(\mathcal{C}^{5}(\tau_{1,2,3,4};P_{0},\dotsc,P_{4})\cap\Lambda_{2}\right)\geq
12-7\geq 5,$

and we can determine a rational point $P_{5}$ of
$\mathcal{C}^{5}(\tau_{1,2,3,4};P_{0},\dotsc,P_{4})\setminus
L(P_{0},\dotsc,P_{4})$ over an extension $L_{4}/L_{3}$ of degree at most 2916.
By construction, $(P_{0},\dotsc,P_{5})$ is a 5-polar point of
$\tau_{1,2,3,4}$.

#### Case: $k=6$.

Recall that $\rho(6)=8905$ and thus we work in $\mathbb{P}_{K_{n}}^{8903}$.
Using the $k=5$ case, we pass to an extension $L_{1}/K_{n}$ of resolvent
degree at most $\text{RD}(2916)$ and assume that we have a 5-polar point
$(P_{0},\dotsc,P_{5})$. The sixth polar cone
$\mathcal{C}^{6}(\tau_{1,2,3,4};P_{0},\dotsc,P_{5})$ has type
$\left[\begin{matrix}4&3&2&1\\\ 1&7&28&84\end{matrix}\right]$. It follows that
$\mathcal{C}^{6}(\tau_{1,2,3,4};P_{0},\dotsc,P_{5})_{1}$ is an intersection of
84 hyperplanes, so
$\mathcal{C}^{6}(\tau_{1,2,3,4};P_{0},\dotsc,P_{5})_{1}\cong\mathbb{P}_{L_{1}}^{8819}$.
Note that
$\mathcal{C}^{6}(\tau_{1,2,3,4};P_{0},\dotsc,P_{5})_{2}\cap\mathcal{C}^{6}(\tau_{1,2,3,4};P_{0},\dotsc,P_{5})_{1}$
has type $\left[\begin{matrix}2\\\ 28\end{matrix}\right]$ in
$\mathcal{C}^{6}(\tau_{1,2,3,4};P_{0},\dotsc,P_{5})_{1}$ and $2^{13}=8192$ is
the largest power of 2 less than $4\cdot 3^{7}=8748$. Consequently, we split
$\mathcal{C}^{6}(\tau_{1,2,3,4};P_{0},\dotsc,P_{5})_{2}$ into three
intersections of quadrics:

$\mathcal{C}^{6}(\tau_{1,2,3,4};P_{0},\dotsc,P_{5})_{2}=W_{1}\cap W_{2}\cap
W_{3},$

where $W_{1}$ has type $\left[\begin{matrix}2\\\ 2\end{matrix}\right]$ and
both $W_{2},W_{3}$ have type $\left[\begin{matrix}2\\\ 13\end{matrix}\right]$.
We have that

$8819=(2939+1)(2)+2939,$

and so by applying Proposition 2.27 (with $k=2939$ and $\ell=2$), we can
determine a 2939-plane $\Lambda_{1}\subseteq
W_{1}\cap\mathcal{C}^{6}(\tau_{1,2,3,4};P_{0},\dotsc,P_{5})_{1}$ over an
extension $L_{2}/L_{1}$ of degree at most 4. Additionally,
$W_{2}\cap\Lambda_{1}$ has type $\left[\begin{matrix}2\\\
13\end{matrix}\right]$ inside $\Lambda_{1}$ and

$2939=(209+1)(13)+209.$

We then apply Proposition 2.27 (with $k=209$ and $\ell=13$) to determine a
209-plane $\Lambda_{2}\subseteq W_{2}\cap\Lambda_{1}$ over an extension
$L_{3}/L_{2}$ of degree at most 8192. Similarly, $W_{3}\cap\Lambda_{2}$ has
type $\left[\begin{matrix}2\\\ 13\end{matrix}\right]$ in $\Lambda_{2}$ and

$209=(14+1)(13)+14,$

and thus we apply Proposition 2.27 (with $k=14$ and $\ell=13$) to determine a
14-plane $\Lambda_{3}\subseteq W_{3}\cap\Lambda_{2}$ over an extension
$L_{4}/L_{3}$ of degree at most 8192. Observe that
$\mathcal{C}^{6}(\tau_{1,2,3,4};P_{0},\dotsc,P_{5})\cap\Lambda_{3}$ has type
$\left[\begin{matrix}4&3\\\ 1&7\end{matrix}\right]$ inside $\Lambda_{3}$ and

$\dim\left(\mathcal{C}^{6}(\tau_{1,2,3,4};P_{0},\dotsc,P_{5})\cap\Lambda_{3}\right)\geq
14-8\geq 6.$

As a consequence, we can determine a rational point $P_{6}$ of
$\mathcal{C}^{6}(\tau_{1,2,3,4};P_{0},\dotsc,P_{5})\setminus
L(P_{0},\dotsc,P_{5})$ over an extension $L_{5}/L_{4}$ of degree at most 8748.
It follows that $(P_{0},\dotsc,P_{6})$ is a 6-polar point of $\tau_{1,2,3,4}$.

#### Case: $k=7$.

Note that $\rho(7)=34546$, so we work in $\mathbb{P}_{K_{n}}^{34544}$. By the
$k=6$ case, we pass to an extension $L_{1}/K_{n}$ of resolvent degree at most
$\text{RD}(8748)$ and assume that we have a 6-polar point
$(P_{0},\dotsc,P_{6})$. Observe that the seventh polar cone
$\mathcal{C}^{7}(\tau_{1,2,3,4};P_{0},\dotsc,P_{6})$ has type
$\left[\begin{matrix}4&3&2&1\\\ 1&8&36&120\end{matrix}\right]$ and so
$\mathcal{C}^{7}(\tau_{1,2,3,4};P_{0},\dotsc,P_{6})_{1}$ is an intersection of
120 hyperplanes. Thus,
$\mathcal{C}^{7}(\tau_{1,2,3,4};P_{0},\dotsc,P_{6})_{1}\cong\mathbb{P}_{L_{1}}^{34424}$.
Inside $\mathcal{C}^{7}(\tau_{1,2,3,4};P_{0},\dotsc,P_{6})_{1}$,
$\mathcal{C}^{7}(\tau_{1,2,3,4};P_{0},\dotsc,P_{6})_{2}\cap\mathcal{C}^{7}(\tau_{1,2,3,4};P_{0},\dotsc,P_{6})_{1}$
has type $\left[\begin{matrix}2\\\ 36\end{matrix}\right]$. Note that
$2^{14}=16384$ is the largest power of 2 less than $4\cdot 3^{8}=26244$.
Consequently, we split
$\mathcal{C}^{7}(\tau_{1,2,3,4};P_{0},\dotsc,P_{6})_{2}$ into three
intersections of quadrics:

$\mathcal{C}^{7}(\tau_{1,2,3,4};P_{0},\dotsc,P_{6})_{2}=W_{1}\cap W_{2}\cap
W_{3},$

where $W_{1}$ has type $\left[\begin{matrix}2\\\ 8\end{matrix}\right]$ and
both $W_{2},W_{3}$ have type $\left[\begin{matrix}2\\\ 14\end{matrix}\right]$.
We have that

$34424=(3824+1)(8)+3824,$

and apply Proposition 2.27 (with $k=3824$ and $\ell=8$) to determine a
3824-plane $\Lambda_{1}\subseteq
W_{1}\cap\mathcal{C}^{7}(\tau_{1,2,3,4};P_{0},\dotsc,P_{6})_{1}$ over an
extension $L_{2}/L_{1}$ of degree at most 256. Additionally,
$W_{2}\cap\Lambda_{1}$ has type $\left[\begin{matrix}2\\\
14\end{matrix}\right]$ inside $\Lambda_{1}$ and

$3824=(254+1)(14)+254.$

Consequently, we can apply Proposition 2.27 (with $k=254$ and $\ell=14$) to
determine a 254-plane $\Lambda_{2}\subseteq W_{2}\cap\Lambda_{1}$ over an
extension $L_{3}/L_{2}$ of degree at most 16384. Observe that
$W_{3}\cap\Lambda_{2}$ has type $\left[\begin{matrix}2\\\
14\end{matrix}\right]$ in $\Lambda_{2}$ and

$254=(16+1)(14)+16,$

and so, by Proposition 2.27 (with $k=16$ and $\ell=14$) to determine a
16-plane $\Lambda_{3}\subseteq W_{3}\cap\Lambda_{2}$ over an extension
$L_{4}/L_{3}$ of degree at most 16384. Observe that
$\mathcal{C}^{7}(\tau_{1,2,3,4};P_{0},\dotsc,P_{6})\cap\Lambda_{3}$ has type
$\left[\begin{matrix}4&3\\\ 1&8\end{matrix}\right]$ inside $\Lambda_{3}$ and

$\dim\left(\mathcal{C}^{7}(\tau_{1,2,3,4};P_{0},\dotsc,P_{6})\cap\Lambda_{3}\right)\geq
16-9\geq 7.$

It follows that we can determine a rational point $P_{7}$ of
$\mathcal{C}^{7}(\tau_{1,2,3,4};P_{0},\dotsc,P_{6})\setminus
L(P_{0},\dotsc,P_{6})$ over an extension $L_{5}/L_{4}$ of degree at most
26244. Note that $(P_{0},\dotsc,P_{7})$ is a 7-polar point of
$\tau_{1,2,3,4}$.

#### Case: $k=8$.

Recall that $\rho(8)=77040$ and thus we work in $\mathbb{P}_{K_{n}}^{77038}$.
By the $k=7$ case, we pass to an extension $L_{1}/K_{n}$ of resolvent degree
at most $\text{RD}(26244)$ and assume that we have a 7-polar point
$(P_{0},\dotsc,P_{7})$. Observe that the eighth polar cone
$\mathcal{C}^{8}(\tau_{1,2,3,4};P_{0},\dotsc,P_{7})$ has type
$\left[\begin{matrix}4&3&2&1\\\ 1&9&45&165\end{matrix}\right]$ and so
$\mathcal{C}^{8}(\tau_{1,2,3,4};P_{0},\dotsc,P_{7})_{1}$ is an intersection of
165 hyperplanes. Thus,
$\mathcal{C}^{8}(\tau_{1,2,3,4};P_{0},\dotsc,P_{7})_{1}\cong\mathbb{P}_{L_{1}}^{76873}$
and
$\mathcal{C}^{8}(\tau_{1,2,3,4};P_{0},\dotsc,P_{7})_{2}\cap\mathcal{C}^{8}(\tau_{1,2,3,4};P_{0},\dotsc,P_{7})_{1}$
has type $\left[\begin{matrix}2\\\ 45\end{matrix}\right]$ inside
$\mathcal{C}^{8}(\tau_{1,2,3,4};P_{0},\dotsc,P_{7})_{1}$. We have that
$2^{16}=65536$ is the largest power of 2 less than $4\cdot 3^{9}=78732$.
Consequently, we split
$\mathcal{C}^{8}(\tau_{1,2,3,4};P_{0},\dotsc,P_{7})_{2}$ into three
intersections of quadrics:

$\mathcal{C}^{8}(\tau_{1,2,3,4};P_{0},\dotsc,P_{7})_{2}=W_{1}\cap W_{2}\cap
W_{3},$

where $W_{1}$ has type $\left[\begin{matrix}2\\\ 13\end{matrix}\right]$ and
both $W_{2},W_{3}$ have type $\left[\begin{matrix}2\\\ 16\end{matrix}\right]$.
Observe that

$76873=(5490+1)(13)+5490.$

Thus, we can apply Proposition 2.27 (with $k=5490$ and $\ell=13$) to determine
a 5490-plane $\Lambda_{1}\subseteq
W_{1}\cap\mathcal{C}^{8}(\tau_{1,2,3,4};P_{0},\dotsc,P_{7})_{1}$ over an
extension $L_{2}/L_{1}$ of degree at most 8192. We also have that
$W_{2}\cap\Lambda_{1}$ has type $\left[\begin{matrix}2\\\
16\end{matrix}\right]$ inside $\Lambda_{1}$ and

$5490=(322+1)(16)+322.$

As a result, we apply Proposition 2.27 (with $k=322$ and $\ell=16$) to
determine a 322-plane $\Lambda_{2}\subseteq W_{2}\cap\Lambda_{1}$ over an
extension $L_{3}/L_{2}$ of degree at most 65536. Note that
$W_{3}\cap\Lambda_{2}$ has type $\left[\begin{matrix}2\\\
16\end{matrix}\right]$ in $\Lambda_{2}$ and

$322=(18+1)(16)+18,$

and so Proposition 2.27 (with $k=18$ and $\ell=16$) allows us to determine an
18-plane $\Lambda_{3}\subseteq W_{3}\cap\Lambda_{2}$ over an extension
$L_{4}/L_{3}$ of degree at most 65536. Observe that
$\mathcal{C}^{8}(\tau_{1,2,3,4};P_{0},\dotsc,P_{7})\cap\Lambda_{3}$ has type
$\left[\begin{matrix}4&3\\\ 1&9\end{matrix}\right]$ inside $\Lambda_{3}$ and

$\dim\left(\mathcal{C}^{8}(\tau_{1,2,3,4};P_{0},\dotsc,P_{7})\cap\Lambda_{3}\right)\geq
18-10\geq 8.$

Consequently, we can determine a rational point $P_{8}$ of
$\mathcal{C}^{8}(\tau_{1,2,3,4};P_{0},\dotsc,P_{7})\setminus
L(P_{0},\dotsc,P_{7})$ over an extension $L_{5}/L_{4}$ of degree at most
78732. By design, $(P_{0},\dotsc,P_{8})$ is an 8-polar point of
$\tau_{1,2,3,4}$.

#### Case: $k=9$.

Recall that $\rho(9)=612581$ and thus we work in
$\mathbb{P}_{K_{n}}^{612579}$. By the $k=8$ case, we pass to an extension
$L_{1}/K_{n}$ of resolvent degree at most $\text{RD}(78732)$ and assume that
we have a 8-polar point $(P_{0},\dotsc,P_{8})$. Note that the ninth polar cone
$\mathcal{C}^{9}(\tau_{1,2,3,4};P_{0},\dotsc,P_{8})$ has type
$\left[\begin{matrix}4&3&2&1\\\ 1&10&55&220\end{matrix}\right]$. So,
$\mathcal{C}^{9}(\tau_{1,2,3,4};P_{0},\dotsc,P_{8}))_{1}$ is an intersection
of 220 hyperplanes and
$\mathcal{C}^{9}(\tau_{1,2,3,4};P_{0},\dotsc,P_{8})_{1}\cong\mathbb{P}_{L_{1}}^{612359}$.
Observe that
$\mathcal{C}^{9}(\tau_{1,2,3,4};P_{0},\dotsc,P_{8})_{2}\cap\mathcal{C}^{9}(\tau_{1,2,3,4};P_{0},\dotsc,P_{8})_{1}$
has type $\left[\begin{matrix}2\\\ 55\end{matrix}\right]$ inside
$\mathcal{C}^{9}(\tau_{1,2,3,4};P_{0},\dotsc,P_{8})_{1}$ and $2^{17}=131072$
is the largest power of 2 less than $4\cdot 3^{10}=236196$. Correspondingly,
we split $\mathcal{C}^{9}(\tau_{1,2,3,4};P_{0},\dotsc,P_{8})_{2}$ into four
intersections of quadrics:

$\mathcal{C}^{9}(\tau_{1,2,3,4};P_{0},\dotsc,P_{8})_{2}=W_{1}\cap W_{2}\cap
W_{3}\cap W_{4},$

where $W_{1}$ has type $\left[\begin{matrix}2\\\ 4\end{matrix}\right]$ and all
of $W_{2},W_{3},W_{4}$ have type $\left[\begin{matrix}2\\\
17\end{matrix}\right]$. Note that

$612359=(122471+1)(4)+122471,$

and so, by Proposition 2.27 (with $k=122471$ and $\ell=4$), we can determine a
122471-plane $\Lambda_{1}\subseteq
W_{1}\cap\mathcal{C}^{9}(\tau_{1,2,3,4};P_{0},\dotsc,P_{8})_{1}$ over an
extension $L_{2}/L_{1}$ of degree at most 16. Additionally,
$W_{2}\cap\Lambda_{1}$ has type $\left[\begin{matrix}2\\\
17\end{matrix}\right]$ inside $\Lambda_{1}$ and

$122471=(6803+1)(17)+6805.$

As a result, we apply Proposition 2.27 (with $k=6805$ and $\ell=17$) to
determine a 6805-plane $\Lambda_{2}\subseteq W_{2}\cap\Lambda_{1}$ over an
extension $L_{3}/L_{2}$ of degree at most 131072. Note that
$W_{3}\cap\Lambda_{2}$ has type $\left[\begin{matrix}2\\\
17\end{matrix}\right]$ in $\Lambda_{2}$ and

$6803=(377+1)(17)+377.$

and so Proposition 2.27 (with $k=18$ and $\ell=16$) allows us to determine an
18-plane $\Lambda_{3}\subseteq W_{3}\cap\Lambda_{2}$ over an extension
$L_{4}/L_{3}$ of degree at most 65536. Also, $W_{4}\cap\Lambda_{3}$ has type
$\left[\begin{matrix}2\\\ 17\end{matrix}\right]$ in $\Lambda_{3}$ and

$377=(20+1)(17)+20.$

From Proposition 2.27 (with $k=18$ and $\ell=16$), we can determine a 20-plane
$\Lambda_{4}\subseteq W_{4}\cap\Lambda_{3}$ over an extension $L_{5}/L_{4}$ of
degree at most 131072. We have that
$\mathcal{C}^{9}(\tau_{1,2,3,4};P_{0},\dotsc,P_{8})\cap\Lambda_{4}$ has type
$\left[\begin{matrix}4&3\\\ 1&10\end{matrix}\right]$ inside $\Lambda_{4}$ and

$\dim\left(\mathcal{C}^{9}(\tau_{1,2,3,4};P_{0},\dotsc,P_{8})\cap\Lambda_{4}\right)\geq
20-11\geq 9.$

As a result, we can determine a rational point $P_{9}$ of
$\mathcal{C}^{9}(\tau_{1,2,3,4};P_{0},\dotsc,P_{8})\setminus
L(P_{0},\dotsc,P_{8})$ over an extension $L_{6}/L_{5}$ of degree at most
236196 and it follows that $(P_{0},\dotsc,P_{9})$ is a 9-polar point of
$\tau_{1,2,3,4}$.

∎

###  3.3 New Bounds from Moduli Spaces

In [Wol2021], Wolfson uses moduli space methods to determine $k$-planes on
intersections of hypersurfaces over extensions of scalars of bounded resolvent
degree. Wolfson then applies these methods to obtain upper bounds on
$\text{RD}(n)$. We will use a similar construction (Theorem 3.24) and thus
begin by giving an example of Wolfson’s process in Example 3.17. To do so, we
must introduce additional language and notation.

######  Definition 3.11.

(Parameter and Moduli Spaces of Hypersurfaces)
Fix $d\geq 2$. The parameter space of degree $d$ hypersurfaces in
$\mathbb{P}_{\mathbb{C}}^{r}$ is

$\displaystyle\mathcal{H}(d;r)\cong\mathbb{P}_{\mathbb{C}}^{\binom{r+d}{d}-1}$

However, there is a natural action of $\text{PGL}(\mathbb{C},r+1)$ on
$\mathbb{P}_{\mathbb{C}}^{r}$ which identifies hypersurfaces which are
projectively equivalent. The parameter space of semi-stable, degree $d$
hypersurfaces in $\mathbb{P}_{\mathbb{C}}^{r}$ is the largest proper Zariski
open $U\subseteq\mathcal{H}(d;r)$ which is
$\text{PGL}(\mathbb{C},r+1)$-invariant; we denote it by $\mathcal{S}(d;r)$.
Consequently, the coarse moduli space of semi-stable, degree $d$ hypersurfaces
in $\mathbb{P}_{\mathbb{C}}^{r}$ is thus

$\displaystyle\mathcal{M}(d;r):=\mathcal{S}(d;r)/\text{PGL}(\mathbb{C},r+1).$

The semi-stable locus $\mathcal{S}(d;r)$ is a dense Zariski open of
$\mathcal{H}(d;r)$ which is $\text{PGL}(\mathbb{C},r+1)$-invariant. It
contains another dense, $\text{PGL}(\mathbb{C},r+1)$-invariant Zariski open
$\mathcal{S}^{\circ}(d;r)\subseteq\mathcal{S}(d;r)$ which parametrizes the
smooth hypersurfaces. The coarse moduli space of smooth, degree $d$
hypersurfaces in $\mathbb{P}_{\mathbb{C}}^{r}$ is

$\displaystyle\mathcal{M}^{\circ}(d;r):=\mathcal{S}^{\circ}(d;r)/\text{
PGL}(\mathbb{C},r+1).$

Each of the above parameter and moduli spaces classifies certain objects. We
will additionally need to refer to the spaces classifying these objects with
an associated choice of $k$-plane.

######  Definition 3.12.

(Parameter and Moduli Spaces of Hypersurfaces with $k$-Planes)
Continuing with the notation of Definition 3.11 and recalling that
$\mathcal{G}\text{r}(k,r)$ is the variety of $k$-planes in
$\mathbb{P}_{\mathbb{C}}^{r}$, we denote the parameter space of degree $d$
hypersurfaces with choice of $k$-plane in $\mathbb{P}_{\mathbb{C}}^{r}$ by
$\mathcal{H}(d;r,k)$; it is the incidence variety

$\displaystyle\mathcal{H}(d;r,k)=\left\\{(V,L)\ |\ L\subseteq
V\right\\}\subseteq\mathcal{H}(d;r)\times\mathcal{G}\text{r}(k,r).$

Similarly, we write $\mathcal{S}(d;r,k),\ \mathcal{M}(d;r,k),\text{ and
}\mathcal{M}^{\circ}(d;r,k)$ for the analogous spaces which additionally
classify a choice of $k$-plane. They similarly arise as incidence varieties or
as quotients of incidence varieties by $\text{PGL}(\mathbb{C},r+1)$.

We now expand these definitions from classifying hypersurfaces to
intersections of hypersurfaces of type $(2,\dotsc,d)$.

######  Definition 3.13.

(Parameter and Moduli Spaces of Intersections of Hypersurfaces)
The parameter space of intersections of hypersurfaces of type $(2,\dotsc,d)$
in $\mathbb{P}_{\mathbb{C}}^{r}$ is

$\displaystyle\mathcal{H}(2,\dotsc,d;r)=\mathcal{H}(2;r)\times\cdots\times\mathcal{H}(d;r).$

The natural action of $\text{PGL}(\mathbb{C},r+1)$ action on
$\mathbb{P}_{\mathbb{C}}^{r}$ induces a diagonal action on
$\mathcal{H}(2,\dotsc,d;r)$ and the parameter space of semi-stable
intersections of hypersurfaces of type $(2,\dotsc,d)$ in
$\mathbb{P}_{\mathbb{C}}^{r}$ is the largest proper Zariski open
$U\subseteq\mathcal{H}(2,\dotsc,d;r)$ which is
$\text{PGL}(\mathbb{C},r+1)$-invariant; it is denoted by
$\mathcal{S}(2,\dotsc,d;r)$.

The moduli space of semi-stable intersections of hypersurfaces of type
$(2,\dotsc,d)$ in $\mathbb{P}_{\mathbb{C}}^{r}$ is

$\displaystyle\mathcal{M}(2,\dotsc,d;r)=\mathcal{S}(2,\dotsc,d;r)/\text{
PGL}(\mathbb{C},r+1).$

In analogy with Definition 3.12, we write $\mathcal{H}(2,\dotsc,d;r,k),\
\mathcal{S}(2,\dotsc,d;r,k),$ and $\mathcal{M}(2,\dotsc,d;r,k)$ for the
respective spaces which additionally classify a choice of $k$-plane; they
analogously arise as incidence varieties or as quotients of incidence
varieties by $\text{PGL}(\mathbb{C},r+1)$, as well.

######  Remark 3.14.

(Parameter and Moduli Spaces as Schemes)
For the interested reader, we note that these spaces can be constructed as
schemes via classical invariant theory, beginning with

$\displaystyle\mathcal{H}(d;r)$
$\displaystyle=\text{Proj}\left(S^{*}\left(\mathbb{C}[x_{0},\dotsc,x_{r}]_{(d)}^{\vee}\right)\right),$
$\displaystyle\mathcal{M}(d;r)$
$\displaystyle=\text{Proj}\left(S^{*}\left(\mathbb{C}[x_{0},\dotsc,x_{r}]_{(d)}^{\vee}\right)^{\text{GL}(\mathbb{C},r+1)}\right).$

The remaining spaces arise analogously from the same constructions as in the
variety case.

We will use the dimension of these parameter and moduli spaces, so we recall
the dimensions in the case of hypersurfaces and give the dimensions in the
case of intersections of hypersurfaces of type $(2,\dotsc,d)$. In the proof of
Proposition 2.26, we observed the combinatorial identity

$\sum\limits_{i=0}^{d}\binom{r+i}{i}=\binom{r+d+1}{d}.$

In Remark 3.15 and Definition 3.21, we will use the slight variation

$\sum\limits_{i=2}^{d}\binom{r+i}{i}=\binom{r+d+1}{d}-(r+2).$ (5)

######  Remark 3.15.

(Dimension of Parameter and Moduli Spaces)
For fixed $d,r\geq 1$, we have

$\dim\left(\mathcal{S}(d;r)\right)=\dim\left(\mathcal{H}(d;r)\right)=\binom{r+d}{d}-1.$

When $\binom{r+d}{d}-(r+1)^{2}<0$, $\mathcal{M}(d;r)$ is empty. When
$\binom{r+d}{d}-(r+1)^{2}\geq 0$, we have

$\dim\left(\mathcal{M}^{\circ}(d;r)\right)=\dim\left(\mathcal{M}(d;r)\right)=\binom{r+d}{d}-(r+1)^{2}.$

Similarly, for $d,r$ for which the following spaces are non-empty, we have

$\displaystyle\dim\left(\mathcal{S}\left(2,\dotsc,d;r\right)\right)=\dim\left(\mathcal{H}\left(2,\dotsc,d;r\right)\right)=\left(\sum\limits_{i=2}^{d}\binom{r+i}{i}\right)-(d-1),$
$\displaystyle\dim\left(\mathcal{M}\left(2,\dotsc,d;r\right)\right)=\left(\sum\limits_{i=2}^{d}\binom{r+i}{i}\right)-(r+1)^{2}-(d-2).$

From equation (5), we re-write these quantities as

$\displaystyle\dim\left(\mathcal{S}\left(2,\dotsc,d;r\right)\right)=\dim\left(\mathcal{H}\left(2,\dotsc,d;r\right)\right)=\binom{r+d+1}{d}-(r+d+1),$
$\displaystyle\dim\left(\mathcal{M}\left(2,\dotsc,d;r\right)\right)=\binom{r+d+1}{d}-(r+1)^{2}-(r+d).$

We now examine how Wolfson’s algorithm determines $k$-planes on hypersurfaces.
A theorem of Waldron (Theorem 3.16) guarantees the existence of a $k$-plane on
a degree $d$ hypersurface once the ambient dimension $r$ is above a given
threshold:

######  Theorem 3.16.

(Theorem 1.6 of [Wal2008])
Fix $d\geq 3$. When $r$ and $k$ are such that

$(k+1)(r-k)-\binom{k+d}{d}\geq 0,$

then the natural maps

$\displaystyle\mathcal{H}(d;r,k)$ $\displaystyle\rightarrow\mathcal{H}(d;r),$
$\displaystyle\mathcal{M}^{\circ}(d;r,k)$
$\displaystyle\rightarrow\mathcal{M}^{\circ}(d;r),$

are surjective.555This is not the entirety of Waldron’s result; see Theorem
1.6 of [Wal2008] for more details.

Definition 2.4 of [Wol2021] shows that the extended Tschirnhaus complete
intersections are defined over $\mathbb{Z}$. Thus, for suitable $d,r,k$,
Waldron’s theorem allows us to determine a $k$-plane on $\tau_{d}$ over an
extension $L/K_{n}$ with

$\text{RD}(L/K_{n})\leq\text{RD}\left(\mathcal{H}(d;r,k)\rightarrow\mathcal{H}(d;r)\right)\leq\dim\left(\mathcal{H}(d;r)\right).$

Moreover, Theorem 2.12 of [Wol2021] shows that $\tau_{1,2,3}$ (along with
certain other extended Tschirnhaus complete intersections) are generically
smooth. It follows that for suitable $r,k$, Waldron’s theorem allows us to
determine a $k$-plane on $\tau_{3}$ over an extension $L^{\prime}/K_{n}$ with

$\text{RD}(L/K_{n})\leq\text{RD}\left(\mathcal{M}^{\circ}(3;r,k)\rightarrow\mathcal{M}^{\circ}(3;r)\right)\leq\dim\left(\mathcal{M}^{\circ}(3;r)\right).$

Note that for a generically finite, dominant, rational map of
$\mathbb{C}$-varieties $Y\dashrightarrow X$, the upper bound

$\text{RD}(Y\dashrightarrow X)\leq\dim(X),$ (6)

follows directly from Definition 2.14; it is also included in Lemma 2.5 of
[FW2019].

######  Example 3.17.

(Wolfson’s Method)
For $n\geq 1559$, we can determine an $8$-plane on $\tau_{1,2,3,4}$ over an
extension $L/K_{n}$ with $\text{RD}(L/K_{n})\leq 78485029$.

###### Proof.

First, $\tau_{1}$ is a hyperplane and so
$\tau_{1}\cong\mathbb{P}_{K_{n}}^{1557}$. Next, $\tau_{2}$ is a quadric
hypersurface in $\tau_{1}$ and thus it is known classically that there is a
778-plane $\Lambda_{2}\subseteq\tau_{1,2}$ over an iterated quadratic
extension $L_{1}/K_{n}$ (see [Wol2021] for details).
$\tau_{1,2,3}\cap\Lambda_{2}$ is a cubic hypersurface in $\Lambda_{2}$ and

$(63+1)(778-63)-\binom{63+3}{3}=0,$

hence we can determine a 63-plane
$\Lambda_{3}\subseteq\tau_{1,2,3}\cap\Lambda_{2}$ over an extension
$L_{2}/L_{1}$ with

$\text{RD}(L_{2}/L_{1})\leq\text{RD}\left(\mathcal{M}^{\circ}(3;778,63)\rightarrow\mathcal{M}^{\circ}(3;778)\right)\leq\dim\left(\mathcal{M}^{\circ}(3;778)\right)=78485029.$

Finally, $\tau_{1,2,3,4}\cap\Lambda_{3}$ is a quartic hypersurface in
$\Lambda_{3}$ and

$(8+1)(63-8)-\binom{8+4}{4}=0,$

hence we can determine an 8-plane
$\Lambda_{4}\subseteq\tau_{1,2,3,4}\cap\Lambda_{3}$ over an extension
$L_{3}/L_{2}$ with

$\text{RD}(L_{3}/L_{2})\leq\text{RD}\left(\mathcal{H}(4;63,8)\rightarrow\mathcal{H}(4;63)\right)\leq\dim\left(\mathcal{H}(4;63)\right)=766479.$

∎

We next provide an example which highlights the current limitations of using
iterated polar cones for determining additional upper bounds on
$\text{RD}(n)$.

######  Example 3.18.

(Wolfson Method for a 9-Plane on $\tau_{1,2,3,4,5}$)
We can determine a 9-plane on $\tau_{1,2,3,4,5}$ over an extension $L/K_{n}$
with

$\text{RD}(L/K_{n})\leq 3298353885918738132194252727911\left(\ \approx 3\cdot
10^{30}\ \right),$

as long as

$\displaystyle r\geq 54097786526\left(\ \approx 5\cdot 10^{10}\right).$

The core construction is the same as Example 3.17. In Wolfson’s notation,
observe that $54097786526=\psi(5,9)_{4}+1$ and

$\dim\left(\mathcal{M}^{\circ}(3;\psi(5,9)_{3})\right)=3298353885918738132194252727911\left(\
\approx 3\cdot 10^{30}\ \right).$

See the proof of Theorem 5.6 of [Wol2021] for details.

######  Remark 3.19.

(Iterated Polar Cone Comparison)
Observe that a $9^{th}$ polar cone
$\mathcal{C}^{9}(\tau_{1,2,3,4,5};P_{0},\dotsc,P_{8})$ of $\tau_{1,2,3,4,5}$
has type $\left[\begin{matrix}5&4&3&2&1\\\
1&10&55&220&715\end{matrix}\right]$. Thus, even for $n$ large enough so that
we may determine a suitable 66-plane $\Lambda$ on
$\mathcal{C}^{9}(\tau_{1,2,3,4,5};P_{0},\dotsc,P_{8})_{1}\cap\mathcal{C}^{9}(\tau_{1,2,3,4,5};P_{0},\dotsc,P_{8})_{2}$
(the intersection of the 220 quadrics and 715 hyperplanes defining
$\mathcal{C}^{9}(\tau_{1,2,3,4,5};P_{0},\dotsc,P_{8})$), it still follows that

$\displaystyle\deg\left(\mathcal{C}^{9}(\tau_{1,2,3,4,5},P_{8})\cap\Lambda\right)$
$\displaystyle=5\cdot 4^{10}\cdot 3^{55},$
$\displaystyle=914616279415496004448658427740160$ $\displaystyle\left(\
\approx 9\cdot 10^{32}\ \right),$
$\displaystyle>3298353885918738132194252727911$ $\displaystyle\left(\ \approx
3\cdot 10^{30}\ \right),$
$\displaystyle=\dim\left(\mathcal{M}^{\circ}(3;\psi(5,9)_{3})\right).$

Indeed, the degree of the analogous intersection grows exponentially in $k$
whereas the required resolvent degree for Wolfson’s method grows polynomially
in $k$. Similar issues arise for $k$-planes on intersections of hypersurfaces
of type (1,2,3,4) for $k\geq 33$.

In [DM1998], Debarre and Manivel give an explicit combinatorial condition for
an intersection of hypersurfaces in $\mathbb{P}_{\mathbb{C}}^{r}$ to contain a
$k$-plane:

######  Theorem 3.20.

(Theorem 2.1 of [DM1998])
Let $V\subseteq\mathbb{P}_{\mathbb{C}}^{r}$ be an intersection of
hypersurfaces of type $\left[\begin{matrix}d&d-1&\cdots&2&1\\\
\ell_{d}&\ell_{d-1}&\cdots&\ell_{2}&\ell_{1}\end{matrix}\right]$ which is not
a quadric hypersurface. When $r$ and $k$ are such that

$\displaystyle(k+1)(r-k)-\sum\limits_{i=1}^{d}\ell_{i}\binom{k+i}{i}\geq 0,$

then the natural maps

$\displaystyle\mathcal{H}\left(2,\dotsc,d;r,k\right)\rightarrow\mathcal{H}\left(2,\dotsc,d;r\right),$
$\displaystyle\mathcal{M}\left(2,\dotsc,d;r,k\right)\rightarrow\mathcal{M}\left(2,\dotsc,d;r\right),$

are surjective. 666This is not the entirety of Debarre and Manivel’s result;
see Theorem 2.1 of [DM1998] for more details.

We will use this result to deal with $k$-planes on intersections of
hypersurfaces directly (instead of using a recursive process on the set of
defining hypersurfaces) to obtain new thresholds for upper bounds on
$\text{RD}(n)$ for $n\geq 348,489,068,134$ in Theorem 3.24.

Similarly to Wolfson, we can use the upper bound

$\text{RD}\left(\mathcal{M}(2,\dotsc,d;r,k)\rightarrow\mathcal{M}(2,\dotsc,d;r)\right)\leq\dim\left(\mathcal{M}(2,\dotsc,d;r)\right),$

once we establish that the Tschirnhaus complete intersections
$\tau_{1,\dotsc,m}$ are semi-stable in Lemma 3.22 and Proposition 3.23.

######  Definition 3.21.

(Notation for Lemma 3.22 )
To succinctly state Lemma 3.22 and its applications, we define a new set
function

$\displaystyle\vartheta:\mathbb{Z}_{\geq 3}\times\mathbb{Z}_{\geq
1}\rightarrow\mathbb{Z}_{\geq 1},$

where $\vartheta(d,k)$ is the smallest positive integer $r$ such that

$\displaystyle(k+1)(r-k)-\sum\limits_{i=2}^{d}\binom{k+i}{i}\geq 0.$

Using the combinatorial identity in equation (5), we can equivalently write

$\displaystyle\vartheta(d,k)=k+\left\lceil\frac{1}{k+1}\left(\binom{k+d+1}{d}-(k+2)\right)\right\rceil.$

######  Lemma 3.22.

($k$-Planes on an Intersection of Hypersurfaces of Type $(2,\dotsc,d)$)
Let $d\geq 3$, and $V\in\mathcal{S}(2,\dotsc,d;r)(K)$ for some
$\mathbb{C}$-field $K$. For any $k\geq 1$, if $r\geq\vartheta(d,k)$, then we
can determine a $k$-plane on $V$ over an extension $L/K$ with

$\displaystyle\text{RD}(L/K)\leq\dim\left(\mathcal{M}(2,\dotsc,d;\vartheta(d,k)\right).$

###### Proof.

First, observe that it suffices to prove the case where $r=\vartheta(d,k)$ by
restriction. Theorem 3.20 then yields that $V$ contains a $k$-plane. We
identify $V$ with
$\mathbb{A}_{K}^{0}\rightarrow\mathcal{M}(2,\dotsc,d;\vartheta(d,k))$ and note
that the resolvent degree of determining a $k$-plane is exactly the resolvent
degree of the map

$\pi_{K}:\mathbb{A}_{K}^{0}\times_{\mathcal{M}(2,\dotsc,d;\vartheta(d,k))}\mathcal{M}(2,\dotsc,d;\vartheta(d,k),k)\rightarrow\mathbb{A}_{K}^{0}$

determined by the pullback square

${\mathbb{A}_{K}^{0}\times_{\mathcal{M}(2,\dotsc,d;\vartheta(d,k))}\mathcal{M}(2,\dotsc,d;\vartheta(d,k),k)}$${\mathcal{M}(2,\dotsc,d;\vartheta(d,k),k)}$${\mathbb{A}_{K}^{0}}$${\mathcal{M}(2,\dotsc,d;\vartheta(d,k)).}$$\scriptstyle{\pi_{K}}$$\scriptstyle{\pi_{\mathcal{M}}}$

However, Lemma 2.5 of [FW2019] yields that

$\displaystyle\text{RD}\left(\pi_{K}\right)\leq\text{RD}\left(\mathcal{M}(2,\dotsc,d;\vartheta(d,k),k)\rightarrow\mathcal{M}(2,\dotsc,d;\vartheta(d,k))\right)$
$\displaystyle\leq\dim\left(\mathcal{M}(2,\dotsc,d;\vartheta(d,k))\right).$

∎

######  Proposition 3.23.

(Semi-Stability of Tschirnhaus Complete Intersections)
For each $d\geq 3$ and $n\geq d+2$,
$\tau_{1,\dotsc,d}\in\mathcal{S}(2,\dotsc,d;n-2)(K_{n})$.

###### Proof.

First, note that $\tau_{1}\subseteq\mathbb{P}_{K_{n}}^{n-1}$ is a hyperplane,
so we can consider
$\tau_{1,\dotsc,d}\subseteq\tau_{1}\cong\mathbb{P}_{K_{n}}^{n-2}$, e.g. as a
$K_{n}$-point of $\mathcal{S}(2,\dotsc,d;n-2)$.

From the definition of $\mathcal{S}(2,\dotsc,d;n-2)$, it suffices to show that
there is some $\text{PGL}(K_{n},n-1)$-invariant polynomial in
$K_{n}[x_{0},\dotsc,x_{n-1}]$ which does not vanish at $\tau_{1,\dotsc,d}$.
Theorem 2.12 of [Wol2021] yields that $\tau_{1,2,3}$ is generically smooth,
hence semi-stable. In particular, there is a $\text{PGL}(K_{n},n-1)$-invariant
polynomial $f(x_{0},\dotsc,x_{n-1})\in K_{n}[x_{0},\dotsc,x_{n-1}]$ which does
not vanish at $\tau_{1,2,3}\in\mathcal{H}(2,3;n-2)(K_{n})$. When pulled back
to $\mathcal{H}(2,\dotsc,d;n-2)(K_{n})$ via the standard projection map,
$f(x_{0},\dotsc,x_{n-1})$ does not vanish at $\tau_{1,\dotsc,d}$ as well. ∎

######  Theorem 3.24.

(Determining a Point on $\tau_{1,\dotsc,d+k}$)
Fix $k,d\geq 1$. For $n\geq\vartheta(d,k)+3$, we can determine a point of
$\tau_{1,\dotsc,d+k}^{\circ}\subseteq\mathcal{T}^{n}$ over an extension
$L/K_{n}$ with

$\displaystyle\text{RD}(L/K_{n})\leq\max\left\\{\dim\left(\mathcal{M}(2,\dotsc,d;\vartheta(d,k))\right),\frac{(d+k)!}{d!}\right\\}.$

###### Proof.

By restriction, it suffices to prove the case where $n=\vartheta(d,k)+3$. As
such, we work in $\mathbb{P}_{K_{n}}^{\vartheta(d,k)+2}$. We then pass to a
hypersurface $H$ which does not contain $[1:0:\cdots:0]$ and $\tau_{1}\cap
H\cong\mathbb{P}_{K_{n}}^{\vartheta(d,k)}$ in this hypersurface. Hence, Lemma
3.22 and Proposition 3.23 yield that we can determine a $k$-plane
$\Lambda\subseteq\tau_{1,\dotsc,d}$ over an extension $L_{1}/K_{n}$ with

$\displaystyle\text{RD}(L_{1}/K_{n})\leq\dim\left(\mathcal{M}(2,\dotsc,d;\vartheta(d,k))\right).$

Thus, $\deg\left(\tau_{1,\dotsc,d+k}\cap\Lambda\right)=\frac{(d+k)!}{d!}$ and
we can determine a point $Q$ of
$\tau_{1,\dotsc,d+k}\cap\Lambda\subseteq\tau_{1,\dotsc,d+k}^{\circ}$ over an
extension of degree at most $\frac{(d+k)!}{d!}$. ∎

######  Remark 3.25.

(Wolfson’s Function $F$)
In Definition 5.4 and Theorem 5.6 of [Wol2021], Wolfson introduces a monotone
increasing function $F(r)$ such that $\text{RD}(n)\leq n-r$ for $n\geq F(r)$;
we will define this function explicitly in Section 4. To remain consistent
with the notation in this paper, we write $F$ as a function of $m$.

Wolfson shows shows that
$\lim\limits_{m\rightarrow\infty}\frac{B(m)}{F(m)}=\infty$, where
$B(m)=(m-1)!+1$ was the previous best bounding function, established in
[Bra1975]. We construct a similar function $G(m)$ and prove that $G$ similarly
improves on $F$ (Theorem 4.1).

######  Definition 3.26.

(Defining the Function $G(m)$)
We define $\varphi:\mathbb{Z}_{\geq 15}\times\mathbb{Z}_{\geq
1}\rightarrow\mathbb{Z}_{\geq 1}$ by

$\displaystyle\varphi(d,k)=\max\left\\{\frac{(d+k)!}{d!},\dim\left(\mathcal{M}(2,\dotsc,d;\vartheta(d,k))\right)\right\\}.$

Next, we define $G:\mathbb{Z}_{\geq 1}\rightarrow\mathbb{Z}_{\geq 1}$ for
$2\leq m\leq 14$ by

$m$ | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10
---|---|---|---|---|---|---|---|---|---|---
$G(m)$ | 2 | 3 | 4 | 5 | 9 | 21 | 109 | 325 | 1681 | 15121
$m$ | 11 | 12 | 13 | 14
---|---|---|---|---
$G(m)$ | 151,201 | 1,663,201 | 19,958,401 | 259,459,201

and for $m\geq 15$ by

$\displaystyle G(m)=1+\min\left\\{\varphi(d,m-d-1)\ |\ 4\leq d\leq
m-1\right\\}.$

######  Theorem 3.27.

(Upper Bounds on $\text{RD}(n)$)
For each $m\geq 1$ and all $n\geq G(m)$, $\text{RD}(n)\leq n-m$.

###### Proof.

(Proof of Theorem 3.27)
The claim for $1\leq m\leq 5$ is classical and is covered in [Wol2021]. The
case $m=6$ is Theorem 3.7 and the cases of $7\leq m\leq 14$ are Theorem 3.10.

Now, fix $m\geq 15$. For each $4\leq d\leq m-1$, Theorem 3.24 yields that we
can determine a point of $\tau_{1,\dotsc,m-1}^{\circ}\subseteq\mathcal{T}^{n}$
over an extension of scalars $L/K_{n}$ with
$\text{RD}(L/K_{n})\leq\varphi(d,m-d-1)$ when $n\geq\vartheta(d,m-d-1)+3$.
Note that

$\displaystyle\vartheta(d,m-d-1)+3$
$\displaystyle<\binom{\vartheta(d,m-d-1)+d+1}{d}-(\vartheta(d,m-d-1)+1)^{2}-(\vartheta(d,m-d-1)+d)$
$\displaystyle=\dim\left(\mathcal{M}(2,\dotsc,d;\vartheta(d,m-d-1))\right)$
$\displaystyle\leq\varphi(m,m-d-1).$

It suffices to minimize over all such $d$ and so the definition of $G(m)$ and
Remark 3.5 yield the theorem. ∎

In Section 4, we pin down how $G(m)$ improves on $F(m)$. First, we use $G(m)$
to establish an upper bound on $\text{RD}(n)$ using elementary functions à la
Brauer.

###  3.4 Upper Bounds on the Bounding Function $G(m)$

While $G(m)$ is simpler than $F(m)$, its definition does not immediately yield
a description in terms of elementary functions. We state such a description
now and use the remainder of the subsection to obtain this approximation.

######  Theorem 3.28.

(Upper Bound on the Growth Rate of $\text{RD}(n)$)
For every positive integer $d\geq 4$,
$G(2d^{2}+7d+6)\leq\frac{(2d^{2}+7d+5)!}{d!}$. Hence, for
$n\geq\frac{\left(2d^{2}+7d+5\right)!}{d!}$, it follows that

$\displaystyle\text{RD}(n)\leq n-2d^{2}-7d-6.$

To prove Theorem 3.28, we establish a simple criterion for $m$, in terms of
$d$, so that we can conclude that $G(m)<\frac{(m-1)!}{d!}$ when that criterion
is met.

Recall that $\vartheta(d,k)$ is defined such that an intersection of
hypersurfaces of type $(1,\dotsc,d)$ in $\mathbb{P}^{r}$ contains a $k$-plane
when $r\geq\vartheta(d,k)$. We begin by approximating $\vartheta(d,m-d-1)$
above.

######  Lemma 3.29.

(Upper Bound on $\vartheta$)
Fix $m>d\geq 4$. Then

$\displaystyle\vartheta(d,m-d-1)\leq m-d-2+\binom{m}{d}.$

###### Proof.

We first recall Definition 3.21:

$\displaystyle\vartheta(d,k)=k+\left\lceil\frac{1}{k+1}\left(\binom{k+d+1}{d}-(k+2)\right)\right\rceil.$

By using the identification $k=m-d-1$, we observe that

$\displaystyle\vartheta(d,m-d-1)$
$\displaystyle=(m-d-1)+\left\lceil\frac{1}{(m-d-1)+1}\left(\binom{(m-d-1)+d+1}{d}-((m-d-1)+2)\right)\right\rceil,$
$\displaystyle=(m-d-1)+\left\lceil\frac{1}{m-d}\left(\binom{m}{d}-(m-d+1)\right)\right\rceil,$
$\displaystyle\leq m-d-2+\binom{m}{d}.$

∎

######  Corollary 3.30.

(Upper Bound on Parameter Space Dimension)
Fix $m>d\geq 4$. Then,

$\displaystyle\dim\left(\mathcal{H}(2,\dotsc,d;\vartheta(d,m-d-1))\right)\leq\binom{m-1+\binom{m}{d}}{d}-\left(m-1+\binom{m}{d}\right).$

###### Proof.

Remark 3.15 established that

$\displaystyle\dim\left(\mathcal{H}(2,\dotsc,d;\vartheta(d,m-d-1))\right)\leq\binom{\vartheta(d,m-d-1)+d+1}{d}-(\vartheta(d,m-d-1)+d+1),$

which is non-decreasing in $\vartheta(d,m-d-1)$. Thus, Lemma 3.29 yields

$\displaystyle\dim\left(\mathcal{H}(2,\dotsc,d;\vartheta(d,m-d-1))\right)\leq\binom{m-1+\binom{m}{d}}{d}-\left(m-1+\binom{m}{d}\right).$

∎

We will now introduce a constant $C_{d}$ and give a bound on
$\log\left(\frac{C_{d}}{d+1}\right)$, both of which will be useful in the
proof of Lemma 3.33. We also remind the reader that every use of $\log$ in
this paper refers to the base $e$ logarithm.

######  Definition 3.31.

(Constant for the Proof of Lemma 3.33 )
For each $d\geq 4$, we set

$\displaystyle C_{d}:=\max\left\\{\binom{d+1}{i}\ |\ 0\leq i\leq
d+1\right\\}.$

######  Lemma 3.32.

(Bound on $\log(C_{d})$)
For each $d\geq 1$, it follows that

$\displaystyle\log\left(\frac{C_{d}}{d+1}\right)\leq d+\frac{3}{2}.$

We will frequently use Stirling’s approximations for factorials, including in
the proof of Lemma 3.32, and thus state the version we use explicitly (a
stronger version of which can be found in [Rob1955]):

Let $a$ be an positive integer. Then,

$\sqrt{2\pi}a^{a+\frac{1}{2}}e^{-a}\leq a!\leq a^{a+\frac{1}{2}}e^{1-a}.$ (7)

###### Proof.

(Proof of Lemma 3.32)
Our proof depends on the parity of $d+1$; we begin with the case where
$d+1=2\ell$ is even. Then,
$C_{d}=\binom{2\ell}{\ell}=\frac{(2\ell)!}{(\ell!)^{2}}$ and Stirling’s
approximation yields

$\displaystyle
C_{d}\leq\frac{(2\ell)^{2\ell+\frac{1}{2}}e^{1-2\ell}}{\left(\sqrt{2\pi}\ell^{\ell+\frac{1}{2}}e^{-\ell}\right)^{2}}=\frac{2^{d+\frac{1}{2}}e}{\pi\sqrt{\ell}}.$

Consequently,

$\displaystyle\log\left(\frac{C_{d}}{d+1}\right)\leq\log\left(\frac{2^{d-\frac{1}{2}}e}{\pi\sqrt{\ell}(d+1)}\right)\leq\log\left(\frac{e}{\pi(d+1)\sqrt{\ell}}\right)+\log\left(e^{d+\frac{1}{2}}\right)\leq
d+\frac{1}{2}.$

When $d+1=2\ell+1$ is odd, we observe

$\displaystyle
C_{d}=\binom{2\ell+1}{\ell}\leq\binom{2\ell+2}{\ell+1}\leq\frac{2^{d+\frac{3}{2}}e}{\pi\sqrt{\ell+1}},$

and thus

$\displaystyle\log\left(\frac{C_{d}}{d+1}\right)\leq\log\left(\frac{2^{d+\frac{3}{2}}e}{\pi\sqrt{\ell+1}}\right)\leq
d+\frac{3}{2}.$

∎

Recall that for each $1\leq d\leq m-1$ and when $n$ is large enough, we can
determine an $(m-d-1)$-plane $\Lambda\subseteq\tau_{1,\dotsc,d}^{\circ}$ over
an extension $L_{1}/K_{n}$ with

$\displaystyle\text{RD}(L_{1}/K_{n})\leq\dim\left(\mathcal{M}(2,\dotsc,d;\vartheta(d,m-d-1))\right).$

In such a case, we can determine a point of $\tau_{1,\dotsc,m-1}\cap\Lambda$
over an extension $L_{2}/L_{1}$ of degree $\frac{(m-1)!}{d!}$. Hence, we set

$\displaystyle\varphi(d,m-d-1)=\max\left\\{\frac{(m-1)!}{d!},\dim\left(\mathcal{M}(2,\dotsc,d;\vartheta(d,m-d-1))\right)+m\right\\}$

in Definition 3.26. We next give a condition relating $\varphi(d,m-d-1)$ and
$\varphi(d+1,m-d-2)$.

######  Lemma 3.33.

(The $\varphi$ Condition)
Fix $d\geq 4$. For all $m\geq 2d^{2}+7d+6$, it follows that

$\varphi(d+1,m-d-2)<\varphi(d,m-d-1).$ (8)

###### Proof.

For any such $d$ and $m$, it is always true that

$\displaystyle\frac{(m-1)!}{(d+1)!}<\frac{(m-1)!}{d!}.$

As a result, to conclude (8), we need only show that

$\frac{(m-1)!}{d!}>\binom{m-1+\binom{m}{d+1}}{d+1},$ (9)

since

$\binom{m-1+\binom{m}{d+1}}{d+1}>\dim\left(\mathcal{H}(2,\dotsc,d;\vartheta(d,m-d-1))\right)>\dim\left(\mathcal{M}(2,\dotsc,d;\vartheta(d,m-d-1))\right).$

Observe that

$\binom{m-1+\binom{m}{d+1}}{d+1}=\frac{\left(m-1+\binom{m}{d+1}\right)!}{(d+1)!\left(m-d-2+\binom{m}{d+1}\right)}=\frac{1}{(d+1)!}\prod\limits_{i=1}^{d+1}\left(m-i+\binom{m}{d+1}\right).$
(10)

Next, we approximate $\binom{m}{d+1}$:

$\binom{m}{d+1}=\frac{1}{(d+1)!}\cdot\frac{m!}{(m-d-1)!}=\frac{1}{(d+1)!}\prod\limits_{j=0}^{d+1}(m-j)\leq\frac{1}{(d+1)!}m^{d+2}\leq
m^{d+2}.$ (11)

By substituting the inequality (11) into equation (10) and using that $m-i\leq
m$, we obtain the approximation

$\displaystyle\binom{m-1+\binom{m}{d+1}}{d+1}\leq\frac{1}{(d+1)!}\prod\limits_{i=1}^{d+1}\left(m+m^{d+2}\right),$

which, after substituting into inequality (9) and multiplying both sides by
$d!$, yields the sufficient condition

$(m-1)!>\frac{1}{d+1}\prod\limits_{i=1}^{d+1}\left(m+m^{d+2}\right).$ (12)

Notice that main term of the right side of inequality (12) is of the form
$\prod\limits_{i=1}^{d+1}(a+b)=(a+b)^{d+1}$, hence applying the binomial
theorem yields

$\displaystyle\prod\limits_{i=1}^{d+1}\left(m+m^{d+2}\right)$
$\displaystyle=\sum\limits_{i=0}^{d+1}\binom{d+1}{i}\left(m^{d+2}\right)^{i}m^{d+1-i},$
$\displaystyle=\sum\limits_{i=0}^{d+1}\binom{d+1}{i}m^{di+2i}(m-1)^{d+1-i},$
$\displaystyle=\sum\limits_{i=0}^{d+1}\binom{d+1}{i}m^{(d+1)i+d+1},$
$\displaystyle=m^{d+1}\sum\limits_{i=0}^{d+1}\binom{d+1}{i}\left(m^{d+1}\right)^{i}.$

However, for any positive integer $a$ and $x\geq a$ it follows from induction
$\sum\limits_{i=0}^{a+1}x^{i}\leq 2x^{a+1}.$ Applying this to
$(m-1)\sum\limits_{i=0}^{d+1}\binom{d+1}{i}\left((m-1)^{d}\right)^{i}$ and
recalling $C_{d}=\max\left\\{\binom{d+1}{i}\ |\ 0\leq i\leq d+1\right\\}$
yields

$m^{d+1}\sum\limits_{i=0}^{d+1}\binom{d+1}{i}\left(m^{d+1}\right)^{i}\leq
m^{d+1}C_{d}\left(2\left(m^{d+1}\right)^{d+1}\right).$ (13)

Substituting inequality (13) into inequality (12) and simplifying, we obtain
the condition

$\displaystyle(m-1)!>\frac{2}{d+1}C_{d}m^{d^{2}+3d+2}.$

Next, we apply Stirling’s approximation and re-arrange terms to arrive at the
condition

$\frac{(m-1)^{m-\frac{1}{2}}}{e^{m-1}}>\frac{2C_{d}}{\sqrt{2\pi}(d+1)}m^{d^{2}+3d+2}.$
(14)

Observe that $a^{a-1}<(a-1)^{a-\frac{1}{2}}$ for positive integers $a\geq 8$.
By requiring $m>8$, we need only consider when

$\frac{m^{m-1}}{e^{m-1}}>\frac{2C_{d}}{\sqrt{2\pi}(d+1)}m^{d^{2}+3d+2}.$ (15)

Multiplying both sides of inequality (15) by $e^{d^{2}+3d+2}$, dividing both
sides by $m^{d^{2}+3d+2}$, and simplifying, we arrive at the condition

$\left(\frac{m}{e}\right)^{m-d^{2}-3d-3}>\frac{2C_{d}e^{d^{2}+3d+2}}{\sqrt{2\pi}(d+1)}.$
(16)

We take $\log$ of both sides of inequality (16), which yields

$\left(m-d^{2}-3d-3\right)\log\left(\frac{m}{e}\right)>\log\left(\frac{2C_{d}e^{d^{2}+3d+3}}{\sqrt{2\pi}(d+1)}\right).$
(17)

Requiring $m>e^{2}$, it suffices to consider

$m-d^{2}-3d-3>\log\left(\frac{2}{\sqrt{2\pi}}\right)+\log\left(\frac{C_{d}}{d+1}\right)+\left(d^{2}+3d+2\right).$
(18)

We note $\log\left(\frac{2}{\sqrt{2\pi}}\right)<0$ and apply the bound on
$\log\left(\frac{C_{d}}{d+1}\right)\leq d+\frac{3}{2}$ from Lemma 3.32 to
obtain the sufficient condition

$m-d^{2}-3d-3>\left(d+\frac{3}{2}\right)+\left(d^{2}+3d+2\right),$

which re-arranges to yield our initial supposition

$m\geq 2d^{2}+7d+6.$ (19)

Finally, note that for all $d\geq 4$, $2d^{2}+7d+6>e^{2}>8,$ hence the
requirements $m\geq 8$, $m>e^{2}$ used in the proof are rendered superfluous.
∎

######  Corollary 3.34.

(Upper Bound on the Growth Rate of $G$)
For any $d\geq 4$ and $m\geq 2d^{2}+7d+6$, it follows that

$\displaystyle G(m)<\frac{(m-1)!}{d!}.$

###### Proof.

Recall Definition 3.26; in particular, for $m\geq 15$,

$\displaystyle G(m)=\min\left\\{\varphi(d,m-d-1)\ |\ 15\leq d\leq
m-d-1\right\\}+1.$

Consequently, for any $d\geq 4$ and $m\geq 2d^{2}+3d$, we have

$\displaystyle G(m)\leq\varphi(d+1,m-d-2)+1<\frac{(m-1)!}{d!}$

from Lemma 3.33. ∎

We now prove Theorem 3.28.

###### Proof.

(Proof of Theorem 3.28)
Fix $d\geq 4$. From Corollary 3.34, we have that

$G(m)\leq\frac{(2d^{2}+7d+5)!}{d!}$

for $m\geq 2d^{2}+7d+6$. From Theorem 3.27, it follows that

$\displaystyle\text{RD}(n)\leq n-2d^{2}-7d-6$

for $n\geq\frac{(2d^{2}+7d+5)!}{d!}$. ∎

##  4 Comparison with Prior Bounds

We now give a precise sense of how the bounds from $G(m)$ improve upon the
bounds of $F(m)$.

######  Theorem 4.1.

(Comparison With $F(m)$)
Let $F$ be the function defined in Definition 4.3 (which is originally
Definition 5.4 of [Wol2021]).

  1. 1.

For every $m\geq 1$, $G(m)\leq F(m)$ with equality if and only if
$m=1,2,3,4,5,15,$ or $16$.

  2. 2.

$G(m)$ provides asymptotic improvements on $F(m)$, in the sense that

$\displaystyle\lim\limits_{m\rightarrow\infty}\frac{F(m)}{G(m)}=\infty.$

######  Remark 4.2.

Despite Theorem 4.1, $G(m)$ does not yield a strictly better bound on
$\text{RD}(n)$ for all $n$. As an example,

$\displaystyle G(17)$ $\displaystyle=348,489,068,134,$ $\displaystyle F(17)$
$\displaystyle=871,782,912,001,$ $\displaystyle G(18)$
$\displaystyle=2,964,061,900,801$ $\displaystyle F(18)$
$\displaystyle=14,820,309,504,001,$

so $\frac{F(17)}{G(17)}\sim 2.502$ and $\frac{F(18)}{G(18)}\sim 5.000$.
However, for any integer $n$ between $F(17)$ and $G(18)$, $F$ and $G$ yield
the same upper bound; namely, $\text{RD}(n)\leq n-17$.

The remainder of the section is spent proving Theorem 4.1. We will now define
Wolfson’s function $F(m)$. Our presentation will vary slightly from Wolfson’s
and we refer the reader to Section 5 of [Wol2021] for details of the
construction.

######  Definition 4.3.

(Wolfson’s Functions)
Given $(d,k)\in\mathbb{Z}_{\geq 3}\times\mathbb{Z}_{\geq 1}$, set
$\psi(d,k)_{0}=k$. For $0\leq i<d-2$, set

$\displaystyle\psi(d,k)_{i+1}:=\psi(d,k)_{i}+\left\lceil\frac{1}{\psi(d,k)_{i}+1}\cdot\binom{\psi(d,k)_{i}+d-i}{d-i}\right\rceil,$

along with $\psi(d,k)_{d-1}:=2\psi(d,k)_{d-2}+1$.

Additionally, define

$\displaystyle\Phi(d,k):=\max\left\\{\frac{(d+k)!}{d!},\dim\left(\mathcal{M}\left(3;\psi(d,k)_{d-2}\right)\right)+d+k+1\right\\}.$

Finally, for $m\leq 3$, set $F(m):=m+1$ and for $m\geq 4$, set

$\displaystyle F(m):=2\left\lfloor\frac{1}{2}\left(\min\limits_{1\leq d\leq
m-2}\Phi(d,m-d-1)\right)\right\rfloor+1.$

######  Remark 4.4.

(Outline for Section 4)
Our goal is to establish a criterion for $m$, in terms of $d$, so that we can
conclude $F(m)>\frac{(m-1)!}{d!}$ when this criterion is met. We do this by
examining when

$\displaystyle\Phi(d,m-d-1)<\Phi(d+1,m-d-2).$

We first show that for any $m\geq d+3$,

$\dim\left(\mathcal{M}\left(3;\psi(d,m-d-1)_{d-2}\right)\right)\leq\dim\left(\mathcal{M}\left(3;\psi(d+1,m-d-2)_{d-1}\right)\right),$
(20)

which will then leave us to consider when

$\frac{(m-1)!}{d!}<\dim\left(\mathcal{M}\left(3;\psi(d+1,m-d-2)_{d-1}\right)\right).$
(21)

We begin by introducing auxiliary functions which will be useful for proving
inequality (20) holds for $m\geq d+3$.

######  Definition 4.5.

(Auxiliary Functions $\Psi(d;i)$)
For any $d\geq 2$, any $d-2\geq i\geq 1$ and $x\in\mathbb{Z}_{\geq 1}$, we set

$\displaystyle\Psi(d;i)(x)=x+\left\lfloor\frac{1}{x+1}\cdot\binom{x+d-i}{d-i}\right\rfloor.$

######  Remark 4.6.

(Key Properties of $\Psi(d;i)$)
The functions $\Psi(d,i)$ satisfy

  1. 1.

$\Psi(d;i+1)\left(\psi(d,m-d-1)_{i}\right)=\psi(d,m-d-1)_{i+1}$ for $i\leq
d-3$, and

  2. 2.

$\Psi(d+1;i+1)=\Psi(d,i)$.

######  Lemma 4.7.

($\Psi(d;i)$ are Non-Decreasing)
For each $m\geq d+1$ with $d\geq 2$ and each $d-2\geq i\geq 1$, the function
$\Psi(d;i)(x)$ is non-decreasing.

###### Proof.

First, observe that

$\displaystyle\Psi(d;i)(x+1)-\Psi(d;i)(x)=1+\left\lfloor\frac{1}{x+2}\cdot\binom{x+1+d-i}{d-i}\right\rfloor-\left\lfloor\frac{1}{x+1}\cdot\binom{x+d-i}{d-i}\right\rfloor,$

thus

$\displaystyle\Psi(d;i)(x+1)-\Psi(d;i)(x)\geq\frac{1}{x+2}\cdot\binom{x+1+d-i}{d-i}-\frac{1}{x+1}\cdot\binom{x+d-i}{d-i}.$

Now, observe that

$\displaystyle\frac{1}{x+2}\cdot\binom{x+1+d-i}{d-i}$
$\displaystyle=\frac{(x+1+d-i)!}{(x+2)(d-i)!(x+1)!}=\frac{(x+d-i)!}{(d-i)!(x+1)!}\left(\frac{x+d-i+1}{x+2}\right),$

and

$\displaystyle\frac{1}{x+1}\cdot\binom{x+d-i}{d-i}$
$\displaystyle=\frac{(x+d-i)!}{(x+1)(d-i)!x!}=\frac{(x+d-i)!}{(d-i)!(x+1)!}.$

Hence,

$\displaystyle\Psi(d;i)(x+1)-\Psi(d;i)(x)\geq\frac{(x+d-i)!}{(d-i)!(x+1)!}\left(\frac{x+d-i+1}{x+2}-1\right),$

and the right side is positive when $i\leq d-2$. ∎

######  Lemma 4.8.

($\psi(d,m-d-1)_{d-2}$ is Non-Decreasing in $d$)
Fix $m\geq 4$. Then,

$\displaystyle\psi(2,m-3)_{0}\leq\psi(3,m-4)_{1}\leq\cdots\leq\psi(m-3,2)_{m-5}\leq\psi(m-2,1)_{m-4}.$

Furthermore, for each $2\leq d\leq m-3$,

$\displaystyle\dim\left(\mathcal{M}\left(3;\psi(d,m-d-1)_{d-2}\right)\right)\leq\dim\left(\mathcal{M}\left(3;\psi(d+1,m-d-2)_{d-1}\right)\right).$

###### Proof.

In light of Remark 4.6 and Lemma 4.7, it suffices to show

$\displaystyle\psi(d+1,m-d-2)_{1}\geq\psi(d,m-d-1)_{0}$

to prove the claim. However,

$\displaystyle\psi(d+1,m-d-2)_{1}$
$\displaystyle=m-d-2+\left\lceil\frac{1}{m-d-1}\cdot\binom{m-2}{d-1}\right\rceil\geq
m-d-1=\psi(d,m-d-1)_{0}.$

∎

######  Remark 4.9.

Having proved Lemma 4.8, we now begin to work towards the second task outlined
in Remark 4.4: establishing a criterion for $m$ in terms of $d$, which, when
met, implies

$\displaystyle\frac{(m-1)!}{d!}\leq\dim\left(\mathcal{M}\left(3;\psi(d+1,m-d-2)_{d-1}\right)\right).$

Next, we establish simple functions which we use to approximate
$\psi(d,m-d-1)_{d-2}$ from below.

######  Definition 4.10.

(Auxiliary Functions $\Omega$ and $\omega(d,i)$)
For each $d\geq 4$ and $d-3\geq i\geq 1$, define
$\omega(d,i):\mathbb{R}_{>0}\rightarrow\mathbb{R}_{>0}$ by

$\displaystyle\omega(d,i)(x)=\frac{1}{(d-i)!}x^{d-i-1}.$

Similarly, for each pair $m\geq d$ with $d\geq 4$, define the function
$\Omega$ by

$\displaystyle\Omega(d,m)=\left(\omega(d,d-3)\circ\cdots\circ\omega(d,1)\right)(m-d-1).$

######  Remark 4.11.

(Bounding Properties of $\Omega$ and $\omega(d,i)$)
Observe that for each $i$,

$\displaystyle\omega(d,i+1)(\psi(d,m-d-1)_{i})\leq\psi(d,m-d-1)_{i+1}.$

In particular,

$\displaystyle\Omega(d,m)\leq\psi(d,m-d-1)_{d-2}.$

######  Example 4.12.

Consider the cases when $d=5$ and $m=10,100$. Then,

$\displaystyle\Omega(5,10)$ $\displaystyle\sim 1.185,$
$\displaystyle\Omega(5,100)$ $\displaystyle\sim 1.996\times 10^{8},$
$\displaystyle\psi(5,10)_{3}$ $\displaystyle=133,$
$\displaystyle\psi(5,100)_{3}$ $\displaystyle\sim 3.633\times 10^{8}.$

######  Lemma 4.13.

(Explicit Form of $\Omega$)
For any $d\geq 4$, set

$\mathfrak{C}_{d}:=\prod\limits_{i=3}^{d-1}\frac{1}{(i!)^{(i-2)!}}.$

For all $m\geq d+2$,

$\Omega(d,m)=\mathfrak{C}_{d}(m-d-1)^{(d-2)!},$

###### Proof.

We proceed by induction on $d$. When $d=4$, we have

$\displaystyle\Omega(4,m)$
$\displaystyle=\rho(4,1)(m-5)=\frac{1}{3!}(m-5)^{2}=\mathfrak{C}_{2}(m-5)^{2!}.$

For arbitrary $d$, recall

$\displaystyle\Omega(d,m)$
$\displaystyle=\left(\omega(d,d-3)\circ\cdots\circ\omega(d,2)\circ\omega(d,1)\right)(m-d-1),$

and

$\displaystyle\Omega(d-1,m-1)$
$\displaystyle=\left(\omega(d-1,d-4)\circ\cdots\circ\omega(d-1,2)\circ\omega(d-1,1)\right)(m-d-1).$

By definition, however, $\omega(d+1,i+1)=\omega(d,i)$. Therefore,

$\displaystyle\Omega(d,m)$
$\displaystyle=\left(\omega(d,d-3)\circ\cdots\circ\omega(d,2)\circ\omega(d,1)\right)(m-d-1),$
$\displaystyle=\left(\omega(d,d-3)\circ\cdots\circ\omega(d,2)\right)\left(\omega(d,1)(m-d-1)\right),$
$\displaystyle=\left(\omega(d-1,d-4)\circ\cdots\circ\omega(d-1,1)\right)\left(\omega(d,1)(m-d-1)\right),$
$\displaystyle=\Omega(d-1,\omega(d,1)(m-d-1)+d+1).$

By induction, we know that

$\displaystyle\Omega(d-1,\omega(d,1)(m-d-1)+d+1)$
$\displaystyle=\mathfrak{C}_{d-1}\left(\omega(d,1)(m-d-1)\right)^{(d-3)!},$
$\displaystyle=\mathfrak{C}_{d-1}\left(\frac{1}{(d-1)!}(m-d-1)^{d-2}\right)^{(d-3)!},$
$\displaystyle=\mathfrak{C}_{d-1}\left(\frac{1}{(d-1)!}\right)^{(d-3)!}\left((m-d-1)^{d-2}\right)^{(d-3)!},$
$\displaystyle=\mathfrak{C}_{d}(m-d-1)^{(d-2)!}.$

Consequently,

$\Omega(d,m)=\Omega(d-1,\omega(d,1)(m-d-1)+d+1)=\mathfrak{C}_{d}(m-d-1)^{(d-2)!}.$

∎

In the following lemma, we prove an inequality that will be useful in the
proof of Proposition 4.15, the proposition which establishes the criterion we
seek.

######  Lemma 4.14.

(Bounding $\log\left(\mathfrak{C}_{d}\right)$)
For each $d\geq 4$,

$\log\left(\mathfrak{C}_{d}\right)\geq
2(d-2)!-2(d-2)!\log(d-1)-2(d-3)!\log(d-1).$

###### Proof.

Observe

$\log\left(\mathfrak{C}_{d}\right)=-\sum\limits_{i=3}^{d-1}(i-2)!\log(i!)\geq-\sum\limits_{i=3}^{d-1}(i-2)!\log\left(e^{1-i}i^{i}\right),$

where the approximation is due to Stirling’s approximation (equation 7).
Hence,

$\displaystyle\log\left(\mathfrak{C}_{d}\right)$
$\displaystyle\geq-\sum\limits_{i=3}^{d-1}(i-2)!\left(1-i+i\log(i)\right),$
$\displaystyle=\sum\limits_{i=3}^{d-1}(i-1)!-\sum\limits_{i=3}^{d-1}(i-2)!(i)\log(i),$
$\displaystyle=\sum\limits_{j=1}^{d-3}(j+1)!-\sum\limits_{j=1}^{d-3}j!(j+2)\log(j+2),$
$\displaystyle=\sum\limits_{j=1}^{d-3}(j+1)!-\sum\limits_{j=1}^{d-3}j!(j+1)\log(j+2)-\sum\limits_{j=1}^{d-3}j!\log(j+2),$
$\displaystyle\geq\sum\limits_{j=1}^{d-3}(j+1)!-\log(d-1)\sum\limits_{j=1}^{d-3}(j+1)!-\log(d-1)SL_{j=1}^{d-3}j!,$
$\displaystyle=(1-\log(d-1))\sum\limits_{j=1}^{d-3}(j+1)!-\log(d-1)\sum\limits_{j=1}^{d-3}j!.$

Recall that for any positive integer $a$, $\sum\limits_{i=1}^{a}i!\leq 2(a!)$.
Since $d\geq 4$, it follows that $1-\log(d-1)$ and $-\log(d-1)$ are negative,
hence

$\displaystyle\log\left(\mathfrak{C}_{d}\right)$ $\displaystyle\geq
2(1-\log(d-1))(d-2)!-2\log(d-1)(d-3)!,$
$\displaystyle=2(d-2)!-2(d-2)!\log(d-1)-2(d-3)!\log(d-1).$

∎

######  Proposition 4.15.

(The $\Omega$ Condition)
Fix $d\geq 6$. For any $m\geq d^{2}-d+4$ such that

$\displaystyle
m^{2}-\frac{5}{2}m+\frac{1}{2}<(d+1)+\log(d)\left(d+\frac{1}{2}\right)+6(d-3)!(d-2-\log(d-1)),$

it follows that

$\displaystyle\Phi(d,m-d-1)<\Phi(d+1,m-d-2).$

###### Proof.

In light of Lemma 4.8, it suffices to have

$\frac{(m-1)!}{d!}<\dim\left(\mathcal{M}(3,\psi(d+1,m-d-2)_{d-1}\right)+m.$
(22)

First, we approximate $\psi(d+1,m-d-2)_{d-1}$ below by
$\left\lceil\Omega(d+1,m-d-2)\right\rceil$ to get

$\dim\left(\mathcal{M}\left(3,\psi(d+1,m-d-2)_{d-1}\right)\right)+m\\\
\geq\dim\left(\mathcal{M}\left(3,\left\lceil\Omega(d+1,m-d-2)\right\rceil\right)\right)+m.$
(23)

Observe that

$\displaystyle\dim\left(\mathcal{M}\left(3,\left\lceil\Omega(d+1,m-d-2)\right\rceil\right)\right)+m$
$\displaystyle=\frac{1}{6}\left(\left\lceil\Omega(d+1,m-d-2)\right\rceil^{3}+6\left\lceil\Omega(d+1,m-d-2)\right\rceil^{2}+11\left\lceil\Omega(d+1,m-d-2)\right\rceil+6\right)$
$\displaystyle\geq\frac{1}{6}\left(\Omega(d+1,m-d-2)^{3}+6\Omega(d+1,m-d-2)^{2}+11\Omega(d+1,m-d-2)+6\right)$
$\displaystyle\ -\left(\Omega(d+1,m-d-2)+1\right)^{2}+m,$
$\displaystyle=\frac{1}{6}\Omega(d+1,m-d-2)^{3}-\frac{1}{6}\Omega(d+1,m-d-2)+m.$

Since $d\geq 6$ and $m\geq d^{2}-d+4\geq 34$, it follows that

$\displaystyle\Omega(d+1,m-d-2)\geq\Omega(6,m-8)\geq\Omega(6,26)>5.6\times
10^{12},$

and consequently

$\frac{1}{6}\Omega(d+1,m-d-2)^{3}-\frac{1}{6}\Omega(d+1,m-d-2)+m>\frac{1}{7}\Omega(d+1,m-d-2)^{3}.$
(24)

Substituting inequality (24) into inequality (22) and re-arranging, we have
the sufficient criterion

$\displaystyle\frac{7(m-1)!}{d!}<\Omega(d+1,m-d-2)^{3}=\mathfrak{C}_{d}^{3}(m-d-2)^{3(d-2)!}.$

Next, we apply Stirling’s approximations (equation 7)and obtain

$\displaystyle\frac{7(m-1)^{m-\frac{1}{2}}e^{2-m}}{\sqrt{2\pi}d^{d+\frac{1}{2}}e^{-d}}<\mathfrak{C}_{d}^{3}(m-d-2)^{3(d-2)!},$

which we re-arrange as

$\frac{(m-1)^{m-\frac{1}{2}}}{e^{m}(m-d-2)^{3(d-2)!}}<\frac{\sqrt{2\pi}}{7}\cdot
e^{d-2}\cdot d^{d+\frac{1}{2}}\cdot\mathfrak{C}_{d}^{3}.$ (25)

We take $\log$ of both sides of inequality (25) and examine them individually.
First, the right side of inequality (25) becomes

$\log\left(\frac{\sqrt{2\pi}}{7}\right)+(d+2)+\left(d+\frac{1}{2}\right)\log(d)+3\log\left(\mathfrak{C}_{d}\right).$
(26)

By applying Lemma 4.14 to $\log\left(\mathfrak{C}_{d}\right)$, observing
$\log\left(\frac{\sqrt{2\pi}}{7}\right)\geq-1$, and simplifying, it suffices
to replace expression (26) with

$(d+1)+\left(d+\frac{1}{2}\right)\log(d)+6(d-2)!-6(d-2)!\log(d-1)-6(d-3)!\log(d-1).$
(27)

The left side of inequality (25) becomes

$\left(m-\frac{1}{2}\right)\log(m-1)-m-3(d-2)!\log(m-d-2).$ (28)

For $m\geq d^{2}-d+4$,

$\log(m-d-2)>\log\left(d^{2}-2d+1\right)=2\log(d-1),$

and multiplying by $3(d-2)!$ yields

$3(d-2)!\log(m-d-2)>6(d-2)!\log(d-1).$ (29)

By combining expressions (27) and (28) with inequality (29), we obtain

$\displaystyle\left(m-\frac{1}{2}\right)\log(m-1)-m<(d+1)+\left(d+\frac{1}{2}\right)\log(d)+6(d-2)!-6(d-3)!\log(d).$

Using the approximation $\log(m-1)<m-1$, we finally arrive at the condition

$\displaystyle
m^{2}-\frac{5}{2}m+\frac{1}{2}<(d+1)+\left(d+\frac{1}{2}\right)\log(d)+6(d-3)!\left(d-2-\log(d-1)\right),$

as claimed above. ∎

Using the simple approximations $\log(d)\geq\log(6)>1$ and $d-2-\log(d-1)>1$,
we arrive at a simplified condition.

######  Corollary 4.16.

(The Simplified $\Omega$ Condition)
Fix $d\geq 6$. For any $m\geq d^{2}-d+4$ such that

$\displaystyle m^{2}-\frac{5}{2}m\leq 6(d-3)!+2d+1,$

it follows that

$\displaystyle\Phi(d,m-d-1)<\Phi(d+1,m-d-2).$

Moreover,

$\displaystyle F(m)\geq\frac{(m-1)!}{d!}.$

###### Proof.

Together, Lemma 4.8 and Proposition 4.15 yield that

$\displaystyle\Phi(d,m-d-1)<\Phi(d^{\prime},m-d^{\prime}-1)$

for each $d<d^{\prime}<m-2$. For any $d^{\prime\prime}<d$, we have

$\displaystyle\frac{(m-1)!}{d!}<\frac{(m-1)!}{(d^{\prime\prime})!}\leq\Phi(d^{\prime\prime},m-d^{\prime\prime}-1).$

Hence,

$\displaystyle F(m)\geq
2\left\lfloor\frac{1}{2}\Phi(d,m-d-1)\right\rfloor+1\geq\frac{(m-1)!}{d!}.$

∎

We state and prove a corollary, then we recall and prove Theorem 4.1.

######  Corollary 4.17.

(Bounding the Ratio $\frac{F(m)}{G(m)})$
For $d\geq 11$ and $m\geq 2d^{2}+11d+15$, $\frac{F(m)}{G(m)}>d+1.$

######  Remark 4.18.

We expect that better estimates of $\frac{F(m)}{G(m)}$ could reasonably be
obtained. However, Corollary 4.17 suffices to prove Theorem 4.1, which
establishes that $G(m)$ is the better bounding function and thus we do not
need additional data on the growth rate of $F(m)$.

###### Proof.

(Proof of Corollary 4.17)
Let $d\geq 11$. Recall that Corollary 3.34 applies for $m\geq 2d^{2}+7d+6$ and
so we set $m_{d}=2d^{2}+7d+6$. Similarly, Corollary 4.16 applies for $m\geq
d^{2}-d+4$ such that

$\displaystyle m^{2}-\frac{5}{2}m\leq 6(d-3)!+2d+1.$

Correspondingly, we set

$\displaystyle M_{d}=\max\left\\{m\in\mathbb{Z}\ |\ m^{2}-\frac{5}{2}m\leq
6(d-3)!+2d+1\right\\}.$

Observe that $2d^{2}+7d+6\geq d^{2}-d+4$ and, since $d\geq 11$, we have

$\displaystyle m_{d+1}=2d^{2}+11d+15<M_{d}.$

Corollary 3.34 yields $G(m_{d+1})<\frac{(m_{d}-1)!}{(d+1)!}$ and Corollary
4.16 yields that $F(m_{d+1})\geq\frac{(m_{d}-1)!}{d!}$. As a result, we have

$\displaystyle\frac{F(m_{d+1})}{G(m_{d+1})}>\frac{\left(\frac{(m_{d}-1)!}{d!}\right)}{\left(\frac{(m_{d}-1)!}{(d+1)!}\right)}=\frac{(d+1)!}{d!}=d+1.$

In fact,

$\displaystyle\left(6(d-3)!+2d+1\right)-\left(m_{d+2}^{2}-\frac{5}{2}m_{d+2}\right)$

is positive and strictly increasing for $d\geq 11$, so $m_{d+2}<M_{d}$. Hence,
Corollaries 3.34 and 4.16 yield that

$\displaystyle\frac{F(m)}{G(m)}>\frac{\left(\frac{(m-1)!}{d!}\right)}{\left(\frac{(m-1)!}{(d+1)!}\right)}=d+1.$

for all $m\geq m_{d}=2d^{2}+7d+6$. ∎

###### Proof.

(Proof of Theorem 4.1)
First, observe that Corollary 4.17 implies that

$\displaystyle\lim\limits_{m\rightarrow\infty}\frac{F(m)}{G(m)}=\infty,$

and that $G(m)\leq F(m)$ for

$\displaystyle m\geq 2\left(11^{2}\right)+11(11)+15=378.$

The verification of that $G(m)\leq F(m)$ for $1\leq m\leq 59$ comes from
explicit computation and the relevant data is provided in Appendices 5.1 and
5.2.

Finally, we address the cases of $60\leq m\leq 377$. Recall that Lemma 3.29
yields

$\displaystyle\vartheta(d,m-d-1)\leq m-d-2+\binom{m}{d}.$

For a fixed $d$, $\vartheta(m,d-m-1)$ is bounded above by a polynomial of
degree $d$ in $m$ with positive coefficients. Remark 3.15 yields that
$\dim\left(\mathcal{M}(2,\dotsc,d;\vartheta(d,m-d-1)\right)$ is also a
polynomial in $m$, hence there is a polynomial $p_{d}(m)$ with positive
coefficients that bounds
$\dim\left(\mathcal{M}(2,\dotsc,d;\vartheta(d,m-d-1)\right)$ above. Hence,
there is a minimal positive integer $a_{d}$ such that
$\frac{(a_{d}-1)!}{d!}>p_{d}(a_{d})$. Moreover, $\frac{m!}{d!}>p_{d}(m)$ for
all $m\geq a_{d}$. In particular,

$\displaystyle G(m)\leq\frac{(m-1)!}{d!}$

for all $m\geq a_{d}$.

We compute explicitly that for all $m\geq 57$,
$\frac{(m-1)!}{9!}>\dim\left(\mathcal{M}(2,\dotsc,9;\vartheta(9,m-10)\right)$
and thus

$\displaystyle G(m)\leq\frac{(m-1)!}{9!}.$

Additionally, we explicitly compute that for $m\leq 377$,

$\displaystyle\frac{(m-1)!}{6!}<\dim\left(\mathcal{M}(3;\psi(7,m-d-1)_{5}\right).$

and the same argument proving Corollary 4.16 yields that

$\displaystyle F(m)\geq\frac{(m-1)!}{6!}.$

As a consequence,

$\displaystyle\frac{F(m)}{G(m)}\geq\frac{\frac{(m-1)!}{6!}}{\frac{(m-1)!}{9!}}\geq\frac{9!}{6!}=504$

for all $60\leq m\leq 376$, which yields the theorem. ∎

##  5 Appendices

###  5.1 Explicit Bounds on $\text{RD}(n)$

Here we provide initial data on the behavior of $G(m)$ and provide data about
$F(m)$ for comparison.

Table 1: Upper Bounds on $\text{RD}(n)$ $m$ $G(m)$ $F(m)$ $F(m)/G(m)$ First
Established by 2 3 3 1 Babylonians & Egyptians 3 4 4 1 Ferrari 4 5 5 1 Bring,
in [Bri1786] 5 9 9 1 Segre, in [Seg1945] 6 21 41 1.952 Theorem 3.7, fixing the
gap in the proof in [Che1954] 7 109 121 1.175 8 325 841 2.645 9 1681 6721
3.998 Theorem 3.10 10 15121 60481 4.000 11 151,201 604,801 4.000 12 1,663,201
6,652,801 4.000 13 19,958,401 78,485,043 3.932 Theorem 3.10 / Theorem 3.24 14
259,459,201 320,082,459 1.234 15 3,632,428,801 3,632,428,801 1 Wolfson, in
[Wol2021] 16 54,486,432,001 54,486,432,001 1 17 348,489,068,134
871,782,912,001 2.502 Theorem 3.24 18 2,964,061,900,801 14,820,309,504,001 5

When $m=12,13,14$, the methods of Theorems 3.10 and 3.24 both obtain the new
bound on resolvent degree.

###  5.2 Explicit Approximations of $F(m)/G(m)$

Here we provide additional data about the ratio $\frac{F(m)}{G(m)}$. In
particular, the behavior exhibited from $m=58$ to $m=59$ illustrates why the
ratio $\frac{F(m)}{G(m)}$ is not always non-decreasing.

Table 2: $F(m)/G(m)$ for $19\leq m\leq 59$

$m$ | $F(m)/G(m)$ |  $G(m)$ given by determining an |  $F(m)$ given by determining an
---|---|---|---
19 | 5.000 |  |
20 | 5.000 |  |
21 | 5.000 |  $(m-6)$-plane on $\tau_{1,\dotsc,5}$ |  $(m-5)$-plane on $\tau_{1,2,3,4}$
22 | 5.000 |  |
23 | 5.000 |  |
24 | 5.000 |  |
25 | 29.930 |  |
26 | 30.000 |  |
27 | 30.000 |  |
28 | 30.000 |  |
29 | 30.000 |  $(m-7)$-plane on $\tau_{1,\dotsc,6}$ |  $(m-5)$-plane on $\tau_{1,2,3,4}$
30 | 30.000 |  |
31 | 30.000 |  |
32 | 30.000 |  |
33 | 30.000 |  |
34 | 146.129 |  |
35 | 210.000 |  |
36 | 210.000 |  |
37 | 210.000 |  |
38 | 210.000 |  $(m-8)$-plane on $\tau_{1,\dotsc,7}$ |  $(m-5)$-plane on $\tau_{1,2,3,4}$
39 | 210.000 |  |
40 | 210.000 |  |
41 | 210.000 |  |
42 | 210.000 |  |
43 | 210.000 |  |
44 | 294.103 |  |
45 | 1680.000 |  |
46 | 1680.000 |  |
47 | 1680.000 |  |
48 | 1680.000 |  |
49 | 1680.000 |  $(m-9)$-plane on $\tau_{1,\dotsc,8}$ |  $(m-5)$-plane on $\tau_{1,2,3,4}$
50 | 1680.000 |  |
51 | 1680.000 |  |
52 | 1680.000 |  |
53 | 1680.000 |  |
54 | 1680.000 |  |
55 | 1680.000 |  |
56 | 2613.173 |  |
57 | 15120.000 |  $(m-10)$-plane on $\tau_{1,\dotsc,9}$ |  $(m-5)$-plane on $\tau_{1,2,3,4}$
58 | 15120.000 |  |
59 | 3024.000 |  $(m-10)$-plane on $\tau_{1,\dotsc,9}$ |  $(m-6)$-plane on $\tau_{1,\dotsc,5}$

###  5.3 Proof of Technical Lemma

We now recall the statement of Lemma 2.8 and provide a proof. We will continue
to use the notation established in Definition 2.1.

Lemma 2.8. (Technical Lemma)
Let $P,Q\in\mathbb{P}^{r}(K)$ and $f\in K[x_{0},\dotsc,x_{r}]$ be a
homogeneous polynomial of degree $d$. Applying a projective change of
coordinates as necessary, we assume that

$\displaystyle P$ $\displaystyle=[1:p_{1}:\cdots:p_{r}],$ $\displaystyle Q$
$\displaystyle=[1:q_{1}:\cdots:q_{r}],$

so that the line determined by $P$ and $Q$ is

$\displaystyle L(P,Q)(K)=\left\\{[1:\lambda p_{1}+\mu q_{1}:\cdots:\lambda
p_{r}+\mu q_{r}]\ |\ [\lambda:\mu]\in\mathbb{P}^{1}(K)\right\\}.$

For any point $R_{\lambda:\mu}=[1:\lambda p_{1}+\mu q_{1}:\cdots:\lambda
p_{r}+\mu q_{r}]\in L(P,Q)(K)$,

$f\left(R_{\lambda:\mu}\right)=f(\lambda P)+f(\mu
Q)+\sum\limits_{k=1}^{d-1}\frac{1}{k!}t(d-k,f,\lambda P)(\mu Q).$

###### Proof.

Set $p_{0}=1$ and $q_{0}=1$. Let $f\in K[x_{0},\dotsc,x_{r}]$ be a homogeneous
polynomial of degree $d$. When $r=1$, the claim follows immediately from the
binomial formula. We thus assume $r\geq 2$. As partial derivatives are linear,
it suffices to consider the case where $f$ is a monomial:

$\displaystyle f(x_{0},\dotsc,x_{r})=ax_{0}^{i_{0}}x_{1}^{i_{1}}\cdots
x_{r}^{i_{r}}.$

Note that

$\displaystyle f\left(R_{\lambda:\mu}\right)$
$\displaystyle=a\prod\limits_{j=0}^{r}\left(\lambda p_{j}+\mu
q_{j}\right)^{i_{j}}$
$\displaystyle=a\prod\limits_{j=0}^{r}\sum\limits_{\ell_{j}=0}^{i_{j}}\binom{i_{j}}{\ell_{j}}(\lambda
p_{j})^{\iota(j)-\ell_{j}}(\mu q_{j})^{\ell_{j}}$
$\displaystyle=a\sum\limits_{\ell_{0}=0}^{i_{0}}\cdots\sum\limits_{\ell_{r}=0}^{i_{r}}\left(\prod\limits_{j=0}^{r}\binom{i_{j}}{\ell_{j}}(\lambda
p_{j})^{i_{j}-\ell_{j}}(\mu q_{j})^{\ell_{j}}\right).$

To simplify notation, we denote an indexing set

$\displaystyle I=\left\\{(k_{0},\dotsc,k_{r})\in\mathbb{Z}^{r+1}\ |\ 0\leq
k_{j}\leq i_{j}\right\\}$

and, in accordance with the notation of Definition 2.1, partition it into
subsets

$\displaystyle I_{k}=\left\\{(k_{0},\dotsc,k_{r})\in\mathbb{Z}^{r+1}\ |\
k_{0}+\cdots+k_{r}=k\right\\}$

for $0\leq k\leq d$. Thus, we write

$\displaystyle f\left(R_{\lambda:\mu}\right)$
$\displaystyle=a\sum\limits_{I}\left(\prod\limits_{j=0}^{r}\binom{i_{j}}{k_{j}}(\lambda
p_{j})^{\iota(j)-k_{j}}(\mu q_{j})^{k_{j}}\right)$
$\displaystyle=a\sum\limits_{k=0}^{d}\sum\limits_{I_{k}}\left(\prod\limits_{j=0}^{r}\binom{i_{j}}{k_{j}}(\lambda
p_{j})^{i_{j}-k_{j}}(\mu q_{j})^{k_{j}}\right)$

Note that

$\displaystyle f(\lambda P)$
$\displaystyle=a\left(\prod\limits_{j=0}^{r}(\lambda
p_{j})^{i_{j}}\right)=a\sum\limits_{I_{0}}\left(\prod\limits_{j=0}^{r}\binom{i_{j}}{0}(\lambda
p_{j})^{i_{j}-0}(\mu q_{j})^{0}\right),$ $\displaystyle f(\mu Q)$
$\displaystyle=a\left(\prod\limits_{j=0}^{r}(\mu
q_{j})^{i_{j}}\right)=a\sum\limits_{I_{d}}\left(\prod\limits_{j=0}^{r}\binom{i_{j}}{i_{j}}(\lambda
p_{j})^{i_{j}-i_{j}}(\mu q_{j})^{i_{j}}\right).$

Thus,

$\displaystyle f\left(R_{\lambda:\mu}\right)-f(\lambda P)-f(\mu Q)$
$\displaystyle=a\sum\limits_{k=1}^{d-1}\sum\limits_{I_{k}}\left(\prod\limits_{j=0}^{r}\binom{i_{j}}{l_{j}}(\lambda
p_{j})^{i_{j}-k_{j}}(\mu q_{j})^{k_{j}}\right).$

Consequently, it suffices to show that

$\displaystyle\frac{1}{k!}t(d-k,f,\lambda P)(\mu
Q)=a\sum\limits_{I_{k}}\left(\prod\limits_{j=0}^{r}\binom{i_{j}}{l_{j}}(\lambda
p_{j})^{i_{j}-k_{j}}(\mu q_{j})^{k_{j}}\right)$

for each $1\leq k\leq d-1$. Recall that

$\displaystyle[k]^{*}$ $\displaystyle=\left\\{1,\dotsc,k\right\\},$
$\displaystyle[r]$ $\displaystyle=\left\\{0,1,\dotsc,r\right\\},$
$\displaystyle I_{k}^{*}$
$\displaystyle=\text{Hom}_{{\textbf{Set}}}([k]^{*},[r])$

and for $\iota\in I_{k}^{*}$, we set $|\iota|(j)=|\iota^{-1}(j)|$. Hence,

$\displaystyle\ \frac{1}{k!}t(d-k,f,\lambda P)(\mu Q)$
$\displaystyle=\frac{1}{k!}\sum\limits_{\iota\in
I_{k}^{*}}\left(\partial_{0}^{|\iota|(0)}\cdots\partial_{r}^{|\iota|(0)}f\right)\biggr{\rvert}_{\lambda
P}\ (\mu q_{0})^{|\iota|(0)}\cdots(\mu q_{r})^{|\iota(r)|}$
$\displaystyle=\frac{1}{k!}\sum\limits_{\iota\in
I_{k}^{*}}a\frac{i_{0}}{(i_{0}-|\iota|(0))!}\cdots\frac{i_{r}}{(i_{r}-|\iota|(r))!}(\lambda
p_{0})^{i_{0}-|\iota|(0)}\cdots(\lambda p_{r})^{i_{r}-|\iota|(r)}(\mu
q_{0})^{|\iota|(0)}\cdots(\mu q_{r})^{|\iota(r)|}$
$\displaystyle=\frac{a}{k!}\sum\limits_{\iota\in
I_{k}^{*}}\prod\limits_{j=0}^{r}\frac{i_{j}}{(i_{j}-|\iota|(j))!}(\lambda
p_{j})^{i_{j}-|\iota|(j)}(\mu q_{j})^{|\iota|(j)}$

Note that for any $\iota,\iota^{\prime}\in I_{k}^{*}$ whose fibers at every
point of $[r]$ have the same cardinality, we have

$\displaystyle\partial_{0}^{|\iota|(0)}\cdots\partial_{r}^{|\iota|(0)}f=\partial_{0}^{|\iota^{\prime}|(0)}\cdots\partial_{r}^{|\iota^{\prime}|(0)}f$

as partial derivatives of polynomials commute. Now, given any $\iota\in
I_{k}^{*}$, we can apply a permutation of the symmetric group $S_{k}$ to
$[k]^{*}$ and not change the number of fibers of each cardinality. However, to
get a unique representative of each unordered class of partial derivative, we
must identify any permutations which fix all of the fibers, but permute the
elements within any such fiber. Thus, to a given $\iota$, there are

$\displaystyle\frac{\left|S_{k}\right|}{\left|S_{|\iota|(0)}\right|\cdots\left|S_{|\iota(r)|}\right|}=k!\prod\limits_{l=0}^{r}\frac{1}{(|\iota|(j))!}$

permutations which give the same unordered partial derivative. Hence,

$\displaystyle\frac{1}{k!}t(d-k,f,\lambda P)(\mu Q)$
$\displaystyle=\frac{a}{k!}\sum\limits_{\iota\in
I_{k}^{*}}\prod\limits_{j=0}^{r}\frac{i_{j}}{(i_{j}-|\iota|(j))!}(\lambda
p_{j})^{i_{j}-|\iota|(j)}(\mu q_{j})^{|\iota|(j)}$
$\displaystyle=\frac{a}{k!}\sum\limits_{I_{k}}\left(k!\prod\limits_{l=0}^{r}\frac{1}{(|\iota|(j))!}\right)\prod\limits_{j=0}^{r}\frac{i_{j}}{(i_{j}-k_{j})!}(\lambda
p_{j})^{i_{j}-k_{j}}(\mu q_{j})^{k_{j}}$
$\displaystyle=a\sum\limits_{I_{k}}\prod\limits_{j=0}^{r}\binom{i_{j}}{k_{j}}(\lambda
p_{j})^{i_{j}-k_{j}}(\mu q_{j})^{k_{j}}$

which yields the lemma. ∎

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