# Schmidt rank of quartics over perfect fields

David Kazhdan  and  Alexander Polishchuk  Einstein Institute of Mathematics,
The Hebrew University of Jerusalem, Jerusalem 91904, Israel
<EMAIL_ADDRESS>Department of Mathematics, University of Oregon,
Eugene, OR 97403, USA; National Research University Higher School of
Economics; and Korea Institute for Advanced Study<EMAIL_ADDRESS>
###### Abstract.

Let ${\mathbf{k}}$ be a perfect field of characteristic $\neq 2$. We prove
that the Schmidt rank (also known as strength) of a quartic polynomial $f$
over ${\mathbf{k}}$ is bounded above in terms of only the Schmidt rank of $f$
over $\overline{{\mathbf{k}}}$, an algebraic closure of ${\mathbf{k}}$.

A.P. is partially supported by the NSF grant DMS-2001224, and within the
framework of the HSE University Basic Research Program and by the Russian
Academic Excellence Project ‘5-100’.

##  1\. Introduction

Recall that the Schmidt rank (also known as strength) of a homogeneous
polynomial $f\in{\mathbf{k}}[x_{1},\ldots,x_{n}]$ (see [1], [2] and references
therein) is defined as the minimal number $r$ such that $f$ admits a
decomposition $f=g_{1}h_{1}+\ldots+g_{r}h_{r}$, with $\deg(g_{i})$ and
$\deg(h_{i})$ smaller than $\deg(f)$. We denote the Schmidt rank of $f$ as
${\operatorname{rk}}^{S}_{{\mathbf{k}}}(f)$.

It is conjectured in [1] that for a homogeneous polynomial $f$ of degree $d$
over a non-closed field ${\mathbf{k}}$ one has

${\operatorname{rk}}^{S}_{{\mathbf{k}}}(f)\leq\kappa_{d}\cdot{\operatorname{rk}}^{S}_{\overline{{\mathbf{k}}}}(f),$

where $\overline{{\mathbf{k}}}$ is an algebraic closure of ${\mathbf{k}}$.
This is known to be true for $d\leq 3$ with $\kappa_{d}=d$ since in this case
the Schmidt rank is equal to the slice rank defined as the minimal $r$ such
that $f\in(l_{1},\ldots,l_{r})$, where $\deg(l_{i})=1$ (see [4, Thm. A] for
the case of cubics).

One can also ask a weaker question whether there exists a function $c(r,d)$
such that

${\operatorname{rk}}^{S}_{{\mathbf{k}}}(f)\leq
c({\operatorname{rk}}^{S}_{\overline{{\mathbf{k}}}}(f),d).$

Our main result is that this weaker question has a positive answer in the case
of quartic polynomials.

Theorem A. Assume that the ground field ${\mathbf{k}}$ is perfect of
characteristic $\neq 2$. Then there exists a function $r\mapsto c(r)$ such
that for any homogeneous quartic polynomial $f(x_{1},\ldots,x_{n})$, such that
${\operatorname{rk}}^{S}_{\overline{{\mathbf{k}}}}(f)=r$, one has
${\operatorname{rk}}^{S}_{{\mathbf{k}}}(f)\leq c(r)$.

We find it convenient to use of the following refined version of Schmidt rank.

######  Definition 1.1.

For a collection of nonnegative integers $(r_{k},\ldots,r_{1})$ and a
homogeneous polynomial $f$ of degree $d$, we say that the refined Schmidt rank
of $f$ is at most $(r_{k},\ldots,r_{1})$, and write
${\operatorname{rk}}^{S}_{\mathbf{k}}(f)\leq(r_{k},\ldots,r_{1})$, if there
exists a decomposition over ${\mathbf{k}}$,

$f=\sum_{j=1}^{k}\sum_{i=1}^{r_{j}}f_{ij}g_{ij},$

where $\deg(f_{ij})=d-\deg(g_{ij})=j$ for $j=1,\ldots,k$.

Thus, for a quartic polynomial $f$, we have
${\operatorname{rk}}^{S}_{\mathbf{k}}(f)\leq(r_{2},r_{1})$ if there exist
quadrics $q_{1},\ldots,q_{r_{2}}$ and linear forms $l_{1},\ldots,l_{r_{1}}$,
such that $f\in(q_{1},\ldots,q_{r_{2}},l_{1},\ldots,l_{r_{1}})$. In our proof
of Theorem A we show the existence of functions
$(c_{2}(r_{2},r_{1}),c_{1}(r_{2},r_{1}))$, such that if
${\operatorname{rk}}^{S}_{\overline{{\mathbf{k}}}}(f)\leq(r_{2},r_{1})$ then
${\operatorname{rk}}^{S}_{{\mathbf{k}}}(f)\leq(c_{2}(r_{2},r_{1}),c_{1}(r_{2},r_{1}))$.
Then one can set

$c(r)=\max_{r_{1}+r_{2}=r}(c_{2}(r_{2},r_{1})+c_{1}(r_{2},r_{1})).$

One can work through our proof of Theorem A and get explicit formulas for
$(c_{2}(r_{2},r_{1}),c_{1}(r_{2},r_{1}))$. As an illustration of this, we give
formulas for $(c_{2}(1,r_{1}),c_{1}(1,r_{1}))$.

Theorem B. Let $f$ be a homogeneous quartic polynomial defined over a perfect
field ${\mathbf{k}}$ with ${\operatorname{char}}({\mathbf{k}})\neq 2$. Assume
that ${\operatorname{rk}}^{S}_{\overline{{\mathbf{k}}}}(f)\leq(1,r)$. Then
${\operatorname{rk}}^{S}_{{\mathbf{k}}}(f)\leq(2,C(r))$, where

$C(r)=8r(41+20\cdot(10r+1)^{10r+1}).$

The main idea of the proof of Theorem A is to study decompositions of a
quartic polynomial $f$ of the form

$f=\sum_{i=1}^{r}q_{i}q^{\prime}_{i}\mod(P),$

where $q_{i}$, $q^{\prime}_{i}$ are of degree $2$ and $P$ is a subspace of
linear forms. The main result about such decomposition is that if the rank of
any linear combination of $(q_{\bullet},q^{\prime}_{\bullet})$ is sufficiently
large then the above decomposition is essentially unique (possibly after
enlarging $P$): the only way to get a new decomposition is by making an
orthogonal change of basis in the linear space with the basis
$(q_{\bullet},q^{\prime}_{\bullet})$. We then apply this result to the
decompositions obtained from a given one over $\overline{{\mathbf{k}}}$ by
applying the Galois group action. If the rank of
$(q_{\bullet},q^{\prime}_{\bullet})$ is sufficiently large, then we obtain a
$1$-cocycle of the Galois group with values in the orthogonal group measuring
how the decomposition transforms under the Galois action. We can assume that
this $1$-cocycle is trivial (after passing to an extension of ${\mathbf{k}}$
of small degree). We then use a certain linear algebra result from [3] to
prove the existence of a decomposition with
$(q_{\bullet},q^{\prime}_{\bullet})$ defined over ${\mathbf{k}}$. Furthermore,
we have a bound on the slice rank of $f-\sum_{i=1}^{r}q_{i}q^{\prime}_{i}$
over $\overline{{\mathbf{k}}}$, and hence over ${\mathbf{k}}$ by [4, Thm. A].
This gives the required bound on the Schmidt rank of $f$ over ${\mathbf{k}}$.

##  2\. Preliminaries

###  2.1. Criterion for an ideal generated by quadrics and linear forms to be
prime

In this subsection we fix a ground field ${\mathbf{k}}$ (and omit it from the
notation). By a quadric we mean an element of ${\mathbf{k}}[V]_{2}$, i.e., a
quadratic form.

######  Definition 2.1.

For a subspace of quadrics $Q$, and a collection of quadrics
$q_{1},\ldots,q_{r}$, we define ${\operatorname{srk}}(q_{1},\ldots,q_{r},Q)$
as the minimum of ${\operatorname{srk}}(\sum_{i}c_{i}q_{i}+q)$, where $q\in Q$
and $c_{i}$ are constants, such that either $q\neq 0$ or $(c_{\bullet})\neq
0$. In particular, ${\operatorname{srk}}(Q)$ is the minimal slice rank of a
nonzero element of $Q$.

We denote by ${\operatorname{rk}}(q)$ the usual rank of a quadric $q$. It is
easy to see that the rank and the slice rank of a quadric $q$ are related by

$2{\operatorname{srk}}(q)-1\leq{\operatorname{rk}}(q)\leq
2{\operatorname{srk}}(q).$

In other words, we have

${\operatorname{srk}}(q)=\lceil\frac{{\operatorname{rk}}(q)}{2}\rceil.$

######  Lemma 2.2.

Let $q_{1},\ldots,q_{r}$ be quadratic forms such that

$R=\min_{(c_{1},\ldots,c_{r})\neq 0}{\operatorname{rk}}(c_{1}\nu_{1}+\ldots
c_{r}\nu_{r})\geq 2r+1.$

Then the subscheme $q_{1}=\ldots=q_{r}=0$ is normal of codimension $r$.

###### Proof.

Consider the Jacobian matrix $J(q_{1},\ldots,q_{r})$. The locus
$S(q_{1},\ldots,q_{r})$ where $J(q_{1},\ldots,q_{r})$ has rank $<r$ coincides
with the union of kernels of $\sum c_{j}\nu_{j}$ over
$(c_{1},\ldots,c_{r})\neq 0$. Hence,

${\operatorname{codim}}_{V}S(q_{1},\ldots,q_{r})\geq R-(r-1)\geq r+2.$

But this implies that $(q_{1}=\ldots=q_{r}=0)$ is a complete intersection,
nonsingular in codimension $1$, hence normal. ∎

######  Proposition 2.3.

(i) Let $Q$ be a subspace of quadratic forms such that
${\operatorname{srk}}(Q)\geq\dim Q+1$. Then the ideal $(Q)$ is prime.

(ii) Let $L$ be a subspace of linear forms, $Q$ a subspace of quadratic forms.
Assume that ${\operatorname{srk}}(Q)\geq\dim Q+\dim L+1$. Then the ideal
$(Q,L)$ is prime.

###### Proof.

(i) For any $q\in Q$ we have ${\operatorname{rk}}(q)\geq
2{\operatorname{srk}}(q)-1\geq 2\dim Q+1$. Hence, by Lemma 2.2, the ideal
$(Q)$ is prime.

(ii) Consider the quotient $\overline{S}=S/(L)$ of the algebra of polynomials
$S$ by the ideal $(L)$. For any $q\in Q$ we have
${\operatorname{srk}}(\overline{q})\geq{\operatorname{srk}}(q)-\dim L$, where
$\overline{q}$ is the image of $q$ in $\overline{S}$. Hence, the image
$\overline{Q}$ of $Q$ in $\overline{S}$ satisfies the assumptions of (i), so
the ideal $(\overline{Q})$ in $\overline{S}$ is prime. Therefore, its preimage
in $S$, namely $(Q,L)$, is also prime. ∎

###  2.2. Almost invariant quadratic forms

We will use the following result from [3].

######  Theorem 2.4.

Let $E/{\mathbf{k}}$ be a finite Galois extension with the Galois group $G$,
and let $V_{0},V^{\prime}_{0}$ be finite dimensional ${\mathbf{k}}$-vector
spaces. Let us set $V=V_{0}\otimes_{{\mathbf{k}}}E$,
$V^{\prime}=V^{\prime}_{0}\otimes_{{\mathbf{k}}}E$. Suppose $T:V\to
V^{\prime}$ is an $E$-linear operator such that for any $\sigma\in G$, one has

${\operatorname{rk}}_{E}(\sigma(T)-T)\leq r,$

for some $r\geq 0$. Then there exists a ${\mathbf{k}}$-linear operator
$T_{0}:V_{0}\to V^{\prime}_{0}$, such that

${\operatorname{rk}}_{E}(T-(T_{0})_{E})\leq r(2+(r+1)^{r+1}),$

where $(T_{0})_{E}:V\to V^{\prime}$ is obtained from $T_{0}$ by the extension
of scalars.

We need the following consequence of this theorem for quadratic forms.

######  Corollary 2.5.

Let $E/{\mathbf{k}}$ be a finite Galois extension with the Galois group $G$,
where ${\operatorname{char}}(2)\neq 2$, and let $V_{0}$ be a finite
dimensional ${\mathbf{k}}$-vector space, $V=V_{0}\otimes_{{\mathbf{k}}}E$.
Assume that $q$ is a quadratic form on $V$ such that for any $\sigma\in G$,
one has ${\operatorname{rk}}_{E}(\sigma(q)-q)\leq r$ for some $r\geq 0$, where
${\operatorname{rk}}_{E}$ is the usual rank of the quadratic form. Then there
exists a quadratic form $q_{0}$ on $V_{0}$ such that
${\operatorname{rk}}_{E}(q-q_{0})\leq 2r(2+(r+1)^{r+1})$.

###### Proof.

Let $T:V\to V^{*}$ be the symmetric linear map associated with $q$. Our
assumption implies that ${\operatorname{rk}}_{E}(\sigma(T)-T)\leq r$ for any
$\sigma\in G$. By Theorem 2.4, there exists an operator $T_{0}:V_{0}\to
V_{0}^{*}$ such that ${\operatorname{rk}}_{E}(T-T_{0})\leq r(2+(r+1)^{r+1})$.
Let $T_{0}^{*}:V_{0}\to V_{0}^{*}$ be the dual operator. Then

${\operatorname{rk}}_{E}(T-\frac{1}{2}(T_{0}+T_{0}^{*}))\leq
2r(2+(r+1)^{r+1}),$

so we can let $q_{0}$ be the quadratic form corresponding to
$\frac{1}{2}(T_{0}+T_{0}^{*})$. ∎

##  3\. Schmidt rank for quartics

From now on we assume that the ground field ${\mathbf{k}}$ is perfect and has
characteristic $\neq 2$.

###  3.1. Case $r_{2}=1$

We start with a proof of Theorem B dealing with the case $r_{2}=1$, since it
is simpler but still shows the main idea.

######  Lemma 3.1.

Let ${\mathbf{k}}^{\prime}/{\mathbf{k}}$ be a quadratic extension, and let $f$
be a homogeneous polynomial over ${\mathbf{k}}$ such that
${\operatorname{rk}}^{S}_{{\mathbf{k}}^{\prime}}(f)\leq(r_{2},r_{1})$. Then
${\operatorname{rk}}^{S}_{{\mathbf{k}}}(f)\leq(2r_{2},2r_{1})$.

###### Proof.

By assumption $f\in(Q,P)$, where $Q$ is a subspace of quadrics and $P$ is a
subspace of linear forms, both defined over ${\mathbf{k}}^{\prime}$. Hence,
$f\in(Q+\sigma(Q),P+\sigma(P))$, where $\sigma$ is the generator of the Galois
group of ${\mathbf{k}}^{\prime}/{\mathbf{k}}$. Since the subspaces $Q$ and $P$
are defined over ${\mathbf{k}}$, this implies the assertion. ∎

###### Proof of Theorem B.

We have to check that if $E/{\mathbf{k}}$ is a finite Galois extension, and

$f\equiv qq^{\prime}\mod(P),$

where $q,q^{\prime}$ are quadratic forms over $E$ and $P$ is an
$r$-dimensional subspace of linear forms defined over $E$, then
${\operatorname{rk}}^{S}_{{\mathbf{k}}}(f)\leq(2,C(r))$.

If ${\operatorname{srk}}_{E}(q)\leq 9r$ or ${\operatorname{srk}}_{E}(q)\leq
9r$ then ${\operatorname{srk}}_{E}(f)\leq 10r$, and so
${\operatorname{srk}}_{\mathbf{k}}(f)\leq 40r\leq C(r)$. Thus, we can assume
that ${\operatorname{srk}}_{E}(q)>9r$ and
${\operatorname{srk}}_{E}(q^{\prime})>9r$. Let $G$ be the Galois group of
$E/{\mathbf{k}}$. For any $\sigma\in G$ we have

$f\equiv\sigma(q)\sigma(q^{\prime})\mod(\sigma(P)).$

By assumption, the slice rank of $\overline{q}=q\mod(P+\sigma(P))$ is $\geq
2$, hence the quadric $\overline{q}$ is irreducible. In other words, the ideal
$(q,P+\sigma(P))$ is prime. Since $qq^{\prime}\in(\sigma(q),P+\sigma(P))$, we
have either $q\in(\sigma(q),P+\sigma(P))$ or
$q^{\prime}\in(\sigma(q),P+\sigma(P))$. Since the slice ranks of $q$ and
$q^{\prime}$ are $>2r$, this means that either

$\sigma(q)\equiv c(\sigma)\cdot q\mod(P+\sigma(P)),\ \sigma(q^{\prime})\equiv
c(\sigma)^{-1}\cdot q^{\prime}\mod(P+\sigma(P)),\text{ or }$ $\sigma(q)\equiv
c(\sigma)\cdot q^{\prime}\mod(P+\sigma(P)),\ \sigma(q^{\prime})\equiv
c(\sigma)^{-1}\cdot q\mod(P+\sigma(P)),$

for some $c(\sigma)\in E^{*}$.

Let $H\subset G$ be the set of $\sigma\in G$ for which the first possibility
holds. Let us consider separately two cases.

Case ${\operatorname{srk}}(q,q^{\prime})>3r$. Assume first that
$\sigma_{1},\sigma_{2}\in H$. Then we have

$\sigma_{1}\sigma_{2}(q)\equiv\sigma_{1}(c(\sigma_{2}))\cdot\sigma_{1}(q)\equiv
c(\sigma_{1})\sigma_{1}(c(\sigma_{2}))\cdot
q\mod(P+\sigma_{1}(P)+\sigma_{1}\sigma_{2}(P)).$

If $\sigma_{1}\sigma_{2}\not\in H$, we would get that a nontrivial linear
combination of $q$ and $q^{\prime}$ is in
$P+\sigma_{1}(P)+\sigma_{1}\sigma_{2}(P)$, contradicting the assumption
${\operatorname{srk}}(q,q^{\prime})>3r$. Hence, $\sigma_{1}\sigma_{2}\in H$.
Furthermore, since a nonzero multiple of $q$ cannot be contained in
$P+\sigma_{1}(P)+\sigma_{1}\sigma_{2}(P)$, we have

$c(\sigma_{1})\sigma_{1}(c(\sigma_{2}))=c(\sigma_{1}\sigma_{2}),$

i.e., $c(\sigma)$ is a $1$-cocycle of $H$. A similar argument shows that if
exactly one of $\sigma_{1},\sigma_{2}$ belongs to $H$ then
$\sigma_{1}\sigma_{2}\not\in H$, and if $\sigma_{1},\sigma_{2}\not\in H$ then
$\sigma_{1}\sigma_{2}\in H$, proving that $H$ is a subgroup of index $\leq 2$
in $G$.

Case ${\operatorname{srk}}(q,q^{\prime})\leq 3r$. In this case we have
$q^{\prime}\equiv cq\mod(P_{0})$ for some subspace of linear forms $P_{0}$ of
dimension $\leq 3r$ (defined over $E$) and some $c\in E^{*}$. This implies
that for $\sigma\not\in H$ we have

$\sigma(q)\equiv c(\sigma)\cdot q^{\prime}\equiv c(\sigma)c\cdot
q\mod(P+\sigma(P)+P_{0}).$

Redefining $c(\sigma)$ for $\sigma\not\in H$ we obtain that

$\sigma(q)\equiv c(\sigma)\cdot q\mod(P+\sigma(P)+P_{0}),\ \ \sigma(q)\equiv
c(\sigma)\cdot q\mod(P+\sigma(P)+P_{0})$ (3.1)

for all $\sigma\in G$. Now for $\sigma_{1},\sigma_{2}\in H$ we have

$\sigma_{1}\sigma_{2}(q)\equiv\sigma_{1}(c(\sigma_{2}))\cdot\sigma_{1}(q)\equiv
c(\sigma_{1})\sigma_{1}(c(\sigma_{2}))\cdot
q\mod(P+\sigma_{1}(P)+\sigma_{1}\sigma_{2}(P)+P_{0}+\sigma_{1}(P_{0})).$

On the other hand,

$\sigma_{1}\sigma_{2}(q)\equiv c(\sigma_{1}\sigma_{2})\cdot
q\mod(P+\sigma_{1}\sigma_{2}(P)+P_{0}).$

Since ${\operatorname{srk}}(q)>9$ this implies that $c(\sigma)$ is a
$1$-cocycle of $H$.

In either case we obtain that for a subgroup $H\subset G$ of index $\leq 2$
and a subspace of linear forms $P_{0}$ of dimension $\leq 3r$, the congruences
(3.1) hold for some $1$-cocycle $c:H\to E^{*}$.

Let ${\mathbf{k}}^{\prime}/{\mathbf{k}}$ be the subextension of $E$
corresponding to the subgroup $H\subset G$, so that the extension
$E/{\mathbf{k}}^{\prime}$ is Galois with the Galois group $H$. By Hilbert’s
Theorem 90, the cocycle $c(\sigma)$ of $H$ is trivial, so rescaling $q$ and
$q^{\prime}$, we can assume that

$\sigma(q)\equiv q\mod(P+\sigma(P)+P_{0}),\ \ \sigma(q^{\prime})\equiv
q^{\prime}\mod(P+\sigma(P)+P_{0})$

for all $\sigma\in H$. But this implies that
${\operatorname{srk}}_{E}(\sigma(q)-q)\leq 5r$ and
${\operatorname{srk}}_{E}(\sigma(q^{\prime})-q^{\prime})\leq 5r$. Hence, we
obtain

${\operatorname{rk}}_{E}(\sigma(q)-q)\leq 10r,\ \
{\operatorname{rk}}_{E}(\sigma(q^{\prime})-q^{\prime})\leq 10r$

for all $\sigma\in H$. By Corollary 2.5, there exist quadrics $q_{0}$ and
$q^{\prime}_{0}$ defined over ${\mathbf{k}}^{\prime}$ such that

$\max({\operatorname{rk}}_{E}(q-q_{0}),{\operatorname{rk}}_{E}(q^{\prime}-q^{\prime}_{0}))\leq
20r(2+(10r+1)^{10r+1}).$

Hence,

$\max({\operatorname{srk}}_{E}(q-q_{0}),{\operatorname{srk}}_{E}(q^{\prime}-q^{\prime}_{0}))\leq
10r(2+(10r+1)^{10r+1}).$

It follows that

${\operatorname{srk}}_{E}(f-q_{0}q^{\prime}_{0})\leq{\operatorname{srk}}_{E}(f-qq^{\prime})+{\operatorname{srk}}_{E}(qq^{\prime}-q_{0}q^{\prime}_{0})\leq
r+20r(2+(10r+1)^{10r+1})=r(41+20\cdot(10r+1)^{10r+1}).$

By [4, Thm. A], this implies that

${\operatorname{srk}}_{{\mathbf{k}}^{\prime}}(f-q_{0}q^{\prime}_{0})\leq
4r(41+20\cdot(10r+1)^{10r+1}),$

hence,
${\operatorname{rk}}^{S}_{{\mathbf{k}}^{\prime}}(f)\leq(1,4r(41+20\cdot(10r+1)^{10r+1}))$.
Since the extension ${\mathbf{k}}^{\prime}/{\mathbf{k}}$ is either trivial or
quadratic, applying Lemma 3.1 we get the result. ∎

###  3.2. Quadratic decompositions of quartics

######  Lemma 3.2.

Let $Q$ be a subspace of quadrics,

$q_{1},\ldots,q_{r},q^{\prime}_{1},\ldots,q^{\prime}_{r},p_{1},\ldots,p_{s},p^{\prime}_{1},\ldots,p^{\prime}_{s}$

quadratic forms, where $r>s$, such that

$\sum_{i=1}^{r}q_{i}q^{\prime}_{i}\equiv\sum_{i=1}^{s}p_{i}p^{\prime}_{i}\mod(Q).$

Then for some constants
$a_{1},\ldots,a_{r},a^{\prime}_{1},\ldots,a^{\prime}_{r}$ such that
$\sum_{i}a_{i}a^{\prime}_{i}=0$, we have

${\operatorname{srk}}(\sum_{i}(a_{i}q_{i}+a^{\prime}_{i}q^{\prime}_{i}),Q)\leq
c(r,s,\dim Q):=2^{s}(r+\dim Q)+2^{s-1}(s-2).$

###### Proof.

We use the induction on $s$. In the case $s=0$ we have to prove that

${\operatorname{srk}}(\sum_{i}(a_{i}q_{i}+a^{\prime}_{i}q^{\prime}_{i}),Q)\leq
c(r,0,\dim Q)=r+\dim Q-1$

for some isotropic $(a_{\bullet},a^{\prime}_{\bullet})$. Indeed, assume this
is not true. Then ${\operatorname{srk}}(q_{1},\ldots,q_{r-1},Q)\geq r+\dim Q$,
so the ideal $(q_{1},\ldots,q_{r-1},Q)$ is prime. But we have

$q_{r}q^{\prime}_{r}\in(q_{1},\ldots,q_{r-1},Q).$

Hence, swapping $q_{r}$ with $q^{\prime}_{r}$ if necessary, we deduce that
$q_{r}\in(q_{1},\ldots,q_{r-1},Q)$ which is impossible since
${\operatorname{srk}}(q_{\bullet},Q)\geq r+\dim Q\geq r>0$.

Assume the assertion holds for $s-1$. We have

$\sum_{i=1}^{r}q_{i}q^{\prime}_{i}\equiv\sum_{i=2}^{s}p_{i}p^{\prime}_{i}\mod(p_{1},Q).$

Hence, by the induction assumption,

${\operatorname{srk}}(\sum_{i}(a_{i}q_{i}+a^{\prime}_{i}q^{\prime}_{i}),p_{1},Q)\leq
c(r,s-1,\dim Q+1)$

for some isotropic $(a_{\bullet},a^{\prime}_{\bullet})$. Changing the basis in
$(q_{\bullet})$, we can assume that either ${\operatorname{srk}}(q_{1},Q)\leq
c(r,s-1,\dim Q+1)$, or there exists a subspace $L$ of linear forms of
dimension $\leq c(r,s-1,\dim Q+1)$ such that $p_{1}\in(q_{1},Q,L)$. In the
former case we are done since $c(r,s,\dim Q)\geq c(r,s-1,\dim Q+1)$. In the
latter case, we have

$\sum_{i=2}^{r}q_{i}q^{\prime}_{i}\equiv\sum_{i=2}^{s}p_{i}p^{\prime}_{i}\mod(q_{1},Q,L).$

Applying the induction assumption we obtain that there exists an isotropic
vector $(a_{>1},a^{\prime}_{>1})$ such that

${\operatorname{srk}}_{L}(\sum_{i=2}^{r}(a_{i}q_{i}+q^{\prime}_{i}q^{\prime}_{i}),q_{1},Q)\leq
c(r-1,s-1,\dim Q+1),$

hence

$\displaystyle{\operatorname{srk}}(\sum_{i=2}^{r}(a_{i}q_{i}+q^{\prime}_{i}q^{\prime}_{i}),q_{1},Q)\leq
c(r-1,s-1,\dim Q+1)+\dim L\leq$ $\displaystyle c(r-1,s-1,\dim
Q+1)+c(r,s-1,\dim Q+1)=c(r,s,\dim Q).$

Since any linear combination of
$\sum_{i=2}^{r}(a_{i}q_{i}+q^{\prime}_{i}q^{\prime}_{i})$ with $q_{1}$ will
correspond to an isotropic vector, the assertion follows. ∎

######  Proposition 3.3.

Let $Q$ be a subspace of quadrics,

$q_{1},\ldots,q_{r},q^{\prime}_{1},\ldots,q^{\prime}_{r},p_{1},\ldots,p_{r},p^{\prime}_{1},\ldots,p^{\prime}_{r}$

quadratic forms, such that

$\sum_{i=1}^{r}q_{i}q^{\prime}_{i}\equiv\sum_{i=1}^{r}p_{i}p^{\prime}_{i}\mod(Q).$
(3.2)

Assume that for any constants
$a_{1},\ldots,a_{r},a^{\prime}_{1},\ldots,a^{\prime}_{r}$ such that
$\sum_{i}a_{i}a^{\prime}_{i}=0$, we have

${\operatorname{srk}}(\sum_{i}(a_{i}q_{i}+a^{\prime}_{i}q^{\prime}_{i}),Q)\geq
C(r,\dim Q):=2^{r}(r+\dim Q)+2^{r-1}(r-2)+1.$

Then there exists a subspace of linear forms $L$ of dimension at most

$D(r,\dim Q):=(2^{r}-1)(r+\dim Q-1)+r\cdot 2^{r-1}$

and a linear transformation $A:{\mathbf{k}}^{2r}\to{\mathbf{k}}^{2r}$
preserving the quadratic form $\sum_{i=1}^{r}x_{i}y_{i}$, such that for the
linear operator $\phi$ from ${\mathbf{k}}^{2r}$ to the space of quadrics
sending the standard basis $(e_{\bullet},f_{\bullet})$ to
$(q_{\bullet},q^{\prime}_{\bullet})$, we have

$p_{i}\equiv\phi(Ae_{i}),\ \ p^{\prime}_{i}\equiv\phi(Af_{i})\mod(Q,L).$

###### Proof.

We use induction on $r$. In the case $r=0$ we can take $L=0$, $D(0,\dim Q)=0$.

Assume the assertion holds for $r-1$. We have

$\sum_{i=1}^{r}q_{i}q^{\prime}_{i}\equiv\sum_{i=2}^{r}p_{i}p^{\prime}_{i}\mod(p_{1},Q).$

Hence, by Lemma 3.2, changing $(q_{\bullet},q^{\prime}_{\bullet})$ by an
orthogonal transformation, we can achieve that

${\operatorname{srk}}(q_{1},p_{1},Q)\leq c(r,r-1,\dim Q+1).$

Since ${\operatorname{srk}}(q_{1},Q)\geq C(r,\dim Q)\geq c(r,r-1,\dim Q+1)+1$,
this implies that there exists a subspace of linear forms $L$ of dimension
$\leq c(r,r-1,\dim Q+1)$, such that

$p_{1}\in(q_{1},Q,L).$

Note that if $p_{1}\in(Q,L)$ then we get

$\sum_{i=1}^{r}q_{i}q^{\prime}_{i}\equiv\sum_{i=2}^{r}p_{i}p^{\prime}_{i}\mod(Q,L).$

Hence, by Lemma 3.2, we would get

$\displaystyle{\operatorname{srk}}(\sum_{i}(a_{i}q_{i}+a^{\prime}_{i}q^{\prime}_{i}),Q)\leq{\operatorname{srk}}_{L}(\sum_{i}(a_{i}q_{i}+a^{\prime}_{i}q^{\prime}_{i}),Q)+\dim
L\leq$ $\displaystyle c(r,r-1,\dim Q)+c(r,r-1,\dim Q+1)\leq C(r,\dim Q)-1,$

which is a contradiction. Hence, rescaling $q_{1}$ and $q^{\prime}_{1}$, we
can assume that

$p_{1}\equiv q_{1}\mod(Q,L).$

Also, from $p_{1}\in(q_{1},Q,L)$ we deduce that

$\sum_{i=2}^{r}q_{i}q^{\prime}_{i}\equiv\sum_{i=2}^{r}p_{i}p^{\prime}_{i}\mod(q_{1},Q,L).$

Since for any isotropic vector $(a_{>1},a^{\prime}_{>1})$ one has

$\displaystyle{\operatorname{srk}}_{L}(\sum_{i=2}^{r}(a_{i}q_{i}+a^{\prime}_{i}q^{\prime}_{i}),q_{1},Q)\geq{\operatorname{srk}}(\sum_{i=2}^{r}(a_{i}q_{i}+a^{\prime}_{i}q^{\prime}_{i}),q_{1},Q)-\dim
L\geq C(r,\dim Q)-\dim L\geq$ $\displaystyle C(r,\dim Q)-c(r,r-1,\dim Q+1)\geq
C(r-1,\dim Q+1),$

we can apply the induction hypothesis and deduce that for some subspace of
linear forms $L^{\prime}\supset L$ of dimension

$D(r-1,\dim Q+1)+\dim L\leq D(r-1,\dim Q+1)+c(r,r-1,\dim Q+1)=D(r,\dim Q),$

after changing the basis
$(q_{2},\ldots,q_{r},q^{\prime}_{2},\ldots,q^{\prime}_{r})$ by an orthogonal
transformation, we have

$p_{i}\equiv q_{i},\ \ p^{\prime}_{i}\equiv
q^{\prime}_{i}\mod(q_{1},Q,L^{\prime}),$

for $i\geq 2$. This means that we have

$p_{i}\equiv q_{i}+c_{i}q_{1},\ \ p^{\prime}_{i}\equiv
q^{\prime}_{i}+c^{\prime}_{i}q_{1}\mod(Q,L^{\prime}),$

for $i\geq 2$. Substituting this into (3.2) and recalling that $p_{1}\equiv
q_{1}\mod(Q,L)$, we get

$q_{1}q^{\prime}_{1}\equiv
q_{1}\cdot[p^{\prime}_{1}+\sum_{i=2}^{r}(c_{i}q^{\prime}_{i}+c^{\prime}_{i}q_{i})+(\sum_{i=2}^{r}c_{i}c^{\prime}_{i})q_{1}]\mod(Q,L^{\prime}).$

Since

${\operatorname{srk}}(Q)\geq C(r,\dim Q)\geq\dim Q+D(r,\dim Q)+1\geq\dim
Q+\dim L^{\prime}+1,$

the ideal $(Q,L^{\prime})$ is prime, so we get

$p^{\prime}_{1}\equiv
q^{\prime}_{1}-\sum_{i=2}^{r}(c_{i}q^{\prime}_{i}+c^{\prime}_{i}q_{i})-(\sum_{i=2}^{r}c_{i}c^{\prime}_{i})q_{1}\mod(Q,L^{\prime}).$

It remains to observe that the linear transformation

$\displaystyle Ae_{1}=e_{1},\ \
Af_{1}=f_{1}-\sum_{i=2}^{r}(c_{i}f_{i}+c^{\prime}_{i}e_{i})-(\sum_{i=2}^{r}c_{i}c^{\prime}_{i})e_{1},$
$\displaystyle Ae_{i}=e_{i}+c_{i}e_{1},\ \
Af_{i}=f_{i}+c^{\prime}_{i}e_{1},\text{ for }i\geq 2,$

preserves the quadratic form $\sum x_{i}y_{i}$. ∎

We are mainly interested in the case $Q=0$ in the above proposition (the case
of general $Q$ was introduced in order for the inductive argument to work).

######  Corollary 3.4.

(i) Assume that

$\sum_{i=1}^{r}q_{i}q^{\prime}_{i}=\sum_{i=1}^{r}p_{i}p^{\prime}_{i},$

where $q_{i}$, $q^{\prime}_{i}$, $p_{i}$, $p^{\prime}_{i}$ are quadrics and

${\operatorname{srk}}(\sum_{i}(a_{i}q_{i}+a^{\prime}_{i}q^{\prime}_{i}))\geq
C(r,0)=(r-1)\cdot 2^{r}+r\cdot 2^{r-1}+1$

for any isotropic $(a_{\bullet},a^{\prime}_{\bullet})$. Then there exists a
subspace of linear forms $L$ of dimension at most

$D(r,0)=(r-1)\cdot(2^{r}-1)+r\cdot 2^{r-1}\leq C(r,0)$

and a linear transformation $A:{\mathbf{k}}^{2r}\to{\mathbf{k}}^{2r}$
preserving the quadratic form $\sum_{i=1}^{r}x_{i}y_{i}$, such that for the
linear operator $\phi$ from ${\mathbf{k}}^{2r}$ to the space of quadrics
sending the standard basis $(e_{\bullet},f_{\bullet})$ to
$(q_{\bullet},q^{\prime}_{\bullet})$, we have

$p_{i}\equiv\phi(Ae_{i}),\ \ p^{\prime}_{i}\equiv\phi(Af_{i})\mod(L).$

(ii) Assume that

$q_{0}^{2}+\sum_{i=1}^{r}q_{i}q^{\prime}_{i}=p_{0}^{2}+\sum_{i=1}^{r}p_{i}p^{\prime}_{i},$

where $q_{i}$, $q^{\prime}_{i}$, $p_{i}$, $p^{\prime}_{i}$ are quadrics and
for any constants $a_{0},\ldots,a_{r},a^{\prime}_{1},\ldots,a^{\prime}_{r}$
such that $a_{0}^{2}+4\sum_{i}a_{i}a^{\prime}_{i}=0$, one has
${\operatorname{srk}}(a_{0}q_{0}+\sum_{i=1}^{r}(a_{i}q_{i}+a^{\prime}_{i}q^{\prime}_{i})\geq
c(r,r-1,0)+C(r,0)+1$. Then there exists a subspace of linear forms $L$ of
dimension at most $D(r,0)+c(r+1,r,0)$, such that after making a linear change
in $(q_{0},\ldots,q_{r},q^{\prime}_{1},\ldots,q^{\prime}_{r})$ preserving the
quadratic form $a_{0}^{2}+4\sum_{i=1}^{r}a_{i}a^{\prime}_{i}$, one has

$q_{0}\equiv p_{0},\ \ q_{i}\equiv p_{i},\ \ q^{\prime}_{i}\equiv
p^{\prime}_{i}\mod(L).$

###### Proof.

(i) This is the case $Q=0$ of Proposition 3.3.

(ii) We have

$(q_{0}-p_{0})(q_{0}+p_{0})+\sum_{i=1}^{r}q_{i}q^{\prime}_{i}\equiv\sum_{i=1}^{r}p_{i}p^{\prime}_{i}.$

Hence, by Lemma 3.2, there exists a nonzero vector
$(a_{0},\ldots,a_{r},a^{\prime}_{1},\ldots,a^{\prime}_{r},b)$ such that

$(a_{0}-b)(a_{0}+b)+\sum_{i=1}^{r}a_{i}a^{\prime}_{i}=0$

and

${\operatorname{srk}}((a_{0}-b)(q_{0}+p_{0})+(a_{0}+b)(q_{0}-p_{0})+\sum_{i=1}^{r}(a_{i}q_{i}+a^{\prime}_{i}q^{\prime}_{i}))\leq
c(r+1,r,0).$

We can rewrite these conditions as

$a_{0}^{2}+\sum_{i=1}^{r}a_{i}a^{\prime}_{i}=b^{2}$

and
${\operatorname{srk}}(a_{0}q_{0}+\frac{1}{2}\sum_{i=1}^{r}(a_{i}q_{i}+a^{\prime}_{i}q^{\prime}_{i})-bp_{0},Q)\leq
c(r+1,r,0)$. Note that by the assumption, we necessarily have $b\neq 0$. Thus,
after making an orthogonal transformation of
$(q_{0},\ldots,q_{r},q^{\prime}_{1},\ldots,q^{\prime}_{r})$, we can assume
that ${\operatorname{srk}}(q_{0}-p_{0})\leq c(r+1,r,0)$. Thus, we have
$q_{0}\equiv p_{0}\mod(L_{0})$,

$\sum_{i=1}^{r}q_{i}q^{\prime}_{i}\equiv\sum_{i=1}^{r}p_{i}p^{\prime}_{i}\mod(L_{0}),$

for some subspace of linear forms $L_{0}$ of dimension $\leq c(r+1,r,0)$.

Now applying part (i), we find a subspace of linear forms $L\supset L_{0}$ of
dimension $\leq D(r,0)+c(r+1,r,0)$, such that after an orthogonal change of
$(q_{1},\ldots,q_{r},q^{\prime}_{1},\ldots,q^{\prime}_{r})$, we have

$p_{i}\equiv q_{i}\mod(L),\ \ p^{\prime}_{i}\equiv q^{\prime}_{i}\mod(L).$

∎

###  3.3. Proof of Theorem A

We will prove the existence of functions $c_{1}(r_{2},r_{1})$ and
$c_{2}(r_{2},r_{1})$ such that if a quartic $f$ satisfies
${\operatorname{rk}}^{S}_{\overline{{\mathbf{k}}}}(f)\leq(r_{2},r_{1})$ then
${\operatorname{rk}}^{S}_{{\mathbf{k}}}(f)\leq(c_{2}(r_{2},r_{1}),c_{1}(r_{2},r_{1}))$.

We use the induction on $r_{2}$. In the case $r_{2}=0$ we just have the slice
rank, so by [4, Thm. A], we can set

$c_{1}(0,r_{1})=4r_{1},\ \ c_{2}(0,r_{1})=0.$

Now assume that the functions $c_{1}(r_{1},r_{2})$ and $c_{2}(r_{1},r_{2})$
are already constructed for $r_{2}<r$. Let $f$ be a quartic over
${\mathbf{k}}$, and $E/{\mathbf{k}}$ is a finite Galois extension such that
${\operatorname{rk}}^{S}_{E}(f)\leq(r,p)$, i.e., over $E$ we have a
decomposition

$f\equiv\sum_{i=1}^{r}q_{i}q^{\prime}_{i}\mod(P),$ (3.3)

where $(q_{i},q^{\prime}_{i})$ are quadrics, and $P$ is a subspace of linear
forms of dimension $p$.

Below we will use constants $C(r,0)$, $D(r,0)$ and $c(r,s,0)$ introduced in
Sec. 3.2.

Case 1. Assume first that
${\operatorname{srk}}(q_{\bullet},q^{\prime}_{\bullet})>6p+3C(r,0)$.

Note that for any element $\sigma$ of the Galois group
$\operatorname{Gal}(E/{\mathbf{k}})$ we have

$\sum_{i=1}^{r}q_{i}q^{\prime}_{i}\equiv\sum_{i=1}^{r}\sigma(q_{i})\sigma(q^{\prime}_{i})\mod(P+\sigma(P)).$

Since
${\operatorname{srk}}_{P+\sigma(P)}(\sum_{i}(a_{i}q_{i}+a^{\prime}_{i}q^{\prime}_{i}))>4p+3C(r,0)\geq
C(r,0)$ for any nonzero $(a_{\bullet},a^{\prime}_{\bullet})$, by Corollary
3.4(i), there exists a subspace of linear forms $L_{\sigma}\supset
P+\sigma(P)$ of dimension $\leq N=2p+C(r,0)$ and an orthogonal transformation
$A_{\sigma}$ of the $2r$ dimensional space with the basis
$(q_{\bullet},q^{\prime}_{\bullet})$ such that

$\sigma(q_{i})\equiv A_{\sigma}(q_{i}),\ \ \sigma(q^{\prime}_{i})\equiv
A_{\sigma}(q^{\prime}_{i})\mod(L_{\sigma})$

(note that here $A_{\sigma}(q_{i})$ and $A_{\sigma}(q^{\prime}_{i})$ are
linear combinations of $(q_{\bullet},q^{\prime}_{\bullet})$).

We claim that $\sigma\mapsto A_{\sigma}$ defines a cocycle with values in the
orthogonal group. Indeed, we have

$\sigma_{1}(A_{\sigma_{2}})A_{\sigma_{1}}(q_{i})\equiv\sigma_{1}(A_{\sigma_{2}})(\sigma_{1}(q_{i}))\equiv\sigma_{1}\sigma_{2}(q_{i})\mod(\sigma_{1}(L_{\sigma_{2}})+L_{\sigma_{1}}),$

hence,

$q_{i}\equiv
A_{\sigma_{1}\sigma_{2}}^{-1}\sigma_{1}(A_{\sigma_{2}})A_{\sigma_{1}}(q_{i})\mod(\sigma_{1}(L_{\sigma_{2}})+L_{\sigma_{1}}+L_{\sigma_{1}\sigma_{2}}).$

Similarly,

$q^{\prime}_{i}\equiv
A_{\sigma_{1}\sigma_{2}}^{-1}\sigma_{1}(A_{\sigma_{2}})A_{\sigma_{1}}(q^{\prime}_{i})\mod(\sigma_{1}(L_{\sigma_{2}})+L_{\sigma_{1}}+L_{\sigma_{1}\sigma_{2}}).$

Since
${\operatorname{srk}}(q_{\bullet},q^{\prime}_{\bullet})>3N\geq\dim(\sigma_{1}(L_{\sigma_{2}})+L_{\sigma_{1}}+L_{\sigma_{1}\sigma_{2}})$,
this implies that
$A_{\sigma_{1}\sigma_{2}}=\sigma_{1}(A_{\sigma_{2}})A_{\sigma_{1}}$.

Recall that $H^{1}$ of the Galois group with values in the orthogonal group
classifies nondegenerate quadratic forms on ${\mathbf{k}}^{2r}$, and any such
quadratic form becomes equivalent to $\sum_{i}x_{i}y_{i}$ over an extension
${\mathbf{k}}^{\prime}$ obtained by adjoining at most $r$ square roots to
${\mathbf{k}}$. Indeed, this follows from the identity

$\lambda_{1}x^{2}+\lambda_{2}y^{2}=\lambda_{1}(x+\sqrt{-\lambda_{2}/\lambda_{1}}y)(x-\sqrt{-\lambda_{2}/\lambda_{1}}y).$

In our case we have the class in $H^{1}$ of the Galois group of
$E/{\mathbf{k}}$, hence we can choose ${\mathbf{k}}^{\prime}$ to be a
subextension of $E$, and the cocycle $\sigma\mapsto A_{\sigma}$ becomes a
coboundary when restricted to $\operatorname{Gal}(E/{\mathbf{k}}^{\prime})$.
Thus, we can make an orthogonal change of basis in
$(q_{\bullet},q^{\prime}_{\bullet})$ such that

$\sigma(q_{i})\equiv q_{i},\ \ \sigma(q^{\prime}_{i})\equiv
q^{\prime}_{i}\mod(L_{\sigma})$

for any $\sigma$ in the Galois group
$\operatorname{Gal}(E/{\mathbf{k}}^{\prime})$. By Corollary 2.5, there exist
quadratic forms $\overline{q}_{\bullet}$, $\overline{q}^{\prime}_{\bullet}$
defined over ${\mathbf{k}}^{\prime}$, such that

${\operatorname{srk}}(q_{i}-\overline{q}_{i})\leq N^{\prime},\ \
{\operatorname{srk}}(q^{\prime}_{i}-\overline{q}^{\prime}_{i})\leq
N^{\prime},$

with $N^{\prime}=N(2+(2N+1)^{2N+1})$.

Thus, we have

$f-\sum_{i=1}^{r}\overline{q}_{i}\overline{q}^{\prime}_{i}\in(P^{\prime})$

for some subspace of linear forms $P^{\prime}$ over $E$ of dimension $\leq
N^{\prime\prime}=p+2rN^{\prime}$. In other words, the slice rank of
$\widetilde{f}=f-\sum_{i=1}^{r}\overline{q}_{i}\overline{q}^{\prime}_{i}$ over
$E$ is $\leq N^{\prime\prime}$. By [4, Thm. A], this implies that the slice
rank of $\widetilde{f}$ over ${\mathbf{k}}^{\prime}$ is $\leq
4N^{\prime\prime}$. This means that

${\operatorname{rk}}^{S}_{{\mathbf{k}}^{\prime}}(f)\leq(r,4N^{\prime\prime}).$

By Lemma 3.1, it follows that

${\operatorname{rk}}^{S}_{{\mathbf{k}}^{\prime}}(f)\leq(2^{r}\cdot
r,2^{r}\cdot 4N^{\prime\prime}).$

Case 2. Next, assume that
${\operatorname{srk}}(q_{\bullet},q^{\prime}_{\bullet})\leq 3N$, so there
exists a nontrivial linear combination
$\sum_{i}a_{i}q_{i}+\sum_{i}a^{\prime}_{i}q^{\prime}_{i}$ which has slice rank
$\leq 3N$. The restriction of the quadratic form $\sum_{i=1}^{r}x_{i}y_{i}$ to
the linear subspace
$\sum_{i}a_{i}x_{i}+\sum_{i}a^{\prime}_{i}x^{\prime}_{i}=0$ has rank $2r-1$ or
$2r-2$. So enlarging $E$ if necessary we can find linear combinations
$\overline{q}_{0},(\overline{q}_{i},\overline{q}^{\prime}_{i})_{i=1}^{r-1}$ of
$(q_{\bullet},q^{\prime}_{\bullet})$ (where possibly $\overline{q}=0$) such
that

$\sum_{i=1}^{r}q_{i}q^{\prime}_{i}\equiv\overline{q}_{0}^{2}+\sum_{i=1}^{r-1}\overline{q}_{i}\overline{q}^{\prime}_{i}\mod(\sum_{i}a_{i}q_{i}+\sum_{i}a^{\prime}_{i}q^{\prime}_{i}).$

Hence, renaming $\overline{q}_{i}$, $\overline{q}^{\prime}_{i}$ by $q_{i}$,
$q^{\prime}_{i}$, we obtain

$f\equiv q_{0}^{2}+\sum_{i=1}^{r-1}q_{i}q^{\prime}_{i}\mod(P^{\prime}),$

where $P$ is a space of linear forms of dimension $\leq p^{\prime}=\dim P\leq
p+3N$.

Furthermore, we claim that we can assume that
${\operatorname{srk}}(q_{0},q_{\bullet},q^{\prime}_{\bullet})>3N^{\prime}$,
where $N^{\prime}=D(r-1,0)+c(r,r-1,0)$. Indeed, otherwise arguing as above we
obtain a decomposition (3.3) with $r$ replaced by $r-1$ and $p$ replaced by
$p^{\prime\prime}=p^{\prime}+3N^{\prime}$. In other words, we would have
${\operatorname{rk}}^{S}_{E}(f)\leq(r-1,p^{\prime\prime})$, so by the
induction assumption we would deduce that

${\operatorname{rk}}^{S}_{{\mathbf{k}}}(f)\leq(c_{2}(r-1,p^{\prime\prime}),c_{1}(r-1,p^{\prime\prime})).$

Since for every $\sigma\in\operatorname{Gal}(E/{\mathbf{k}})$, one has

$q_{0}^{2}+\sum_{i=1}^{r-1}q_{i}q^{\prime}_{i}\equiv\sigma(q_{0})^{2}+\sum_{i=1}^{r-1}\sigma(q_{i})\sigma(q^{\prime}_{i})\mod(P^{\prime}+\sigma
P^{\prime}),$

by Corollary 3.4(ii), we get that

$\sigma(q_{0})\equiv A_{\sigma}q_{0},\ \ \sigma(q_{i})\equiv A_{\sigma}q_{i},\
\ \sigma(q^{\prime}_{i})\equiv A_{\sigma}q^{\prime}_{i}\mod(L_{\sigma}),$

for some orthogonal transformation $A_{\sigma}$ and a subspace of linear forms
$L_{\sigma}\supset P^{\prime}+\sigma P^{\prime}$ of dimension $\leq
N^{\prime}$ (here we use the inequality

$3D(r-1,0)+3c(r,r-1,0)\geq C(r-1,0)+c(r-1,r-2,0)$

which is easy to check). Since
${\operatorname{srk}}(q_{0},q_{\bullet},q^{\prime}_{\bullet})>3N^{\prime}$, as
in Case 1, this implies that $\sigma\mapsto A_{\sigma}$ is a $1$-cocycle.

Arguing as in Case 1, we find a subextension ${\mathbf{k}}^{\prime}\subset E$
obtained by adjoing at most $r-1$ square roots to ${\mathbf{k}}$, such that
after making a change of basis in $(q_{\bullet},q^{\prime}_{\bullet})$, we get

$\sigma(q_{i})\equiv q_{i},\ \sigma(q^{\prime}_{i})\equiv
q^{\prime}_{i}\mod(L_{\sigma})$

for any $\sigma\in\operatorname{Gal}(E/{\mathbf{k}}^{\prime})$. Hence, by
Corollary 2.5, there exist quadratic forms $\overline{q}_{\bullet}$,
$\overline{q}^{\prime}_{\bullet}$ defined over ${\mathbf{k}}^{\prime}$, such
that

${\operatorname{srk}}(q_{i}-\overline{q}_{i})\leq N^{\prime\prime}\text{ for
}i=0,\ldots,r-1,\ \
{\operatorname{srk}}(q^{\prime}_{i}-\overline{q}^{\prime}_{i})\leq
N^{\prime\prime}\text{ for }i=1,\ldots,r-1,$

with $N^{\prime\prime}=N^{\prime}(2+(2N^{\prime}+1)^{2N^{\prime}+1})$, and we
get

${\operatorname{srk}}_{E}(f-\overline{q}_{0}^{2}-\sum_{i=1}^{r-1}\overline{q}_{i}\overline{q}^{\prime}_{i})\leq
M=p^{\prime}+(2r-1)N^{\prime\prime}.$

By [4, Thm. A], this implies that the slice rank of $\widetilde{f}$ over
${\mathbf{k}}^{\prime}$ is $\leq 4M$, so we get

${\operatorname{rk}}^{S}_{{\mathbf{k}}^{\prime}}(f)\leq(r,4M),$

and so by Lemma 3.1,

${\operatorname{rk}}^{S}_{{\mathbf{k}}^{\prime}}(f)\leq(2^{r}\cdot
r,2^{r}\cdot 4M).$

## References

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  * [2] E. Ballico, A. Bik, A. Oneto, E. Ventura, Strength and slice rank of forms are genericaly equal, arXiv:2102.11549.
  * [3] D. Kazhdan, A. Polishchuk, Almost invariant subspaces and operators, arXiv:2107.08085.
  * [4] D. Kazhdan, A. Polishchuk, Linear subspaces in hypersurfaces over perfect fields, arXiv:2107.08080.