# Finding nontrivial zeros of quadratic forms over rational function fields of
characteristic 2

Tímea Csahók, Péter Kutas, Mickaël Montessinos, Gergely Zábrádi111Supported by
MTA Rényi Institute Lendület Automorphic Research Group, by the Thematic
Excellence Programme, Industry and Digitization Subprogramme, NRDI Office,
2020, and by the NKFIH Research grants FK-127906 and K-135885.

###### Abstract

We propose polynomial-time algorithms for finding nontrivial zeros of
quadratic forms with four variables over rational function fields of
characteristic 2. We apply these results to find prescribed quadratic
subfields of quaternion division division algebras and zero divisors in
$M_{2}(D)$, the full matrix algebra over a division algebra, given by
structure constants. We also provide an implementation of our results in MAGMA
which shows that the algorithms are truly practical.

##  1 Introduction

The theory of quadratic spaces has a long history in mathematics and has
applications in topology, number theory, algebraic geometry and in many other
areas of mathematics. Two quadratic forms are equivalent if there is an
invertible linear change of variables transforming one form into the other one
(or alternatively, there exists an vector space isomorphism between their
corresponding quadratic spaces that also respects the quadratic structure).
The theory of quadratic forms is vastly different in characteristic 2 and in
any other characteristic. Nevertheless, the concept of equivalence is key in
both cases. A quadratic form is called isotropic if it admits a nontrivial
zero and is called anisotropic otherwise. Quadratic spaces have a well-known
decomposition into the direct sum of special quadratic subspaces containing
isotropic vectors (called hyperbolic planes) and an anisotropic part. This
somehow motivates the fact that computing isotropic vectors is useful in
determining equivalence of quadratic forms.

Let $K=\mathbb{F}_{q}(t)$ be the rational function field in one variable,
where $q$ is an odd prime power. In [10] the authors describe a polynomial-
time algorithm that decides whether two quadratic forms over $K$ are
equivalent, and if so, finds an explicit equivalence between them. The key
tool is a subroutine that finds isotropic vectors of the form. The algorithm
doesn’t naturally generalize to field extensions and doesn’t work when $q$ is
a power of 2. Computing isotropic vectors of quadratic forms in odd
characteristic function fields (i.e., finite extensions of
$\mathbb{F}_{q}(t)$) is considered in [12]. The algorithm works for any
extensions but is not claimed to run in polynomial time. Furthermore, as
demonstrated in [10], quadratic form algorithms can be used to find zero
divisors in quaternion algebras over quadratic field extensions.

The only known quadratic form algorithm in the characteristic 2 case comes
from the well-known correspondence between quaternion algebras and ternary
quadratic forms. Since in [9] the main algorithm can find zero divisors in
quaternion algebras it can be used to find nontrivial zeros of ternary
quadratic forms. In this work we consider the algorithmic problem of finding
nontrivial zeros of quadratic forms over $\mathbb{F}_{2^{k}}(t)$ in 4
variables. Our contributions are the following:

  * •

We propose a polynomial-time algorithm that decides whether a four-variable
form is isotropic or not. If it is, it also outputs a nontrivial zero.

  * •

We provide a Magma implementation for finding zeros of ternary quadratic
forms. Even though the algorithm is not novel it hasn’t been implemented
before.

  * •

An implementation of our main algorithm in Magma [2].

The paper is structured as follows. In Section 2 we recall theoretical and
algorithmic preliminaries. In Section 3 we describe our algorithm for finding
nontrivial zeros. We also provide some applications of this result, such
constructing quaternion algebras with prescribed Hasse invariants, finding
zero divisors in $M_{2}(D)$ where $D$ is a quaternion algebra over
$\mathbb{F}_{2^{k}}(t)$ and finding prescribed maximal subfields in quaternion
algebras. In Section 4 we provide details about our Magma implementation of
our main algorithm. For the implementation, see
https://github.com/Char2QuadForms/Char2QuadForms. In Appendix A we give
pseudo-code algorithms for the main subroutines necessary for algorithm 1.

##  2 Preliminaries

###  2.1 Number theory background

We end this section by stating some classical results independent of the
characteristic (even though we only use them in characteristic $2$).

In this section we collect the background we need from Number theory. The
following discussion is independent of the characteristic.

We are going to use the following higher dimensional variant of Hensel’s
lemma. Let $O$ be a complete discrete valuation ring with maximal ideal $P$.
Given a multivariate polynomial $f(x_{1},\dots,x_{n})\in O[x_{1},\dots,x_{n}]$
such that the gradient $(\frac{\partial f}{\partial
x_{1}},\dots,\frac{\partial f}{\partial x_{n}})$ is nonzero modulo $P$ at a
modulo $P$ solution $(\overline{u_{1}},\dots,\overline{u_{n}})$ then this
lifts to a solution in $O$. However, the lift is not unique in general: using
the one-variable Hensel’s lemma one can even choose an arbitrary lift of
$\overline{u_{1}},\dots,\overline{u_{j-1}},\overline{u_{j+1}},\dots,\overline{u_{n}}$
for any $1\leq j\leq n$ with $\frac{\partial f}{\partial
x_{j}}(u_{1},\dots,u_{n})\not\equiv 0\pmod{P}$.

We state a variant of the Hasse–Minkowski theorem over the field
$\mathbb{F}(t)$ of rational functions over a finite field $\mathbb{F}$ [13,
Chapter VI, 3.1]. It was proved by Hasse’s doctoral student Herbert Rauter in
1926 [15].

######  Theorem 2.1.

A non-degenerate quadratic form over $\mathbb{F}(t)$ is isotropic over
$\mathbb{F}(t)$ if and only if it is isotropic over every completion of
$\mathbb{F}(t)$.

For ternary quadratic forms there exists a slightly stronger version of this
theorem which is a consequence of the product formula for quaternion algebras
or Hilbert’s reciprocity law [13, Chapter IX, Theorem 4.6]:

######  Theorem 2.2.

Let $Q$ be a ternary non-degenerate quadratic form over $\mathbb{F}(t)$. Then
if it is isotropic in every completion except maybe one then it is isotropic
over $\mathbb{F}(t)$.

Finally, we need the following version (extract) of the local reciprocity law
for function fields.

######  Theorem 2.3 (Thm. I.1.1, Cor. I.1.2, Prop. III.1.2 in [14]).

Let $K$ be a nonarchimedean local field. Then the map $L\mapsto
N_{L/K}(L^{\times})$ is a bijection from the set of finite abelian extensions
$L$ of $K$ to the norm subgroups in $K^{\times}$. Further, for any abelian
extension $L|K$ we have $\operatorname{Gal}(L/K)\cong
K^{\times}/N_{L/K}(L^{\times})$. If $L/K$ is unramified then we have
$N_{L/K}(\mathcal{O}_{L}^{\times})=\mathcal{O}_{K}^{\times}$. Here
$\mathcal{O}_{K}$ (resp. $\mathcal{O}_{L}$) denotes the valuation ring in $K$
(resp. in $L$).

###  2.2 Algorithmic preliminaries

##### Quadratic form algorithms

Quadratic forms over fields of characteristic different from 2 have a long
algorithmic history. When $char(K)\neq 2$, then the theory of ternary
quadratic forms has a close connection to quaternion algebras over
$\mathbb{Q}$. Namely finding a nontrivial zero of the quadratic form
$ax^{2}+by^{2}-cz^{2}$ (where $c\neq 0$) is equivalent to finding a zero
divisor in the quaternion algebra $(\frac{a}{c},\frac{b}{c})$. This
algorithmic correspondance is exploited in [11] to provide an algorithm for
finding zeros of indefinite rational ternary quadratic forms. In [5] a more
direct approach is followed which is also used by [6] in the case where $K$ is
a rational function field. Every approach uses lattice reduction in some
fashion. None of these approaches generalize to extension fields (even
quadratic extensions).

In [16] Simon proposes an algorithm which finds nontrivial zeros of quadratic
forms in four or more variables. The main idea of the algorithm is the
following. Let $Q_{1}$ be a quadratic form in 4 variables. Then one first
finds a quadratic form $Q_{2}$ of dimension 2 such that the orthogonal sum of
the corresponding quadratic spaces results ina hyperbolic space (direct sum of
hyperbolic planes). Then one can use the algorithm from [17] to compute a
maximal isotropic subspace of the new quadratic space (which will have
dimension 3 in this case). This will have a nontrivial intersection with the
original 4-dimensional quadratic space and the intersection can be computed
efficiently. Any nonzero element in the intersection corresponds to nontrivial
zero. The main algorithmic tool in finding the suitable form $Q_{2}$ is the
computation of the 2-Sylow part of a certain class group of an imaginary
quadratic field. The algorithm requires an oracle for factoring the
discriminant of the form which was known to be necessary for forms with 4
variables. Interestingly, under GRH, Castel [3] showed that when the number of
variables is at least 5, then one can adapt Simon’s algorithm in a way that a
factoring oracle is no longer necessary. The case of finding nontrivial zeros
over rational function fields with arbitrary many variables was considered in
[10]. The main idea of this algorithm is quite simple: split the 4-variable
form into two binary forms and find a common value they both represent. The
main theoretical tool here is the local global principle and the following
efficient formula for the number of monic irreducible polynomials in a given
residue class of a given degree [19]:

######  Lemma 2.4.

Let $a,m\in\mathbb{F}_{q}[t]$ be such that $deg(m)>0$ and the $gcd(a,m)=1$.
Let $N$ be a positive integer and let

$S_{N}(a,m)=\\#\\{f\in\mathbb{F}_{q}[t]\leavevmode\nobreak\ \text{monic
irred.}\leavevmode\nobreak\ |\leavevmode\nobreak\ f\equiv
a\leavevmode\nobreak\ (mod\leavevmode\nobreak\ m),\leavevmode\nobreak\
deg(f)=N\\}.$

Let $M=deg(m)$ and let $\Phi(m)$ denote the number of polynomials in
$\mathbb{F}_{q}[t]$ relative prime to $m$ whose degree is smaller than M. Then
we have the following inequality:

$|S_{N}(a,m)-\frac{q^{N}}{\Phi(m)N}|\leq\frac{1}{N}(M+1)q^{\frac{N}{2}}.$

The algorithm could be adapted to the rational setting but it will become
heuristic as there is no such efficient formula in the rational case.

##### Splitting quaternion algebras in characteristic 2

In characteristic 2 (to the best of our knowledge) there is no direct
algorithm for finding nontrivial zeros of ternary quadratic forms. However,
there is a similar relation between split quaternion algebras and quadratic
forms with nontrivial zeros. In [9] the authors study the problem of finding
primitive idempotents in full matrix algebras over $\mathbb{F}_{q}(t)$ given
by a structure constant representation. In particular this encompasses the
case of quaternion algebras over function fields of characteristic 2. The main
idea of the algorithm is the following. One computes two maximal orders, one
over $\mathbb{F}_{q}[t]$ and one over the ring of rational functions whose
denominator has degree larger than the degree of the numerator (the maximal
order corresponding to the degree valuation). This intersection can be
computed using lattice reduction techniques. The intersection will be finite
algebra over the base field $\mathbb{F}_{q}$ which contains a rank 1 element
from the large algebra which can be retrieved by computing the structure of
this algebra. This algorithm runs in polynomial-time but has not been
implemented so far.

##  3 Finding nontrivial zeros of quadratic forms over $\mathbb{F}_{2^{k}}(t)$

In this section we concentrate on the case of characteristic $2$. In
subsection 3.1 we recall some basic facts on quadratic forms and quaternion
algebras in characteristic $2$. Then subsection 3.2 is devoted to developing
local criteria for the existence of nontrivial zeros of quadratic forms using
Hensel’s lemma. In section 3.3 with the help of these criteria we propose an
algorithm that decides whether or not a quadratic form is isotropic globally
and if so the algorithm finds a nontrivial zero. We give an example in
subsection 3.3.1. Finally, in section 3.4 we apply the results in the previous
section to finding a zero divisor in split quaternion algebras defined over
quadratic extensions. Using the construction of quaternion algebras with given
local splitting conditions this leads to finding, in polynomial time, zero
divisors in degree two matrix rings over nonsplit quaternion algebras defined
over the ground field.

###  3.1 Quadratic forms and quaternion algebras in characteristic 2

In this subsection we recall important facts about quadratic forms and
quaternion algebras in characteristic 2. Our main source is [18, Chapter 6].
From here on $F$ will always denote a field with characteristic $2$.

######  Lemma 3.1.

[18, Chapter 6] For every quaternion algebra $A$ over $F$ there exists an
$F$-basis $1,i,j,k$ of $A$ such that

$i^{2}+i=a,\,j^{2}=b,\quad\text{and}\quad k=ij=j(i+1)$

where $a,b\in F$.

We denote the quaternion algebra over $F$ with parameters $a,b$ as
$\left[\frac{a,b}{F}\right)$. We recall some facts about quadratic forms over
fields of characteristic 2.

######  Definition 3.2.

A quadratic form over $F$ is a homogeneous polynomial $Q$ of degree two in $n$
variables $x_{1},\ldots,x_{n}$ for some $n$. We say that $Q$ is isotropic if
there exist $a_{1},\ldots,a_{n}\in F$ not all zero such that
$Q(a_{1},\ldots,a_{n})=0$. If $Q$ is not isotropic, we say that $Q$ is
anisotropic.

We can also view a quadratic form $Q$ with $n$ variables over $F$ as a
$Q:F^{n}\to F$ function. This motivates the following definition.

######  Definition 3.3.

We say that two quadratic forms $Q_{1}$ and $Q_{2}$ are isometric if there
exists a $\varphi:F^{n}\to F^{n}$ invertible linear map such that
$Q_{1}\circ\varphi=Q_{2}$.

######  Definition 3.4.

Let $Q_{1}$ and $Q_{2}$ be diagonal quadratic forms in $n$ variables. We call
$Q_{1}$ and $Q_{2}$ similar if there exist a quadratic form $Q^{\prime}$ that
is isometric to $Q_{2}$ and such that $Q^{\prime}$ can be obtained from
$Q_{1}$ by multiplication of $Q_{1}$ by a non-zero $g\in F$.

Even though if $\text{char}\,F=2$, not all quadratic forms can be diagonalized
(we get $ax^{2}+axy+by^{2}$ as the general form), the following can be said
about quadratic forms in four variables.

######  Lemma 3.5.

[7, Cor. 7.32] Every regular quadratic form in four variables over $F$ is
equivalent to a quadratic form in the form of

$a_{1}x_{1}^{2}+x_{1}x_{2}+b_{1}x_{2}^{2}+a_{3}x_{3}^{2}+x_{3}x_{4}+b_{2}x_{4}^{2}$

where $a_{1},a_{3},b_{1},b_{2}\in F$.

######  Corollary 3.6.

Every regular quadratic form in four variables over $F$ is equivalent to a
quadratic form in the form of

$a_{1}x_{1}^{2}+a_{1}x_{1}x_{2}+a_{1}a_{2}x_{2}^{2}+a_{3}x_{3}^{2}+a_{3}x_{3}x_{4}+a_{3}a_{4}x_{4}^{2}$

where $a_{1},a_{2},a_{3},a_{4}\in F$.

###### Proof.

We start from the canonical form described in Lemma 3.5.After substituting
$x_{2}\leftarrow a_{1}x_{2}$ and $x_{4}\leftarrow a_{3}x_{4}$, we get that
$a_{1}x_{1}^{2}+a_{1}x_{1}x_{2}+a_{1}^{2}b_{1}x_{2}^{2}+a_{3}x_{3}^{2}+a_{3}x_{3}x_{4}+a_{3}^{2}b_{2}x_{4}^{2}$.
After setting $a_{2}=a_{1}b_{1}$ and $a_{4}=a_{3}b_{2}$ we arrive to the form
$a_{1}x_{1}^{2}+a_{1}x_{1}x_{2}+a_{1}a_{2}x_{2}^{2}+a_{3}x_{3}^{2}+a_{3}x_{3}x_{4}+a_{3}a_{4}x_{4}^{2}$.
∎

The following lemma [18, Theorem 6.4.11] highlights a connection between the
isotropy of quadratic forms and the splitting of quaternion algebras:

######  Lemma 3.7 (Hilbert equation).

A quaternion algebra $\left[\frac{a,b}{F}\right)$ is split if and only if
$bx^{2}+bxy+aby^{2}=1$ has a solution with $x,y\in F$.

Now if $X^{2}+X+a$ has a solution in $F$ (in this case put $K_{a}:=F$) then
the form $x^{2}+xy+ay^{2}$ is equivalent to $x^{2}+xy$ which represents all
elements in $F$. Otherwise let $\alpha$ be a root of the polynomial
$X^{2}+X+a$ in a quadratic extension $K_{a}/F$. Then
$x^{2}+xy+ay^{2}=N_{K/F}(x+y\alpha)$ is the norm form. Therefore in case $F$
is a local field of characteristic $2$, we may apply Thm. 2.3 to deduce

######  Lemma 3.8.

Assume $F$ is a local field of characteristic $2$. Then we have
$F^{\times}/N_{K_{a}/F}(K_{a}^{\times})\cong\operatorname{Gal}(K_{a}/F)$ is
cyclic of order at most $2$. The subgroup $N_{K_{a}/F}(K_{a}^{\times})\leq
F^{\times}$ is uniquely determined by the extension $K_{a}/F$. In particular,
the regular quadratic form

$a_{1}x_{1}^{2}+a_{1}x_{1}x_{2}+a_{1}a_{2}x_{2}^{2}+a_{3}x_{3}^{2}+a_{3}x_{3}x_{4}+a_{3}a_{4}x_{4}^{2}$

has no nontrivial solutions in $F$ if and only if $K_{a_{2}}=K_{a_{4}}$ is a
quadratic extension of $F$ such that exactly one of $a_{1}$ and $a_{3}$ is
represented by the form $x^{2}+xy+a_{2}y^{2}$. We have $K_{a_{2}}=K_{a_{4}}$
if and only if $x^{2}+x+a_{2}+a_{4}$ splits in $F$.

###### Proof.

If $K_{a_{2}}\neq K_{a_{4}}$ then the intersection
$a_{1}N_{K_{a_{2}}/F}(K_{a_{2}}^{\times})\cap
a_{3}N_{K_{a_{4}}/F}(K_{a_{4}}^{\times})$ is a full coset of
$N_{K_{a_{2}}/F}(K_{a_{2}}^{\times})\cap N_{K_{a_{4}}/F}(K_{a_{4}}^{\times})$
and is therefore nonempty. The last statement follows from Artin–Schreier
theory. ∎

In order to handle quadratic forms, just like in odd characteristics, we will
need to introduce a quadratic residue symbol. If $\mathbb{F}$ is a finite
field of characteristic $2$ and $\pi$ is an irreducible polynomial in
$\mathbb{F}[t]$, then every element in $\mathbb{F}[t]/(\pi)$ will be a square
(as the factor ring is a finite field of characteristic 2), so the definition
will need to differ slightly. The following definition and lemma with proof
can be found in [4].

######  Definition 3.9.

For a monic irreducible $\pi$ in $\mathbb{F}[t]$ and any $f\in\mathbb{F}(t)$
that has no pole at $\pi$, let

$[f,\pi):=\begin{cases}0,&\text{if}\ f\equiv x^{2}+x\pmod{\pi}\ \text{for
some}\ x\in\mathbb{F}[t]\\\ 1,&\text{otherwise}\end{cases}$

If $[f,\pi)=0$, we say that $f$ is a quadratic residue modulo $\pi$.
Similarly, for the place at $\infty$ we define

$[f,\infty):=\begin{cases}0,&\text{if}\ f\equiv x^{2}+x\pmod{t^{-1}}\
\text{for some}\ x\in\mathbb{F}[t^{-1}]\\\ 1,&\text{otherwise}\end{cases}$

whenever $f\in\mathbb{F}(t)$ has no pole at $\infty$ (ie. $\deg(f)\leq 0$). If
$f$ has a pole at the (finite or infinite) place $\pi$ then $[f,\pi)$ has no
meaning.

######  Lemma 3.10.

The symbol $[f,\pi)$ has the following properties:

  1. 1.

if $f_{1}\equiv f_{2}\,\pmod{\pi}$, then $[f_{1},\pi)=[f_{2},\pi)$,

  2. 2.

$[f,\pi)\equiv f+f^{2}+\ldots+f^{q^{\text{deg}\pi}/2}\,\pmod{\pi}$, where
$q=|F|$,

  3. 3.

$[f_{1}+f_{2},\pi)=[f_{1},\pi)+[f_{2},\pi)$,

  4. 4.

$[f^{2}+f,\pi)=0$.

###  3.2 Local lemmas

We denote by $v_{f}$ the $f$-adic valuation on $\mathbb{F}_{2^{k}}(t)$ for a
(finite or infinite) prime $f\in\mathbb{F}_{2^{k}}(t)$, by
$\mathbb{F}_{2^{k}}(t)_{(f)}$ the $f$-adic completion, and by
$\mathbb{F}_{2^{k}}(t)_{(f)}^{+}:=\\{u\in\mathbb{F}_{2^{k}}(t)_{(f)}\mid
v_{f}(u)\geq 0\\}$ its valuation ring.

We are interested in the range of the quadratic form $x^{2}+xy+ay^{2}$ for
some $a\in\mathbb{F}_{2^{k}}(t)$. Note that this is the norm form of the
quadratic Artin–Schreier extension adjoining the root of $X^{2}+X+a$.

######  Definition 3.11.

For $a\in\mathbb{F}_{2^{k}}(t)$ we call the quadratic form $x^{2}+xy+ay^{2}$
_minimal_ if all the poles of $a$ (including $\infty$) have odd multiplicity.

Note that for a finite prime $f$ the multiplicity of the the pole of $a$ is by
definition the exponent of $f$ in the denominator of $a$. The multiplicity of
the pole of $a$ at $\infty$ is the degree of $a$ if it is positive and $0$
otherwise. The only elements $a\in\mathbb{F}_{2^{k}}(t)$ that have no poles
are the constants $a\in\mathbb{F}_{2^{k}}$ in which case the form
$x^{2}+xy+ay^{2}$ is minimal. The fact that each norm form is equivalent to a
minimal follows easily from Artin–Schreier theory. We include an algorithmic
proof as we need its running time.

######  Lemma 3.12.

Any quadratic form $x^{2}+xy+ay^{2}$ with is equivalent to a minimal form. The
equivalent minimal form can be found in polynomial time.

###### Proof.

Assume $a=\frac{g_{1}}{f^{2r}h_{1}}$ with $f\nmid g_{1},h_{1}$ for some finite
prime $f$. Since $\mathbb{F}_{2^{k}}[t]/(f)$ is a finite field of
characteristic $2$, the $2$-Frobenius is bijective on
$\mathbb{F}_{2^{k}}[t]/(f)$. In particular, there exists a polynomial
$g\in\mathbb{F}_{2^{k}}[t]$ such that $f\mid g^{2}h_{1}+g_{1}$ (one can find
$g$ by squaring $\frac{g_{1}}{h_{1}}$ $k\deg(f)-1$-times modulo $f$). So we
may replace the variable $x$ by $x_{1}=x+\frac{gy}{f^{r}}$ to obtain

$\displaystyle
x^{2}+xy+ay^{2}=x_{1}^{2}+\frac{g^{2}y^{2}}{f^{2r}}+x_{1}y+\frac{gy^{2}}{f^{r}}+\frac{g_{1}y^{2}}{f^{2r}h_{1}}=$
$\displaystyle=x_{1}^{2}+x_{1}y+\frac{g^{2}h_{1}+g_{1}+f^{r}h_{1}g}{f^{2r}h_{1}}y^{2}$

and $a^{\prime}:=\frac{g^{2}h_{1}+g_{1}+f^{r}h_{1}g}{f^{2r}h_{1}}$ has one
less $f$ in the denominator. Repeating the process for all finite primes in
the denominator of $a$ we are reduced to handle the case of the infinite
prime. This is entirely analogous: assume we have $a=\frac{g_{1}}{h_{1}}$ with
$2r:=\deg g_{1}-\deg h_{1}$ even and positive. Since the leading coefficient
of $a$ is a square in $\mathbb{F}_{2^{k}}$, there exists $0\neq
c\in\mathbb{F}_{2^{k}}$ such that $\deg(g_{1}+c^{2}t^{2r}h_{1})<\deg g_{1}$.
Therefore putting $x_{1}=x+ct^{r}y$ we obtain the form

$\displaystyle
x^{2}+xy+ay^{2}=x_{1}^{2}+c^{2}t^{2r}y^{2}+x_{1}y+ct^{r}y^{2}+\frac{g_{1}}{h_{1}}y^{2}=$
$\displaystyle=x_{1}^{2}+x_{1}y+\frac{h_{1}ct^{r}+h_{1}c^{2}t^{2r}+g_{1}}{h_{1}}y^{2}$

such that
$a^{\prime}=a+ct^{r}+c^{2}t^{2r}=\frac{h_{1}ct^{r}+h_{1}c^{2}t^{2r}+g_{1}}{h_{1}}$
has smaller degree than $a$. Repeating this step several times we deduce the
statement. ∎

######  Remark 3.13.

The above proof also shows that the minimal form of $x^{2}+xy+ay^{2}$ is
unique up to an additive constant of the form $\alpha^{2}+\alpha$ with
$\alpha\in\mathbb{F}_{2^{k}}$.

By the local-global principle (Theorem 2.1) we are reduced to identifying the
range of a minimal quadratic form $x^{2}+xy+ay^{2}$ locally at each place $f$
of $\mathbb{F}_{2^{k}}(t)$. We may apply Lemma 3.8, however, for our purposes
we need to identify explicit congruence conditions on $c$ being in the range.
Put $K_{a,f}$ for the splitting field of the polynomial $X^{2}+X+a$ over
$\mathbb{F}_{2^{k}}(t)_{f}$. We distinguish two cases whether or not $a$ has a
pole at $f$, ie. whether or not the splitting field of $X^{2}+X+a$ ramifies at
$f$. At first we treat the case when $a$ is an $f$-adic integer.

######  Lemma 3.14.

Assume $v_{f}(a)\geq 0$.

  1. 1.

If $v_{f}(c)$ is even then the equation $x^{2}+xy+ay^{2}=c$ has a solution in
$\mathbb{F}_{2^{k}}(t)_{(f)}$.

  2. 2.

If $v_{f}(c)$ is odd then the equation $x^{2}+xy+ay^{2}=c$ has a solution in
$\mathbb{F}_{2^{k}}(t)_{(f)}$ if and only if $[a,f)=0$.

###### Proof.

Note that by Hensel’s lemma the extension $K_{a,f}/\mathbb{F}_{2^{k}}(t)_{f}$
is unramified. Therefore by Thm. 2.3 the image of the norm map
$N_{K_{a,f}/\mathbb{F}_{2^{k}}(t)_{f}}\colon
K_{a,f}^{\times}\to\mathbb{F}_{2^{k}}(t)_{f}^{\times}$ contains the group of
units $(\mathbb{F}_{2^{k}}(t)_{f}^{+})^{\times}$ in the ring of integers.
Further, $N_{K_{a,f}/\mathbb{F}_{2^{k}}(t)_{f}}$ is onto if and only if
$K_{a,f}=\mathbb{F}_{2^{k}}(t)_{f}$. The latter is equivalent to $[a,f)=0$. ∎

######  Lemma 3.15.

Let $a_{1},a_{3}\in\mathbb{F}_{2^{k}}[t]$ be square-free polynomials with
$\gcd(a_{1},a_{3})=1$ and $a_{2},a_{4}\in\mathbb{F}_{2^{k}}(t)$. Let $f$ be a
place, ie. either a monic irreducible polynomial or $f=\infty$ such that
$v_{f}(a_{1}a_{3})$ is odd. Assume that neither $a_{2}$ nor $a_{4}$ has a pole
at $f$. Then the equation
$a_{1}x_{1}^{2}+a_{1}x_{1}x_{2}+a_{1}a_{2}x_{2}^{2}+a_{3}x_{3}^{2}+a_{3}x_{3}x_{4}+a_{3}a_{4}x_{4}^{2}=0$
has a nontrivial solution in $\mathbb{F}_{2^{k}}(t)_{(f)}$ if and only if at
least one of the two conditions holds:

  1. 1.

$[a_{2},f)=0$

  2. 2.

$[a_{4},f)=0$

###### Proof.

This is a combination of Lemmas 3.8 and 3.14. ∎

Now we turn our attention to the case when $a$ has a pole at $f$ (ie.
$K_{a,f}$ ramifies). Note that unlike in the case of characteristic $0$ there
exist infinitely many ramified quadratic extensions of local fields of
characteristic $2$. By Lemma 3.12 the pole must be of odd degree $2r+1$
therefore the following lemma is relevant. In this case it is more convenient
to multiply by $f^{2r+1}$ and put $b=af^{2r+1}$ which is an $f$-adic unit.
Note that $c$ is in the range of the quadratic form $x^{2}+xy+ay^{2}$ if and
only if so is $cd^{2}$ for all $0\neq d\in\mathbb{F}_{2^{k}}(t)$ therefore we
may rescale $c$ by a square element as convenient.

######  Lemma 3.16.

Let $b,c$ be in $\mathbb{F}_{2^{k}}(t)_{(f)}$ such that $v_{f}(b)=0$ (ie. $b$
is an $f$-adic unit) and $v_{f}(c)=0$ or $1$. Then the equation

$f^{2r+1}x^{2}+f^{2r+1}xy+by^{2}=cf^{2r}$

has a solution in $\mathbb{F}_{2^{k}}(t)_{(f)}$ if and only if it has a
solution modulo $f^{4r+3}$. All such solutions lie in the valuation ring
$\mathbb{F}_{2^{k}}(t)_{(f)}^{+}$.

###### Proof.

$\Rightarrow:$ Suppose we have a solution
$(u,v)\in\mathbb{F}_{2^{k}}(t)_{(f)}$. Assume for contradiction that one of
$u$ and $v$ is not in $\mathbb{F}_{2^{k}}(t)_{(f)}^{+}$. Multiplying by the
square of the common denominator $f^{l}$ of $u$ and $v$ we obtain
$u_{1}=f^{l}u,v_{1}=f^{l}v\in\mathbb{F}_{2^{k}}(t)_{(f)}^{+}$ such that
$f^{2r+2l}\mid f^{2r+1}u_{1}^{2}+f^{2r+1}u_{1}v_{1}+bv_{1}^{2}$ but $f$ does
not divide at least one of $u_{1}$ and $v_{1}$. Since $f\nmid b$ we obtain
$f^{2r+1}\mid v_{1}^{2}$ whence $f^{r+1}\mid v_{1}$. So we deduce
$f^{2r+2}\mid f^{2r+1}u_{1}v_{1}+bv_{1}^{2}$ and $f^{2r+2}\mid
f^{2r+1}u_{1}^{2}$ contradicting to $f\nmid u_{1}$. Hence we may reduce the
equality $f^{2r+1}u^{2}+f^{2r+1}uv+bv^{2}=cf^{2r}$ modulo $f^{4r+3}$.

$\Leftarrow:$ Assume we have $u_{0},v_{0}\in\mathbb{F}_{2^{k}}(t)_{(f)}^{+}$
such that

$c_{0}f^{2r}:=f^{2r+1}u_{0}^{2}+f^{2r+1}u_{0}v_{0}+bv_{0}^{2}\equiv
cf^{2r}\pmod{f^{4r+2}}\ .$

Then we must have $f^{r}\mid v_{0}$ and put $v_{0}=f^{r}v_{1}$ so dividing by
$f^{2r}$ we deduce

$c_{0}=fu_{0}^{2}+f^{r+1}u_{0}v_{1}+bv_{1}^{2}\equiv c\pmod{f^{2r+2}}$

Since $f^{2}\nmid c$ at least one of $u_{0}$ and $v_{1}$ is not divisible by
$f$. Putting $c_{1}:=\frac{c-c_{0}}{f^{2r+2}}$, we look for the solution of
the original equation in the form $x=u_{0}+f^{r+1}x_{1}$, and
$y=v_{0}+f^{2r+1}y_{1}$. So we are reduced to solving the equation

$\displaystyle
f^{2r+1}(u_{0}+f^{r+1}x_{1})^{2}+f^{2r+1}(u_{0}+f^{r+1}x_{1})(f^{r}v_{1}+f^{2r+1}y_{1})+$
$\displaystyle+b(f^{r}v_{1}+f^{2r+1}y_{1})^{2}=f^{2r}(c_{0}+f^{2r+2}c_{1})\ .$

Using the equation for $c_{0}$ and dividing by $f^{4r+2}$ we obtain the
equivalent equation

$\displaystyle
fx_{1}^{2}+x_{1}v_{1}+u_{0}y_{1}+f^{r+1}x_{1}y_{1}+by_{1}^{2}=c_{1}\ .$ (1)

Now note that Hensel’s lemma applies to (1) since the gradient

$\left(\frac{\partial}{\partial
x_{1}}(fx_{1}^{2}+v_{0}x_{1}+u_{0}y_{1}+fx_{1}y_{1}+by_{1}^{2}-c_{1}),\right.$
$\left.\frac{\partial}{\partial
y_{1}}(fx_{1}^{2}+v_{0}x_{1}+u_{0}y_{1}+fx_{1}y_{1}+by_{1}^{2}-c_{1})\right)=$
$=(v_{1}+f^{r+1}y_{1},u_{0}+f^{r+1}x_{1})\equiv(v_{1},u_{0})\pmod{f}$

is nonzero modulo $f$. Therefore $(u_{0},v_{1})$ lifts to a solution modulo
$f^{4r+3}$ $\Leftrightarrow$ (1) has a solution modulo $f$
$\overset{\text{Hensel}}{\Leftrightarrow}$ (1) has a solution in
$\mathbb{F}_{2^{k}}(t)_{(f)}$ $\Leftrightarrow$ $(u_{0},v_{0})$ lifts to a
solution of $f^{2r+1}x^{2}+f^{2r+1}xy+by^{2}=cf^{2r}$ in
$\mathbb{F}_{2^{k}}(t)_{(f)}$. ∎

###  3.3 Finding nontrivial zeros

Let
$Q(x_{1},x_{2},x_{3},x_{4})=a_{1}x_{1}^{2}+a_{1}x_{1}x_{2}+a_{1}a_{2}x_{2}^{2}+a_{3}x_{3}^{2}+a_{3}x_{3}x_{4}+a_{3}a_{4}x_{4}^{2}$
where $a_{i}\mathbb{F}_{2^{k}}(t)$. In this section we provide an algorithm
for deciding whether $Q$ admits a nontrivial zero and if so, returns a
nontrivial zero $(x_{1},x_{2},x_{3},x_{4})$. The main idea is similar to the
main algorithm of [10]. We replace $Q$ with a similar form $Q^{\prime}$ and
then decide whether $Q^{\prime}$ has a nontrivial zero using the local-global
principle. If so, then we look for a common $c\in\mathbb{F}_{2^{k}}(t)$ which
is represented by both $a_{1}x_{1}^{2}+a_{1}x_{1}x_{2}+a_{1}a_{2}x_{2}^{2}$
and $a_{3}x_{3}^{2}+a_{3}x_{3}x_{4}+a_{3}a_{4}x_{4}^{2}$ and then solve the
equations $a_{1}x_{1}^{2}+a_{1}x_{1}x_{2}+a_{1}a_{2}x_{2}^{2}=c$ and
$a_{3}x_{3}^{2}+a_{3}x_{3}x_{4}+a_{3}a_{4}x_{4}^{2}=c$ separately. Solving
these equations is equivalent to finding zero divisors in quaternion algebras
over $\mathbb{F}_{2^{k}}(t)$. This is a special case of the main algorithm
from [9, Section 4].

######  Theorem 3.17.

Let
$Q(x_{1},x_{2},x_{3},x_{4})=a_{1}x_{1}^{2}+a_{1}x_{1}x_{2}+a_{1}a_{2}x_{2}^{2}+a_{3}x_{3}^{2}+a_{3}x_{3}x_{4}+a_{3}a_{4}x_{4}^{2}$
where $a_{i}\in\mathbb{F}_{2^{k}}(t)$. Then there exists a polynomial-time
algorithm which decides whether $Q$ is isotropic and if so it finds a
nontrivial zero of $Q$.

###### Proof.

By rescaling and dividing by common factors we can assume that
$a_{1},a_{3}\in\mathbb{F}_{2^{k}}[t]$ and $gcd(a_{1},a_{3})=1$. We look for a
common $c\in\mathbb{F}_{2^{k}}[t]$ which is represented by both
$a_{1}x_{1}^{2}+a_{1}x_{1}x_{2}+a_{1}a_{2}x_{2}^{2}$ and
$a_{3}x_{3}^{2}+a_{3}x_{3}x_{4}+a_{3}a_{4}x_{4}^{2}$. Note that $c$ is
represented by both these forms if and only if it is represented by both forms
locally at each place $f$. By Lemma 3.12 we may assume that both
$a_{1}x_{1}^{2}+a_{1}x_{1}x_{2}+a_{1}a_{2}x_{2}^{2}$ and
$a_{3}x_{3}^{2}+a_{3}x_{3}x_{4}+a_{3}a_{4}x_{4}^{2}$ are minimal (in the sense
of Definition 3.11). Denote by $S$ the set of places where at least one of the
following holds:

  1. 1.

$a_{2}$ has a pole at $f$;

  2. 2.

$a_{4}$ has a pole at $f$;

  3. 3.

$v_{f}(a_{1}a_{3})$ is odd.

We look for $c$ in the form $c=f_{1}f_{2}\cdots f_{m}h$ where
$f_{1},\dots,f_{m}\in S$ are monic irreducible polynomials and $h$ is
irreducible. If $f\notin S$, then $v_{f}(a_{1})$ and $v_{f}(a_{3})$ have the
same parity. Since $a_{1}$ and $a_{3}$ are coprime polynomials, their
valuations must actually be even. If $v_{f}(c)=0$ the forms represent $c$
locally at $f$ by Lemma 3.14(1). On the other hand, if $f\in S$ then we
distinguish two cases.

First assume that neither $a_{2}$ nor $a_{4}$ has a pole at $f$ (whence
$v_{f}(a_{1}a_{3})$ is odd). Then whether or not a square-free polynomial $c$
is represented by the form
$a_{1}x_{1}^{2}+a_{1}x_{1}x_{2}+a_{1}a_{2}x_{2}^{2}$ (resp.
$a_{3}x_{3}^{2}+a_{3}x_{3}x_{4}+a_{3}a_{4}x_{4}^{2}$) depends only on the
class of $c$ modulo $f^{2}$. So we may decide by checking all the residue
classes modulo $f^{2}$ whether there is a common value $c$ of the two forms.
If there is no common value then we are done (the $4$-variable form is not
isotropic). By Lemma 3.15 this happens if and only if $[a_{2},f)=[a_{4},f)=1$.
We put $f$ among $f_{1},\dots,f_{m}$ if all the common square-free values of
the two forms are divisible by $f$. Either way, there possibly appears a
condition on $c$ modulo $f^{2}$ (which we shall encode in the choice of $h$).

Now assume that either $a_{2}$ or $a_{4}$ has a pole at $f$. We use Lemma 3.8
in order to decide whether there is a common value of the forms
$a_{1}x_{1}^{2}+a_{1}x_{1}x_{2}+a_{1}a_{2}x_{2}^{2}$ and
$a_{3}x_{3}^{2}+a_{3}x_{3}x_{4}+a_{3}a_{4}x_{4}^{2}$ locally at $f$: If
$a_{2}+a_{4}$ has a pole of odd degree at $f$ then $x^{2}+x+a_{2}+a_{4}$ does
not split in $\mathbb{F}_{2^{k}}(t)_{(f)}$ hence there exist common values. We
reduce the poles at $f$ in $a_{2}+a_{4}$ of even degree by adding elements of
the form $\frac{d^{2}}{t^{2s}}+\frac{d}{t^{s}}$ ($d\in\mathbb{F}_{2^{k}}[t]$).
If all the poles are removed then we check whether $x^{2}+x+a_{2}+a_{4}$
splits by computing the symbol $[a_{2}+a_{4},f)$ as in the previous case.
Finally, for finding common values we use random values $c$ modulo $f^{4r+3}$
by Lemma 3.16 where $2r+1:=\max(-v_{f}(a_{2}),-v_{f}(a_{4}))$ is the bigger
order of the pole at $f$ of $a_{2}$ and $a_{4}$. Again, if $v_{f}(c)$ is odd
then we put $f$ into the finite set $\\{f_{1},\dots,f_{m}\\}$.

Finally, if $f=\infty\in S$ then the congruence condition on $c$ involves a
condition on the parity of the degree of $c$, as well as a condition modulo a
power of $t$. Even if $\infty\notin S$ then the condition $v_{\infty}(c)=0$
means the degree of $c$ must be even.

Now if none of the above congruence conditions were contradictory then we
deduce that the $4$-variable form is isotropic by Theorem 2.1. So we proceed
with finding a nontrivial zero looking for $c=f_{1}\cdots f_{m}h$ where the
monic irreducible polynomials $f_{1},\dots,f_{m}\in S$ are determined above
and we choose $h$ irreducible satisfying all the above congruence conditions
(including possibly a condition at $\infty$ if it belongs to $S$). This is
possible by Lemma 2.4. By construction, $c$ is a common value of
$a_{1}x_{1}^{2}+a_{1}x_{1}x_{2}+a_{1}a_{2}x_{2}^{2}$ and
$a_{3}x_{3}^{2}+a_{3}x_{3}x_{4}+a_{3}a_{4}x_{4}^{2}$ locally at all places in
$S$. Further, if $g\neq h$ is a (finite or infinite) place not in $S$ then $c$
is also a common value locally at $g$, so the only exception could be at $h$.
However, by Hilbert’s reciprocity law (Theorem 2.2) $c$ is also a common value
locally at $h$.

Input: $a_{1},a_{3}\in\mathbb{F}_{2^{k}}[t]$ and
$a_{2},a_{4}\in\mathbb{F}_{2^{k}}(t)^{\times}$ such that $a_{1}$ and $a_{3}$
are coprime, and every pole of $a_{2}$ (resp. $a_{4}$) has an odd order.

Output: $h\in\mathbb{F}_{2^{k}}[t]$ such that $h$ is represented by both
binary quadratic forms $Q_{1}(x,y)=a_{1}(x^{2}+xy+a_{2}y^{2})$ and
$Q_{2}(x,y)=a_{3}(x^{2}+xy+a_{4}y^{2})$, or $\bot$ if no such $h$ exists.

$Conds\leftarrow[\ ]$;

$c\leftarrow 1$;

for  _$g\in\mathrm{Poles}(a_{2})\cup\mathrm{Poles}(a_{4})$_ do

/* Apply lemmas 3.16 and 3.14 to find $h_{g}$ */

_ _ if  _$\exists
h_{g}\in\mathbb{F}_{2^{k}}[t],N\in\mathbb{N}\mid\nu_{g}(h_{g})\leq 1$ and
$\forall h\in\mathbb{F}_{2^{k}}(t)_{(g)},h$ is represented by $Q_{1}$ and
$Q_{2}$ if_

 __ $h=h_{g}\mod g^{N}$ then

$\mathrm{Append}(\mathrm{Conditions},(h_{g},g^{N}))$;

$c\leftarrow c\times g^{\nu_{g}(h_{g})\mod 2}$;

end if

else

return _$\bot$_

end if

end for

for  _$g\in\mathbb{F}_{2^{k}}(t)$ irreducible, such that $\nu_{g}(a_{1}a_{3})$
is odd and $g\notin\mathrm{Poles}(a_{2})\cup\mathrm{Poles}(a_{4})$_ do

/* Apply lemmas 3.14 and 3.15 to find $\nu_{g}$ */

if  _$\exists\nu\in\\{0,1\\}\mid\forall h\in\mathbb{F}_{2^{k}}(t)_{(g)},h$ is
represented by $Q_{1}$ and $Q_{2}$ if $\nu_{g}(h)=\nu\mod 2$_ then

$c\leftarrow c\times g^{\nu}$;

end if

else

return _$\bot$_

end if

end for

$i\leftarrow\mathbb{F}_{2^{k}}$-automorphism of $\mathbb{F}_{2^{k}}(t)$
sending $t$ to $1/X$;

$\nu_{\infty}\leftarrow\max(\deg(a_{1}),\deg(a_{3}))$;

/* Use remark 3.18 to find $h_{\infty}$ and $N_{\infty}$ */

if  _$\exists
h_{\infty}\in\mathbb{F}_{2^{k}}[t],N_{\infty}\in\mathbb{N}\mid\forall
h\in\mathbb{F}_{2^{k}}(t)_{(t)},h$ is represented by $Q_{1}$ and $Q_{2}$ if
$\exists N\in\mathbb{N},t^{2N+\nu_{\infty}}i(h)=h_{\infty}\mod
t^{N_{\infty}}$_ then

return _$cg:g\in\mathbb{F}_{2^{k}}[X]$ is irreducible,
$\deg{(cg)}=\nu_{\infty}+\deg{(h_{\infty})}\mod 2$,
$\forall(h,f)\in\mathrm{Conditions},cg=h\mod f$ and $\exists
n\in\mathbb{Z}\mid t^{2n+\nu_{\infty}}i(cg)=h_{\infty}\mod t^{N_{\infty}}$_

end if

else

return _$\bot$_

end if

Algorithm 1 SplitQuaternaryForm

∎

######  Remark 3.18.

The case $f=\infty$ can be treated like the case $f=t$ after applying the
automorphism fixing $\mathbb{F}_{2^{k}}$ and sending $t$ to $\frac{1}{t}$. It
is necessary to multiply the equation by a power of $t$ to normalize it, and
to check that if the new $a_{2}$ or $a_{4}$ have a pole at $t$, it has an odd
order (otherwise one may apply again the algorithm from lemma 3.12). When
checking a candidate polynomial for this condition, one should be careful to
normalize it with a power of $t$ of the same parity as the one used to derive
the condition. The condition on the degree of $h$ then is the one that allows
$h$ to have the prescribed valuation at the place $t$ after applying the
automorphism and normalizing.

####  3.3.1 An example

We give a short example of how the algorithm works. Let $K=\mathbb{F}_{2}(t)$
and consider the form
$a_{1}x_{1}^{2}+a_{1}x_{1}x_{2}+a_{1}a_{2}x_{2}^{2}+a_{3}x_{3}^{2}+a_{3}x_{3}x_{4}+a_{3}a_{4}x_{4}^{2}$
where $a_{1}=t^{2}+t+1,\leavevmode\nobreak\ a_{2}=t,\leavevmode\nobreak\
a_{3}=1,\leavevmode\nobreak\ a_{4}=1$. We have to look at $f$-adic solvability
for $f=t^{2}+t+1$ and $f=\infty$ as these are the places for which the
$f$-adic valuation of $a_{1}a_{3}$ is odd or $a_{2}$ has pole ($a_{4}$ is
regular everywhere).

One has $[a_{2},t^{2}+t+1)=1$ and $[a_{4},t^{2}+t+1)=0$ which implies that the
$(t^{2}+t+1)$-adic valuation of a common value $c$ of
$(t^{2}+t+1)(x_{1}^{2}+x_{1}x_{2}+tx_{2}^{2})$ and
$x_{3}^{2}+x_{3}x_{4}+x_{4}^{2}$ must be odd. Therefore we need to look for
$c$ in form of $c=(t^{2}+t+1)h$ where $h$ is an irreducible polynomial over
$\mathbb{F}_{2}$. Further, $h=x_{1}^{2}+x_{1}x_{2}+tx_{2}^{2}$ admits a
nonzero solution modulo $(t^{2}+t+1)$ if $h\equiv t\pmod{t^{2}+t+1}$. On the
other hand, $v_{\infty}((t^{2}+t+1)h)=\deg((t^{2}+t+1)h)$ must be even in
order for $x_{3}^{2}+x_{3}x_{4}+x_{4}^{2}=(t^{2}+t+1)h$ to be solvable at
$\infty$ since the extension by a root of $X^{2}+X+1$ is unramified at
$\infty$. Putting $z=1/t$ this boils down to the solvability of
$zx_{1}^{2}+zx_{1}x_{2}+x_{2}^{2}=zh$ in $\mathbb{F}_{2}[[z]]$ which is
equivalent to $h\equiv 1\pmod{z^{2}}$ by Lemma 3.16. After rescaling $h$ so
that it is a polynomial in $t$ our condition is that the coefficient of
$t^{2n-1}$ is zero in $h$ where $\deg(h)=2n$.

Finally, a little computation shows that $h(t)=t^{6}+t+1$ satisfies

  1. 1.

$h\equiv t\pmod{t^{2}+t+1}$

  2. 2.

$\deg(h)=2n$ is even

  3. 3.

the coefficient of $t^{2n-1}$ in $h$ is zero

So we are reduced to finding a solution to the following two equations
globally:

  1. 1.

$x_{1}^{2}+x_{1}x_{2}+tx_{2}^{2}=(t^{2}+t+1)(t^{6}+t+1)$

  2. 2.

$x_{3}^{2}+x_{3}x_{4}+x_{4}^{2}=t^{6}+t+1$

Running the known algorithms for binary forms we find $x_{1}=t^{3}+t^{2}+1$,
$x_{2}=t$, $x_{3}=t^{4}+1$, $x_{4}=t^{3}$.

###  3.4 Applications

In this subsection we give two applications of our results and methods. One is
to finding separable quadratic extensions $L$ of $\mathbb{F}_{2^{k}}(t)$
inside a quaternion algebra that is split by $L$. The other is constructing
quaternion algebras over $\mathbb{F}_{2^{k}}(t)$ with prescribed Hasse
invariants.

######  Theorem 3.19.

Let $L$ be a separable quadratic extension of $\mathbb{F}_{2^{k}}(t)$ and let
$B$ be a quaternion algebra over $\mathbb{F}_{2^{k}}(t)$ which is split by
$L$. Then there exists a polynomial-time algorithm which finds a subfield of
$B$ isomorphic to $L$.

###### Proof.

If $B$ is split, then one can find a explicit isomorphism between $B$ and
$M_{2}(\mathbb{F}_{2^{k}}(t))$ in polynomial time using the main algorithm
from [9, Section 4] (the algorithm also decides whether $B$ is split or not).
From such an isomorphism a suitable maximal subfield can be constructed easily
(by constructing a matrix whose minimal polynomial corresponds to a defining
polynomial of $L$).

Now suppose that $B$ is a division algebra. In that case $B$ contains a
maximal subfield isomorphic to $L$ [18, Lemma 6.4.12]. Let
$L=\mathbb{F}_{2^{k}}(t)(s)$ where $s^{2}+s=c$ and
$c\in\mathbb{F}_{2^{k}}(t)$. If we find an element $u\in B$ such that
$u^{2}+u=c$, then $u+s$ is a zero divisor as $u$ is not in the center. Suppose
that $B$ has the following quaternion basis:

$i^{2}+i=a$ $j^{2}=b$ $ij=j(i+1)$

Let us look for $u$ in the form of
$u=\lambda_{1}+\lambda_{2}i+\lambda_{3}j+\lambda_{4}ij$, where
$\lambda_{i}\in\mathbb{F}_{2^{k}}(t)$.

$\displaystyle
u^{2}+u=\lambda_{1}^{2}+\lambda_{2}^{2}a+\lambda_{3}^{2}b+\lambda_{4}^{2}ab+\lambda_{3}\lambda_{4}b+$
$\displaystyle+\lambda_{1}+i(\lambda_{2}^{2}+\lambda_{2})+j(\lambda_{2}\lambda_{3}+\lambda_{3})+ij(\lambda_{2}\lambda_{4}+\lambda_{4})$

For this to be in $\mathbb{F}_{2^{k}}[t]$, $\lambda_{2}=1$ must hold. Now we
investigate if the following equation has a non-trivial solution:

$\lambda_{1}^{2}+\lambda_{3}^{2}b+\lambda_{4}^{2}ab+\lambda_{3}\lambda_{4}b+\lambda_{1}+a+c=0$
(2)

Let $\mu_{2}$ equal to the product of the denominators of all $\lambda_{i}$,
$\mu_{1}:=\lambda_{1}\mu_{2}$, let us introduce new variables
$\mu_{3}:=\lambda_{3}\mu_{2}$ and $\mu_{4}:=\lambda_{4}\mu_{2}$. Then
multiplying (2) by $\mu_{2}^{2}$ gives

$\mu_{1}^{2}+\mu_{1}\mu_{2}+(a+c)\mu_{2}^{2}+b\mu_{3}^{2}+b\mu_{3}\mu_{4}+ab\mu_{4}^{2}=0$
(3)

where $\mu_{1},\mu_{2},\mu_{3},\mu_{4}\in\mathbb{F}_{2^{k}}[t]$. Now we find a
solution to the above equation using the algorithm from Theorem 3.17 ([18,
Lemma 6.4.12] guarantees the existence of a solution) which returns $u$. The
algorithm of Theorem 3.17 runs in polynomial time which implies the statement
of the theorem. ∎

######  Remark 3.20.

The main motivation behind studying this algorithm is that in can be used to
find zero divisors in split quaternion algebras over $L$. Namely, one frst
constructs a subalgebra $B$ of the large algebra $A$ that is a quaternion
algebra over $\mathbb{F}_{2^{k}}(t)$. If this algebra is not split, then it
contains a subfield isomorphic to $L$ which is generated by some quaternion
element $x$. Let $L=\mathbb{F}_{2^{k}}(t)(s)$. Then $x+s$ will be a zero
divisor.

The next proposition shows how to construct a quaternion division algebra with
given Hasse invariants.

######  Proposition 3.21.

Let $v_{1},\dots,v_{l}$ be places of $\mathbb{F}_{2^{k}}(t)$ such that $l$ is
even. Then there exists a polynomial-time algorithm which constructs a
quaternion algebra over $\mathbb{F}_{2^{k}}(t)$ which is ramified exactly at
$v_{1},\dots,v_{l}$.

###### Proof.

Let $f_{1},\dots f_{m}$ be the finite places amongst the $v_{i}$. First we
find a monic irreducible polynomial in $b\in\mathbb{F}_{2^{k}}[t]$ such that
$[b,f_{i})=1$. This can be accomplished in the following way. One finds
quadratic non-square $r_{i}$ modulo every $f_{i}$ ($\mathbb{F}_{q}[t]/(f_{i})$
is finite field of cardinality $2^{\deg(f_{i})k}$) and then obtains a residue
class $r$ modulo $f_{1}\cdots f_{m}$ such that $r\equiv r_{i}\pmod{f_{i}}$ by
Chinese remaindering. Then using Lemma 2.4 one finds an irreducible polynomial
of suitably large degree which is congruent to $r$ mod $f_{1}\cdots f_{m}$ by
choosing random elements from the residue class until an irreducible is found.

Let $a=f_{1}\cdots f_{m}$. We show that the quaternion algebra $A=[a,b)$
ramifies at every $f_{i}$. The algebra $A$ ramifies at $f_{i}$ if and only if
the quadratic form $ax^{2}+axy+aby^{2}+z^{2}$ has a nontrivial zero in
$\mathbb{F}_{2^{k}}(t)_{(f_{i})}$. Since the form is homogeneous, it is enough
to show that it does not admit an integral zero. The variable $z$ must be
divisible by $f_{i}$ since $a$ is divisible by $f_{i}$. Now setting
$z=f_{i}z^{\prime}$ and dividing by $f_{i}$ we get the following equation:

$a/f_{i}x^{2}+a/f_{i}xy+a/f_{i}by^{2}+f_{i}z^{\prime 2}=0$

Suppose this equation has a nontrivial solution $(x_{0},y_{0},z_{0})$. One may
assume that $f_{i}$ does not divide $x_{0},y_{0}$ and $z_{0}$ simultaneously.
Then the following congruence condition holds:

$a/f_{i}x_{0}^{2}+a/f_{i}x_{0}y_{0}+a/f_{i}by_{0}^{2}\equiv 0\pmod{f_{i}}$

Since $a/f_{i}$ is coprime to $f_{i}$ one can divide the congruence by
$a/f_{i}$. If $y_{0}$ is not divisible by $f_{i}$, then $b$ is a quadratic
residue mod $f_{i}$ which is a contradiction. If $y_{0}$ is divisible by
$f_{i}$, then so is $x_{0}$. However, if $x_{0}$ and $y_{0}$ are both
divisible by $f_{i}$, then $z_{0}$ is not divisible by $f_{i}$ and then
$a/f_{i}x_{0}^{2}+a/f_{i}x_{0}y_{0}+a/f_{i}by_{0}^{2}+f_{i}z_{0}^{2}$ is not
divisible by $f_{i}^{2}$ which is a contradiction.

The algebra $A$ is split at $b$ since the equation
$ax^{2}+axy+aby^{2}+z^{2}=0$ has a solution modulo $b$ (setting $z=0$ and
$x=y=1$) which can be lifted by Hensel’s lemma. $A$ is clearly split at all
the other finite places and has the required splitting condition at $\infty$
by Hilbert reciprocity (Theorem 2.2). ∎

######  Corollary 3.22.

Let $D$ be a quaternion division algebra over $\mathbb{F}_{2^{k}}(t)$ and let
$A$ be an algebra isomorphic to $M_{2}(D)$ given by structure constants. Then
one can find a zero divisor in $A$ in polynomial time.

###### Proof.

We compute the local indices of $A$ using the algorithm [8, Proposition
6.5.3.] and then use Proposition 3.21 to compute a division quaternion algebra
$D_{0}$ with those exact invariants. Since we have constructed a structure
constant representation of $D_{0}$, we can construct a structure constant
representation of $M_{2}(D_{0})$ by considering the basis where the matrix has
one nonzero entry and that runs through the basis of $D_{0}$. Then as stated
previously, one can construct an explicit isomorphism between $A$ and
$M_{2}(D_{0})$ from an explicit isomorphism between $A^{op}\otimes
M_{2}(D_{0})$ and $M_{16}(\mathbb{F}_{2^{k}}(t))$ in polynomial time using the
main algorithm from [9, Section 4]. Finally, the preimage of the matrix
$\begin{pmatrix}1&0\\\ 0&0\end{pmatrix}$ is a zero divisor. ∎

##  4 Implementation

In this section, we give details about our
implementation222https://github.com/Char2QuadForms/Char2QuadForms of the
algorithm in the Magma language [2]. We then provide details on the practical
efficiency of our implementation and discuss the computational bottlenecks.

###  4.1 Implementation details

The core of our code is a practical implementation of algorithm 1. Our first
step is to take as an input a quaternary quadratic form as a degree $4$ square
matrix with coefficients in $\mathbb{F}_{2^{k}}(t)$. If $Q$ is the input
quadratic form, we apply successively the reductions from lemma 3.5, corollary
3.6 and lemma 3.12 to obtain coefficients
$a_{1},a_{2},a_{3},a_{4}\in\mathbb{F}_{2^{k}}(t)$ which follow the hypotheses
of theorem 3.17, and such that the quadratic form
$a_{1}(x_{1}^{2}+x_{1}x_{2}+a_{2}x_{2}^{2})+a_{3}(x_{3}^{2}+x_{3}x_{4}+a_{4}x_{4}^{2})$
is similar to $Q$. In addition, we make coefficients $a_{1}$ and $a_{3}$
square-free, as it simplifies computation and does not affect the place of the
zeros.

Our implementation of algorithm 1 follows the structure of the pseudo-code
representation. See appendix A for more details on the subroutines for each
case. Once all the conditions for a common value have been established, we
randomly generate polynomials that satisfy said conditions until we find a
prime polynomial. Because of the bound given in lemma 2.4, finding one such a
polynomial can be done in probabilistic polynomial time. Once we find a prime
polynomial which satisfies every condition, we move to the last step of the
implementation.

We independently solve equations
$a_{1}(x_{1}^{2}+x_{1}x_{2}+a_{2}x_{2}^{2})=ch$ and
$a_{3}(x_{1}^{2}+x_{1}x_{2}+a_{3}x_{2}^{2})=ch$. Both these equations directly
reduce to a Hilbert equation, and then by lemma 3.7 the problem reduces to
finding an explicit isomorphism between a given quaternion algebra and the
degree $2$ matrix algebra. In practice, we solve the equation
$a_{1}(x_{1}^{2}+x_{1}x_{2}+a_{2}x_{2}^{2})=c$ by computing an explicit
isomorphism between
$A=\left[\frac{a_{2},a_{1}/c}{\mathbb{F}_{2^{k}}(t)}\right)$ and
$M_{2}\left(\mathbb{F}_{2^{k}}(t)\right)$. It follows from [18, equation
6.4.5] that in $A$, $nrd(x+yj+zk)=x^{2}+\frac{a_{1}}{c}(y^{2}+yz+a_{2}z^{2})$.
We therefore find a singular matrix which pulls back to a quaternion of the
form $x+yj+zk$ with $x\neq 0$, and then we set $x_{1}=\frac{y}{x}$ and
$x_{2}=\frac{z}{x}$.

We use the main algorithm from [9] to compute the explicit isomorphism between
$A$ and $M_{2}\left(\mathbb{F}_{2^{k}}(t)\right)$. Since, to the best of our
knowledge, it has not been implemented yet, we provide an implementation in
Magma which may be of independent interest.

###  4.2 Computational data

In table 1 we show the running time for some executions of algorithm 1. This
running time does not include solving the resulting ternary forms, which we
will discuss separately. The tests were executed on the online Magma
calculator333http://magma.maths.usyd.edu.au/magma/ with randomly generated
polynomials. The degree of the input polynomials were not randomly chosen, but
they were affected by the steps of minimization of the coefficients. We give
the degrees of the coefficients after the minimisation steps. By the degree of
a rational function we mean the maximal of the degrees of its numerator and of
its denominator. The column $\deg h$ refers to the degree of the value
represented by both binary forms, that is the output of algorithm 1. The
column $q$ refers to the cardinal of the finite field underlying our rational
function field. The running times are given in seconds. We note that all
running times here are given for input corresponding to an isotropic quadratic
form. In general, it is faster for the algorithm to recognize an anisotropic
form than to split an isotropic one.

$\deg a_{1}$ | $\deg a_{2}$ | $\deg a_{3}$ | $\deg a_{4}$ | $\deg h$ | $q$ | Running time
---|---|---|---|---|---|---
$83$ | $71$ | $7$ | $93$ | $384$ | $2$ | $0.570$
$204$ | $211$ | $1048$ | $211$ | $604$ | $2$ | $4.680$
$21$ | $25$ | $121$ | $25$ | $75$ | $2^{10}$ | $6.090$
$15$ | $21$ | $102$ | $21$ | $64$ | $2^{20}$ | $29.520$
Table 1: Running times of algorithm 1

The last part of our implementation is the main algorithm from [9] (see the
discussion in subsection 4.1). A subroutine for this algorithm is the
computation of a maximal order in a quaternion algebra. Since this subroutine
was not implemented in Magma for algebras over fields of characteristic 2, we
gave our own implementation using the polynomial time algorithm given in [9,
subsection 3.2]. Our implementation of this algorithm runs significantly
slower than the Magma built-in function for maximal order computation in odd
characteristic. As a result, we do not draw conclusions regarding the running
time for this part of the implementation.

However, the implementation is still practical for small input. In table 2 we
give running time for our function solving equations of the form
$x^{2}+xy+ay^{2}=h$. Every line refers to a computation done over
$\mathbb{F}_{2}(t)$.

$\deg a$ | $\deg h$ | Running time
---|---|---
$1$ | $4$ | $2.390$
$1$ | $8$ | $20.360$
$5$ | $14$ | $311.460$
Table 2: Running time for solving $x^{2}+xy+ay^{2}=h$

## References

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  * Bosma et al. [1997] Wieb Bosma, John Cannon, and Catherine Playoust. 1997.  The Magma algebra system. I. The user language.  _J. Symbolic Comput._ 24, 3-4 (1997), 235–265.  https://doi.org/10.1006/jsco.1996.0125 Computational algebra and number theory (London, 1993).
  * Castel [2013] Pierre Castel. 2013\.  Solving quadratic equations in dimension 5 or more without factoring.  _The Open Book Series_ 1, 1 (2013), 213–233.
  * Conrad [2010] Keith Conrad. 2010\.  Quadratic reciprocity in characteristic 2.  _Unpublished notes, available athttp://www.math.uconn.edu/~kconrad/blurbs/ugradnumthy/QRchar2.pdf_ (2010).
  * Cremona and Rusin [2003] John Cremona and David Rusin. 2003.  Efficient solution of rational conics.  _Math. Comp._ 72, 243 (2003), 1417–1441.
  * Cremona and van Hoeij [2006] John Cremona and Mark van Hoeij. 2006.  Solving conics over function fields.  _Journal de théorie des nombres de Bordeaux_ 18, 3 (2006), 595–606.
  * Elman et al. [2008] Richard S Elman, Nikita Karpenko, and Alexander Merkurjev. 2008\.  _The algebraic and geometric theory of quadratic forms_. Vol. 56.  American Mathematical Soc.
  * Ivanyos [1996] Gábor Ivanyos. 1996\.  _Algorithms for algebras over global fields_.  Ph.D. Dissertation. Hungarian Academy of Sciences.
  * Ivanyos et al. [2018] Gábor Ivanyos, Péter Kutas, and Lajos Rónyai. 2018\.  Computing Explicit Isomorphisms with Full Matrix Algebras over $\mathbb{F}_{q}(x)$.  _Foundations of Computational Mathematics_ 18, 2 (2018), 381–397.
  * Ivanyos et al. [2019] Gábor Ivanyos, Péter Kutas, and Lajos Rónyai. 2019\.  Explicit equivalence of quadratic forms over $\mathbb{F}_{q}(t)$.  _Finite Fields and Their Applications_ 55 (2019), 33–63.
  * Ivanyos and Szántó [1996] Gábor Ivanyos and Ágnes Szántó. 1996.  Lattice basis reduction for indefinite forms and an application.  _Discrete Mathematics_ 153, 1-3 (1996), 177–188.
  * Koprowski [2021] Przemysław Koprowski. 2021\.  Isotropic vectors over global fields.  _arXiv preprint arXiv:2111.08569_ (2021).
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  * Milne [1997] James S Milne. 1997\.  Class field theory.  _lecture notes available at http://www. math. lsa. umich. edu/jmilne_ (1997).
  * Rauter [1926] H Rauter. 1926.  _über die Darstellbarkeit durch quadratische Formen im Körper der rationalen Funktionen einer Unbestimmten über dem Restklassenkörper mod p_.  Ph.D. Dissertation. Halle.
  * Simon [2005a] Denis Simon. 2005a.  Quadratic equations in dimensions 4, 5 and more.  _preprint_ (2005).
  * Simon [2005b] Denis Simon. 2005b.  Solving quadratic equations using reduced unimodular quadratic forms.  _Math. Comp._ 74, 251 (2005), 1531–1543.
  * Voight [2021] John Voight. 2021\.  _Quaternion algebras_. Graduate texts in mathematics, Vol. 288.  Springer,Cham.
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##  Appendix A Algorithm subroutines

Input:
$a_{1},a_{3},f\in\mathbb{F}_{2^{k}}[t],a_{2},a_{4}\in\mathbb{F}_{2^{k}}(t)$
such that $a_{1}$ and $a_{3}$ are nonzero coprime polynomials, each pole of
$a_{2}$ and $a_{4}$ has an odd multiplicity, and $f$ is an irreducible
polynomial which is a pole of $a_{2}$ or $a_{4}$.

Output: $h_{f}\in\mathbb{F}_{2^{k}}[t],n\in\mathbb{N}$ such that for all
$h\in\mathbb{F}_{2^{k}}[t]$, if $h=h_{f}\mod f^{n}$, then $h$ is represented
by both binary forms $a_{1}(x^{2}+xy+a_{2}y^{2})$ and
$a_{3}(x^{2}+xy+a_{4}y^{2})$ over $\mathbb{F}_{2^{k}}(t)_{(f)}$. Outputs
$\bot$ instead if such a tuple does not exist.

$N\leftarrow 2\max(-\nu_{f}(a_{2}),-\nu_{f}(a_{4}))+1$;

$K_{2}\leftarrow\mathbb{F}_{2^{k}}(t)/(t^{2}+t+a_{2})$;

$K_{4}\leftarrow\mathbb{F}_{2^{k}}(t)/(t^{2}+t+a_{4})$;

$K_{6}\leftarrow\mathbb{F}_{2^{k}}(t)/(t^{2}+t+a_{2}+a_{4})$;

if  _$(\mathrm{not}\ K_{6}=\mathbb{F}_{2^{k}}(t))\ \mathrm{or}\
\frac{a_{1}}{a_{3}}\in N_{K_{2}/\mathbb{F}_{2^{k}}(t)}(K_{2}^{\times})$_ then

repeat

$c\xleftarrow{\mathrm{Random}}\mathbb{F}_{2^{k}}[t]/(f^{N})$;

until  _$\nu_{f}(c)\leq 1\ \mathrm{and}\ c\in
N_{K_{2}/\mathbb{F}_{2^{k}}(t)}(K_{2}^{\times})\cap
N_{K_{4}/\mathbb{F}_{2^{k}}(t)}(K_{4}^{\times})$_ ;

return _c,N_

end if

else

return _$\mathrm{\bot}$_

end if

Algorithm 2 CommonValuePole

Input:
$a_{1},a_{3},f\in\mathbb{F}_{2^{k}}[t],a_{2},a_{4}\in\mathbb{F}_{2^{k}}(t)$
such that $a_{1}$ and $a_{2}$ are nonzero coprime polynomials, each pole of
$a_{2}$ and $a_{4}$ has an odd multiplicity, and $f$ is an irreducible
polynomial which is not a pole of $a_{2}$ or $a_{4}$ but such that
$\nu_{f}(a_{1}a_{3})$ is odd.

Output: $\nu\in\\{0,1\\}$ such that a polynomial $h\in\mathbb{F}_{2^{k}}[t]$
is represented by both binary quadratic forms $a_{1}(x^{2}+xy+a_{2}y^{2})$ and
$a_{3}(x^{2}+xy+a_{4}y^{2})$ over $\mathbb{F}_{2^{k}}(t)_{(f)}$ if
$\nu_{f}(h)=\nu\mod 2$. Outputs $\bot$ if there is no such $\nu$.

if  _$[a_{1},f)=[a_{3},f)=1$_ then

return _$\bot$_

end if

else if  _$[a_{1},f)=\nu_{f}(a_{1})\mod 2$_ then

return _$1$_

end if

else

return _$0$_

end if

Algorithm 3 CommonValueOdd

Input:
$a_{1},a_{3}\in\mathbb{F}_{2^{k}}[t],a_{2},a_{4}\in\mathbb{F}_{2^{k}}(t)$ such
that $a_{1}$ and $a_{3}$ are nonzero coprime polynomials, and each pole of
$a_{2}$ and $a_{4}$ has an odd multiplicity.

Output: $h_{\infty}\in\mathbb{F}_{2^{k}}[t],N_{\infty}\in\mathbb{N}$ such that
a polynomial $h\in\mathbb{F}_{2^{k}}[t]$ is represented by both binary forms
$a_{1}(x^{2}+xy+a_{2}y^{2})$ and $a_{3}(x^{2}+xy+a_{4}y^{2})$ over the
completion of $\mathbb{F}_{2^{k}}(t)$ at the infinite place if
$\deg(h)=\nu_{t}(h_{\infty})\mod 2$ and $\exists n\in\mathbb{Z}\mid
x^{2n+\nu_{\infty}}i(cg)=h_{\infty}\mod N_{\infty}$, with
$\nu_{\infty}=\max{(\deg{a_{1}},\deg{a3})}$.

$i\leftarrow\mathbb{F}_{2^{k}}$-automorphism of $\mathbb{F}_{2^{k}}(t)$
sending $t$ to $1/t$;

$\nu_{\infty}\leftarrow-\max_{i\in\\{1,3\\}}\deg(a_{i})$;

$ia_{1}\leftarrow i(x^{\nu_{\infty}}a_{1}),ia_{3}\leftarrow
i(x^{\nu_{\infty}}a_{3})$;

$ia_{2}\leftarrow i(a_{2}),ia_{4}\leftarrow i(a_{4})$;

/* By minimalise we mean to apply the algorithm described in the proof of
lemma 3.12 */

minimalise(ia2);

minimalise(ia4);

if  _$t\in\mathrm{Poles}(ia_{2})\cup\mathrm{Poles}(ia_{4})$_ then

$\mathrm{Res}\leftarrow\mathrm{CommonValuePole}(ia_{1},ia_{2},ia_{3},ia_{4},t)$;

if  _$\mathrm{Res}=\bot$_ then

return _$\bot$_

end if

else

$(h_{\infty},N_{\infty})\leftarrow Res$;

end if

end if

else if  _$\nu_{t}(ia_{1}ia_{3})=1\mod 2$_ then

$\mathrm{Res}\leftarrow\mathrm{CommonValueOdds}(a_{1},a_{2},a_{3},a_{4},f)$;

if  _$\mathrm{Res}=\bot$_ then

return _$\bot$_

end if

else

$h_{\infty}\leftarrow t^{\mathrm{Res}}$;

$N_{\infty}\leftarrow 0$;

end if

end if

else

$h_{\infty}\leftarrow 1$;

$N_{\infty}\leftarrow 0$;

end if

return _$(h_{\infty},N_{\infty})$_

Algorithm 4 CommonValueInf


# Dynamics of the fall of a quantum 4He crystal in superfluid liquid

V. L. Tsymbalenko<EMAIL_ADDRESS>NRC Kurchatov Institute, 123182
sq.Kurchatov 1
Institute for Physical Problems RAS, 119334 Kosigin st.2
Moscow, Russia

###### Abstract

The motion of helium crystals has been experimentally studied when the
crystals fall in the superfluid liquid owing to gravity at temperatures above
the roughening transitions where the whole crystal surface is in the
atomically rough state. The rate of crystal fall at $T=1.25$K is higher than
at $T=1.54$K. This is proof of the essential role of the normal component of
superfluid helium in the deceleration of crystal motion. The pressure
measurements in the container have shown the effect of surface kinetics on the
motions of the crystal and its size. The fall of crystals with the low surface
mobility at $T=1.54$K does not change the pressure significantly. The high
surface mobility at $T=1.25$K results in decreasing the pressure in the
container in the course of the fall of a crystal. The pressure drop exceeds
the difference in the hydrostatic pressure for the initial and final positions
of the crystal. After the stop the pressure in the container relaxes to the
difference mentioned above. This fact demonstrates an additional growth of the
crystal in the flow of a superfluid liquid.

###### pacs:

67.80. -s, 68.08. -p

##  I Introduction

The fast kinetics of the atomically rough surfaces of 4He crystals AP
displays a remarkable interplay between the fluid flow and the superfluid-
solid interface. This fact is mentioned by Nozieres, Uwaha NU , and M.Kagan MK
, who theoretically studied the tangential instability of a flat liquid-solid
interface. It is found that the instability of the cylindrical crystal shape
is associated with the fast interface kinetics. The fluid flow in the
direction of the cylindrical axis of a 4He crystal leads to developing the
instability on the smaller scales MaxVLT . As it concerns the faceted
interfacial segments with low mobility, the fluid flow has no observable
effect. However, there exist conditions when the crystal facets acquire high
mobility comparable with that of the atomically rough surface. Provided that a
large overpressure is applied to the crystalline facet, the growth rate of the
facet increases drastically by 2 - 3 orders of magnitude in jump-like manner
VLT2015 . At the initial stage the 4He crystal grows at high growth rates and
giant accelerations of the interface. This situation is theoretically analyzed
within the framework of the Rayleigh-Taylor and Richtmyer-Meshkov
instabilities BDT .

The experimental test of theoretical conclusions is complicated by the
requirement to prepare the fluid flow with controlled parameters while the 4He
crystal is in the container. It has been observed that the fluid jet created
with the heat flux BKP , the oscillating loop VLT2013  or the motion of
charges MaxVLT  distorts the crystalline surface. Observations of the crystal
shapes at the stage of fast growth demonstrate the instability due to
acceleration of liquid-solid interface BDT . All these observations have been
performed with immobile crystals under the fluid flow induced by the methods
mentioned above.

Since the direct measurement of the velocity of fluid flow is difficult, the
alternative method is to set a 4He crystal into motion in the immobile fluid.
The velocity of the crystal can be directly measured using video recording of
the crystal motion. The force exerted to the crystal is determined by the
method of setting the crystal into motion. In a series of experiments a
superconducting loop is used for forcing the crystal to oscillate. The crystal
is either pierced by the crossbar or is fixed on the crossbar VLT2013 . The
oscillation frequency of such system is determined by the crystal mass and
force arising from the fluid flow. The maximum velocity of the crystal, which
one can reach, is as large as 3 cm/s. In the first case, a further enhancement
of the oscillation amplitude results in the fast remelting of the crystal
before its detaching from the crossbar. In the second case the crystal escapes
from the crossbar.

Another method is the study of free fall of a 4He crystal in the field of
gravity. Here the motive force is the weight of the crystal, compensated
partially by the force of buoyancy. A direct observation of the crystal shape
transformation during the fall of the crystal has been performed by the
Japanese group at 0.3K using a high-speed camera NOTY . At these temperatures
the shape of the 4He crystal is governed by the facets with low growth rates.
As can be seen from this study, the effect of facet interface kinetics upon
the motion of the 4He crystal is insignificant.

In experiments which continue the study of burst-like growth effects within
the 0.1 – 0.2K range we have noticed a curious feature VLT2019 . As the
crystal stops to grow, the mobility of crystalline facets remains very high
VLT2004 . The crystal starts to remelt in the hydrostatic pressure gradient
and then detaches from its nucleation site at the wall in the upper part of
the container. The fall of the crystal is accompanied by a decreasing pressure
in the container. It is unexpected and intriguing that the pressure changes
non-monotonically. At first, the pressure drops below the difference in
hydrostatic pressures between the crystal nucleation site and the container
bottom. Then the pressure relaxes to the pressure difference indicated. Such
behavior of the pressure implies two essential facts. Firstly, the pressure
decrease in the course of fall is possible only provided that the crystal is
growing and its volume increases. In other words, as the crystal stops to grow
in the burst-like growth regime, the crystal facets keep their high mobility.
Secondly, the extra pressure drop and the following pressure relaxation prove
unambiguously the increase of the crystal volume during the fall of the
crystal not only due to hydrostatic pressure but also due to an additional
factor.

In Ref. VLT2019  the influence of fluid flow is involved as a factor
explaining such an effect. The simplified model is suggested for the fall of
an isotropic sphere with a mobile interfacial boundary which is encircled by
the fluid flow. The fluid flow is supposed to be laminar as in all theoretical
works known so far NU ; MK ; MaxVLT ; BDT . The model has explained the
pressure behavior and given the estimate for the magnitude of the averaged
kinetic growth coefficient $K=0.22$ s/m, $V=K\Delta\mu$. Here $V$ is the
interface growth rate and $\Delta\mu$ is the difference in chemical
potentials. However, the time behavior of pressure $p(t)$ measured
experimentally in the course of the fall differs noticeably from the
prediction of the model. The reasons may be as follows: growth anisotropy of
crystal facets, specific trajectory of the fall and orientation of the
crystal, specific features of fluid flow at such velocities of encircling and
so on.

Many of these difficulties can be avoided by observing the fall of helium
crystals at 1.2 - 1.6K. Within this temperature range all the liquid-solid 4He
interface is in the atomically rough state. The kinetic growth coefficient is
isotropic. The crystals have approximately spherical shape VLT1995 . The
numerical magnitude of the kinetic growth coefficient for such temperatures
lies within the range $K\sim 0.05-0.3$ s/cm, i.e. it is of the order of the
magnitude given in Ref. VLT2019 . Comparing the dynamics of the fall of a
crystal in the superfluid with different normal and superfluid densities, we
can clarify and understand the role and contribution of normal and superfluid
components to the dynamics of motion of 4He crystals.

##  II Experimental methods and results

The crystals are grown in optical container similar to those used in
experiments previously VLT1996 ; VLT1997 . The inner diameter of the container
is 20 mm and the volume is 10 cm3. A capacitive pressure sensor is fixed on
the upper flange of the container. The pressure is measured at 2 ms time
intervals. The crystal nucleates at a tungsten tip located at 10.5 mm distance
from the bottom of the container. A high voltage is applied to the tip. After
nucleation and growth, the crystal has 1-2 mm diameter. The hydrostatic
pressure gradient leads to a remelting of the crystal, while its center shifts
down and it subsequently separates from the tip VLT1995 . The fall of the
crystal is recorded by a television system with a ccd-matrix. The crystal is
illuminated by a light pulse of an infrared LED lasting 5 $\mu s$ and
synchronized with the frame. An example of shooting the fall of a crystal at
two temperatures is shown in Fig.1.

Figure 1: The crystal fall at two temperatures. The upper series is made at T
= 1.54K, and the lower series at T = 1.25K. The intervals between the frames
from left to right: 60ms, 40ms and 40ms. The light pulse duration is 5 $\mu
s$.

The relaxation time of the pressure in the container to the pressure of the
external system is $\sim$4 s, i.e., the crystals fall over an interval
$\sim$160 ms at almost constant mass of helium in the container. Note that
when the crystal is separated from the tip, a protrusion remains on its
surface, as was previously observed in Ref. VLT1995 . Then, the protrusion
flattens and at lower temperature this occurs faster due to the higher kinetic
growth coefficient. The observation for the protrusion reveals that the
crystal rotates in the course of its fall.

The experiments were carried out at two temperatures. At T=1.54K, the crystal
has the bcc structure, and at T=1.25K the crystal has the hcp structure. The
kinetic growth coefficient of an atomically rough surface in this temperature
range is determined by the energy dissipation processes in the liquid. These
processes are insensitive to the crystal structure MaxVLT2 . The surface
growth rate in both cases is linearly dependent on supersaturation and
isotropic for small deviations from equilibrium. The kinetic growth
coefficients measured from surface relaxation BLN  and from pressure
relaxation MaxVLT2  are consistent and equal to $K=0.063\pm 0.012m/s$ at 1.54K
and $K=0.23\pm 0.03m/s$ at 1.25K.

Fig.2 shows the results of measurements on the position of the crystal center
$H$ as a function of the fall time. It can be seen that the temperature growth
decelerates during the fall.

Figure 2: The position of crystal center versus the square of the fall time.
The ordinate corresponds to the distance from the center of the crystals to
the bottom of the container. The squares refer to the crystals grown at 1.25K
and circles at 1.54K

This can also be seen in the video of Fig.1, where the crystals starting from
the same position are at different heights after 140 ms. In Fig.3, top panel,
we see the time behavior of the fall rate $V(t)$ of crystals, obtained by
differentiating the smoothed data $H(t)$.

Figure 3: Above: The time behavior of the fall rate of crystals showing the
velocity $V(t)$ as a function of time. Below: the change in the acceleration
of fall over time. The squares refer to the crystals grown at 1.25K and
circles at 1.54K The solid curves refer to 1.54K and the dashed lines to
1.25K.

Crystals fall with acceleration. Lowering the temperature increases the fall
rate of crystals. The wave on the dependence $V(t)$ at T = 1.54K is probably
caused by the rotation of the crystal and the non-sphericity of its shape. The
Fig. 3, bottom panel, shows the change in acceleration in the process of fall,
obtained by numerical differentiation of the upper graph. Acceleration is
determined by the points included within the 20 ms time interval. To avoid
dispersion in the velocity graph at T = 1.52K, the time interval is enlarged
to $\pm 30$ ms for the point at 130 ms. The low accuracy in determining the
acceleration allows us to draw qualitative conclusions only. The acceleration
of the crystal motion decreases with time at both temperatures. The magnitudes
of acceleration are within the range of $30-100cm/s^{2}$. It is possible that
the acceleration at $\sim 1.25K$ becomes constant after $\sim 60ms$.

Averaged over all measurements the results for the pressure variations are
shown in Fig.4. To reduce noise, the records of $\Delta p(t)$ in each series
are combined by the time of separating the crystal from the tip and
summarized. One can see that the pressure variation differs drastically for
two temperatures.

Figure 4: Upper curve: the pressure change in the container during the fall of
the crystal at 1.54K. The pressure difference between the beginning and the
end of the fall is 0.026mbar. This value lies within the measurement noise
magnitude of 0.03mbar. Bottom curve: $\Delta p(t)$ during the fall of the
crystal at 1.25K. The arrows show the time of crystal fall. The dashed line
corresponds to the difference in the hydrostatic pressure between the initial
and final positions of the crystals.

At high temperature, the change in pressure during the fall, if it occurred,
does not exceed the measurement noise of $\sim 0.03$mbar. At low temperatures,
a pressure drop is observed similar to that observed at 106 mK for crystals
whose surface retains mobility after nucleation in the burst-like growth mode
VLT2019 . As in that case, the pressure drops below the difference in the
hydrostatic pressure between the place of crystal nucleation and the bottom of
the container. Then, pressure relaxation to this magnitude is observed, which
in our case is $\Delta p_{0}=0.16$ mbar. The same kinetic growth coefficients
for the crystals that grow at different temperatures in the completely
different states lead to a qualitatively identical change in pressure upon
falling.

The observations on the fall of the crystals at 1.16K show that the crystals
have their flat facets facing down, resulting in floating during the motion,
see Fig.5. The lower facet is oriented perpendicular to the direction of fall.
The moment of forces acting on the body from the side by the liquid flow turns
the crystal in this way LAMB .

Figure 5: The crystal fall illuminated with stroboscopic light: pulse duration
5 $\mu$s, pulse interval 20 ms, 1.16K

Below 0.6 K, the crystal facets have such low growth rate that the nucleated
crystals remain hanging on the tip for tens of minutes, as was previously
observed VLT1995 .

##  III Discussion of the results

The crystal falls at temperatures where the normal component density of
superfluid helium is significant. Since the rates of fall are small and the
fluid can be considered as incompressible, the equations of hydrodynamics are
separated into the equations for the superfluid and normal components LL . The
equation for the normal component reads

$\frac{\partial{\vec{v}_{n}}}{\partial
t}+(\vec{v}_{n}\nabla)\vec{v}_{n}=-\frac{1}{\rho_{n}}\nabla
p_{n}+\frac{\eta_{n}}{\rho_{n}}\triangle\vec{v}_{n}$ (1)

Here $\vec{v_{n}}$, $p_{n}$, $\eta_{n}$ and $\rho_{n}$ are the velocity,
pressure, viscosity and density of the normal component, respectively. The
equation can be transformed to the dimensionless representation similar to the
classical Navier-Stokes equation:

$\displaystyle t\rightarrow t\frac{R}{U},r\rightarrow rR,v_{n}\rightarrow
v_{n}U,p_{n}\rightarrow\rho_{n}U^{2}p_{n}$
$\displaystyle\frac{\partial{v_{n}}}{\partial t}+(v_{n}\nabla)v_{n}=-\nabla
p_{n}+\frac{1}{Re_{n}}\triangle\vec{v}_{n},$ (2) $\displaystyle
Re_{n}=UR\frac{\rho_{n}}{\eta_{n}}=\frac{UR}{\nu_{n}}.$

Here $U$ is the velocity of the body, $R$ is its radius and $Re_{n}$ is the
Reynolds parameter for the normal component flow. The normal component density
is $0.41\rho$ at 1.54K, $0.14\rho$ at 1.25K and $0.08\rho$ at 1.16K Maynard .
The viscosity of the normal component is 20 $\mu P$ at 1.54K and 23 $\mu P$ at
1.25K EHH . For the crystal radius of $\sim 1mm$ and falling rates of 1-10
cm/sec, the Reynolds parameter varies within Ren = 400 – 4000 at 1.54K and Ren
= 100 – 1000 at 1.25K. There exist a lot of studies on the motion of spheres
in helium at saturated vapor pressure. Without attempting to provide a
complete description of these results, we use only a few works for
illustrative purposes. The resistance to the motion of a ball in normal helium
is proportional to velocity squared and estimated by the relation BHH ; DHH ;
LR

$F_{n}=Cx_{n}\frac{1}{2}\rho_{n}U^{2}\pi R^{2}.$ (3)

The parameter $Cx_{n}$ is 0.2-0.5 in the indicated range of the Reynolds
numbers.

The superfluid component flow is described by the ideal fluid equation

$\triangle\varphi=0,\vec{v}_{s}=\nabla\varphi.$ (4)

The experimental studies of the motion of spherical bodies in superfluid
helium show that resistance to motion is observed starting at velocities of
$\sim 0.1$ cm/s A . Feynman explained this phenomenon by the production of
vortices during helium flow around an object and gave an estimate for the
critical velocity of such process, $v_{crit}=(h/m_{4}R)ln(R/A)$ F , where $h$
is the Plank constant, $m_{4}$ is the helium atom mass and $A$ is the
interatomic distance. The resistance to the motion of the sphere is
approximately described by a relation similar to expression (3) A ; Vinen ;
Sch

$F_{s}=Cx_{s}\frac{1}{2}\rho_{s}U^{2}\pi R^{2}.$ (5)

The parameter $Cx_{s}$ in the indicated range of the Reynolds numbers is
0.3-0.5. A number of authors use the Reynolds parameter
$Re_{s}=Rv_{s}/\nu_{s}$ to evaluate the nature of superfluid flow and
turbulence. The value $\nu_{s}~{}=h/m_{4}~{}\simeq~{}10^{-3}$ cm2/s is taken
as the kinematic viscosity. For the rates of crystal fall in these
experiments, this corresponds to the values $Re_{s}=100-1000$.

The equations (III) and (4) are related via the boundary condition on the
crystal surface. Andreev and Knizhnik AK  proposed the condition
$v_{surf}=v_{n}$, that is, the mass flow through the phase boundary is
provided only by the superfluid component. This statement is consistent with
the experimental fact that near $\lambda$-transition, when
$\rho_{s}\rightarrow 0$, the kinetic growth coefficient also tends to zero
MaxVLT2 . Therefore, the motion of a sphere in the normal component (III)
occurs as the motion of an impenetrable ball, whose size and shape, generally
speaking, change during the motion. The superfluid flow around a sphere with
the mass flow through its surface is considered in Ref. VLT2019  under the
assumption that the superfluid flow is laminar. In this case, the crystal
grows mainly in the direction perpendicular to the flow of the liquid. This
change in the shape leads to increasing the added mass and, as a result, to
decreasing the acceleration of movement. In our experiments, as is seen in
Fig.1, the crystals rotate during the fall. This leads to more uniform surface
growth and small change in the shape of the crystal. It follows that the added
mass of the superfluid liquid also changes insignificantly, and does not lead
to reducing the acceleration during a fall within the framework of the model
in Ref. VLT2019 .

In our experiments, the flow around a crystal differs significantly from that
considered in Ref. VLT2019 . The rotation of the crystal leads to the
appearance of a Magnus force perpendicular to the direction of motion. The
formation of vortices in the normal and superfluid components creates a
complex picture of the pressure distribution on the crystal surface. For these
reasons, the flow of liquid around the crystal is very complicated. The
problem of crystal growth taking into account these factors has not yet been
solved.

For the values of the Reynolds parameter in these experiments (see above), the
crystal falls down with the formation of a vortex pattern in the normal
component and the creation of quantum vortices in the superfluid component.
Combining formulas (3) and (5), we obtain an expression for the total force
$F$ acting on the crystal

$F=(Cx_{n}\kappa_{n}+Cx_{s}\kappa_{s})\frac{1}{2}\rho U^{2}\pi
R^{2},\kappa_{n}+\kappa_{s}=1.$ (6)

Here the coefficients $\kappa_{s}$ and $\kappa_{n}$ are the fractions of the
superfluid and normal components, respectively. The equation of motion of the
crystal reads

$\rho^{\prime}\alpha\frac{4\pi}{3}R^{3}\dot{U}=-(Cx_{n}\kappa_{n}+Cx_{s}\kappa_{s})\frac{1}{2}\rho
U^{2}\pi R^{2}+\Delta\rho\frac{4\pi}{3}R^{3}g,$ (7)

where $\rho^{\prime}$ is the density of the crystal, $\Delta\rho$ is the
difference in density between the solid and liquid phases, and $\alpha$ takes
the added mass of the liquid into account. For a sphere, the parameter
$\alpha$ is equal to $\alpha=1+\frac{\rho}{2\rho^{\prime}}\simeq 1.45$. The
solution of equation (7) has the form

$\displaystyle
U(t)=U_{0}\tanh\left(\frac{t}{\tau}\right),<Cx>=Cx_{n}\kappa_{n}+Cx_{s}\kappa_{s},$
(8) $\displaystyle
U_{0}=\left(\frac{8}{3}\frac{\Delta\rho}{\rho}\frac{gR}{<Cx>}\right)^{\frac{1}{2}}\simeq
0.54\sqrt{\frac{gR}{<Cx>}},$
$\displaystyle\tau=\alpha\left(\frac{8}{3}\frac{(\rho^{\prime})^{2}}{\rho\Delta\rho}\frac{1}{<Cx>}\frac{R}{g}\right)^{\frac{1}{2}}\simeq
7.93\sqrt{\frac{R}{g<Cx>}}.$

From the form of equation (7) it follows that if the condition $Cx_{n}>Cx_{s}$
is fulfilled, then the lower rate of crystal fall at higher temperatures is
explained by increasing the normal component fraction. The values of the
$Cx_{n,s}$ coefficients at a liquid pressure 25 bar are unknown. Assuming
these values to be of the order of $Cx_{n,s}$ at the saturated vapor pressure
for solid balls A , we obtain an estimate for the parameters $U_{0}$ and
$\tau$. For $<Cx>=0.5$, $U_{0}\approx 8$ cm/s and $\tau\approx 120$ ms. For
$<Cx>=0.1$, $U_{0}\approx 17$ cm/s and $\tau\approx 260$ ms. As can be seen
from Fig. 2, the second pair of parameters is better consistent with the
experimental data. These estimates should be considered as qualitative since
the essential features of the process, for example, such as the rotation of
the crystal and the possible dependence of the Reynolds parameter on velocity,
are not taken into account when deriving equation (7).

The decrease in pressure during the fall of the crystal is clear evidence for
the growth of its volume. The change in crystal volume is due to two factors.
First, due to growth in the denser liquid layers at the bottom of the
container, where the pressure exceeds that for the liquid in the initial
position of the crystal by $\Delta p_{0}=\rho g\Delta h=0.16$mbar. The second
source of the bulk growth is the effect of liquid flow. The qualitative change
in the shape of a crystal is clear from the general picture of the liquid flow
around the crystal. The flow rate of the liquid is maximum for the cross
section which gives the maximum difference in chemical potentials and the
maximum growth rate VLT2019 . The crystal starts to increase in transverse
dimension and volume. We estimate its value by the magnitude of pressure
relaxation after the motion stops, $\Delta p_{rel}\simeq 0.15mbar$. The total
increase in the crystal volume with an initial size of $\sim 1mm$ will be
$\sim 30\%$. The transverse crystal size will increase by about 0.2 mm. The
video recording has a resolution of 0.04 mm per pixel, which prevents from
measuring this effect with the necessary accuracy. Owing to the rotation of
the crystals, it is also impossible to check carefully the model VLT2019
predicting an increase in the crystal size mainly in the direction
perpendicular to the liquid. A comparison of the measurements of $p(t)$ at two
temperatures shows, see Fig. 4, that a decrease in the fraction of the
superfluid component leads to a practically constant volume of the crystal
during the fall. This results from the small pressure variation in the
container. As is noted above, this conclusion agrees with the observation that
the crystal grows from the superfluid component.

##  IV Conclusion

The study of falling the crystals in superfluid helium has shown a significant
effect of the normal component on the resistance to motion. This is clearly
seen in the dependencies $V(t)$ in Fig.3. For the fluid of the 41% normal
component fraction, the velocity of motion is noticeably lower than that in a
fluid with the 14% normal component concentration. The influence of the
interface growth kinetics is clearly shown by the pressure measurements during
the fall. The low kinetic coefficient of interface growth at 1.54K does not
result in a noticeable crystal growth either due to gradient of hydrostatic
pressure or pressure of the liquid flow. The increase in the interface
mobility at 1.25K demonstrates these both effects. When falling, the pressure
decreases due to crystal growth under influence of both effects, and then
increases after stopping the crystal at the bottom of the container due to
melting the excess crystal volume as a result of the liquid flow effect. The
experimental results have confirmed the hypothesis that the nonmonotonic
behavior pressure observed for the fall of crystals nucleated in the burst-
like growth mode at $\sim 0.1$ K is associated with the remaining high
kinetics of crystal facets after their growth. The growth rate of crystal
facets for $\sim 0.6$s after growing a crystal in the burst-like growth mode
VLT2019  exceeds the growth rate of crystal facets in the normal state by
$\sim 20$ times Balibar .

###### Acknowledgements.

The author is grateful to V. V. Dmitriev for the possibility of performing
these experiments at Kapitza Institute for Physical Problems RAS. The author
is also grateful to V. V. Zavyalov for supporting this work, S.N.Burmistrov
for helpful comments and V. S. Kruglov for interest to the work.

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