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# Simultaneous diagonal equations over P-adic fields

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## Abstract

Consider a system of R diagonal forms of degree k in N variables over a P-adic field K. Write k = pτm where p is the residue field characteristic and (m, p) = 1. We prove that this system has a non-trivial P-adic solution if N exceeds (Rk)2τ+5, a bound independent of the field K. We prove the same if N exceeds 4nR2k2 where n = [K : Qp]. Both results improve previously known bounds.
ﬁelds
D. Brink, H. Godinho, P. H. A. Rodrigues
February 2008
Abstract. Consider a system of R diagonal forms of degree k in N
variables over a p-adic ﬁeld K. Write k = p
τ
m where p is the residue
ﬁeld characteristic and (m, p) = 1. We prove that this system has a
non-trivial p-adic solution if N exceeds (Rk)
2τ+5
, a bound independent
of the ﬁeld K. We prove the same if N exceeds 4nR
2
k
2
where
n = [K : Q
p
]. Both results improve previously known bounds.
Let K be a ﬁnite extension of the ﬁeld of p-adic numbers Q
p
. Let O be the
ring of integers in K and let p be O’s unique maximal ideal. We say that
Consider R simultaneous diagonal equations
a
11
X
k
1
+ · · · + a
1N
X
k
N
= 0
.
.
.
.
.
.
.
.
.
a
R1
X
k
1
+ · · · + a
RN
X
k
N
= 0
()
with coeﬃcients a
ij
in O. Write the degree as k = p
τ
m with p - m. A
solution x = (x
1
, . . . , x
N
) K
N
is called non-trivial if at least one x
j
is
non-zero. It is a special case of a conjecture of Emil Artin that () has a
non-trivial solution whenever N > Rk
2
. This conjecture has been veriﬁed
by Davenport and Lewis for a single diagonal equation over Q
p
and for a
pair of equations of odd degree over Q
p
(see [3] and [4]), but the general
case remains open.
The main results of the present paper are the following two theorems.
Theorem 1. The system () has a non-trivial solution if the number of
variables N exceeds (Rk)
2τ+5
.
Theorem 2. Let n be the degree of the ﬁeld extension K/Q
p
. Then ()
has a non-trivial solution if N exceeds 4nR
2
k
2
.
Theorem 1 has the virtue of being independent of K and can be compared
with Skinner [11] where the bound N > k
6τ+4
is given for a single diagonal
equation. Theorem 2 is a natural generalisation of Knapp [7, Theorem 1]
1
and improves Dodson [6, Theorem 1] and Knapp [7, Theorem 3]. See also
Skinner [11] for other references.
Deﬁne the integer Γ(R, k) as minimal with the property that any system
() with N > Γ(R, k) has a non-trivial solution over K. Then Theorems
1 and 2 can be restated as Γ(R, k) (Rk)
2τ+5
and Γ(R, k) 4nR
2
k
2
,
respectively. The idea of the proof of the theorems is to ﬁrst solve () in
the ﬁnite residue ring O/p
γ
(for a suitable exponent γ), and then lift this
solution to K via a version of Hensel’s lemma.
A solution x O
N
is called primitive if at least one coordinate x
j
is a
unit in O. Deﬁne the integer Φ(R, k, ν) as minimal with the property that
any system () with N > Φ(R, k, ν) has a primitive solution modulo p
ν
.
The Chevalley-Warning theorem (see [2, Lemma 4]) states that any sys-
tem of homogeneous polynomials over a ﬁnite ﬁeld has a non-trivial zero
if the number of variables exceeds the sum of the polynomials’ degrees. In
the special case of systems of diagonal equations, the Chevalley-Warning
theorem gives
Φ(R, k, 1) Rk. (1)
For general moduli a, b 1 one has the relation
Φ(R, k, a + b) + 1 (Φ(R, k, a) + 1) (Φ(R, k, b) + 1) . (2)
This is shown using a well-known ”contraction” argument (see examples in
([4] and [11]). The idea is to construct a primitive solution modulo p
a+b
in
N = (Φ(R, k, a) + 1) · (Φ(R, k, b) + 1) variables as follows: First divide the
left hand side of () into Φ(R, k, a) + 1 subsystems of diagonal forms, each
in Φ(R, k, b) + 1 variables, and solve each system primitively modulo p
b
.
Then multiply each of these solutions by a new variable to form a system
of diagonal forms in Φ(R, k, a) + 1 variables. Since every coeﬃcient is a
multiple of p
b
, to solve this new system primitively modulo p
a+b
is basically
to solve it modulo p
a
. This results in a primitive solution modulo p
a+b
to
() which proves (2).
Let A = (a
ij
) be the coeﬃcient matrix of (). A solution x O
N
is called
non-singular if the matrix (a
ij
x
k
j
) has rank R modulo p, or, equivalently, if
the columns of A corresponding to the indices j with x
j
6≡ 0 (mod p) have
rank R modulo p.
The following strong version of Hensel’s lemma is a natural generalisa-
tion of [5, Lemma 9], from p-adic to p-adic ﬁelds. The deﬁnition of γ here
is somewhat better than the value 2 + 1 often found in the literature (al-
though Alemu [1] has a result for one equation similar to the lemma below).
Lemma 1. Let e be the ramiﬁcation index of K over Q
p
and deﬁne
γ :=
1 for τ = 0,
e(τ + 1) for τ > 0 and p 6= 2,
e(τ + 2) for τ > 0 and p = 2.
The system () then has a non-trivial solution in K if it has a non-singular
solution modulo p
γ
.
2
Proof. We ﬁrst show that a unit u O
is a kth power if u ξ
k
(mod p
γ
)
for some ξ O
. This is the standard Hensel’s lemma for τ = 0, so we may
assume τ > 0. Then multiplication x 7→ k·x maps p
e
onto k·p
e
= p
+e
= p
γ
for p 6= 2, and p
2e
onto k · p
2e
= p
γ
for p = 2. For any n > e/(p 1), the p-
n
and the the multiplicative group 1 + p
n
([9,
Kapitel II, Satz 5.5]). It follows that exponentiation x 7→ x
k
maps 1 + p
e
(for p 6= 2) and 1 + p
2e
(for p = 2) onto 1 + p
γ
. The diagram shows the
situation for p 6= 2:
1 + p
e
x7→x
k
// //
log
1 + p
γ
p
e
x7→k·x
// //
p
γ
exp
OO
Therefore, the elements of the set ξ
k
·(1+ p
γ
) = ξ
k
+p
γ
, to which u belongs,
are all kth powers.
Now let x = (x
1
, . . . , x
N
) be a non-singular solution to () modulo p
γ
.
We may assume x
1
, . . . , x
R
6≡ 0 (mod p) and that the ﬁrst R columns of
A have rank R modulo p, i.e. form a non-singular matrix modulo p. Row
operations on A will not change the solution set, so we may assume
A =
a
11
0 a
1,R+1
. . . a
1N
.
.
.
.
.
.
.
.
.
0 a
RR
a
R,R+1
. . . a
RN
with a
11
, . . . , a
RR
6≡ 0 (mod p). For each i = 1, . . . , R we have x
k
i
u
i
(mod p
γ
) with u
i
= (a
i,R+1
x
k
R+1
+ · · · + a
iN
x
k
N
)/a
ii
. By the above,
the equation X
k
= u
i
has a solution x
0
i
because it has the solution x
i
mod-
ulo p
γ
. We conclude that (x
0
1
, . . . , x
0
R
, x
R+1
, . . . , x
N
) solves ().
The notion of a p-normalised system of diagonal equations over Q
p
was
introduced in [5]. It is shown there that any system of the form () over
Q
p
has a non-trivial solution provided that any p-normalised system has a
non-trivial solution. All of this is easily generalised to π-normalised systems
with p-adic coeﬃcients (see [7] for details).
Let µ(d) be the maximal number of columns of the coeﬃcient matrix A
which, when considered modulo p, lie in a d-dimensional subspace of F
N
q
.
The key property of π-normalised systems is the inequality
µ(d) N (R d)N/Rk for d = 0, . . . , R 1. (3)
This is [5, Lemma 11] combined with [2, eq. (9)]. An equivalent statement
of this inequality is that any matrix having (R d) rows which are linear
combinations of the rows of A, independent modulo p, contains at least
(R d)N/Rk columns which are nonzero modulo p.
The following slight strengthening of [2, Lemma 2] essentially gives one
extra non-singular submatrix.
3
Lemma 2. Suppose () is π-normalised and has more than k(tR 1)
variables, where t is arbitrary. Then the coeﬃcient matrix A contains t
disjoint R × R submatrices which are non-singular modulo p.
Proof. For every d = 0, . . . , R 1, the assumption N > k(tR 1) combined
with (3) implies µ(d) N (R d)t since µ(d) is integral. Now the
conclusion follows by a combinatorial result of Aigner (see [8, Lemma 1] or
the comment before [2, Lemma 2]).
Next, we extend and improve [2, Lemma 5] using the same idea of proof.
Lemma 3. Suppose () is π-normalised and has more than Rk ·Φ(R, k, ν)
k(R 1)
2
variables, where ν is arbitrary. Then () has a non-singular
solution modulo p
ν
.
Proof. Suppose ﬁrst () has N = k(tR 1) + 1 variables for some t to be
deﬁned later. Then, by Lemma 2, A has t disjoint R×R submatrices which
are non-singular modulo p. Discard all variables not belonging to one of
these t submatrices. Then we have tR variables left. In each of all but one
of the t submatrices, replace all R variables by one new variable. Then we
have a new system with t 1 + R variables. This system, by deﬁnition,
has a primitive solution modulo p
ν
if t 1 + R > Φ(R, k, ν), hence if
t = Φ(R, k, ν) R + 2. Not all the new variables of this solution can be
zero modulo p since the columns corresponding to the old variables form a
non-singular submatrix modulo p and so are linearly independent modulo p.
Therefore, ”inﬂating” the new variables again gives a non-singular solution
to our original system () in N = Rk · Φ(R, k, ν) k(R 1)
2
+ 1 variables,
and the lemma is proved.
Recall that Γ(R, k) is the minimal integer such that any system () with
N > Γ(R, k) has a non-trivial solution. From Lemmas 1 and 3 it follows
that
Γ(R, k) Rk · Φ(R, k, γ) k(R 1)
2
(4)
since any bound on Γ(R, k) may be proved under the assumption that ()
is π-normalised. For degree k not divisible by p, (4) and (1) give
Γ(R, k) (Rk)
2
k(R 1)
2
, (5)
which extends [2, Theorem 3].
Now, Theorem 2 follows from (4) and the following lemma.
Lemma 4. With γ deﬁned as in Lemma 1, we have
Φ(R, k, γ)
(
p(p 1)
1
nRk for p > 2,
4nRk for p = 2.
Proof. To bound Φ(R, k, γ), we must ﬁnd a primitive solution modulo p
γ
4
to (). The additive group of the ﬁnite residue ring O/p
γ
is equal to the
direct sum of n cyclic subgroups of order p
γ/e
,
O/p
γ
= Zλ
1
· · · Zλ
n
.
This can be seen for example by counting the number of elements of any
given order in both groups and noting that these numbers are the same (see
also [1] for a diﬀerent proof and a more general statement). Writing each
coeﬃcient a
ij
of () as a Z-linear combination of the λ
i
’s, we see that it
suﬃces to solve nR congruences
c
i1
X
k
1
+ · · · + c
iN
X
k
N
0 (mod p
γ/e
), i = 1, . . . , nR (6)
with coeﬃcients c
ij
Z. We shall only look for solutions x T
N
where
T = {x Q
p
| x
p
= x} is the set of Teichm¨uller representatives. Since
{x
k
| x T} = {x
(k,p1)
| x T}, we may in (6) replace the exponent
k by (k, p 1). Now, by a theorem of Schanuel [10], the system (6) has a
non-trivial solution x T
N
if N > nR(k, p1)(p
γ/e
1)(p1)
1
. Recalling
k = p
τ
m, we see that (k, p 1) divides m and conclude that Φ(R, k, γ) is
bounded by nR(k, p 1)p
τ+1
(p 1)
1
p(p 1)
1
nRk for p 6= 2, and by
4nRk for p = 2.
The next two lemmas and the ﬁnal proof of Theorem 1 are much inspired
by the ideas presented in Skinner [11].
Lemma 5. Any a O can be written as
a c
p
τ
0
+ πc
p
τ
1
+ π
2
c
p
τ
2
+ · · · + π
p
τ
1
c
p
τ
p
τ
1
(mod p)
with c
j
O and π being a prime element of O.
Proof. If R O is a set of representatives for O/p, then so is {r
p
τ
| r R},
because the map x 7→ x
p
τ
is a bijection F
q
F
q
. Hence, with suitable
r
n
R, we can write
a =
X
n=0
r
p
τ
n
π
n
=
p
τ
1
X
j=0
π
j
X
i=0
r
p
τ
j+ip
τ
π
ip
τ
p
τ
1
X
j=0
π
j
X
i=0
r
j+ip
τ
π
i
!
p
τ
(mod p),
which proves the lemma.
Lemma 6. Φ(R, k, e) Φ(Rp
τ
, m, e).
Proof. We have to ﬁnd a primitive solution x O
N
to the R congruences
a
i1
X
k
1
+ · · · + a
iN
X
k
N
0 (mod p), i = 1, . . . , R.
Write each polynomial in this system as a sum of p
τ
polynomials using the
above lemma on each coeﬃcient a = a
ij
. Thus it suﬃces to ﬁnd a primitive
solution to Rp
τ
congruences
c
p
τ
i1
X
k
1
+ · · · + c
p
τ
iN
X
k
N
0 (mod p), i = 1, . . . , Rp
τ
.
5
Since
c
p
τ
i1
X
k
1
+ · · · + c
p
τ
iN
X
k
N
(c
i1
X
m
1
+ · · · + c
iN
X
m
N
)
p
τ
(mod p),
it suﬃces to ﬁnd a primitive solution to the Rp
τ
congruences
c
i1
X
m
1
+ · · · + c
iN
X
m
N
0 (mod p), i = 1, . . . , Rp
τ
.
Such a solution exists by deﬁnition for N > Φ(Rp
τ
, m, e).
We can ﬁnally prove Theorem 1. Clearly, Φ(Rp
τ
, m, e) is bounded by
Γ(Rp
τ
, m) which is in turn bounded by (Rk)
2
m(Rp
τ
1)
2
by (5) since
m is not divisible by p. For τ = 0 we already have the bound (5) which is
superior to the one given in Theorem 1. So assume τ > 0. Then Lemma 6
implies
Φ(R, k, e) < (Rk)
2
. (7)
From (4), (2), and (7) it now follows that
Γ(R, k) Rk · Φ(R, k, γ)
Rk · (Φ(R, k, e) + 1)
γ/e
(Rk)
2γ/e+1
(Rk)
2τ+5
.
This concludes the proof of Theorem 1.
References
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6
[8] L. Low, J. Pitman & A. Wolﬀ, Simultaneous diagonal congruences, J.
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7
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