SOLUTION OF DIOPHANTINE EQUATIONS IN p-ADIC
INTEGERS.
IAN KIMING
Setup: K algebraic number field, O = OK the ring of integers of K.
Let p be a prime ideal of O and νp the attached exponential valuation. So we
have the completion Kp of K w.r.t. νp . One often refers to Kp as a field of ‘p-adic
numbers’. Similarly, elements of Op — the valuation ring in Kp — are often called
‘p-adic integers’.
We wish to give a complement to section 4.4 of [2] in the case of polynomials of
several variables. We give only the most basic statements: we will make a remark
to the effect that a Diophantine equation with coefficients in Op has a (non-trivial)
solution if and only if it has a (non-trivial) solution modulo pm for any m ∈ N.
Here, the notion of a congruence modulo pm between elements of Op has the obvious
meaning: α ≡ β (pm ) if and only if α − β ∈ pm . Remember here that the ideal ‘p’ is
really the maximal ideal pOp of Op . So we are talking about congruences modulo
(pOp )m . Clearly, (pOp )m = pm Op . It is useful to note the following:
Lemma 1. The canonical injection O ,→ Op induces an isomorphism:
O/pm ∼
= Op /pm Op .
Proof. pm Op is the set of elements of Op of valuation ≥ m so we see that the natural
homomorphism
O ,→ Op −→ Op /pm Op
m
has kernel p . Surjectivity follows thus: Choose a prime element π ∈ p, i.e., an
element with valuation 1. If now R is a system of representatives of O/p then every
element α ∈ Op has a (unique) representation of form:
α=
∞
X
ai · π i ,
i=0
where ai ∈ R ⊆ O. Then
P∞
i=m
ai · π
β := π m ·
i−m
∞
X
∈ Op and hence
ai · π i−m ∈ pm Op ;
i=m
thus α ≡
Pm−1
i=0
i
m
ai · π (p Op ). We have
Pm−1
i=0
ai · π i ∈ O.
Before we turn to Diophantine equations we will make some additional observations concerning p-adic convergence:
Proposition 1. (i). Let (αn ) be a sequence in Kp . Then (αn ) converges if and
only if the sequence (αn − αn+1 ) converges to 0.
(ii). Let (αn ) be a sequence in Kp such that the sequence (νp (αn )) is bounded from
below. Then (αn ) has a convergent subsequence.
1
2
IAN KIMING
Proof. (i). ‘Only if’ is clear as a convergent sequence is a Cauchy sequence. Suppose
conversely that (αn −αn+1 ) converges to 0. Let t ∈ R be given. There is then N ∈ N
such that
νp (αn − αn+1 ) ≥ t whenever n ≥ N .
If then m, n ≥ N , n > m, we obtain:
νp (αm − αn )
= νp ((αm − αm+1 ) + . . . + (αn−1 − αn ))
≥ min{νp (αm − αm+1 ), . . . , νp (αn−1 − αn )}
≥ t.
So, (αn ) is a Cauchy sequence and hence convergent.
(ii). Choose a prime element π ∈ p. Now, if M ∈ N is such that (νp (αn )) ≥ −M for
all n we have π M αn ∈ Op for all n. And if (π M αn ) has a convergent subsequence
then so does (αn ) (with limit π −M α if the subsequence of (π M αn ) has limit α).
Thus it suffices to show that any sequence (αn ) of elements in Op has a convergent
subsequence. So let αn ∈ Op be arbitrary.
Since Op /pOp ∼
= O/p is finite we see that the sequence (αn mod p) in Op /pOp
(1)
has a constant subsequence. Hence there is β1 ∈ Op and a subsequence (αn ) of
(αn ) such that
αn(1) ≡ β1 (p)
for all n.
Since Op /pm Op ∼
= O/pm is finite for any m ∈ N we can repeat the process and
(m)
obtain a sequence of elements β1 , β2 , . . . ∈ Op and subsequences (αn ) of (αn ) such
(m)
(m+1)
) is a subsequence of (αn ) for m ∈ N, such that
that (αn
βm+1 ≡ βm
(pm )
for all m ∈ N, and such that for all m ∈ N we have:
αn(m) ≡ βm
(pm )
for all n ∈ N .
Then
(n+1)
αn+1 − αn(n) ≡ βn+1 − βn ≡ 0
(n+1)
(pn )
(n)
(n)
so that the sequence (αn+1 − αn ) converges to 0. By (i) we then have that (αn )
(n)
converges. But (αn ) is a subsequence of the original sequence (αn ).
Let now F (x1 , . . . , xn ) be a polynomial in x1 , . . . , xn and coefficients in Op .
Theorem 1. (i). Let α1 , . . . , αn ∈ Op be arbitrary and let s ∈ N.
Then the equation F (x1 , . . . , xn ) = 0 has a solution x1 , . . . , xn ∈ Op satisfying
xi ≡ αi (ps ) for i = 1, . . . , n, if and only if for each m ∈ N the congruence
F (x1 , . . . , xn ) ≡ 0
(pm )
has a solution x1 , . . . , xn ∈ Op with xi ≡ αi (ps ) for i = 1, . . . , n.
(ii). The equation F (x1 , . . . , xn ) = 0 has a solution x1 , . . . , xn ∈ Op if and only if
for each m ∈ N the congruence
F (x1 , . . . , xn ) ≡ 0
has a solution x1 , . . . , xn ∈ Op .
(pm )
SOLUTION OF DIOPHANTINE EQUATIONS IN p-ADIC INTEGERS.
3
(iii). Suppose that F is a form, i.e., a homogeneous polynomial. Then the equation
F (x1 , . . . , xn ) = 0 has a non-trivial solution x1 , . . . , xn ∈ Op , i.e., a solution with
(x1 , . . . , xn ) 6= (0, . . . , 0) if and only if for each m ∈ N the congruence
(pm )
F (x1 , . . . , xn ) ≡ 0
has a solution x1 , . . . , xn ∈ Op with (x1 , . . . , xn ) 6≡ (0, . . . , 0) (p).
Proof. (i). The ‘only if’ part is clear (consider the reduction mod pm of a solution).
So assume that for each m ∈ N the congruence F (x1 , . . . , xn ) ≡ 0 (pm ) has
(m)
(m)
(m)
a solution (β1 , . . . , βn ) ∈ (Op )n with βi
≡ αi (ps ) for i = 1, . . . , n. Using
(m)
(m)
Proposition 1 (i) repeatedly n times we see that the sequence ((β1 , . . . , βn ))
(m)
(m)
(m)
has a subsequence ((γ1 , . . . , γn )) such that the sequence (γi ) is convergent
for each i = 1, . . . , n. Now, since for each i = 1, . . . , n and each m ∈ N the number
(m)
(m0 )
γi equals βi
for some m0 ≥ m, we have for each m ∈ N that
(m)
F (γ1
(m)
and γi
, . . . , γn(m) ) ≡ 0
(pm )
≡ αi (ps ) for each i = 1, . . . , n.
(m)
Put γi := limm γi for i = 1, . . . , n. Clearly, γi ≡ αi (ps ) for i = 1, . . . , n.
We claim that F (γ1 , . . . , γn ) = 0. To see this, let t ∈ N be arbitrary. Since
(m)
γi → γi for each i there is M ∈ N such that
(m)
νp (γi
− γi ) ≥ t
for all m ≥ M .
for each i. In particular,
(t+M )
γi
≡ γi (pt )
for each i. Since F is a polynomial with coefficients in Op this implies
(t+M )
F (γ1 , . . . , γn ) ≡ F (γ1
(t+M )
But we also have F (γ1
(t+M )
, . . . , γn
, . . . , γn(t+M ) )
(pt ) .
) ≡ 0 (pt+M ). We deduce
F (γ1 , . . . , γn ) ≡ 0
(pt ) ,
i.e., νp (F (γ1 , . . . , γn )) ≥ t. Since t ∈ N was arbitrary we deduce F (γ1 , . . . , γn ) = 0
as desired.
(ii), (iii). Again the ‘only if’ part is immediate under the assumptions of (ii).
Assume then case (iii) and that F (α1 , . . . , αn ) = 0 where αi ∈ Op with α1 , . . . , αn
not all 0. Then
r := min{νp (αi )}
i
is not ∞. Choosing a prime element π ∈ p we then have π −r αi ∈ Op for all i, and
π −r αi 6≡ 0 (p) for at least one i. But as F is homogeneous we also have:
F (π −r α1 , . . . , π −r αn ) = F (α1 , . . . , αn ) = 0 .
For the ‘if’ parts of both (ii) and (iii), note that the assumptions and the finiteness of Op /pOp ∼
= O/p together imply the existence of α1 , . . . , αn ∈ Op such that
for each m ∈ N the congruence
F (x1 , . . . , xn ) ≡ 0
(pm )
4
IAN KIMING
has a solution x1 , . . . , xn ∈ Op with xi ≡ αi (p) for i = 1, . . . , n. In case (iii)
we may additionally assume (α1 , . . . , αn ) 6≡ (0, . . . , 0) (p). Now apply part (i) for
s = 1.
The theorem tells us that we can deduce solvability of the equation
F (x1 , . . . , xn ) = 0
in Op from the solvability of the infinitely many necessary congruences
(∗)
F (x1 , . . . , xn ) ≡ 0
(pm )
where m runs through N.
Since Op /pm Op ∼
= O/pm is not only a finite ring, but can also be explicitly
described by a finite amount of computation — cf. the proof of Lemma 1 — it is
clear that solvability of an individual congruence (i.e., for a fixed m) can be decided
by a finite amount of computation.
Hence the natural question remains whether it is possible to ‘reduce’ the infinitely
many congruences (∗) to checking only finitely many of them. This is a delicate
question. We will be content with giving an example of such a ‘reduction’. The
following theorem is from the book [1].
Theorem 2. Let F (x1 , . . . , xn ) be a polynomial with coefficients in Op .
Suppose that there is s ∈ N, i ∈ {1, . . . , n}, and α1 , . . . , αn ∈ Op such that
νp (
∂F
(α1 , . . . , αn )) = s − 1
∂xi
and
F (α1 , . . . , αn ) ≡ 0 (p2s−1 ) .
Then F (x1 , . . . , xn ) = 0 has a solution x1 , . . . , xn ∈ Op with xj ≡ αj (ps ) for
j = 1, . . . , n.
Proof. We show by induction on m ≥ 0 that the congruence
(†)
F (x1 , . . . , xn ) ≡ 0
(p2s−1+m )
has a solution with xj ≡ αj (ps ) for j = 1, . . . , n. The statement then follows from
Theorem 1 (i).
For m = 0 we obviously take xj = αj , j = 1, . . . , n. For the induction step
m → m + 1 (m ≥ 0) let β1 , . . . , βn be a solution to (†) with βj ≡ αj (ps ). Consider
then the polynomial:
f (x) := F (β1 , . . . , βi−1 , βi + x, βi+1 , . . . , βn ) = a0 + a1 x + a2 x2 + . . .
where aj ∈ Op .
By assumptions we have now f (0) ≡ 0 (p2s+m−1 ) and νp (f 0 (0)) = s − 1 so that
we can write:
a0 = π 2s+m−1 · a , a1 = π s−1 · u
where a ∈ Op and u is a unit of Op ; here, π is of course a prime element of p.
We find then that
f (π s+m x) ≡ π 2s+m−1 (a + ux)
Hence, if we define γj := βj for j 6= i, and
γi := βi − au−1 π s+m
(p2s+2m ) .
SOLUTION OF DIOPHANTINE EQUATIONS IN p-ADIC INTEGERS.
5
then γj ≡ αj (ps ) for j = 1, . . . , n, and:
F (γ1 , . . . , γn ) ≡ 0
(p2s+m ) .
Example/Exercise 1: (For those with a little background in algebraic geometry).
Suppose that C is a projective curve over Q given by an equation
F (x1 , . . . , xn ) = 0
with a homogeneous polynomial F with coefficients in Z. As usual we may assume
that the gcd of the coefficients is 1. For any prime p we can then speak of the
reduction C̄p of C modulo p, i.e., C̄p is the projective curve given by F considered
as a (non-zero) homogeneous polynomial over Fp .
Now theorem 2 implies (take s = 1): If P̄ is a non-singular Fp -rational point of
C̄p then P̄ lifts to a Zp -rational point of C (i.e., if we consider C as a curve over
Qp then there is a rational point that reduces to P̄ modulo p).
Example/Exercise 2: Consider a quadratic form of shape
F (x1 , . . . , xn ) = a1 x21 + . . . + an x2n
where the ai are integers and (say) gcd(a1 , . . . , an ) = 1.
You will be able to formulate a criterion that can be checked via a finite amount
of computation for the existence of a non-trivial solution in p-adic integers (i.e., in
Zp ) to the equation:
(])
F (x1 , . . . , xn ) = 0 .
This is of interest for instance for the following reason: One can show that given
F (i.e., given the ai ) a finite set S of primes can be explicitly determined so that
(]) necessarily has a non-trivial solution in Zp whenever p 6∈ S. For each of the
finitely many primes p ∈ S the non-trivial solvability in Zp can be decided via
a finite amount of computation by the above indicated method. Hence there is
an algorithm that decides whether (]) has a non-trivial solution in Qp for every p
(note that — as F is homogeneous — non-trivial solvability in Qp is the same as
non-trivial solvability in Zp ). Now, by the famous Hasse-Minkowski theorem the
non-trivial solvability of (]) in ordinary rational numbers is equivalent to it being
non-trivially solvable in Qp for every p.
The conclusion is then that the question of non-trivial solvability of (]) in rational
numbers can be decided be a finite amount of computation.
References
[1] Z. I. Borevich, I. R. Shafarevich: ‘Number Theory’. Academic Press, 1966.
[2] H. Koch: ‘Number Theory. Algebraic Numbers and Functions’. Graduate Studies in Mathematics 24, AMS 2000.
Department of Mathematics, University of Copenhagen, Universitetsparken 5, DK2100 Copenhagen Ø, Denmark.
E-mail address: [email protected]