Download Report

Seminar on p-adic analysis, talk 4
Patrik Lengacher, Linda Raabe, Nicolas Wider
01.11.2011
Example
√
√
√
We want to compute 6 and 7 in Z5 . We start with 6. Let a := a0 +5a1 +52 a2 +...
∈
√
Z5 . Then a2 = (a0 + 5a1 + 52 a2 + ...)2 = a20 + 5(2a0 a1 ) + 52 (a11 + 2a0 a2 ) + .... If a = 6,
its square has to be equal to 6: a2 = a20 + 5(2a0 a1 ) + 52 (a11 + 2a0 a2 ) + ... = 6. We can
look at these equations modulo 5k and get:
a20 =
a20 + 5(2a0 a1 ) +
a20 + 5(2a0 a1 )
52 (a11 + 2a0 a2 )
6
mod 5
=
6
mod 52
=
6
mod 53
...
Then we get a0 = 1 or 4 mod 5. Let's choose 1 (4 would lead to another square root
of 6 in Z5 ). Then we get a1 = 3 and a2 =
√. We could go on and compute the innitely
many ai and get the 5-adic expansion of 6, but the principle should be clear so I stop
here.
Let's have a look at
√
7 in Z5 . With the same algorithm as above, we get:
a20 =
a20 + 5(2a0 a1 ) +
a20 + 5(2a0 a1 )
52 (a11 + 2a0 a2 )
7
mod 5
=
7
mod 52
=
7
mod 53
...
But the rst equation does not have√a solution as every square is either 1 or -1 modulo
5. So we don't get any solution and 7 ∈/ Z5 .
A general statement of this is Hensel's Lemma.
Theorem 1.39 (Hensel's Lemma)
Let F (x) = c0 + c1 x + ... + cn xn be a polynomial with coecients ci ∈ Zp . Let F 0 (x) =
c1 + 2c2 x + ... + ncn xn−1 be its derivative. Suppose a0 ∈ Zp satises F (a0 ) = 0 mod p
and F 0 (a0 ) 6= 0 mod p. Then there exists a unique p-adic integer a ∈ Zp such that
F (a) = 0 and a = a0 mod p.
1
Proof. We will inductively construct a p-adic integer ak = b0 + b1 p + ... + bk pk that
satises the conditions F (ak ) = 0 mod pk+1 and ak = a0 mod p. To satisfy the second
condition we choose b0 = a0 mod p. Then for any k ≥ 1, ak = a0 mod p. The rst
condition is satised as well: F (a0 ) = F (b0 ) = F (a0 ) = 0 mod p0+1=1
Assume we already found b0 , ..., bk−1 . Then we have to uniquely compute bk . We know
that ak = ak−1 + bk pk , so we have
F (ak ) = F (ak−1 + bk pk )
n
X
=
ci (ak−1 + bk pk )i
i=0
= c0 +
n
X
ci (ak−1 + bk pk )i
i=1
= c0 +
n
X
i−1
ci (aik−1 + iak−1
bk pk + somethingpk+1 )
i=1
= c0 +
n
X
i=1
ci aik−1 +
n
X
k
ici ai−1
k−1 bk p
mod p
i=1
= F (ak−1 ) + bk pk F 0 (ak−1 )
We want F (a0 ) to be zero mod pk+1 , so we have F (ak−1 ) + bk pk F 0 (ak−1 ) = 0 mod p.
)
Since F (ak−1 ) = 0 mod pk , we can divide by pk . That gives us bk F 0 (ak−1 ) = − F (apk−1
k
mod p. Since ak−1 = a0 mod p and F 0 (a0 ) 6= 0 mod p, also F 0 (ak−1 ) 6= 0 mod p and
we can divide by F 0 (ak−1 ).
F (a
)
⇒ bk = − F 0 (a k−1)pk mod p.
k−1
Now let a := limi→∞ aki for a convergent subsequence (aki )i of (ak )k . Then a is the root
of F we were looking for.
Theorem 1.42
A polynomial F with integer coecients has a root in Zp ⇔ it has an integer root
mod pk for any k ≥ 1.
Proof. ⇒: Let ak = b0 + ... + bk pk be the kth p-adic expansion of a. Then ak = a
mod pk+1 and hence we have F (ak ) = F (a) = 0 mod pk+1 for any k ≥ 1 as well.
⇐: Let {a1 , ..., ak , ...} be the roots of F mod pk+1 . By theorem 1.34 the sequence has a
convergent subsequence {aki }. Let a := limi→∞ aki . Since polynomials are continuous,
we have F (a) = F (limi→∞ aki ) = limi→∞ F (aki ). And since F (aki ) = 0 mod pki +1 for
any ki , F (a) must be zero.
Proposition 1.43
A rational integer a ∈ Z not divisible by p has a square root in Zp for p 6= 2 ⇔ it has a
square root mod p.
2
Proof. ⇒: Assume a has no square root mod p. By theorem ?? it cannot have a square
root in Zp .
⇐:Let a20 = a mod p the square root mod p. Then for P (x) = x2 −a we have P (a0 ) = 0
mod p and P 0 (a0 ) = 2a0 6= 0 mod p. So by Hensel's Lemma a has a square root in Zp
(the root of P ).
As we have seen so far the p-adic integers Zp are quite dierent from the ordinary integers
Z. But nevertheless the algebraic properties are similar in each cases. Before we start
we recall the set of p-adic integers:
Zp =
(∞
X
)
i
ai p |ai ∈ {0, . . . , p − 1}
i=0
Now we give an alternative denition for the order of a p-adic integer:
Denition
For a p-adic integer a = i≥0 ai pi with a 6= 0 we dene the order of a p-adic integer as
the rst index ν = ν (a) such that aν 6= 0. We denote this by ordp (a) = ν = ν(a).
P
Of course the order ordp (a) dened in this way is the same as the largest power of p
such that pordp (a) divides a. Therefore this denition is equivalent to the one given in
the previous talks. The order is a mapping from Zp \{0} to N. We normally extend this
denition by the convention ν(0) = ∞.
Remark
The terminology of the order comes from a formal analogy to the ring of holomorphic
functions. If f (z) ia nonzero holomorphic function of a complex variable z ∈ C then we
can write the Taylor series of f in the neighborhood of a point a ∈ C in the following
way:
X
an (z − an )n ,
f (z) =
(am 6= 0, kz − ak ≤ )
n≥m
We also call the index m of the rst nonzero coecient the order (of vanishing) of f at
a.
Now we will specify the algebraic structure of the p-adic integers.
Proposition 1.44
The ring Zp of p-adic integers is an integral domain.
Proof. We already know that Zp is a commutative ring and not equal to {0}. It remains to
show that it contains no zero divisor. If we recall the Qp itself is a eld hence contains no
zero devisors we can directly follow that there are also no zero divisors in Zp . Otherwise
we continue as following. Consider two p-adic integers:
a=
X
ai pi 6= 0,
b=
i≥0
X
i≥0
3
bi pi 6= 0
Dene ν = ν(a) = ordp (a) and ω = ν(b) = ordp (b) to be the order of a respectively b.
So aν and bω are the rst nonzero coecients of a and b with 0 < aν , bω < p. Then the
product of a and b looks like the following:
a · b = aν bω pν+ω + (aν+1 bω + aν bω+1 )pν+ω+1 + . . .
Hence by the denition of multiplication the rst nonzero coecient of the product is
cν+ω .
cν+ω ≡ aν bω
0 < cν+ω < p,
mod p
This implies that ab can not be equal to zero unless a or b is equal to zero, therefore Zp
contains no zero divisors.
The calculation in the preceding proof gives rise to a Corollary.
Corollary
The order ν : Zp → N ∪ {∞} satises
ν(ab) = ν(a) + ν(b),
ν(a + b) ≥ min(ν(a), ν(b))
for a, b and a + b unequal to zero.
Proof. The statement follows directly by investigation of the product expansion as done
in the previous proposition.
Next we look at the nite eld with p elements Fp = Z/pZ. We can construct a ring
homomorpism : Zp → Fp called the reduction mod p as follows:
a=
X
ai pi 7→ a0
mod p
i≥0
Obviously this mapping is surjective and the kernel is
{a ∈ Zp |a0 = 0} =

X

i≥1




 X

i
j
= pZp
ai p = p
aj+1 p
 

j≥0
where pZp is a subgroup of index p in Zp . Because the quotient Zp /pZp through is
isomorphic to the eld Fp , pZp is a maximal ideal.
Now we formulate a proposition about the invertible elements Z×
p in the ring of p-adic
integers.
Proposition
The group Z×
p in Zp consists of p-adic integers of order zero, namely
Z×
p =

X

i≥0


ai pi |a0 6= 0

4
Proof. First we remark that if a p-adic integer a is invertible then so must be its reduction
(a) in Fp . Therefore we have a rst inclusion



X
i
a
p
|a
=
6
0
Z×
⊂
i
0
p


i≥0
Conversely we need to show that a p-adic integer a with order ν(a) = 0 is invertible.
Then the P
reduction (a) ∈ Fp is not zero and therefore invertible in this eld. So for
given a = i≥0 ai pi we choose 0 < b0 < p such that a0 b0 ≡ 1 mod p. Then we can write
a0 b0 = 1 + kp. From a = a0 + pα we can follow
a · b0 = 1 + kp + pαb0 = 1 + tp
for some p-adic integer t. If the p-adic integer 1 + tp is invertible then a is invertible
because we can write
a · b0 (1 + tp)−1 = 1,
a−1 = b0 (1 + tp)−1
Hence it is enough to look at the case where a0 = 1 and a = 1 + tp. The p-adic expansion
gives us
(1 + tp)−1 = 1 − tp + (tp)2 − · · · = 1 + c1 p + c2 p2 + · · ·
with ci ∈ {0, 1, . . . , p − 1}
The second equality is a reformulation which can be achieved by applying the rules of
multiplication for p-adic integers. Therefore (1 + tp) is invertible and so is a. Hence we
get the desired result.
From this we can follow a corollary.
Corollary 1.45
The ring Zp has a unique maximal ideal, namely,
pZp = Zp \Z×
p
Proof. It only remains to proof that it is a unique maximal ideal. Therefore suppose I
is a maximal ideal of Zp . Because pZp is also a maximal ideal, the ideal I has to contain
−1 ∈ I and
an element from its complement, say a ∈ Z×
p . Because I is an ideal 1 = a · a
therefore I = Zp .
In Z the maximal ideals are mZ for m prime. The previous corollary also corresponds
to a partition Zp = Z×
p t pZp . In fact we even have
Zp \{0} =
G
pk Z×
p
k≥0
We can write down two additional Corollaries which easily follow.
5
Corollary 1.46
Every nonzero p-adic integer a ∈ Zp has canonical representation a = pν u, where ν =
ν(a) is the p-adic order of a and u ∈ Z×
p is a p-adic unit.
Proof. Consider the p-adic integer a = i≥ν ai pi with ν = ν(a) the order of a. Then we
can write
X
X
X
P
j≥0
i≥ν
i≥ν
bj pj = pν u
ai pi−ν = pν
ai pi = pν
a=
with bi = ai+ν and u ∈ Z×
p is a p-adic unit.
Corollary 1.47
The rational integer a ∈ Z that are invertible in Zp are the integers coprime to p. The
quotients of integers m/n ∈ Q with n 6= 0 that are p-adic integers are those that have a
denominator n coprime to p.
Proof. First we make the following observation. For any p-adic integer a =
p·
X
ai p i =
i≥0
X
ai pi+1 =
i≥0
X
P
i≥0
and p.
bj pj = v
j≥1
with bi = ai−1 . And v will never going to be a unit or even 1. Therefore p is not
invertible in Zp . Of course the same holds for all power or multiple of p. Now by the
previous corollary we have Z×
p = Zp \pZp and therefore exactly the integers coprime to p
are invertible. Each integer m ∈ Z is also a p-adic integer. Now by the rst part exactly
the n coprime to p are invertible in Zp . And therefore m · n1 = mn is a p-adic integer if
and only if n is coprime to p.
This corollaries showed us that Zp in comparison to Z even contains multiplicative inverses for some elements. But because not all elements are invertible it is still not a eld.
Let us now look at the principal ideals of the ring Zp .
pk = pk Zp = {a ∈ Zp |ordp (a) ≥ k}
The intersection is equal to {0}:
Zp ⊃ pZp ⊃ · · · pk Zp ⊃ · · · ⊃
\
pk Zp = {0}
k≥0
Indeed, for any a 6= 0 with order ν(a) = k we get a ∈/ pk+1 . Even more we can state
the following proposition.
Proposition 1.46
The ring Zp is a principal ideal domain. More precisely, its ideals are the principal ideals
{0} and pk Zp , ∀k ∈ N.
6
Proof. We start with a nonzero ideal I of Zp . Furthermore let 0 6= a ∈ I be an element
of minimal order k = ν(a) < ∞. (This is possible because the order only takes a discrete
k
−1
set of values). We can write a = pk u for some unit u ∈ Z×
p . Hence p = u a ∈ I and
pk = pk Zp ⊂ I . This gives the rst inclusion. Conversely, for any b ∈ I , with order
ω = ν(b) ≥ k we can write
b = pω v = pk · pω−k v ∈ pk Zp
k
for v ∈ Z×
p an appropriate unit. Hence I ⊂ p Z. Therefore for any ideal I of Zp we get
k
I = p Zp for some k ∈ N.
Finally Zp gives rise to the same algebraic structure as Z which is itself also an principal
ideal domain
Proposition 1.47
For p ∈ N prime and m ∈ N coprime to p. There exists a primitive mth root of unity in
Qp if and only if m|p − 1.Moreover if m|(p − 1) every mth root of unity is also a (p − 1)
root of unity. The set of (p−1)th roots of unity is a cyclic subgroup of Z×
p of order (p−1).
Proof. ⇐ Let α be a mth root and k ∈ N. It holds that
αm ≡ αkm ≡ 1 and αp−1 ≡ 1
mod p
and therefore p − 1 = km and m|p − 1.
⇒ Let m|(p − 1), hence p − 1 = km for k ≥ 1. Therefore any mth of unity is also a
(p − 1)th root of unity. Let f (x) = xp−1 − 1 and f 0 (x) = (p − 1)xp−2 . Take x0 ∈ Z×
p
to be any rational integer satisfying 1 ≤ x0 ≤ p − 1. Then f (x0 ) ≡ 0 mod p and
f 0 (x0 ) 6= 0 mod p since |f 0 (x0 )|p = 1. We apply Hensel's Lemma, therefore there
are exactly p − 1 solutions which are (p − 1) roots of unity. The rst digits of these
roots are 1, . . . , p − 1. Let α ∈ Qp is an mth root of unity. Therefore, since αm = 1,
we must have |α|p = 1. i.e. α ∈ Zp . If α0 is it's rst digit then α ≡ 1 mod p
hence m divides p − 1, the order of (Z/pZ)× . Since a polynomial with coecients
in a eld can only have as many roots as it's degree, the polynomial xp−1 cannot
have more than p − 1 roots and these roots must be all the roots of unity in Q. It
is clear that the roots form a group under multiplication. Finally since any nite
subgroup of the multiplicative group of any eld is cyclic, the group of (p − 1)th
roots is a cyclic subgroup of Z×
p of order (p − 1)
Remark
The pth root cannot be handled by means of Hensel's Lemma, since if f (x) = xp − 1 then
f 0 (x) = pxp−1 − 1 which is 0 for every x.
Remark
Let us show that the elds Qp and Qq , for q 6= p prime, are not isomorphic.
7
Proof. Let us choose an m ∈ N such that m|p − 1 and m - q − 1. By the proposition ??
there is a mth root of unity in Qp . If there would exist such an isomorphism it would
send the mth roots of unity form one eld into the other. Since there are no mth roots
of unitiy in Qq , Qq and Qp cannot be isomorphic.
Remark
When p is odd, p − 1 is even and −1 belongs the a cyclic subgroup of order p − 1. The
number −1 will have a square root in Qp precisley when (p − 1)/2 is still even, namley
when p ≡ 1 mod 4. We have that
√
−1 ∈ Qp
⇔
4|p − 1
⇔
p≡1
mod 4
√
A number i = −1 can thus be found in Q5 , Q13 , ...
The (p − 1)th roots of unity are related to the signumfunction sgnp (x) introduced in the
next theorem.
Theorem 1.48
For any x ∈ Zp the limit lim xp exists. This limit is denoted by sgnp (x)and has the
n→∞
following properties:
n
1. sgnp (x) depends only on the rst digit x0 in the canonical p-adic expansion of x.
2. sgnp (xy) = sgnp (x) · sgnp (y).
3. sgnp (x) = 0 if x0 = 0, and it is a (p − 1)th root of unity if x 6= 0.
n
Proof. Let x0 ∈ {1, 2, . . . , p − 1}. First we show that the sequence {xp0 } converges. By
Euler's Theorem,
ϕ(pn )
{x0
mod pn ,
}≡1
where ϕ us Euler's ϕ-function: for a positive integer m, ϕ(m) is equal to the number
of integers smaller than m and coprime to m. Observe that since p is a prime, we have
ϕ(pn ) = pn − pn−1 . Thus,
xp0
and hence
n −pn−1
≡1
n
xp0 ≡ x0p
mod pn ,
n
|xp0 − x0p
n−1
|p ≤
n−1
mod pn ,
1
.
pn
n
Since 1/pn → 0 as n → ∞, the sequence {xp0 } is Cauchy, and by the completeness of
Zp , it converges to a limit in Zp , which we denote by
sgnp (x0 ) = lim xp0
n
n→∞
The limit obviously exists for x0 = 0, so sgnp (x) is dened for x0 ∈ {1, 2, . . . , p − 1}, and
sgnp (0) = 0.
Next we show that the limit exists for all x ∈ Zp and is dened by the rst digit x0 of x.
For this we will need the following lemma.
8
Lemma 1.49
Suppose x ∈ Zp with the rst digit x0 . Then we have |xp − xp0 |p ≤ p−1 |x − x0 |p .
Proof. Let x = x0 + α, with |α|p ≤ p−1
- Then xp − xp0 = (x − x0 ) · f (x, α), where f (x, α)
is a polynomial. Since every factor | pj xp−j αj−1 |p ≤ p−1 for j ≥ 1, by the strong triangle
inequality we obtain |xp − xp0 |p ≤ p−1 |x − x0 |p .
Applying the lemma, we obtain
n
n
|xp − xop |p ≤ p−n |x − xo |p
which implies that lim xp exists and is equal to the limit of x0 . Thus we have dened
n→∞
sgnp (x0 ) for all x ∈ Zp , and property (??) of the proposition is satised. Property (??)
follows form the multiplicative property of limits.
It remains to show that if x0 ∈ {1, 2, . . . , p − 1}, then sgnp (x0 ) is a (p − 1)th root of unity.
Using property (??) and Fermat's Little theorem, we obtain
n
p−1
sgnp−1
p (x0 ) = sgnp (x0 ) = sgnp (1) = 1.
The values of sgnp (x0 ) are thus solutions of the equation y p − y = 0. Since Qp is a eld,
this equation cannot have more than p solutions in Qp , and hence in Zp . Consequently,
the only solutions of this equation are the values o the signum function.
Suppose we want to nd a root of a polynomial in Q. Evidently if there are roots in Q,
then there are roots in R and in all Qp . Hence we can certainly conclude that there are
no rational roots if there is some p ≤ ∞ for which there are no p-adic roots. A converse
statement would be more interesting, but is it true? If polynomial has p-adic roots for
all p including ∞, does it follow that is has a rational root? Here is a simple example
when such a converse statement holds.
Proposition 1.51
A number x ∈ Q is a square if and only if it is a square in every Qp , p ≤ ∞.
Proof. For any x ∈ Q× we have
x=±
Y
pordp (x) .
p<∞
Notice that x is a square in R if and only if it is positive. In Qp we can write x = pordp (x) u,
where u ∈ Zp . Then x is a square in Qp if and only if ordp (x) is even and u = v 2 for
some unit v ∈ Z×
p . If we write out the factorization, we see that x is a square in Q if and
only if it is a square in each Qp .
This manifestation of the so-called Local-to-Global Principle, which asserts that the existence or nonexistence of solutions in Q (global solutions) of a Diophantine equation can
be detected by studying, for each p ≤ ∞, the solutions in Qp (local solutions). Unfortunately, this principle is not universal, but it holds in some important cases, for instance
for quadratic forms in severals variables.
9
Example
√
Let us show that 2 is irrational. Consider the polynomial x2 − 2 = 0. The goal is to
show that it has no sloution
√ in Q5 . With the local-to-global principle it follows that it
has no solution in Q i.e. 2 is irrational. The solution is a 5-adic integer i.e. |2|5 = 1.
Let α be a solution and
|α2 |5 = |α|5 |α|5 = 1.
Therefore |α|5 = 1 and it has to hold that α2 ≡ 2 mod 5k for k > 0. Now let a0 to be
the rst coecient of the 5-adic expansion. We need that
a20 ≡ 2
mod 5
But 2 is not a quadratic residue mod 5 and hence there is no solution in Q5 .
10