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Valuations
Dave Mendes da Costa
November 18, 2009
1 Valuations on number fields
Everyone knows about the absolute value |x|∞ of a rational number x ∈ Q. This is a very useful notion, giving us a measure of the size of a number, a metric and associated topology on Q and ulimately a way of completing Q to form R. And we all like R.
What is perhaps less well known is that the abolute value, |·|∞, is not the only way to proceed. To see this we first need to define what we want a valuation to do:
Defintion : A valution (or norm) on a ring k is a funtion |·| : k → R such that: (1) |a| ≥ 0 ∀x ∈ k |x| = 0 ⇔ x = 0 (2) |ab| = |a| |b| ∀x, y ∈ k (3) ∃C ∈ R such that |a| ≤ 1 ⇒ |a + 1| ≤ C
Examples :
- The trivial valuation, |x| 0 =
1 if x 6 = 0 0 if x = 0
. If k is finite then this is the only valuation on k. In
this case we may take C = 1.
- The absolute value, |·|∞. Here we have C = 2.
- If |·| 1 is a valuation on k and λ > 0 then |·| 2 := |·|λ 1 is also a valuation on k. In this case we say that |·| 1 and |·| 2 are equivalent. Equivalence is, of course, an equivalence relation and the equivalence classes are known as places.
This definition is due to Artin. It is sometimes formulated with (3) replaced by the more familiar Tri- angle Inequality:
(3’) |a + b| ≤ |a| + |b| ∀a, b ∈ k.
It may be shown that (3’) holds if and only if (3) holds with C = 2. Therefore every valution is equivalent to one satisfying the triangle inequality. In the case that (3) holds with C = 1 we get the stronger condition:
|a + b| ≤max{|a| , |b|} (Ultrametric Inequality)
If |a| 6 = |b| then we get an equality. We call valuations with C = 1 non-achimedian and the others are called archimedian.
So far the only example of a non-arch valuation on Q is |·| 0 , but we can do better than that. Let p be a prime number in Z and a = peb ∈ Q where b = mn with gcd(m, n, p) = 1. Putting |a|p = (^) p^1 e and setting | 0 |p = 0 yields a non-arch valuation on Q called the p-adic valuation.
These are essentially the only non-trivial non-arch valuations of Q:
Theorem(Ostrowski): Every non-trivial valuation on Q is equivalent to either |·|p for some prime p or |·|∞.
We may generalise the case of Q to a number field k. The arch valuations come from the real em- beddings and the pairs of complex embeddings with the usual valuations on R and C respectively. For the non-arch (non trivial) valuations we proceed as follows: Let P be a prime ideal of k and let x ∈ k∗. Then by the theorem on uniqueness of factorisation of ideals we get that the fraction ideal (x) may be expressed as (x) = Pe^
Pe ii
where the product is over the remaining prime ideals of k. We may then form the P-adic valuation by putting |x|P = (^) N (^1 P)e and | 0 |P = 0 where N (P) is the norm of P.
An analogue of Ostrowski’s Theorem holds for number fields. We denote the places of k by Mk, the arch places by M (^) k∞ and the non-arch places by M (^) k^0. It is worth noting that since non-arch (resp. arch) places come from prime ideals they are referred to as prime (resp. infinite) places.
Finally, since Q ⊂ k we get a valuation on Q from each valuation on k by restriction. It is easy to see that the arch places on k descend to the arch place on Q and that if P is a prime ideal lieing over the rational prime p then the P-adic valuation on k restricts to a valuation equivalent to the p-adic valuation on Q.
2 Completions
Recall that a complete metric space is one where every Cauchy sequence converges. We all know that Q is not complete with respect to |·|∞ and nor is it with repect to any of the p-adic valuations. How- ever, every metric space has a completion (a complete metric space containing k as a dense subspace) which is essentially the space of equivalence classes of Cauchy sequences under the equivalence relation (an) ≡ (bn) ⇔ |an − bn| → 0 as n → ∞.
The completion of Q with repect to |·|∞ is R and with respect to |·|p the completion is denoted Qp and is called the space of p-adic numbers. Note that we usualy write R as Q∞ and refer to ∞ as the infinite prime.
We can also complete a number field k with respect a valuation ν, and we denote the completion by kν.
Theorem: Let ν ∈ M (^) k∞ , then kν is isomorphic to either R or C.
3 The structure of (discrete) complete non-arch fields
In this section k will denote an arbitary complete non-arch field with vaulation |·|.
The p-adic numbers were invented/discovered by Hensel in order to bring the might of power series to bear in number theory. He defined them as formal Laurent series in p and later the valuation ap- proach which we are using here developed. We shall now recover the Hensellian picture.
Lemma: Let (an) ⊂ k be a sequence. Then
n=0 an^ converges if and only if^ an^ →^ 0 as^ n^ → ∞.
Definition: Let R = {x ∈ k : |x| ≤ 1 } and I = {x ∈ R : |x| < 1 }. We call R the ring of integers of k.
Since I is a maximal ideal in R we can define the Residue Class Field R/I.
Consider the sequence (inverse system) of abelian groups:
ϕn+ → Z/pn+1Z ϕn → Z/pnZ ϕn− 1 → · · · ϕ 0 → Z/pZ
with ϕn : Z/pn+1Z → Z/pn+1Z being the homomorphism which takes α ∈ Z/pn+1Z to its residue mod pn.
We define the inverse limit of this sequence to be the set of all sequences (... , a 2 , a 1 , a 0 ) with an ∈ Z/pn+1Z and ϕn(an) = an− 1 and denote it by lim←−−Z/pnZ.
From our previous comments we see that Zp ∼= lim←−−Z/pnZ. The algebraic construction takes the in- verse limit as the definition of Zp and defines Qp to be its field of fractions.
5 Hasse’s Principle
The p-adic numbers have a key application to the study of Diophantine equations. This is due to the fact that Qp contains a copy of Q and so a polynomial f ∈ Q[X] can be seen as a polynomial in Qp[X]. If there are no solutions in Qp, then there cannot be any solutions in Q. Handy. An obvious question is whether we can go the other way, that is does the existence of a solution to a Diophantine equation in every completion of Q (including R) imply a solution in Q itself? Such equations satisfy a local-global principle, or Hasse’s Principle. A useful tool in the study of this problem is the following:
Theorem (Hensel’s Lemma): Let f ∈ Z[X], 2 ≤ k ∈ Z, p a rational prime. Suppose that we have r ∈ Z such that f (r) ≡ 0(mod pk−^1 ). Then if f ′(r) 6 ≡ 0(mod p) we have that there exists r such that f (r) ≡ 0(mod pk) and with r ≡ r(modpk−^1 ).
The importance of this theorem to us is that it allows us to lift solutions to equations in finite fields to solutions in Zp. Indeed, if we look at the inverse limit definition of Zp we see that a solution in Zp corresponds to a solution in every ring Z/pnZ.
Hasse’s Principle holds for quadratic forms (Hasse-Minkowski Theorem) but fails in general for poly- nomials of higher degree. A famous example is
3 x^3 + 4y^3 + 5z^3 = 0
which has a non-trivial solution in R and Qp for every prime p but has no non-trivial solution in Q.
6 Adeles`
We have seen that we can glue together the rings Z/pnZ to form Zp. This is a dandy thing to do since we can talk about all of the rings at once in a meaningful way. But why stop there?
We define the (ring of ) Ad`eles of a number field k to be Ak =
ν kν^ where^
means that for each adle a = (aν )ν ∈ Ak all but finitely many of the entries of aν for ν ∈ M (^) k^0 are integers (this is a restricted topological product). This restriction is to ensure that the adeles form a locally compact topological group which is a key property if we want to do abstract Fourier analysis on the adeles (`a la Tate).
So why do we like the ad`eles? Well, an immediate reason is that since we can embed k diagonally into Ak we can study k in all of its completions simultaneously and thus without taking any of them to be of special importance over the others. This socialist view is necessary in order to gain a unified sense algebraic number theory.
Since we can view k as lieing inside Ak and since k is dense in each of its completions (by defini- tion) it seems natural to ask about the density of k in Ak. The two key results in this area are Weak
and Strong Approximation:
Theorem (Weak Approximation): Let |·|j for 1 ≤ j ≤ n be pairwise inequivalent valuations on k. Let a 1... , an be arbitrary elements in k and let > 0. Then there exists an x ∈ k such that |x − ai|i < for 1 ≤ i ≤ n.
Weak Approximation means that if we restrict Ak to finitely many places then k is dense in the re- striction. This may be alternatively stated as: k is dense is
ν kν^ with the (unrestricted) product topology.
This seems to be strong evidence that k is dense in Ak however this is not the case. Let us look at the case k = Q. Consider (ap)p = (1, 2 , 3 , 5 , 7 , 11 , 13 ,.. .) ∈ AQ. Now we cannot have b ∈ Q is such that |b − ap|p < 1 for every p ∈ MQ since |b − ap|p = max{|ap|p , |b|p} = 1 for almost all p. A similar argument shows in fact that Q is discrete in AQ. The best we can get is:
Theorem (Strong Approximation): Let P be a finite set of non-arch places on a number field k. Let > 0 and aν ∈ kν for each ν ∈ P be arbitrarily chosen. Then there is an x ∈ k such that |x − aν |ν < for each ν ∈ P and |x|ν ≤ 1 for each non-arch place ν /∈ P. Further, if ai ∈ Oν for each ν ∈ P then x ∈ Ok.
Corollary (Alternate Statement of S.A.): Let Ak,ν 0 be Ak restricted to all places except ν 0. Then k is dense in Ak,ν 0.
So we have weak and strong approximation for number fields k. The real meat of these notions comes though when we look at varieties defined over k. Our discussion so far can be interpretted as looking at
the affine line X(k) = {(x, y) ∈ k
2 : y = 0} and the question of the density of X(k) in X(Ak). In this case we get both weak and strong approximation.
Weak approximation and the Hasse principle are related: if X is a variety defined over k and X(kν ) 6 = ∅ for all ν ∈ Mk then the density of X(k) in X(
ν kν^ ) (where^
ν kν^ has the unrestricted product topol- ogy) implies that X(k) 6 = ∅ since the empty set is not dense in a non-empty set. The failure of the Hasse Principle on some varieties imples that weak approximation is not a trivial concept.
7 Extending valuations
It is clear that if L/k is an extension of number fields then every valuation on L restricts to a valuation on k, but given a valuation on k how far can we extend it to a valuation on L? It is obvious how the arch valuations extend so let us concentrate on the non-arch ones.
Proposition: Let L/k be a finite field extension on degree n and suppose theat k is equipped with a non-arch valuation |·| 1 with respect to which k is complete. Then |·| 1 extends to a unique valuation |·| 2 on L given by
|x| 2 =
NL/k(x)
∣∣ 1 n 1
This proposition clearly fails when k is not complete since if k = Q then L = Q(i) is a finite extension of Q but we have (2) = (1 + i)(1 − i) in Z[i] and thus both the (1 + i) − adic and (1 − i) − adic valuation extend the 2 − adic valuation.
Let us suppose that k is equipped with a non-arch valuation |·| which does not make it complete. Denote the completion of k by ̂k. Then since L and ̂k both contain k we may consider L ⊗k ̂k. Now there is an a ∈ L such that L = k(a) and this has minimum polynomial f (x) ∈ k[x]. Thus we have L = (^) (fk ([xx])).
This polynomial may well become irreducible over ̂k so that f = g 1... gr with gi ∈ ̂k[x]. In which case we have that
