TANT 4 - Complete fields

Hello there! These are notes for the fourth class of the course "Topics in algebra and number theory" held in Block 4 of the academic year 2017/18 at the University of Copenhagen.

In the previous lecture we have struggled with non-Archimedean places of a number field K and of a function field F(t). In particular we have proved that for every number field K the maps ΣKSpec(OK){0}[ϕ]{xOKϕ(x)<1}andSpec(OK){0}ΣKp[||p] are one the inverse of the other. Now we want to prove a similar statement for the Archimedean places ΣK,. Recall that the Archimedean world is the world of R and C, i.e. it is the world of the analysis that we are accustomed to.

For this reason to understand Archimedean places we need to start building an analytic theory of valued fields. In the second lecture we have defined absolute values over a field and we have seen that they carry with them a canonical topology. Now we want to define a category of valued fields.

 Definition 1  The category of valued fields is the category whose objects are couples (K,ϕ), where K is a field and ϕ:KR0 is an absolute value. A morphism (K,ϕ)(L,ψ) in this category is given by any homomorphism of fields f:KL such that ψ(f(x))=ϕ(x) for all xK.

Now it is natural to look for special objects in this category, e.g. for objects with universal properties. In the following section we will define such an object, the completion of a valued field (K,ϕ), whose definition is related to the analytic properties of ϕ.


Analytic properties of absolute values

Since we are interested in the analytic theory of valued fields the first place to start from will be the concept of limit. The definition looks familiar, at least to anyone who was born after Weierstrass...

 Definition 2  Let (K,ϕ) be a valued field. We say that a sequence {xn}nNK converges to xK (with respect to ϕ) if for every εR>0 there exists nεN such that for all nnε we have that ϕ(xnx)<ε.

 Exercise 3  Prove that if {xn}nNK converges to xK and to xK then x=x.

In view of the previous exercise we will say that a sequence {xn} converges in K if it converges to some (and so, to one) xK. We call x the limit of the sequence {xn}.

 Exercise 4  Prove that convergence respects sums and products. More precisely, prove that if {xn}nNK converges to xK and {yn}nNK converges to yK then {xn+yn}nN converges to x+y and {xnyn}nN converges to xy.

 Exercise 5  Prove that every morphism of valued fields f:(K,ϕ)(L,ψ) is sequentially continuous. More precisely prove that if {xn}K converges to xK then {f(xn)}L converges to f(x).

As we all learned some time ago, if a sequence {xn}nN converges then the xn's get "closer and closer" as n+, i.e. the sequence {xn}nN is a Cauchy sequence.

 Definition 6  Let (K,ϕ) be a valued field. We say that a sequence {xn}nNK is a Cauchy sequence if for every εR>0 there exists nεN such that for all n,mnε we have that ϕ(xnxm)<ε.

 Exercise 7  Prove that Cauchy sequences respect sums and products. More precisely, prove that if {xn}nNK and {yn}nNK are Cauchy sequences then {xn+yn}nN  and {xnyn}nN are also Cauchy sequences.

Testing whether a sequence is Cauchy or not can be hard but it is doable in principle, whereas testing if a sequence has a limit or not can be really tough. For example we know that the sequences of rational numbers xn:=nk=0(1)klog2(k+2)k+2andyn:=nk=0(1)k(2k+1)2 are Cauchy sequences with respect to the canonical Archimedean absolute value ||:QR>0 but no one knows if these sequences converge in Q or not (see here and here).
The problem here is that (Q,||) is a valued field "with many holes" and so not all the Cauchy sequences converge. For example for every αQ>0 we can define a sequence {xn}nN by setting x0:=1andxn+1:=12(xn+αxn). It is not difficult to prove that this sequence is always Cauchy for ||. However, it converges in Q if and only if α is a square (so, for example, it does not converge if α=2).

In order to avoid this sort of problems it would be nice if every Cauchy sequence was convergent. So nice that this property deserves a name.

 Definition 8  A complete field is a valued field (K,ϕ) such that any Cauchy sequence {xn}nNK converges.

 Exercise 9  Prove that the field of real numbers (R,||) is complete with respect to the canonical absolute value defined in Example 9 of the second lecture. Use in particular the definition of R as the set of sequences R:={εanan1a0.a1a2amε{±1}, aj{0,,9}}S9 where S9 is the set of sequences which are definitely equal to 9.

There is another classical Archimedean complete field: the field of complex numbers C with the canonical absolute value. To prove that it is complete starting from the completeness of R we will need the following lemma, which applies also to non-Archimedean absolute values.

 Lemma 10  Let (K,ϕ) be a complete valued field, and suppose that x2+1K[x] is irreducible (i.e. that 1 is not a square in K). Then L:=K[x]/(x2+1)=K(i) is a quadratic field extension of K. L is also complete with respect to the valuation ψ:LR0 defined by ψ(a+ib):=ϕ(a2+b2), and the inclusion KL is a morphism of valued fields.

 Proof  The fact that L is a quadratic field extension of K is part of any course in Galois theory.

It is now trivial to observe that ψ satisfies the first two axioms of the definition of an absolute value, and that ϕ=ψι, where ι:KL. Showing that ψ satisfies the triangle inequality is rather difficult, and we will prove it for a general Galois extension KL in one of the following lectures. If you can't wait you can look at Lemma 2.2 of these notes by Peter Stevenhagen.

Let now {an+ibn}nNL be a sequence such that {an} and {bn} tend to zero with respect to ϕ. Then it is clear from the definition of ψ that {an+ibn} tends to zero with respect to ψ.
Vice versa, let {an+ibn}nNL be a sequence which tends to zero with respect to ψ, and suppose by contradiction that {xn}nN does not tend to zero with respect to ϕ. Then there would exist a number CR>0 such that ϕ(xn)C for all nN and thus the sequence {1+(yn/xn)2}nN would go to zero with respect to ϕ, because ϕ(1+(yn/xn)2)=ϕ(xn)2ϕ(xn+yn).
It is now not difficult to see that this implies that the sequence zn:=yn/xn is a Cauchy sequence. Since K is complete with respect to ϕ this sequence converges to an element zK. Using Exercise 4 we see then that z2=1. But this contradicts the fact that x2+1K[x] is irreducible, and so it follows that {xn}nN actually tends to zero with respect to ϕ. We can make a similar argument to show that yn also tends to zero with respect to ϕ.

Using again Exercise 4 we see now that a sequence {xn+iyn}nNL converges to x+iy with respect to ψ if and only if {xn} converges to x and {yn} converges to y with respect to ϕ. This implies in particular that L is complete with respect to ψ, as we wanted to show. Q.E.D.

Completions

We have just seen how important complete fields are. In particular we can restrict the category of all valued fields to the category of complete valued fields, and we have a forgetful functor (i.e. an inclusion) {complete valued fields}{valued fields}.
A question that now arises automatically is: how many complete fields do exist? The answer is... quite many! Indeed, the following theorem associates to every valued field (K,ϕ) a complete field (KΦ,Φ) which is the complete field "closest" to (K,ϕ).

 Definition 11  Let (K,ϕ) be any valued field. We say that a complete valued field (L,ψ) is a completion of (K,ϕ) if there is a morphism of valued fields ι:(K,ϕ)(L,ψ) such that for every complete valued field (M,ξ) and every morphism of valued fields f:(K,ϕ)(M,ξ) there exists a unique morphism of valued fields f:(L,ψ)(M,ξ) such that f=fι.

 Exercise 12  Prove that if (L1,ψ1) and (L2,ψ2) are two completions of the same valued field (K,ϕ) then there exists a unique isomorphism of valued fields (L1,ψ1)(L2,ψ2).

 Theorem 13  Let (K,ϕ) be a valued field. Then there exists a completion (L,ψ) of (K,ϕ).

 Proof  Define R to be the set of all Cauchy sequences {xn}K. If we define {xn}+{yn}:={xn+yn}and{xn}{yn}:={xnyn} then it is clear from Exercise 7 that R is a ring with respect to these "coordinate-wise" sum and product operations. Observe moreover that the subset m:={{xn}Rlimn+ϕ(xn)=0}R is an ideal, because clearly if {xn}m and {yn}R then ϕ(xnyn)=ϕ(xn)ϕ(yn)0 as n+ because {yn} is a Cauchy sequence and thus {ϕ(yn)}nNR0 is bounded.
Moreover m is a maximal ideal because if {xn}m then there exists n0N such that xn0 for nn0. Thus if we define a sequence {yn}K by putting yn:=1 for all n<n0 and yn:=1/xn for all nn0 we have that the sequence {xnyn1} is definitely zero, and thus it is zero in the quotient ring R/m, i.e. [{xnyn}]=1 in R/m.

What we have shown implies that m is a maximal ideal, i.e. that L:=R/m is a field. Moreover the map KR which sends an element xK to the constant (Cauchy) sequence {xn=x}R induces an inclusion of fields ι:KL.

Let now Ψ:RR0 be defined as Ψ({xn}):=limnϕ(xn), which is well defined because R consists of Cauchy sequences. It is immediate to check that the composition of the inclusion KR and of Ψ is equal to ϕ and it is also easy using the definition of m to show that Ψ descends to a map ψ:LR0 which is an absolute value in view of Exercise 4.

We want now to prove that L is complete with respect to ψ. To do so observe first of all that for every Cauchy sequence {xn}nNK and every ε>0 there exists αK such that limn+ϕ(xnα)<ε. For example we can take α=xnε, where nεN is the integer associated to ε in the definition of a Cauchy sequence.
Let now {{xjn}n}jL be a Cauchy sequence in L, i.e. a sequence of sequences {xjn}nK such that for all ε>0 there exists j0N such that for all j,ln0 we have that limn+ϕ(xjnxln)<ε. Since every sequence xj:={xjn}nK is a Cauchy sequence we can pick elements ajK such that limn+ϕ(xjnaj)<1/j. Observe now that the sequence x:={aj}jNK is a Cauchy sequence because for every ε>0 we can choose j,l,nN big enough so that ϕ(ajal)=ϕ(ajxjn+xlnal+xjnxln)ϕ(ajxjn)+ϕ(xlnal)+ϕ(xjnxln)<ε in view of the definition of the aj's and of the fact that limn+ϕ(xjnxln)<ε/3 for j and l big enough.

We finally have to show the universal property for (L,ψ). Let (M,ξ) be a complete valued field with a map of valued fields (K,ϕ)(M,ξ). If this map is trivial there is nothing to prove, otherwise we have an injection f:KM. Since M is complete with respect to ξ we see immediately that for every Cauchy sequence {xn}K there exists the limit limn+f(xn). This defines a map RM{xn}limn+ι(xn) and it is not difficult to show that this factors through a map f:LM because if two convergent sequences differ by a sequence which goes to zero they have the same limit. It is now immediate to see that f=fι and we only have to prove that f is unique.
Let f:(L,ψ)(M,ξ) be another map of valued fields such that f=fι. Then using Exercise 5 we have for every x=[{xn}]L that f(x)f(x)=limn+f(xn)f(x)=limn+f(ι(xn))f(x)=f(limn+ι(xn)x)=0 and so f=f as we wanted to prove. Q.E.D.

In view of the previous exercise and theorem we can speak of the completion of a valued field (K,ϕ), which we will denote by (Kϕ,Φ).

 Exercise 14  Prove that the completion of (Q,||) is isomorphic to R with the canonical absolute value, where the real numbers are defined as in Exercise 9.

Complete Archimedean fields

We are now ready to prove the main result of this lecture, which will allow us to characterize the set of Archimedean places of a number field. To do so we need firstly a lemma about Archimedean absolute values on Q.

 Lemma 15  Every Archimedean absolute value ϕ:QR0 is equivalent to the canonical (Archimedean) absolute value ||:QR0.

 Proof   Since ϕ is Archimedean we know that there exists nN with ϕ(n)>1, because otherwise ϕ would be bounded on ZQ and thus non-Archimedean (see Exercise 6 of the second lecture).
Let now aN>1 and for every kN write nk=mkj=0cj,kaj where mkN, cj{0,,a1} for all j{0,,m} and cm0. Applying the triangle inequality we get that ϕ(nk)mkj=0ϕ(cj,k)ϕ(a)j(mk+1)max{ϕ(1),,ϕ(a1)}max(ϕ(a)mk,1) and using the fact that mkklog(n)/log(a) we get that ϕ(n)((klog(n)/log(a)+1)max{ϕ(1),,ϕ(a1)}max(ϕ(a)klog(n)/log(a),1))1/k which implies that ϕ(n)max(ϕ(a)log(n)/log(a),1) for every integer aN>1.
Since ϕ(n)>1 this implies that ϕ(a)>1 for all aN>1. Thus we can use the same argument of the previous paragraph to show that for every couple of integers a,bN>1 we have that ϕ(b)ϕ(a)log(b)/log(a)  and ϕ(a)ϕ(b)log(a)/log(b), which implies that ϕ(a)1/log(a)=ϕ(b)1/log(b). Thus α:=ϕ(a)1/log(a)>1 does not depend on a and we have that ϕ(a)=αlog(a)=alog(α) for all aN>1. Using the multiplicativity of ϕ we see that this extends immediately to Q, and thus that ϕ is equivalent to ||. Q.E.D.

 Theorem 16  Let (K,ϕ) be a complete, Archimedean field. Then either (K,ϕ)(R,||) or (K,ϕ)(C,||), up to substituting ϕ with an equivalent absolute value.

 Proof  Recall first of all the K has characteristic zero, because every absolute value on a field of positive characteristic is non-Archimedean (see Exercise 6 of the second lecture). Thus we have that QK. If we can restrict ϕ to Q we get an Archimedean absolute value, which (thanks to the previous lemma) is equivalent to ||. Moreover since K is complete by hypothesis we can use the previous exercise to obtain an inclusion RK.

Suppose now that 1K is a square. Then we get an inclusion ι:CK and it is easy to see that this is automatically an inclusion of valued fields. Recall now that the polynomial tn1C[t] can be written as tn1=n1j=0(tζjn)whereζn:=e2πinC and thus using the inclusion ι we can write xnyn=(xy)n1j=1(ζjnxy) for every x,yK. Thus using the multiplicativity of ϕ we get that the identity ()ϕ(xy)=ϕ((xz)n(yz)n)n1j=1ϕ(ζjn(xz)(yz)) holds for all x,y,zK for which ϕ(ζjn(xz)(yz))0 for all j{1,,n1}.


We want to show now that K=ι(C), and so suppose by contradiction that there exists αKι(C). Define now the continuous function f:CR>0 as f(z):=ϕ(zα) and let z0C be a point of minimum for f, and let r:=f(z0). Take now zC such that ψ(zz0)<r and use the previous identity () to show that f(z)=f(z0)nϕ(1(zz0)n(αz0)n)n1j=1f(ζjn(zz0)+z0)1f(z0)nϕ(1(zz0)n(αz0)n)r1nr(1+(ϕ(zz0)r)n) because ψ(ζn)=1 as it is easy to observe using the fact that ζnn=1. Thus letting n+ and using the fact that f(z)r we get that f(z)=r for all zC with ϕ(zz0)<r. We can now go on by covering C by balls of radius r to show that f is constant. But this is absurd because f(n)ϕ(n)(1ϕ(n)/ϕ(α)) for all nN and we know that ϕ(n)+ as n+. Thus we have proved that K=ι(C), which shows that ι:(C,||)(K,ϕ) is an isomorphism of valued fields.

Suppose now that 1K is not a square. Then we can use what we have just proved and Lemma 10 to show that L:=K[t]/(t2+1) is isomorphic to C as a valued field. But we know that RK which implies that KR as valued fields. Q.E.D.

 Definition 17  Let K be a number field. We say that two embeddings σ,τ:KC are equivalent, and we write στ, if σ(x)=¯τ(x) for all xK. Here ¯z=xiy denotes the complex conjugate of any complex number z=x+iyC.

 Corollary 18  Let K be a number field. Then the map {σ:KC}ΣK,[σ][||σ] is a bijection. Here ||σ is the absolute value defined in Example 11 of the second lecture.

 Proof  It is easy to see that the map is well defined and injective, i.e. that ||σ is equivalent to ||τ if and only if στ. Moreover if ϕ is an Archimedean absolute value on K then, up to substituting ϕ with an equivalent absolute value, we have that KϕR or KϕC as valued fields. Thus the embedding KKϕ is actually an embedding σ:KC and ϕ=||σ because as we have seen ϕ is equal to the absolute value induced on K by the embedding KKϕ. Q.E.D.

Conclusions and references

In this lecture we have:
  • defined the notion of convergence, Cauchy sequence and completeness for a valued field (K,ϕ)
  • associated to any valued field (K,ϕ) a universal, complete valued field (Kϕ,Φ);
  • proved that all complete, Archimedean valued fields are isomorphic either to R or to mathbbC as topological fields;
  • concluded the proof of Ostrowski's theorem which characterizes the set ΣK=ΣKΣK, for any number field K.
In the following lecture we will switch our focus to non-Archimedean complete valued fields, and we will (finally) start to play with the field of p-adic numbers Qp.

Some references for this class are:

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