AMA CSC-AI
A. S. Classification
A.8. Final Comments on Regularity of Haar Measures
3. The Completion Cp of the Field Qp 1. Definition of Cp1.Definition of Cp
Let us define
Cp = Qp = closure of Qp in S2p.
Hence Cp is a completion of Qp:
Cp = Q
.Proposition. The field Cp is a separable metric space.
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3. The Completion Cp of the Field Qp 141
PROOF. The algebraic closure Qp of Qp is a separable metric space (by Corollary 2 in (1.5)) and is dense in Cp. Any countable dense subset of Qp is automatically dense in Cp: For example Q" is dense in Cp.
The universal field Cp is not locally compact: ICp I = pQ = j p' : v E Q} is dense in R,o. We shall use the following notation
Ap = {X E Cp : IXI < 1}: maximal subring of Cp, Mp = {x E Cp : IXI < 1}: maximal ideal of Ap.
Hence MP = Mp, and AP is not a Noetherian ring (2.1).
3.2.
Finite-Dimensional Vector Spaces over a Complete Ultrametric Field
Let us formulate and prove a generalization of (11.3.1) (cf. Theorem 2 in (II.A.6) for the most general version).
Theorem 1. Let K be a complete (nondiscrete) ultrametric field and V a finite- dimensional vector space over K. Then all norms on V are equivalent.
PROOF We use induction on the dimension n of V. Since the property is obvious for n = 1, it is enough to establish it in dimension n assuming that it holds in dimension n - 1. Choose a basis (e,), <,<n of V and consider the vector space isomorphism
go : Kn -* V sending the canonical basis of K" onto the chosen basis of V.
Considering that K" is equipped with the sup norm, we have to show that for any given norm II .II on V, the mapping iP is bicontinuous. First, for x = (x,) E K", we have
5 EIxiIlleill <maxix,l
E11ei11,IIco(x)ll 5 Cllxll (C = Elleill),
which proves the continuity of the map V. Conversely, let F be the subspace of V generated by the lastn - 1 basis vectors. Since the dimension of F is n - 1,
the inductionhypothesis shows that on this subspace, the given norm is equivalent to the supnorm of the components. In particular, F is complete and closed in V.
Sincee = et ¢ F, we can define
d(e,F)=inflle-yll>0
yEF
and put y = d (e F)/ lie 11 < 1. By the induction hypothesis, there is also a constant cF such that
IIYII > cF max IxiI2<i<n (Y= E2<i<nxie, E F).
Al
For each v = ip(x) E E - F, say v = l; e + y ( 0, y E F), we can write
v = (e +
with
IIvII = ICI Ile+y/III = ICI - Ile - y'll
and hence
IIYII = Ilv - dell < max(llvll, max(IIvIl, y-' IIvIU = Ilvll/y (since y < 1). This shows that IIvII > y IIYII- We have thus proved
IIvII >
y Ilyll),
and since Il Y ll > cF maxi>2 I xi l, we have
Ilsv(x)ll = llvll > ymax(l I Hell, CF max>2 ix I)
> cmaxi>i IxiI = c - Ilxll,
with x1 = and c = cv = y min(cF, Ilell). 2
Corollary. If K is a complete (nondiscrete) ultrametric field and L is a finite extension of K, there is at most one extension of the absolute value of K to L.
Any K-automorphism of L is isometric.
PROOF. Same as in (11.3.3). 9
We can now give Krasner's lemma (1.5) in a more general form.
Theorem 2. Let 0 be any algebraically closed extension of Qp and K C §2 any complete subfield. Select an algebraic element a (E S2) over K and denote by as its conjugates over K. Let r = min,. 0a l as - al. Then every algebraic element b over K, b E B<r(a), generates with K an extension containing K(a)-
PROOF. We can proceed as in (1.5), since we now have uniqueness of the extension of absolute values for finite extensions of K. For any algebraic element bsuch that a V K(b), a has a conjugate a° 0 a over K(b) (the automorphism Cr leaves all elements of K(b) fixed), and
lb - a°I = I(b-a)°l = lb-al,
la - a°I < max(la-bl,lb-a°I)=lb- at.
c,, 5;' ..r
fir
I The Completion Cp of the Field Qp 143 Hence
lb-al>la-aaI>r.
Taking the contrapositive, l b - a I < r ==>a E K(b) and K(a) C K(b).
3.3.
The Completion is Algebraically Closed
Theorem. The universal field Cp is algebraically closed.PeooF Let L (C S2p) be a finite - hence algebraic - extension of Cp. We can apply the general form of Krasner's lemma to the extension Cp C S2p, since we already know that
the field Cp is complete,
the field QP is algebraically closed,
he field S2p has an absolute value extending the p-adic one.
Assume that L = Cp(a) is generated by an algebraic element a of degree n > I and let f E Cp[Xj be the monic irreducible polynomial of a. By density of the algebraic closure Qp of Qp in Cp, we can choose (1.5) a polynomial g E Q ,[X]
sufficiently close to f in order to ensure that a root of g generates L over CPI But QP is algebraically closed, so that g has all its roots in Q ,, and this proves that f has degree 1: L= Cp.
Comment. We have not used the possibility of extending the absolute value of Cp to finite extensions of this field, since we work in the field S2p constructed in (2.2). The general possibility of extending absolute values for finite (algebraic)
"tensions-where the base field is not locally compact - involves other algebraic techniques.
3.4.
The Field Cp is not Spherically Complete
Proposition. The universal field Cp is not spherically complete.
PROOF. Hereis an argument showing the existence of strictly decreasing sequences of closedballs of Cp having an empty intersection (without explicitly constructing one suchsequence!).
Let r -p r > 0 be a strictly decreasing sequence of r = pQ = Cp
In theball B = B<,(0) we can choose two closed disjoint balls Bo and Bl with
thesame radiusrl < ro. In each of these we can select two closed disjoint balls of radii rz < ri, say
Bio and Bit closed and disjoint in Bi.
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Continuing these choices, we define sequences of closed balls having decreasing radii given by the sequence (rn) and satisfying in particular
Bi D Bid D ... D Bij ...k D Bij. -kt D ...
(with multi-indices equal to 0 or 1). By construction, two balls having distinct multi- indices of the same length are disjoint. If (i) = (i1, i2, ...) is a binary sequence we can define
B(1) = n Bi,..
n>1
Such an intersection is either empty or is a closed ball of radius r = lim rn having for center any element in it, as always in the ultrametric case. In any case, all B(i) are open subsets of Cp (this is where r = lim rn > 0 is used). If two sequences (i) and (j) are distinct - say i,, jn - then by construction Bi,...i, and Bj,...j. are disjoint, and a fortiori B(i) C Bi,...i,, B(j) C Bj,...j, are disjoint. Since the metric space Cp is separable, the uncountable family of disjoint open sets (B(i)) can only be a countable set of distinct open sets (any countable dense subset must meet all nonempty open sets). This forces most of the B(i) to be empty!
A pictorial representation of the preceding proof is sketched in the exercises.
3.5. The Field Cp is Isomorphic to the Complex Field C
The result of this section will not be used in this book. It gives the answer to a natural question, namely: What is the algebraic structure of the field Cp?
Let us start by the determination of the cardinality of the field Cp.
Lemma. The field Cp has the power of the continuum.
PROOF The unit ball of Qp is Zp - fn>0{0, 1, ... , p- 1),hence has the power of the continuum c: numeration in base p gives a 1-1 correspondence with [0, 11 c R except for contably many overlaps, so these sets have the same cardinality. (In fact, each Zp is homeomorphic to the Cantor set: Exercise 13 of Chapter I.) The field Qp itself has the same power, since it is the countable union of balls pmZp (each having cardinality c). All finite extensions of Qp have the same power. The algebraic closure Qp of Qp still has the same power (the ring of polynomials in one variable over Qp has also the power of the continuum). Finally, a countable product (Q,)N cannot have bigger cardinality. Such a product contains all CauchY sequences of Q,, and
Card(Cp) < Card((Qp)N) = C.
Recall the terminology used for field extensions. A transcendence basis of a field extension K/k is a family (X)1 j in K such that
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3. The Completion Cp of the Field Qp 145
the subfield k(X,),EJ C K is a purely transcendental extension of k, and K/k(X,),EI is an algebraic extension.
Here are some general results of Steimtz concerning field theory:
Two algebraic closures of a field k are k-isomorphic.
Everyfield extension K/k has a transcendence basis.
Two transcendence bases of K/k have the same cardinality.
For example, let Q° be the algebraic closure of Q in Cp and Qb the algebraic closure of Q in C. Then there is an isomorphism
Qa Qb
These fields are countable. But the fields Cp and C have the power of the continuum, hence the same transcendence degree (over the prime field Q or its algebraic closure).
Theorem. The fields C and Cp are isomorphic.
PROOF. Any extension of the rational field Q having the power of the continuum has a transcendence basis having this cardinality. By the above lemma the tran- scendence degrees of C and Cp over Q (or its algebraic closure) are the same, and we can select transcendence bases (X),EI in C and resp. (Y),EI in Cp (indexed by the same set). Now, C isan algebraic closure of Q(X,),E, and C,, is an algebraic closure of Q(Y,),E,. Hence these two algebraic closures are isomorphic.
As a consequence, we can view the field Cp as the complex field C endowed with an exotic topology. But the preceding considerations do not lead to a canonical isomorphism between these universal fields: The axiom of choice has to be used to show the existence of such an isomorphism.
Field J B.,1 D Bpi Residue field Nonzero I.I
Qp D Zp D PZp FP pz
KJRDP=7rR F9(q=Pf)
17riz = Pe!ZQ°pJAOJM° k°=FD=Fpx PQ
Cp 1) Ap D MP FP = Fpx PQ
S2 DA -
p n Al.-. uncountablekn R>oProperties locally compact J of = dimQp K < oc
locally compact Jalgebraically closed
t not locally compact
algebraically closed
complete
J algebraically closed spherically complete
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