Model theory of valued elds Lecture Notes

Model theory of valued fields

Lecture Notes

Lou van den Dries

Fall Semester 2004

Contents

1 Introduction 2

2 Henselian local rings 52.2 Hensel’s Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . 62.3 Completion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92.4 Lifting the residue field . . . . . . . . . . . . . . . . . . . . . . . 132.5 The Ax-Kochen Principle . . . . . . . . . . . . . . . . . . . . . . 15

3 Valuation theory 203.1 Valuation rings and integral closure . . . . . . . . . . . . . . . . . 263.2 Quantifier elimination . . . . . . . . . . . . . . . . . . . . . . . . 303.3 The complete extensions of ACFval . . . . . . . . . . . . . . . . . 37

4 Immediate Extensions 394.1 Pseudoconvergence . . . . . . . . . . . . . . . . . . . . . . . . . . 394.2 Maximal Valued Fields . . . . . . . . . . . . . . . . . . . . . . . . 444.3 Henselization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 484.4 Uniqueness of immediate maximal extensions . . . . . . . . . . . 50

5 The Theorem of Ax–Kochen and Ershov 525.1 Existence of cross-sections in elementary extensions . . . . . . . . 525.2 Extending the value group . . . . . . . . . . . . . . . . . . . . . . 535.3 AKE-theorem with lifting and cross-section . . . . . . . . . . . . 54

6 Unramified Mixed Characteristic 596.1 The Teichmuller Map . . . . . . . . . . . . . . . . . . . . . . . . 596.2 Witt vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 616.3 Coarsening . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 706.4 AKE for unramified mixed characteristic . . . . . . . . . . . . . . 71

1

1 INTRODUCTION 2

1 Introduction

Conventions. Throughout, m,n range over N = {0, 1, 2, . . . }, the set of naturalnumbers. Unless specified otherwise, “ring” means “commutative ring with 1”and given a ring R we let U(R) := {x ∈ R : xy = 1 for some y ∈ R} beits multiplicative group of units, and for a1, . . . , an ∈ R we denote the ideala1R + · · · + anR of R also by (a1, . . . , an)R or just by (a1, . . . , an). A ringis considered as an L-structure for the language L = {0, 1,+,−, ·} of rings.Given a ring R and distinct indeterminates t1, . . . , tn, we let R[t1, . . . , tn] andR[[t1, . . . , tn]] denote the corresponding polynomial ring and formal power seriesring, with

R ⊆ R[t1, . . . , tn] ⊆ R[[t1, . . . , tn]].

Note that every f ∈ R[[t1, . . . , tn]] can be written as

f = a+ t1f1 + · · ·+ tnfn

where a ∈ R and each fi ∈ R[[t1, . . . , ti]]. The element a is uniquely determinedby f and is called the constant term of f , and denoted by f(0). Throughout,k,k1,k2 are fields, and we put k× := k \ {0} = U(k). Given a prime number pthe field Z/pZ of p elements is usually denoted by Fp.

The goal of this course is the famous Ax-Kochen-Ersov Theorem from the 1960’s.We shall prove this result in various stronger forms, and discuss applications.We begin by stating some very special cases of the theorem. Let t be a singleindeterminate.

1.1. Suppose k1 and k2 have characteristic 0. Then

k1[[t]] ≡ k2[[t]] ⇐⇒ k1 ≡ k2.

The direction ⇒ is easy and holds also when k1 and k2 have characteristicp > 0, but the direction ⇐ lies much deeper, and is not yet known to be truewhen k1 and k2 have characteristic p > 0. (This is one of the main openproblems in the subject.) Recall that Qalg ≡ C, as fields, where Qalg is thealgebraic closure of Q, so as a special case of 1.1 we have Qalg[[t]] ≡ C[[t]]. Onthe other hand, Qalg[[t1, t2]] ≡ C[[t1, t2]] and Qalg[t] ≡ C[t]. This highlights thefact that 1.1 is a one-variable phenomenon about power series rings: it fails fortwo or more variables, and also fails for polynomial rings instead of power seriesrings.

A stronger version of 1.1 is in terms of sentences:

1.2. For every sentence σ in the language of rings, there is a sentence σ inthat language, such that for all k of characteristic 0,

k[[t]] |= σ ⇐⇒ k |= σ.

1 INTRODUCTION 3

For the next result, let k[t1, . . . , tn]alg be the subring of k[[t1, . . . , tn]] con-

sisting of all f ∈ k[[t1, . . . , tn]] that are algebraic over k[t1, . . . , tn].

1.3. If k has characteristic 0, then k[t]alg ≼ k[[t]].

For any field k we have a weak version of 1.3 for n variables:

k[t1, . . . , tn]alg ≼1 k[[t1, . . . , tn]].

This is part of the Artin Approximation Theorems. Again, if n ≥ 2, thenk[t1, . . . , tn]

alg is never an elementary substructure of k[[t1, . . . , tn]].Next, let C[[t]]conv be the subring of C[[t]] consisting of the power series

f ∈ C[[t]] that have a positive radius of convergence.

1.4. C[[t]]conv ≼ C[[t]].

We also have C[[t1, . . . , tn]]conv ≼1 C[[t1, . . . , tn]] for any n, but ≼1 cannot bereplaced here by ≼ for n > 1.

A key algebraic property of the power series rings k[[t1, . . . , tn]] is that theyare henselian local rings. This property has a close connection to the theoremsabove, and distinguishes these power series rings from the polynomial ringsk[t1, . . . , tn].

1.5 Definition. A ring R is local if it has exactly one maximal ideal. Givena local ring R, we denote its maximal ideal by m, and we let k = R/m be theresidue field, with residue map

x 7→ x = x+m : R→ k.

Local rings are exactly the models of a (finite) set of axioms in the language ofrings, because a ring R is local iff its set of non-units is an ideal. (Even simpler,a ring R is local iff 1 = 0 and its set of non-units is closed under addition.) Notealso that an element x in a local ring is a unit iff x = 0.

Examples.

1. Fields are exactly the local rings with m = 0.

2. k[[t1, . . . , tn]] is a local ring with m = (t1, . . . , tn). This follows by theabove decomposition of any f ∈ R as a sum f(0) + t1f1 + · · ·+ tnfn, andby noting that f ∈ U(R) iff f(0) = 0. The kernel of the ring morphismf 7→ f(0) : R → k is m, so this morphism induces a field isomorphismR/m→ k. We shall usually identify the residue field R/m with k via thisisomorphism.

3. Let p be a prime number and k a positive integer. Then Z/pkZ is a localring. The integers a0+a1p+ · · ·+ak−1p

k−1 with ai ∈ {0, 1, . . . , p−1} arein one-to-one correspondence with their images in Z/pkZ. The maximalideal of this ring is generated by the image of p. The elements of this ringbehave somewhat like power series in the “variable” p truncated at pk.

1 INTRODUCTION 4

4. Letting k →∞ in (3) we get the ring Zp of p-adic integers. Its elements canbe represented as infinite series

∑∞i=0 aip

i, where all ai ∈ {0, 1, . . . , p− 1}.A precise definition is given in the next section.

All local rings in these examples are henselian, as we shall see in the next section.The localizations

ZpZ = {ab: a, b ∈ Z, b /∈ pZ} ⊆ Q, (p a prime number),

are examples of local rings that are not henselian.

1.6 Definition. Let R be a local ring. We say that R is henselian if for anypolynomial f(X) ∈ R[X] and any α ∈ R such that

f(α) ∈ m, f ′(α) /∈ m,

there is a ∈ R with f(a) = 0 and a ≡ α mod m.

1.7. Remark. By Lemma 2.1 below there can be at most one such a.

The henselian property means that non-singular zeros in the residue field canbe lifted to the ring itself: let R be a local ring, f(X) ∈ R[X], α ∈ R; then

f(α) ∈ m and f ′(α) /∈ m ⇐⇒ f(α) = 0 and f ′(α) = 0

⇐⇒ α is a non-singular zero of f(X).

(Here we let f(X) be the image of f(X) in k[X] obtained by replacing eachcoefficient of f by its residue class.) Note that the henselian local rings areexactly the models of a certain set of sentences in the language of rings: thisset consists of the axioms for local rings and has in addition for each n > 0 anaxiom expressing the henselian property for polynomials of degree ≤ n.The Ax-Kochen-Ersov Theorem concerns a particular class of henselian localrings, namely henselian valuation rings. The rings k[[t]], k[[t]]alg, C[[t]]conv, andZp are indeed valuation rings, unlike k[[t1, . . . , tn]] for n > 1, and Z/pkZ fork > 1.

1.8 Definition. A valuation ring is a domain R whose set of ideals is linearlyordered by inclusion.

Exercises. Prove the following:

1. Each valuation ring is a local ring.

2. Let R be a domain with fraction field K. Then R is a valuation ring ifffor every a ∈ K× either a ∈ R or a−1 ∈ R.

3. k[[t]] is a valuation ring.

4. k[[t1, . . . , tn]] is not a valuation ring if n > 1.

2 HENSELIAN LOCAL RINGS 5

2 Henselian local rings

In this section we establish basic facts about henselian local rings, prove thatcomplete local rings are henselian (Hensel’s Lemma), show that under certainconditions the residue field of a henselian local ring can be lifted, and give ababy version of the Ax-Kochen Principle.

2.1. Lemma. Let R be a local ring, f(X) ∈ R[X], and α ∈ R, such that

f(α) ∈ m, f ′(α) /∈ m.

Then there is at most one a ∈ R such that f(a) = 0 and a ≡ α mod m.

Proof. Suppose a ∈ R satisfies f(a) = 0 and a ≡ α mod m. Then f ′(a) ≡ f ′(α)mod m, hence f ′(a) is a unit. Taylor expansion in x ∈ m around a gives

f(a+ x) = f(a) + f ′(a)x+ bx2 = f ′(a)x+ bx2 (b ∈ R)

= f ′(a)x[1 + f ′(a)

−1bx.

]Since f ′(a)

[1 + f ′(a)

−1bx

]∈ U(R), this gives: f(a+ x) = 0 iff x = 0.

2.2. Lemma. Let R be a local ring. The following are equivalent.

(1) R is henselian.

(2) Each polynomial 1 + X + cd2X2 + · · · + cdnX

n, with n ≥ 2, c ∈ m andd2, . . . , dn ∈ R, has a zero in U(R).

(3) Each polynomial Y n + Y n−1 + cd2Yn−2 + · · ·+ cdn, with n ≥ 2, c ∈ m and

d2, . . . , dn ∈ R, has a zero in U(R).

(4) [Newton Version] Given a polynomial f(X) ∈ R[X], α ∈ R, and c ∈ msuch that f(α) = cf ′(α)2, there is a ∈ R such that f(a) = 0 and a ≡ αmod cf ′(α).

The “Newton version” (4) gives extra precision in the henselian propertywhen f ′(α) /∈ m, but (4) is devised to deal also with the case f ′(α) ∈ m.

Proof. (1)⇒(2). Assume (1) and let f(X) = 1+X + cd2X2+ · · ·+ cdnX

n withc ∈ m and d2, . . . , dn ∈ R. Then for α = −1 we have: f(α) ≡ 0 mod m andf ′(α) ≡ 1 mod m. Thus f(X) has a zero in −1 +m ⊆ U(R), by (1).

(2)⇔ (3). Use the substitution X = 1/Y .(2)⇒(4). For f, α, c as in the hypothesis of (4), let x ∈ R and consider the

expansion:

f(α+ x) = f(α) + f ′(α)x+∑i≥2

bixi

where the bi ∈ R do not depend on x

= cf ′(α)2 + f ′(α)x+∑i≥2

bixi


Set x = cf ′(α)y where y ∈ R. Then

f(α+ cf ′(α)y) = cf ′(α)2

1 + y +∑i≥2

cdiyi

,

where the di ∈ R do not depend on y. Assuming (2), choose y ∈ R such that

1 + y +∑

cdiyi = 0.

This yields an a = α+ cf ′(α)y as required.(4)⇒(1). Clear.

Exercises. Let R be a henselian local ring, m > 0, and suppose chark doesnot divide m (this includes the case chark = 0). Show that for a ∈ U(R):

a is an m-th power in R. ⇐⇒ a is an m-th power in k.

2.2 Hensel’s Lemma

Hensel’s Lemma says that complete local rings are henselian. Here is the idea ofthe proof. Let R be a local ring, and think of the elements in m as infinitesimals(very small). Let f(X) ∈ R[X] and α ∈ R be such that f(α) ∈ m and f ′(α) /∈ m,so f(α) is tiny compared to f ′(α). If f(α) = 0 we can apply Newton’s methodto perturb α by a tiny amount x to make f(α + x) much smaller than f(α):Taylor expansion in x ∈ R around α yields

f(α+ x) = f(α) + f ′(α)x+ terms of higher degree in x

= f ′(α)(f ′(α)

−1f(α) + x+ terms of higher degree in x

)Setting x = −f ′(α)

−1f(α) yields x ∈ m and

f(α+ x) = f ′(α)[multiple of f(α)2]

By the occurrence of f(α)2 as a factor in the above expression, this choice of xwill make f(α+x) much smaller than f(α), provided we have a suitable norm onR by which to measure size. The process above can now be repeated with α+xin the role of α. Iterating this process indefinitely and assuming R is completewith respect to our norm, we can hope to obtain a sequence converging to somea ∈ R such that f(a) = 0 and a ≡ α mod m. A precise version of this limitprocess occurs in the proof of Hensel’s Lemma 2.7.

2.3 Definition. Let R be a ring. A norm on R is a function | | : R → R≥0

such that for all x, y ∈ R:

(i) |x| = 0 ⇐⇒ x = 0, |1| = | − 1| = 1,


(ii) |x+ y| ≤ |x|+ |y|,

(iii) |xy| ≤ |x||y|.

It follows easily that | − x| = |x| for all x ∈ R, and thus we obtain a metricd(x, y) = |x − y| on R; we consider R as a metric space with this metric if thenorm | | is clear from context. Note that then the operations

+, ·,− : R2 → R and | | : R→ R

are continuous. We say that R is complete if it is complete as a metric space.If condition (ii) holds in the strong form

|x+ y| ≤ max{|x|, |y|},

then we call the norm an ultranorm (or ultrametric norm, or nonarchimedeannorm). Note that then |x+ y| = max{|x|, |y|} if |x| = |y|. When our norm is anultranorm, then convergence of infinite series

∑an is an easy matter:

2.4. Lemma. Suppose the ring R is complete with respect to the ultranorm | |.Let (an) be a sequence in R. Then

limn→∞

n∑i=0

ai exists in R ⇐⇒ limn→∞

|an| = 0.

We leave the proof as an exercise. If the limit in this lemma exists, it is denotedby

∑∞n=0 an.

Example. Let A be a ring with 1 = 0, and R := A[[t1, . . . , tn]]. Each f ∈ Rhas a unique representation

f =∑

ai1...inti11 · · · tinn

where the formal sum is over all (i1, . . . , in) ∈ Nn and the coefficients ai1...in liein A. For such f we define ord f ∈ N ∪ {∞} by

ord f = min{i1 + · · ·+ in : ai1...in = 0} if f = 0,

and ord 0 =∞. This order function satisfies:

• ord 1 = 0,

• ord(f + g) ≥ min{ord(f), ord(g)},

• ord(fg) ≥ ord(f) + ord(g).

It follows that |f | := 2−ord f defines an ultranorm on R.

Exercise. Show that R is complete with respect to this norm, and has thepolynomial ring A[t1, . . . , tn] as a dense subring.


2.5. Remark. Let the ring R be complete with respect to the norm | |, and

let x ∈ R satisfy |x| < 1. Then

∞∑i=0

xi := limN→∞

N∑i=0

xi exists in R, and

(1− x)∞∑i=0

xi = 1.

In the next two lemmas and subsequent exercise, R is a ring complete withrespect to the ultranorm | |, and |x| ≤ 1 for all x ∈ R.

2.6. Lemma. The set N = {x ∈ R : |x| < 1} is an ideal. If a ∈ R, then

a ∈ U(R) ⇐⇒ a ∈ U(R), where R := R/N, a := a+N.

Proof. The direction⇒ is obvious. Conversely, let a ∈ R be such that a ∈ U(R).Take b ∈ R such that ab = 1− x with x ∈ N. Then 1− x is a unit in R by theabove remark. Hence a ∈ U(R) and a−1 = b

∑∞i=0 x

i.

2.7. Hensel’s Lemma. Let f(X) ∈ R[X], and let α ∈ R be such that

|f(α)| < 1, f ′(α) ∈ U(R).

Then there is a unique a ∈ R such that f(a) = 0 and |a− α| < 1.

Proof. We shall obtain a as the limit of a sequence {an} in R. Put a0 := α andε := |f(α)| < 1. Suppose an ∈ R is such that

|f(an)| ≤ ε2n

, |an − α| ≤ ε.

(Clearly this is true for n = 0.) Note that an ≡ α mod N, so f ′(an) is a unitbecause f ′(α) is. Now, put

an+1 := an + h, where h = −f ′(an)−1

f(an).

Then:

f(an+1) = f(an) + f ′(an)h+multiple of h2

= 0 +multiple of f(an)2,

hence|f(an+1)| ≤ |f(an)|2 ≤ (ε2

n

)2 = ε2n+1

.

Also,|an+1 − α| ≤ max{|an+1 − an|, |an − α|} ≤ ε,

so the induction step is complete. We have

|an+1 − an| = |h| ≤ |f(an)| ≤ ε2n

,


so {an} is a Cauchy sequence. Let a ∈ R be its limit. Then

f(a) = f(lim an) = lim f(an) = 0.

The condition |an−α| ≤ ε for all n insures that |a−α| ≤ ε < 1. The uniquenessof a follows exactly as in the proof of Lemma 2.1.

Exercise. Let f1, . . . , fN ∈ R[X] where X = (X1, . . . , XN ) is a tuple of distinctindeterminates. Suppose α ∈ RN is a “near zero” of f1, . . . , fN , that is,

|fi(α)| < 1 for i = 1, . . . , N, and det

(∂fi∂xj

(α)

)i,j

is a unit in R.

Then there is a unique a ∈ RN such that f1(a) = · · · = fN (a) = 0, and a ≡ αmod N componentwise.

Hint. Put f = (f1, . . . , fN ) ∈ R[X]N . Check that in R[X]N we have an identity

f(α+X) = f(α) + f ′(α)X + terms of degree ≥ 2 in X,

where f ′(α) is the matrix(

∂fi∂Xj

(α))i,j, and in the matrix product f ′(α)X we

view X as an N × 1 matrix. Verify that the proof of Hensel’s Lemma goesthrough.

2.3 Completion

Let (R, | |) be a normed ring, that is, R is a ring, and | | is a norm on R. As a

metric space, R has a completion R, and the ring operations +,−, · : R2 → Rextend uniquely to continuous operations +,−, · : R2 → R. With these extendedoperations R is again a ring. The norm | | : R → R also extends uniquely to

a continuous function | | : R → R, and | | is a norm on the ring R whose

corresponding metric is the metric of the complete metric space R. We have

now a complete normed ring (R, | |) in which R is a dense subring. If (R′, | |′) isa second complete normed ring extending the normed ring (R, | |) such that R is

dense in R′, then there is a unique normed ring isomorphism (R, | |) ∼= (R′, | |′)that is the identity on R. Thus we have the right to call (R, | |) the completionof the normed ring (R, | |); to keep notations simple we usually write | | for thenorm | | on R.

If |xy| = |x||y| for all x, y ∈ R, then we call | | an absolute value, and in thatcase R is a domain, and | | extends uniquely to an absolute value on the fractionfield. If K is a field with absolute value | |, then the map x 7→ x−1 on K× is

continuous, and the completion K of K with respect to | | is a field, and theextended norm is again an absolute value. In this case, for any subring R ⊆ K,the closure of R in K is a subring, and is thus the completion of R with respectto the restricted norm.

Suppose that R = A[[t1, . . . , tn]] with A a domain. Then its norm given by|f | = 2−ord(f) is an absolute value. Note that R is complete with respect to this


norm. If A = k is a field, then the ring R = k[[t]] is a valuation ring, and itsfraction field K is also complete with respect to the extended absolute value.In this fraction field we have R = {f ∈ K : |f | ≤ 1}. One can identify K in thiscase with the field of Laurent series k((t)).

The field of p-adic numbers. Let p be a prime number. For a ∈ Z we definevp(a) ∈ N ∪ {∞} as follows: if a = 0, then vp(a) is the natural number suchthat a = pvp(a)b where b ∈ Z and p ∤ b, and vp(0) =∞. It is clear that then forall a, b ∈ Z:

1. vp(a+ b) ≥ min{vp(a), vp(b)},

2. vp(ab) = vp(a) + vp(b),

3. vp(1) = 0.

This yields an absolute value | |p on Z by setting

|a|p = p−vp(a).

(The reason we defined |a|p in this way with base p, instead of setting |a|p =2−vp(a) with base 2, is to have |a|

∏p |a|p = 1 for all nonzero a ∈ Z, where now

the product is over all prime numbers p. Taking logarithms this identity becomeslog |a|−

∑p vp(a) log p = 0, which mimicks the identity deg f−

∑P∈C vP (f) = 0

for non-zero f ∈ C[X]. Here vP (f) is the order of vanishing of f at the pointP ; a precise definition is given later. This strenghtens the analogy between thering Z and the polynomial ring C[X]. This analogy can be pushed much furtherand is very productive in number theory.)

The absolute value | |p extends uniquely to an absolute value on Q, which wecall the p-adic absolute value. The completion of Q with respect to the p-adicabsolute value is denoted by Qp, and called the field of p-adic numbers. Theclosure of Z in Qp is denoted by Zp. By previous remarks Zp is a subring ofQp, and is the completion of Z with respect to its p-adic norm. We call Zp thering of p-adic integers. Thus we have the following diagram where all maps areinclusions:

Zp// Qp

Z

OO

// Q

OO

Since |a|p ≤ 1 for a ∈ Z, this remains true for a ∈ Zp. Here are some of themost basic facts about p-adic numbers:

2.8. Proposition. Given any sequence (an) of integers, limn→∞∑n

i=0 aipi

exists in Zp, and for a =∑∞

n=0 anpn we have the equivalences

p|a0 ⇐⇒ a ∈ pZp ⇐⇒ |a|p < 1 ⇐⇒ a /∈ U(Zp).

Moreover:


(i) the map (an) 7→∑∞

n=0 anpn : {0, 1, . . . , p− 1}N → Zp is a bijection;

(ii) Zp is a henselian local ring with maximal ideal pZp;

(iii) for each integer k > 0 the ring morphism Z→ Zp/pkZp is surjective with

kernel pkZ, and thus induces a ring isomorphism

Z/pkZ → Zp/pkZp.

For k = 1 this isomorphism identifies Fp with the residue field of Zp.

(iv) each nonzero x ∈ Qp is of the form x = pku with unique integer exponentk and u ∈ U(Zp).

(v) Zp = {x ∈ Qp : |x|p ≤ 1} = {x ∈ Qp : |x|p < p}.

(vi) Zp is open and closed in Qp, and is compact.

Proof. Let (an) be a sequence in Z. By Lemma 2.4 the series∑∞

n=0 anpn con-

verges, since |anpn|p ≤ p−n → 0 as n→∞. Put a :=∑∞

n=0 anpn. If p|a0, then

a0 = pb with b ∈ Z, so

a = a0 + p∞∑

n=0

an+1pn = p(b+

∞∑n=0

an+1pn) ∈ pZp.

The implications a ∈ pZp ⇒ |a|p < 1, and |a|p < 1 ⇒ a /∈ U(Zp) follow from|p|p < 1 and the fact that |x|p ≤ 1 for all x ∈ Zp. We obtain the implicationa /∈ U(Zp) ⇒ p|a0 by proving its contrapositive: Suppose that p ∤ a0. Takeb ∈ Z with a0b = 1 + pk, k ∈ Z. Then

ab = 1 + p

∞∑n=0

cnpn, c0 := k + a1b, cn = an+1b for n > 0.

Since |p∑∞

n=0 cnpn|p ≤ p−1 < 1, it follows from Remark 2.5 that ab ∈ U(Zp),

hence a ∈ U(Zp).As to (i), to prove injectivity, let (an) and (bn) be two distinct sequences in

{0, 1, . . . , p−1}, and a =∑∞

n=0 anpn, b =

∑∞n=0 bnp

n. Let m be the least n suchthat an = bn. Then a− b = pm(

∑∞i=0(am+i− bm+i)p

i) = 0, since p ∤ (am− bm).For surjectivity, let A be the image of the map in (i). We claim that, givenany k ∈ Z and real ϵ > 0 there is a ∈ A such that |k − a|p < ϵ. To see this,take n such that p−n ≤ ϵ, and take a0, . . . , an−1 ∈ {0, 1, . . . , p − 1} such thatk ≡ a0+a1p+· · ·+an−1p

n−1 mod pn, so a = a0+a1p+· · ·+an−1pn−1+pnb with

b ∈ Z, hence |a−(a0+a1p+ · · ·+an−1pn−1)|p ≤ p−n < ϵ. This proves our claim,

which says that Z is contained in the closure of A. It remains to show that A isclosed in Zp. We equip the finite set {0, 1, . . . , p−1} with the discrete topology,and give {0, 1, . . . , p− 1}N the corresponding product topology, making this setof sequences a compact hausdorff space (by Tychonov). It is easily checked thatthe map in (i) is continuous, so A is compact, hence A is closed in Zp.


Item (ii): By (i) and what we proved before (i), the set of nonunits in Zp

is the ideal {x ∈ Zp : |x|p < 1} = pZp. Hence Zp is a local ring with maximalideal pZp. Since Zp is complete with respect to | |p, it is henselian by Hensel’sLemma.

Item (iii): Let k be a positive integer and a ∈ Zp. By (i) we have a =a0 + a1p+ · · ·+ ak−1p

k−1 + pkb with a0, . . . , ap−1 ∈ {0, . . . , p− 1} and b ∈ Zp,so a+ pkZp = (a0 + a1p+ · · ·+ ak−1p

k−1)+ pkZp. This gives surjectivity of themap in (iv). Suppose ℓ ∈ Z lies in the kernel of this map, that is, ℓ = pkb withb ∈ Zp. To show that ℓ ∈ pkZ we can replace ℓ by ℓ+pkq where q ∈ Z, and thuswe can assume that ℓ = b0 + b1p+ · · · + bk−1p

k−1 where the “base p-digits” bilie in {0, . . . , p− 1}. So pkb = b0 + b1p+ · · ·+ bk−1p

k−1, and by the injectivityin (i) this yields b0 = · · · = bk−1 = 0.

Item (iv): By the above, each nonzero a ∈ Zp has the form pnu with u ∈U(Zp), and we note that this determines n by |a|p = p−n. Using this fact, theset K := {a/pm : a ∈ Zp,m ∈ N} is easily seen to be a subfield of Qp. Note thateach non-zero x ∈ K has the form pku with k ∈ Z and u ∈ U(Zp), and that then|x|p = p−k, so k and u are uniquely determined by x. It follows that if x ∈ Kand |x|p ≤ 1, then x ∈ Zp. It remains to show that K = Qp, and since Qp isthe closure of Q, this will follow by showing that K is closed in Qp. Let (an) bea sequence in K converging to a ∈ Qp. Then the sequence (|an|p) is bounded,so we can take k ∈ N such that |an|p ≤ pk for all n, hence |pkan|p ≤ 1 for all n,and thus pkan ∈ Zp for all n. Since (pkan) is a Cauchy sequence, it convergesto an element b ∈ Zp, hence (an) converges to b/pk ∈ K, that is, a = b/pk ∈ K.

Item (v): The proof of (iv) shows Zp = {x ∈ Qp : |x|p ≤ 1}, and also thatthe p-adic norm of any x ∈ Q×

p is of the form pk with k ∈ Z. So if x ∈ Qp and|x|p < p, then |x|p ≤ 1.

Item (vi): that Zp is open and closed in Qp is immediate from (v), and thatZp is compact follows from (i) and its proof: the map in (i) is a continuousbijection from the compact hausdorff space {0, 1, . . . , p − 1}N onto Zp, so thismap is even a homeomorphism.

Note that the familiar base p representation of a natural number is extendedin (i) to all p-adic integers: the natural numbers are exactly the p-adic integerswhose base p representation

∑anp

n as in (i) has an = 0 for all sufficientlylarge n. In connection with (vi), note that every a ∈ Qp has a compact openneighborhood a+ Zp.

Exercises. Find the base p representation of −1 as in (i). Prove:

1. Zp is a valuation ring;

2. the pnZp are exactly the nonzero ideals of Zp;

3. Zp ∩Q = ZpZ.


2.4 Lifting the residue field

Suppose R is a local ring with a subfield E, that is, E is a subring that happensto be a field. Then E is mapped isomorphically onto a subfield E of the residuefield k by the residue map x 7→ x. (Why?) In case E = k we call E a lift of k.(Example: if R = C[[t]], then the subfield C is such a lift. Non-example: let pbe a prime number; then Z/p2Z is a local ring but has no subfield.)

Suppose E = k. It would be nice if we could extend E to a lift of k. Inwhat follows we indicate how to do this. Let y ∈ k \ E and take x ∈ R suchthat x = y.

Case 1. y is transcendental over E. For f(X) ∈ E[X] \ {0}, we havef(X) ∈ E[X] \ {0}, hence f(x) = f(y) = 0, so f(x) ∈ U(R). In particular, thesubring E[x] of R is mapped isomorphically onto the subring E[y] of k by theresidue map. Thus E[x] is a domain with fraction field E(x) inside R, and E(x)is mapped isomorphically onto the subfield E(y) of k.

In this case we found a subfield E(x) of R that strictly contains E.

Case 2. y is algebraic over E. Take the monic polynomial f(X) ∈ E[X] suchthat its image f(X) ∈ E[X] is the minimum polynomial of y over E. Note thatf(X) ∈ E[X] is irreducible, since f(X) ∈ E[X] is. We have a surjective ringmorphism

a 7→ a : E[x]→ E[y],

where E[x] is a subring of R and E[y] is a subfield of k. If we happened to beso lucky that f(x) = 0, then this map would be injective (why?), and we wouldhave extended E to a strictly larger subfield E[x] of R. In general f(x) = 0,but if we can find an ϵ ∈ m such that f(x+ ϵ) = 0, then we can replace in theabove argument x by x+ ϵ, and E[x+ ϵ] would be a subfield of R that strictlycontains E.

To find such an ϵ we assume at this point that R is henselian and chark = 0.We have f(x) = f(y) = 0, that is, f(x) ∈ m. Since chark = 0, the irreduciblepolynomial f(X) ∈ E[X] is separable, so f ′(y) = 0, that is, f ′(x) /∈ m. Thehenselian assumption on R now yields an ϵ ∈ m such that f(x + ϵ) = 0, asdesired.

This discussion leads to the following important result.

2.9. Theorem. Let R be a henselian local ring with chark = 0. Then theresidue field k can be lifted.

Proof. The unique ring morphism Z→ R is injective, and as usual we identifyZ with its image in R via this morphism. Each non-zero integer in R is a unitsince its residue class is, so Z has a fraction field in R. Thus we have found asubfield of R. By Zorn’s lemma there is a maximal subfield of R; by the earlierdiscussion, a maximal subfield of R is a lift of k.

In some cases this lift allows us to completely describe a local ring R in termsof its residue field. Here is such a case:


2.10 Proposition. Let R be a local ring, with t ∈ R and n such that:

chark = 0, m = tR, tn = 0, tn+1 = 0. (1)

Then we have a ring isomorphism R ∼= k[T ]/(Tn+1).

A first step in proving this proposition is to show that a local ring R as inthe proposition is henselian. This will follow from the next elementary result.

2.11. Lemma. Let R be a local ring and let t ∈ R be such that m = tR. LetA ⊆ R be a set of representatives modulo m, that is, A maps bijectively onto kvia the residue map. Then we can expand in powers of t with coeffficients in A:

(1) For any n, each r ∈ R is of the form

r = a0 + a1t+ · · ·+ antn + stn+1

with a0, . . . , an ∈ A and s ∈ R.

(2) If tn = 0, then tm /∈ tm+1R for m ≤ n, and the tuple (a0, . . . , an) in (1)is uniquely determined by n, r.

(3) If tn = 0 and tn+1 = 0, then the map

(a0, . . . , an) 7→ a0 + a1t+ · · ·+ antn : An+1 → R

is a bijection.

Proof of Lemma. Let r ∈ R. We establish (1) by induction on n. We haver ≡ a0 mod m with a0 ∈ A, that is, r = a0 + st with s ∈ R. This provesthe case n = 0. Suppose r = a0 + a1t + · · · + ant

n + stn+1 as in (1). Sinces = an+1 + ut with an+1 ∈ A and u ∈ R, this gives

r = a0 + a1t+ · · ·+ antn + an+1t

n+1 + utn+2.

For (2), suppose tn = 0, and tm = tm+1r with r ∈ R. Then tm(1 − tr) = 0,hence tm = 0 since 1− tr is a unit. This forces m > n. Next suppose

a0 + a1t+ · · ·+ antn ≡ b0 + b1t+ · · ·+ bnt

n mod tn+1,

where ai, bi ∈ A. Suppose ai = bi for some i ≤ n, and let m be the least iwith this property. After subtracting the first m terms from each side we getamtm ≡ bmtm mod tm+1, hence (am− bm)tm ∈ tm+1R. But am ≡ bm mod m,so am − bm ∈ U(R), and thus tm ∈ tm+1R, a contradiction. Item (3) followsimmediately from (1) and (2).

Proof of Proposition 2.10. Let R, t and n be as in the hypothesis of theproposition. By (2) of the lemma above we have a strictly descending sequenceof ideals

R = t0R ⊃ tR ⊃ · · · ⊃ tnR ⊃ tn+1R = 0.


With this we can define an ultranorm | | on R,

|r| ={

2−m if r = 0, m = max{i : r ∈ tiR} ,0 if r = 0

.

Note that |r| ≤ 1 for all r ∈ R, and |r| < 1 if and only if r ∈ m. Since the normtakes only finitely many values, R is complete in this norm. Thus by Hensel’sLemma R is henselian. By Theorem 2.9 there is a ring embedding j : k → Rsuch that jx = x for all x ∈ k, so A := j(k) is a set of representatives modulom. We extend j to a ring morphism jt : k[T ]→ R by sending T to t. By (1) ofthe above lemma the map jt is surjective, and by (2) we have ker jt = (Tn+1).Thus jt induces a ring isomorphism k[T ]/(Tn+1) ∼= R.

Exercises. Prove the following:

1. Let R := k[T ]/(Tn+1), and let t be the image of T under the canonicalmap k[T ]→ R. Then R is a henselian local ring with maximal ideal tR.

2. Let p be a prime number, and R := Z/(pn+1) and let t be the image of punder the canonical map Z → R. Then R is a henselian local ring withmaximal ideal tR.

2.5 The Ax-Kochen Principle

At this point we shall bring a little logic into the picture. In what follows Lt isthe language of rings augmented by a new constant symbol t, that is Lt = L∪{t}.Here is an easy consequence of Proposition 2.10:

2.12. Corollary. For each sentence σ in the language Lt there is a sentenceσn in L such that for any R, n and t as in Proposition 2.10,

(R, t) |= σ ⇐⇒ k |= σn.

Note the resemblance between this result and 1.2 in the introduction. Thiscorollary yields a “baby” Ax-Kochen principle, but for that we need one morelemma:

2.13. Lemma. For each L-sentence τ there is an L-sentence τ∗ such that, forany local ring R,

R |= τ∗ ⇐⇒ k |= τ.

This follows from the fact that the ideal m is definable in R as the set of allnon-units. Call a sentence τ∗ as in this lemma a lift of τ .

The Ax-Kochen Principle says roughly that as the prime number p goes toinfinity, the rings Zp and Fp[[T ]] are more and more alike. The simple versionof this result below says the same about the rings Z/(pn) and Fp[T ]/(T

n) forany fixed n > 0. For example, for n = 2, the integers

a0 + a1p, (a0, a1 ∈ {0, 1, . . . , p− 1})


are in bijection with their images in Z/(p2), and the polynomials

b0 + b1T ∈ Fp[T ], (b0, b1 ∈ Fp)

are in bijection with their images in Fp[T ]/(T2). These two rings clearly resemble

each other: both are henselian local rings with maximal ideal generated by anelement whose square is 0, and both have exactly p2 elements. But they are verydifferent in one respect: Z/(p2) has no subfield, but Fp[T ]/(T

2) has (the imageof) Fp as subfield. The ring Z/(p2) is an arithmetic object, while Fp[T ]/(T

2)has more a geometric flavour.

2.14. Baby Ax-Kochen Principle. Let σ be any sentence of Lt, and letn > 0. Then (

Z/(pn), pn)|= σ ⇐⇒

(Fp[T ]/(T

n), Tn

)|= σ

for all but finitely many primes p, where pn and Tn are the image of p and Tin Z/(pn) and Fp[T ]/(T

n), respectively.

Before we start the proof, we define for a prime number p the L-sentence chp tobe ∀x(p·x = 1), where p·x denotes the term x+· · ·+x where the logical variablex occurs exactly p times. So for any ring R we have: R |= chp ⇐⇒ p·1 /∈ U(R),and thus for any local ring R we have R |= chp ⇐⇒ char k = p.

Proof. Let Σn be a set of axioms in the language Lt whose models are exactlythe pairs (R, t) such that R is a local ring with char k = 0, m = tR, tn−1 = 0,and tn = 0. By the previous corollary we can take an L-sentence σn−1 suchthat for every model (R, t) of Σn we have

(R, t) |= σ ⇐⇒ k |= σn−1.

Let σ∗n−1 be a lift of σn−1 as in the lemma. Then for every model (R, t) of Σn

we have(R, t) |= σ ⇐⇒ R |= σ∗

n−1.

Thus by the completeness theorem there is a formal proof of σ ↔ σ∗n−1 from

Σn. We can assume that Σn is the union of a finite set with an infinite set{¬ch2,¬ch3,¬ch5, . . . , }. Fix a formal proof of σ ↔ σ∗

n−1 from Σn, and takethe natural number N so large that for p > N the axiom ¬chp is not used inthis proof. Then for p > N , the structures

(Z/(pn), pn

)and

(Fp[T ]/(T

n), Tn

)satisfy all axioms of Σn used in this proof, hence(

Z/(pn), pn)|= σ ⇐⇒ Z/(pn) |= σ∗

n−1

⇐⇒ Fp |= σn−1 by lemma 2.13

⇐⇒ Fp[T ]/(Tn) |= σ∗

n−1

⇐⇒(Fp[T ]/(T

n), Tn

)|= σ


2.15. Remark. The proof as given shows that the equivalence holds for allp > N(σ, n), where (σ, n) 7→ N(σ, n) is a computable function with values in N.

Another (less constructive) argument to obtain this result is to show thatfor each non-principal ultrafilter F on the set of primes,∏

F

(Z/(pn), pn

) ∼= ∏F

(Fp[T ]/(T

n), Tn

)(as Lt-structures).

The existence of the isomorphism follows from proposition 2.10: both sides arelocal rings whose maximal ideal is generated by a single element (the interpre-tation of t) whose (n − 1)th power is not zero, but whose nth power is 0, andboth sides have residue field isomorphic to

∏F Fp, which has characteristic 0.

To formulate other applications of the lifting theorem, we introduce someterminology: Let R be a ring, and X = (X1, . . . , Xn) a tuple of distinct indeter-minates. Given any polynomial f(X) ∈ Z[X], let the polynomial fR(X) ∈ R[X]be obtained from f by replacing each of its coefficients by its image under thering morphism Z→ R, and for a ∈ Rn, put f(a) := fR(a) ∈ R. Given polyno-mials f1, . . . , fk ∈ Z[X] a solution of the system f1(X) = · · · = fk(X) = 0 inR is an n-tuple a ∈ Rn such that f1(a) = · · · = fk(a) = 0. If R is a local ring,then a lift of a tuple (x1, . . . , xn) ∈ kn is a tuple (a1, . . . , an) ∈ Rn such thatai = xi for i = 1, . . . , n.

The next lemma is an immediate consequence of the lifting theorem. In therest of this section X = (X1, . . . , Xn) is a tuple of distinct indeterminates withn > 0.

2.16. Lemma. Let R be a henselian local ring with chark = 0, and f1, . . . , fk ∈Z[X]. Then any solution of f1(X) = · · · = fk(X) = 0 in k can be lifted to asolution in R.

The following corollary answered a question of S. Lang from the 1950’s.

2.17. Corollary (Greenleaf-Ax-Kochen). Let f1, . . . , fk ∈ Z[X]. Then forall but finitely many primes p, every solution of

f1(X) = · · · = fk(X) = 0

in Fp can be lifted to a solution in Zp.

Proof. Since f1, . . . , fk are given by L-terms, we can construct an L-sentenceσf that expresses in any local ring the liftability of solutions in the residue field,that is, for for every local ring R:

R |= σf ⇐⇒ for all x ∈ Rn with f1(x), . . . , fk(x) ∈ m there is y ∈ Rn such

that f1(y) = · · · = fk(y) = 0 and yi − xi ∈ m for i = 1, . . . , n.

Let HLR be a set of axioms in the language L whose models are exactly thehenselian local rings, and put HLR(0) := HLR ∪ {¬ch2,¬ch3,¬ch5, . . . }. So


the models of HLR(0) are exactly the henselian local rings with residue field ofcharacteristic 0. By the lemma we have HLR(0) |= σf , so by compactness thereis N ∈ N such that

HLR ∪ {¬chp : p ≤ N} |= σf .

Now observe that Zp is a model of the lefthandside if p > N .

2.18. Theorem (Chevalley-Warning). Let q = pe, e ∈ N>0, and let f1, . . . , fk ∈Fq[X] \ {0} be such that

∑deg fi < n. Then∣∣{x ∈ Fn

q : f1(x) = · · · = fk(x) = 0}∣∣ ≡ 0 mod p;

in particular, if f1, . . . , fk have constant term 0, then there is a nonzero x ∈ Fnq

such that f1(x) = · · · = fk(x) = 0.

See Serre’s Course of Arithmetic for a proof of the Chevalley-Warning theorem.In combination with the Greenleaf-Ax-Kochen theorem it yields:

2.19. Corollary. Let f1, . . . , fk ∈ Z[X] \ {0} all have constant term 0, andsuppose that

∑deg fi < n. Then for all sufficiently large primes p there exists

a nonzero x ∈ Znp such that f1(x) = · · · = fk(x) = 0.

Later in the course we shall prove:

2.20. Ax-Kochen Principle. Let σ be any L-sentence. Then

Zp |= σ ⇐⇒ Fp[[T ]] |= σ

for all but finitely many primes p.

Here is an application that at the time (mid 1960’s) created a stir:

2.21. Given any positive integer d there is N(d) ∈ N such that for all primenumbers p > N(d): if f(X1, . . . , Xn) ∈ Zp[X1, . . . , Xn] is homogeneous of degreed and n > d2, then there exists a nonzero x ∈ Zn

p with f(x) = 0. (It is enoughto consider the case n = d2 + 1.)

This follows by applying the Ax-Kochen Principle to a theorem of Tsen andLang from the 1950’s which says that if f(X1, . . . , Xn) ∈ R[X1, . . . , Xn] is ho-mogeneous of degree d and n > d2, with R = Fp[[T ]] and p any prime number,then there exists a nonzero x ∈ Rn with f(x) = 0. (This is related to theChevalley-Warning theorem.)

Here is some more history: it was known from the 19th century that aquadratic form (= homogeneous polynomial of degree 2) over a p-adic field in 5variables always has a nontrivial zero in that field. Emil Artin had conjecturedthat this remained true for forms of degree d in d2+1 variables. This conjecturewas proved for d = 3 in the 1950s. Ax and Kochen proved the above asymp-totic form of Artin’s conjecture, and around the same time Terjanian gave acounterexample for d = 4.


More on completion. We still need to explain the term complete local ringthat was used in motivating Hensel’s Lemma. Let R be a ring and I ⊆ Ra proper ideal. (Here proper means that I = R.) Consider the decreasingsequence of ideals:

R = I0 ⊇ I1 ⊇ I2 ⊇ · · · .

Recall that In is the set of R-linear combinations of products a1 · · · an with eachai ∈ I. Suppose

∩n I

n = 0. (This is always true if R is local and noetherian.)Define ordI : R→ N ∪ {∞} by:

ordI(r) =

{max{n : r ∈ In} if r = 0∞ if r = 0

.

Note that then for all r, s ∈ R:

• ordI(r + s) ≥ min{ordI(r), ordI(s)},

• ordI(rs) ≥ ordI(r) + ordI(s),

• ordI > 0 ⇐⇒ r ∈ I.

This defines an ultranorm by |r| = 2−ordI(r); note that for all r ∈ R:

|r| ≤ 1, |r| < 1 ⇐⇒ r ∈ I.

A ring R with proper ideal I is said to be I-adically complete if∩

n In = 0 and

R is complete with respect to this norm.

2.22. Lemma. Let R be a ring with maximal ideal m which is m-adicallycomplete. Then R is a local ring (and thus henselian, by Hensel’s lemma).

Proof. Since m = {x ∈ R : |x| < 1}, Lemma 2.6 yields R \m = U(R).

We define a complete local ring to be a local ring R that is m-adically complete.

Assume the proper ideal I in the ring R is finitely generated and∩

n In = 0.

Let R be the completion of R with respect to the I-adic norm. Then the closureof I in R is IR, also denoted by I. More generally, the closure of In in R isInR, which equals (I)n. Furthermore, the natural map

R/In → R/(I)n

is a ring isomorphism. We leave the proofs of these claims as an exercise.Consider now the ring Z, with ideal I = pZ, where p is a prime number.

Although the p-adic norm is defined with base p instead of 2, this yields thesame topology on Z as the I-adic norm. As rings we can identify Zp with the

I-adic completion Z such that the p-adic norm is a fixed power of the I-adicnorm: |x|p = |x|log2 p for x ∈ Zp.

3 VALUATION THEORY 20

3 Valuation theory

We begin this section with developing basic valuation theory, and end it with aquantifier elimination for algebraically closed valued fields.

A valuation on an (additively written) abelian group A is a map

v : A \ {0} → Γ

into a linearly ordered set Γ, such that v(−x) = vx for nonzero x ∈ A andv(x+ y) ≥ min{vx, vy} for x, y ∈ A such that x, y, x+ y = 0.

By convention we extend v to all of A by setting v0 = ∞, where ∞ isnot in Γ. We extend < to a linear order on Γ∞ = Γ ∪ {∞} by specifyingγ < ∞ for all γ ∈ Γ. Then v(−x) = vx and v(x + y) ≥ min{vx, vy} for allx, y ∈ A without exception. For future use, note that if x, y ∈ A and vx < vy,then v(x + y) = vx. (Use that vx = v(x + y − y) ≥ min{v(x + y), vy}.)More generally, if x1, . . . , xn ∈ A, n ≥ 1, and v(x1) < v(x2), . . . , v(xn), thenv(x1 + · · ·+ xn) = v(x1). Another elementary fact that is often used is that

{(x, y) ∈ A×A : v(x− y) > vx}

is an equivalence relation on A \ {0}. (The reader should check this.)

Our real interest is in valuations on fields, and then Γ is not just a linearlyordered set but has also a compatible group structure:

An ordered abelian group is an abelian (additively written) group Γ equippedwith a translation invariant linear order <. Such Γ is called archimedean if, forany a, b > 0, there is n ∈ N such that na > b. (Examples: The additive groupsof Z,Q,R with the usual ordering are archimedean ordered abelian groups.)Note that ordered abelian groups are torsion-free.

The following result has historic roots in Eudoxos, Euclid, and Archimedes,but its present formulation is due to O. Holder and became only possible afterDedekind’s foundation of the real number system in the 19th century. It saysthat archimedean ordered abelian groups can be embedded into the orderedadditive group of real numbers.

3.1. Lemma. Let Γ be an archimedean ordered abelian group, and γ ∈ Γ>0.Then there is a unique embedding i : Γ→ R of ordered groups such that iγ = 1.If there is no β ∈ Γ with 0 < β < γ, then iΓ = Z.

Proof. If Γ has γ as least positive element, then Γ = Zγ, and the embedding isgiven by i(kγ) = k for k ∈ Z. If Γ has no least positive element, the embeddingis given by iβ := sup{ kn : k ∈ Z, n > 0, nβ < kγ} for β ∈ Γ. The reader shouldfill in the details.

3.2 Definition. A valuation on a domain A is a function v : A \ {0} → Γ,where Γ is an ordered abelian group, such that for all x, y ∈ A \ {0}:

(V1) v(x+ y) ≥ min{vx, vy}, provided x+ y = 0,


(V2) v(xy) = vx+ vy,

Given a domain A we have a bijective correspondence between the set ofultrametric absolute values on A and the set of valuations v : A \ {0} → R,with the ultrametric absolute value | | on A corresponding to the valuationv : A \ {0} → R given by v(a) := − log |a|.

Let v : A \ {0} → Γ be a valuation on the domain A. Note that thenv1 = v(−1) = 0, since v restricted to U(A) is a group morphism. Hencevx = v(−x) for all x ∈ A \ {0}, so v is a valuation on the additive group asdefined earlier. We extend v uniquely to a valuation v : K× → Γ on the fractionfield K of A, by

v(x/y) = vx− vy, x, y ∈ A \ {0}.

Thus v(K×) is a subgroup of Γ. By convention we extend v to all of K bysetting v0 = ∞, where Γ∞ = Γ ∪ {∞} is linearly ordered as before, and wherewe extend + to a binary operation on Γ∞ by γ +∞ = ∞ + γ = ∞ for allγ ∈ Γ∞. Then (V1) and (V2) hold for all x, y ∈ K without the proviso in (V1).

When we refer to a valuation v : K× → Γ on a field K, we shall assumefrom now on that v(K×) = Γ, unless specified otherwise. We call Γ = v(K×)the value group of the valuation.

Motivating Example. Let K = C(t). Each point a ∈ C gives rise to avaluation va : K× → Z by:

va(f(t)) = k if f(t) = (t− a)kg(t)

h(t),

where g, h ∈ C[t] are such that g(a), h(a) = 0. Note that if k > 0, then f(a) = 0and k is the order of vanishing of f at a; if k = 0, then f(a) ∈ C×; and ifk < 0, then f has a pole at a of order −k. Geometrically it is natural toconsider f ∈ K× as defining a meromorphic function on the Riemann sphereC∞ := C ∪ {∞}, and to assign also to the point ∞ on this sphere a valuationv∞ : K× → Z,

v∞(f(t)) = deg h(t)− deg g(t) if f(t) =g(t)

h(t), g, h ∈ C[t].

These valuations are related by the fundamental identity∑a∈C∞

va(f) = 0, f ∈ K×.

(This identity clearly holds for f ∈ C[t] \ {0}, and the general case then followseasily.) This family of valuations is an important tool to understand the fieldK = C(t); the same role is played for other so-called global fields by a familyof valuations satisfying a similar identity. In this course, however, we focus onfields with a single valuation, that is, we study local aspects of fields.

With this example in mind, given a valued field (K, v), we think of K as afield of (meromorphic) functions on some space in which v is associated with a


certain point of the space, where vf > 0 indicates that the “function” f vanishesat that point, vf measuring the order of vanishing. Likewise, vf < 0 meansthat f has a pole at that point (and v(f−1) is the order of vanishing of f−1).Indeed, we shall regard v itself as an abstract point at which the elements ofK can be evaluated. This will become clear in the discussion after the nextdefinition.

3.3 Definition. Let v : K× → Γ be a valuation on a field.

1. Ov := {x ∈ K : vx ≥ 0}, a subring of K.

2. mv := {x ∈ K : vx > 0}, an ideal in Ov.

3. kv := Ov/mv.

Note that vx = 0 ⇐⇒ v(x−1) = 0, for x ∈ K×, so U(Ov) = Ov \mv. ThereforeOv is a local ring with maximal ideal mv, and kv is the residue field.

When thinking of K as a field of functions as above, Ov would be the set offunctions which have no pole at (the point corresponding to) v, and mv wouldbe the set of functions that vanish at v. The residue map Ov → kv is thento be thought of as evaluation at v. We extend the residue map to a mapK → kv ∪ {∞} by sending all elements of K \ Ov to ∞.

Note that we have the following disjoint union in K,

K = mv

�∪ U(Ov)︸︷︷︸Ov

�∪ (mv \ {0})−1.

For each x ∈ K×, either x ∈ Ov or x−1 ∈ Ov, so Ov is a valuation ring in K.We have, therefore, two maps:

Kv //

��

Γ ∪ {∞}

kv ∪ {∞}

The raison d’ etre of valuation theory is in analyzing K in terms of the (usually)simpler objects Γ and kv via these two maps.

Terminology. We call v trivial if Γ = {0}, discrete if Γ ∼= Z as ordered groups,and of rank 1 if Γ = 0 and Γ is archimedean.

If v is discrete, say Γ = Z, take t ∈ K with vt = 1. Then mv = tOv, and eacha ∈ K× can be written uniquely as a = utk with u ∈ O×

v and k ∈ Z. (Simplytake k = v(a), and u = a/tk.) The only non-trivial ideals of Ov are those of theform tnOv. In particular Ov is a principal ideal domain.

Examples.1. Let K = C(t), v = v0 : K× → Z. Then vt = 1 and

Ov =

{f

g: f, g ∈ C[t], g(0) = 0

}= C[t]tC[t].


We have a surjective ring morphism

Ov → C :f(t)

g(t)7→ f(0)

g(0), (f, g ∈ C[t], g(0) = 0)

which has kernel mv, and thus induces a field isomorphism kv = Ov/mv∼= C.

It is traditional to identify kv with C via this isomorphism.2. Let K = Qp. Here we have the p-adic valuation v = vp : K× → Z defined

by vx = k if x = pku, u ∈ U(Zp), which is related to the p-adic absolute valueby |x|p = p−vx. We have Ov = Zp, v(p) = 1, mv = pZp, and kv = Zp/pZp = Fp.

For each ordered abelian group Γ and field k, there is a field K = k((tΓ)) withvaluation v having Γ as the value group and k as the residue field. To constructk((tΓ)) we need some basic facts on well-ordered subsets of Γ.

Recall that a linearly ordered setA is said to be well-ordered if each nonemptysubset of A has a least element, equivalently, there is no strictly decreasing se-quence (αn) in A.

Lemma. Let A,B ⊆ Γ be well-ordered (by the ordering of Γ). Then A ∪ B iswell-ordered, the set A+ B := {α+ β : α ∈ A, β ∈ B} is well-ordered, and foreach γ ∈ Γ there are only finitely many (α, β) ∈ A×B such that α+ β = γ.

Lemma. Let A ⊆ Γ>0 be well-ordered. Then

[A] := {α1 + · · ·+ αn : α1, . . . , αn ∈ A}

is also well-ordered, and for each γ ∈ [A] there are only finitely many tuples(n, α1, . . . , αn) with α1, . . . , αn ∈ A such that γ = α1 + · · ·+ αn.

We now define K = k((tΓ)) to be the set of all formal series f(t) =∑

γ∈Γ aγtγ ,

with coefficients aγ ∈ k, such that the support,

supp(f) = {γ ∈ Γ : aγ = 0},

is a well-ordered subset of Γ. Using the first lemma on well-ordered subsetsof Γ we can define binary operations of addition and multiplication on the setk((tΓ)) as follows:∑

aγtγ +

∑bγt

γ =∑

(aγ + bγ)tγ(∑

aγtγ)(∑

bγtγ)

=∑γ

( ∑α+β=γ

aαbβ

)tγ .

Note that K is a ring, even a domain, with k as a subfield via the identificationa 7→ at0. To show K is a field we shall use the second lemma.

Define v : K \ {0} → Γ by

v(∑

aγtγ)= min{γ : aγ = 0}.


Then v is a valuation on K, and if v(f) > 0, then by the second lemma∑∞

n=0 fn

makes sense as an element of K: for any γ ∈ Γ there are only finitely many nsuch that the coefficient of tγ in fn is not zero. Note that then

(1− f)∞∑

n=0

fn = 1.

It now follows in the usual way that K is a field: write an arbitrary g ∈ K \ {0}as g = ctγ(1− f), with c ∈ k× and v(f) > 0. Then g−1 = c−1t−γ

∑n f

n.The valuation ring is

Ov = {f ∈ K : supp(f) ⊆ Γ≥0}.

For f =∑

aγtγ in K, call a0 the constant term of f . The map sending f to its

constant term sends Ov onto k, and this is a ring homomorphism. Its kernel is

mv = {f ∈ K : supp(f) ⊆ Γ>0},

so this induces an isomorphism Ov/mv → k.For Γ = Z, the field k((tZ)) is the usual field of Laurent series, i.e. k((t)),

with valuation ring k[[t]]. An early study of the fields k((tΓ)) is due to H. Hahn(of “Hahn–Banach” fame and Godel’s thesis advisor), so these valued fields areoften referred to as Hahn fields.

Topology induced by a valuation. An ultrametric space is a set X with amap v : X2 → Γ∞, where Γ∞ is a linearly ordered set with largest element ∞,such that, for all x, y, z ∈ X and γ ∈ Γ∞:

(i) v(x, y) =∞ ⇐⇒ x = y,

(ii) v(x, y) = v(y, x),

(iii) v(x, y), v(y, z) ≥ γ =⇒ v(x, z) ≥ γ.

Example. An abelian group A with valuation v is an ultrametric space withrespect to v(x, y) := v(x− y).

Let X be an ultrametric space as above. For a point a ∈ X and γ ∈ Γ∞, wedefine two kinds of disks:

Da(γ) = {x ∈ X : v(x, a) > γ}, an open disk

Da(γ) = {x ∈ X : v(x, a) ≥ γ}, a closed disk.

We refer to Da as the open disk centered at a with valuation radius γ, andto Da(γ) as the closed disk centered at a with valuation radius γ. The readershould verify that Da(γ) = Db(γ) for each b ∈ Da, and Da(γ) = Db(γ) for eachb ∈ Da, in stark contrast to disks in the euclidean plane which have a uniquecenter. It follows easily that if D,E are disks (of any kind) with non-emptyintersection, then D ⊆ E or E ⊆ D. The open disks form a basis for a topology


on X, the v-topology. It is easy to see that all disks are both open and closed inthe v-topology, and that X with the v-topology is a hausdorff space. A field Kwith valuation v is a topological field with respect to the v-topology, that is, thefield operations +,−, · : K2 → K, and x 7→ x−1 : K× → K× are continuous.Note that then Ov is the closed disk centered at 0 with valuation radius 0, andthat the open disks with valuation radius 0 contained in Ov are exactly thecosets a + mv with a ∈ O. In particular, Ov contains exactly |kv| open disksof valuation radius 0. Indeed, any closed disk of valuation radius γ ∈ v(K×)contains exactly |kv| open disks of valuation radius γ: by a translation one canassume the closed disk contains 0, and then multiplication by an a ∈ K withva = −γ makes the closed disk equal to Ov.

The next result is due to Krull. The short proof below was found by Gravett.

3.4. Proposition. Let v : K× → Γ be a valuation on the field K, and k := kv.Then the cardinality of K is bounded in terms of the cardinalities of k and Γ:|K| ≤ |k||Γ|.

Proof. Let γ ∈ Γ, and define equivalence relations on K by

xγy ⇐⇒ v(x− y) > γ, xγy ⇐⇒ v(x− y) ≥ γ.

Thus the γ-classes are the open disks of valuation radius γ, and the γ-classesare the closed disks of valuation radius γ. Each γ-class D is contained in aunique γ-class D, and each γ-class contains exactly |k|-many γ-classes. Let Dγ

be the set of all γ-classes. Then we have a map fγ : Dγ → k such that for allD,E ∈ Dγ , if D = E and fγ(D) = fγ(E), then D = E.

We now associate to each x ∈ K the function γ 7→ x(γ) : Γ → k by x(γ) =fγ(D) where D ∈ Dγ contains x. Then the map K → kΓ that sends eachx ∈ K to the function γ 7→ x(γ) is injective: if x, y ∈ K and x = y, then forγ := v(x− y) we have x(γ) = y(γ) since x and y are in the same closed disk ofvaluation radius γ but in different open disks of valuation radius γ.

Correspondence between valuation rings and valuations. A valuationring of a field K is a subring A of K such that for each x ∈ K×, either x ∈ Aor x−1 ∈ A. Note that then A is indeed a valuation ring, with K as field offractions.

Let A be a valuation ring of the field K. Recall that we have the disjointunion

K = mA ∪ U(A) ∪ (mA \ {0})−1.

Consider the (abelian) quotient group ΓA = K×/U(A), written additively. Thebinary relation ≥ on ΓA defined by

xU(A) ≥ yU(A) ⇐⇒ x/y ∈ A, (x, y ∈ K×)

makes ΓA into an ordered abelian group, and the natural map

vA : K× → ΓA, vAx = xU(A)


is a valuation. (Check this!) Note that OvA= A.

Every valuation v : K× → Γ with valuation ring Ov = A is equivalent to vAas follows: there is a unique isomorphism of ordered abelian groups i : ΓA → Γsuch that i ◦ vA = v. More generally, two valuations v1 : K× → Γ1 andv2 : K× → Γ2 are said to be equivalent if there is an isomorphism of orderedabelian groups i : Γ1 → Γ2 such that i ◦ v1 = v2; note that such an i is thenuniquely determined. The reader should check that two valuations v1 and v2 ona field are equivalent iff they have the same valuation ring.

A valuation ring A is said to be discrete if the valuation vA on its fractionfield K is discrete. In that case there is a unique valuation v : K× → Z suchthat Ov = A; this v is called the normalized valuation of A. (For example,the value group of a discrete valuation v on a field could be 1

2Z, and then 2vwould be the corresponding normalized valuation.) We also write DVR insteadof discrete valuation ring. Note that Zp and k[[t]] are DVR’s.

3.5 Definition. A valued field is a pair (K,A), where K is a field and A is avaluation ring of K.

If (K,A) is a valued field and E a subfield of K, then A∩E is a valuation ringof E. Note: (E,A ∩ E) ⊆ (K,A) (where “⊆” means “is a substructure of”).

Exercises. Let (K,A) be an algebraically closed valued field. Show that theresidue field kA is algebraically closed and the value group ΓA is divisible.

Let (K,A) be a valued field, and L ⊇ K a field extension. Then a valuation ringB of L lies over A (or dominates A) if A = B∩K, equivalently, (K,A) ⊆ (L,B).Note that then mA = mB∩K, so we have an induced embedding of residue fields,A/mA → B/mB , by means of which kA = A/mA is identified with a subfieldof kB = B/mB . Also U(B) ∩ K = U(A), so we have an induced embeddingvA(x) 7→ vB(x) : ΓA → ΓB (x ∈ K×), of ordered abelian groups, by means ofwhich ΓA is identified with an ordered subgroup of ΓB.

3.1 Valuation rings and integral closure

We begin with recalling some terminology and basic facts concerning rings andideals. (See for example S. Lang’s book “Algebra” for more details.)

Let A ⊆ B be rings. For an ideal I of A we let

IB := {a1b1 + · · ·+ anbn : a1, . . . , an ∈ I, b1, . . . , bn ∈ B}

be the ideal of B generated by I. If p and q are prime ideals of A and Brespectively, we say that q lies over p if q ∩ A = p. In that case we have a ringembedding a+ p 7→ a+ q : A/p→ B/q.

We call an element b ∈ B integral over A if b is a zero of a monic polynomialover A, that is,

bn + a1bn−1 + · · ·+ an = 0

for suitable n > 0 and a1, . . . , an ∈ A. (Then A[b] = A + Ab + · · · + Abn−1, soA[b] is a finitely generated A-module. The converse is also true: if b ∈ B and


A[b] is a finitely generated A-module, then b is integral over A.) The elementsof B that are integral over A form a subring of B, called the integral closure ofA in B. We call A integrally closed in B if each b ∈ B that is integral over Alies already in A.

Given rings A ⊆ B ⊆ C, the ring C is integral over A iff C is integral overB and B is integral over A.

Exercise. Let A ⊆ B be domains, and suppose B is integral over A. Show: Ais a field iff B is a field.

As before, let A ⊆ B be rings, and suppose B is integral over A. Let p be aprime ideal of A. Then there is a prime ideal q of B that lies over p. Moreover,for each such q, the ideal p is maximal iff q is maximal. (Explanation: For suchq we have a natural ring embedding A/p→ B/q, and after identifying A/p withits image in B/q, this last domain is integral over the domain A/p. Hence bythe exercise above, A/p is a field iff B/q is a field.)

We say that a domain A is integrally closed if it is integrally closed in itsfield of fractions.

Given a domain A with fraction field K, and a prime ideal p of A we let

Ap :=

{x

y∈ K : x, y ∈ A, y /∈ p

}be the localization of A with respect to p. It is easy to check that Ap is a localdomain with maximal ideal pAp, and that this maximal ideal lies over p.

3.6. Proposition. Valuation rings are integrally closed.

Proof. Suppose A is a valuation ring of the field K, and let v = vA be thecanonical valuation. To show A is integrally closed, let x ∈ K× be such thatxn + an−1x

n−1 + · · ·+ a0 = 0 with n > 0 and all ai ∈ A. If x /∈ A, then vx < 0,hence

v(xn) = nv(x) < iv(x) + v(ai) = v(aixi), i = 0, . . . , n− 1,

hence v(xn+· · ·+a0) = v(xn) < 0. But v(xn+· · ·+a0) =∞, a contradiction.

3.7. Corollary. Let (K,A) be an algebraically closed valued field. Then A ishenselian.

Proof. Consider a polynomial f(Y ) = Y n + Y n−1 + a2Yn−2 + · · · + an with

n ≥ 2 and a2, . . . , an ∈ A. By Lemma 2.2 it is enough to show that f has a zeroin U(A). By the proposition above, f =

∏ni=1(Y − αi) with all αi ∈ A. Now∑

i αi = −1, so at least one αi must be outside m, and thus be in U(A).

3.8 Definition. Given local rings A,B, we say B dominates A (or that B liesover A) if A ⊆ B and mA ⊆ mB (hence mB ∩A = mA).

One should check that this agrees with our definition of domination forvaluation rings in the previous subsection.

We now state the main results to be proved in this subsection.


3.9 Proposition. Let A be a local subring of the field K. Consider the class ofall local subrings of K that dominate A, partially ordered by domination. Anymaximal element of this class is a valuation ring of K.

In particular, given any local subring A of a field K, there is a valuation ringof K lying above A.

3.10 Proposition. Let A be a local subring of a field K. The integral closureof A in K is the intersection of all valuation rings of K that dominate A.

3.11 Proposition. Let A be a local domain integrally closed in K = Frac(A),and let L be a normal field extension of K. Let B be the integral closure of Ain L. Then, given any maximal ideals n and n′ of B there exists σ ∈ Aut(L|K)such that σ(n) = n′. (Note: all maximal ideals in B lie over mA.)

3.12 Proposition. Let A be a valuation ring of the field K, let L ⊇ K be analgebraic field extension, and let B be the integral closure of A in L. Then everyvaluation ring of L dominating A is of the form Bn for some maximal idealn ⊆ B.

3.13. Corollary. With the same assumption as in the proposition, we have abijection n 7→ Bn from the set of maximal ideals in B onto the set of valuationrings in L dominating A. If in addition L is a normal field extension of K,then for any valuation rings V, V ′ of L dominating A there exists σ ∈ Aut(L|K)such that σ(V ) = V ′.

Proof. We first show that, given any maximal ideal n of B, the localization Bn

is a valuation ring of L dominating A. Note that n lies over mA, hence Bn

dominates A. Since L is the fraction field of B, there are no local subrings of Lthat dominate Bn. Thus by proposition 3.9, Bn is a valuation ring of L.

The surjectivity of n 7→ Bn now follows from Proposition 3.12. Injectivityis left to the reader. The second assertion of the corollary follows from thiscorrespondence and Proposition 3.11.

3.14. Corollary. Let A be a valuation ring of K, let K be an algebraic closureof K and A a valuation ring of K lying over A. Then any valued field embed-ding (K,A)→ (F,C), where F is algebraically closed, extends to an embedding

(K, A)→ (F,C).

Proof. Let j : K → F be a field embedding that extends the field embeddingK → F . Then j−1(C) is a valuation ring of K lying over A. Hence there is

σ ∈ Aut(K|K) such that j−1(C) = σA. Then jσ is a valued field embedding of

(K, A) into (F,C).

It remains to prove 3.9–3.12. For the first two we use the next lemma:

3.15. Lemma. Let A be a local subring of the field K, and suppose x ∈ K× issuch that 1 ∈ mA[x−1] + x−1A[x−1]. Then x is integral over A.


Proof. We have 1 = anx−n + · · ·+ a1x

−1 + a0 with a1, . . . , an ∈ A and a0 ∈ m.Multiplying both sides by xn yields:

xn(1− a0) + (terms of lower degree in x) = 0.

Since 1− a0 is a unit in A, it follows that x is integral over A.

Proof of 3.9. Let V be a maximal element in the class of local subrings of Kthat dominate A, and let x ∈ K×. Put m := mV .

Consider first the case that x is integral over V . Then V [x] is integral overV , so we can take a maximal ideal n of V [x] that lies over m, and then V [x]n isa local subring of K dominating V . By V ’s maximality this gives x ∈ V .

Next, suppose x is not integral over V . Then by the above lemma we have1 /∈ mV [x−1] + x−1V [x−1], so we have a maximal ideal n of V [x−1] such thatn ⊇ m, and thus n ∩ V = m. The local subring V [x−1]n of K dominates V , sothe maximality of V yields x−1 ∈ V .

Proof of 3.10. Any x ∈ K integral over A lies in every valuation ring of Kcontaining A as a subring. Next, suppose x ∈ K× is not integral over A. Bythe lemma above, mA[x−1] + x−1A[x−1] is a proper ideal of A[x−1]. So we cantake a maximal ideal n ⊇ m of A[x−1] that contains x−1. This yields a localsubring A[x−1]n of K that dominates A. Let V be a maximal element of theclass of local subrings of K that dominate A[x−1]n. Then V is a valuation ring(by Proposition 3.9), V dominates A, and x−1 ∈ mV , so x /∈ V .

In the next proof we use the Chinese Remainder Theorem: Let A be an abelian(additive) group with subgroups B1, . . . , Bn (n > 0) such that Bi + Bj = A forall i, j with 1 ≤ i < j ≤ n, and let a1, . . . , an ∈ A. Then there exists x ∈ A suchthat x− ai ∈ Bi for i = 1, . . . , n.

Proof of 3.11. The general case follows from the case [L : K] < ∞ (how?), sowe assume [L : K] <∞ below.

Let n, n′ be two maximal ideals in B, lying over m, and assume towards acontradiction that σn = n′ for all σ ∈ G := Aut(L|K). Thus the two sets ofmaximal ideals of B,

{σn : σ ∈ G}, {σn′ : σ ∈ G}

are disjoint; since [L : K] < ∞, these two sets are also finite. By the ChineseRemainder Theorem there is x ∈ B such that

x ≡ 0 mod σn, x ≡ 1 mod σn′

for all σ ∈ G. Hence σx ∈ n \ n′ for all σ ∈ G. Recall that

NLK(x) := (

∏σ∈G

σx)ℓ

lies inK, where ℓ = 1 if charK = 0, and ℓ = pe for some e ∈ N if charK = p > 0.Each σx is integral over A, so NL

K(x) ∈ A. Also NLK(x) ∈ n \ n′ because n and

n′ are prime ideals, hence NLK(x) ∈ A ∩ n = m ⊆ n′, a contradiction.


Proof of 3.12. Let V be a valuation ring of L lying over A. Valuation rings areintegrally closed, so B ⊆ V . We claim that V = Bn where n := mV ∩B. ClearlyB \ n ⊆ V \ mV , hence Bn ⊆ V . For the other inclusion, let x ∈ V , x = 0.We have a relation anx

n + · · · + a0 = 0, where a0, . . . , an ∈ A and an = 0.Take s ∈ {0, . . . , n} maximal such that vA(as) = mini vA(ai). Put bi := ai/as.Dividing by asx

s yields

(bnxn−s + · · ·+ bs+1x+ 1)︸︷︷︸

y

+x−1 (bs−1 + · · ·+ b0/xs−1)︸︷︷︸

z

= 0.

So y = −(z/x), and thus −xy = z. Since bi ∈ mA for s < i ≤ n, we havey ∈ U(V ), and hence y /∈ n. Thus to get x ∈ Bn it suffices to show thaty, z ∈ B. This will follow if we show that y, z lie in every valuation ring ofL dominating A. If such a ring contains x, it also contains y, hence containsz = −yx. If such a ring contains x−1, it contains z, hence also y = −zx−1.

3.16 Proposition. Let A be a local ring, f(X) ∈ A[X] a monic polynomial ofdegree d > 0, and A[x] := A[X]/(f) with x := X + (f). Let f(X) ∈ k[X] factorin k[X] as

f =n∏

i=1

fiei

where each ei > 0, each fi ∈ A[X] is monic, and f1, . . . , fn are irreducible ink[X] and distinct. Put mi := (m, fi(x))A[x]. Then m1, . . . ,mn are exactly thedistinct maximal ideals of A[x], and A[x]/mi

∼= k[X]/(fi) (as rings) for each i.

Proof. For each i, consider the canonical morphisms,

A[x] = A[X]/(f)π−→ k[X]/(f)

πi−→ k[X]/(fi).

Note that kerπ = mA[x], and kerπi = fi + (f). Therefore

ker(πiπm) = (m, fi(x)) = mi,

since fi(x) is the preimage of fi + (f) in A[x]. Because k[X]/fi is a field,(m, fi(x)) is maximal in A[x]. Note that fi = fj implies fi = 0 under πjπm,since fi ≡ 0 mod fj . Thus m1, . . . ,mn are distinct. It remains to show that themi’s are the only maximal ideals in A[x]. First note that the ring A[x] is integralover A, so each maximal ideal in A[x] lies over m. The polynomial

∏feii − f is

in m[X], and f(x) = 0, so∏

f(x)ei ∈ mA[x]. Thus each maximal ideal of A[x]contains

∏fi(x)

ei , hence contains some fi(x), and thus is one of the mi’s.

3.2 Quantifier elimination

We shall prove quantifier elimination for algebraically closed valued fields usingCorollary 3.14. We also need to know some ways to extend a valuation on afield K to K(x) where x is transcendental over K. This is related to the nextresult, which is a fundamental inequality for valued field extensions. Given a


field extension K ⊆ L we let [L : K] be its degree, that is, the dimension ofL as a vector space over K; to keep things simple we set [L : K] = ∞ if thisdimension is infinite. Likewise, given an extension Γ ⊆ Γ′ of abelian groups welet [Γ′ : Γ] be its index, which by convention is ∞ if Γ′/Γ is infinite.

3.17 Proposition. Let (K,A) ⊆ (L,B) be valued fields, with correspondinginclusions kA ⊆ kB and ΓA ⊆ ΓB between their residue fields and value groups.Then:

[L : K] ≥ [kB : kA] · [ΓB : ΓA].

Proof. Let b1, . . . , bp ∈ B be such that b1, . . . , bp are linearly independent overkA, p ≥ 1. Likewise, let c1, . . . , cq ∈ L× be such that vc1, . . . , vcq lie in distinctcosets of ΓA, q ≥ 1. It is enough to show that then [L : K] ≥ pq; this will followfrom the K-linear independence of the family (bicj), which in turn follows fromthe identity:

v(∑

aijbicj) = mini,j

v(aijbicj) = mini,j{v(aij) + v(cj)} (all aij ∈ K) (2)

First we show that v(a1b1 + · · ·+ apbp) = mini v(ai), where a1, . . . , ap ∈ K. Wecan assume that some ai = 0, and then dividing by an ai of minimum valuation,we can reduce to the case that all ai are in A and v(ai) = 0 for some i. Wemust show that then v(a1b1 + · · ·+ apbp) = 0. We have:

a1b1 + · · ·+ apbp = a1b1 + · · ·+ apbp = 0,

since b1, . . . , bp are linearly independent over kA and ai = 0 for some i. Thisproves what we want. Next, to prove (2) we can assume that for each j there isi such that aij = 0. Then for any j,

v(∑

i

aijbicj)

= v((∑i

aijbi)cj)

= v(∑i

aijbi) + vcj

= mini

v(aij) + vcj

∈ ΓA + vcj .

For j1 = j2 we have ΓA + vcj1 = ΓA + vcj2 , so

v(∑

i

aij1bicj1)= v

(∑i

aij2bicj2),

hence,

v(∑

aijbicj) = minj{min

iv(aij) + vcj}

= mini,j{v(aij) + vcj}


For algebraic extensions this yields:

3.18. Corollary. If [L : K] = n, then [kB : kA] ≤ n (so kB is algebraic overkA), and [ΓB : ΓA] ≤ n (so mΓB ⊆ ΓA for some m ∈ {1, . . . , n}.

If L is an algebraic closure of K, then kB is an algebraic closure of kA, andΓB is a divisible hull of ΓA.

For simple transcendental extensions it is the proof of the fundamental inequal-ity that really matters, as the next two lemmas will show.

3.19. Lemma. Let (K,A) be a valued field, and let L = K(x) be a fieldextension with x transcendental over K. There is a unique valuation ring B ofL lying over A, such that x ∈ B, and x is transcendental over kA. Moreover,this B satisfies kB = kA(x) and ΓB = ΓA.

Proof. Let B be a valuation ring as in the lemma. Then 1, x, x2, . . . are linearlyindependent over kA, so by the proof of Proposition 3.17,

vB(f0 + f1x+ · · ·+ fnxn) = min

ivA(fi) (fi ∈ K).

Thus vB is uniquely determined by vA on K[x], hence on L. In particular, thereis at most one B as in the lemma.

(Existence.) Define v : K[x] \ {0} → ΓA by:

v(f0 + f1x+ · · ·+ fnxn) = min

ivA(fi) (fi ∈ K).

Claim. v is a valuation on K[x] (hence extends to a valuation on L).

It is clear that (V1) is satisfied. We need to check (V2). Let f, g ∈ K[x]\{0},so f =

∑i fix

i, g =∑

j gjxj (fi, gj ∈ K). Take i0 minimal such that v(fi0) =

mini v(fi), and take j0 minimal such that v(gj0) = minj v(gj). Then

fg =∑n

(∑

i+j=n

figj)xn,

and for each n

v(∑

i+j=n

figj) ≥ mini+j=n

{v(fi) + v(gj)} ≥ v(f) + v(g).

Now consider∑i+j=i0+j0

figj = fi0gj0 + terms figj with i < i0 or j < j0︸︷︷︸each has valuation > v(fi0) + v(gj0)

Therefore:v(

∑i+j=i0+j0

figj) = v(fi0) + v(gj0) = v(f) + v(g).

This finishes the proof of the claim. Put B := Ov. We still need to check:


Claim. x ∈ B, and x is transcendental over kA.

To see this, note that v(x) = 0, so x ∈ B. Let a1, . . . , an ∈ A be such thata1, . . . , an are not all zero. Then v(xn + a1x

n−1 + · · · an) = 0, hence

xn + a1xn−1 + · · ·+ an = 0.

This proves that x is transcendental over kA.

Claim. kB = kA(x).

Let b ∈ B be such that v(b) = 0. Write b = f(x)g(x) , with f, g ∈ K[x] \ {0},

so vf = vg. Dividing all coefficients of f and g by a coefficient of g of minimalvaluation, we may assume vg = 0, so vf = 0. Then f(x) = 0, g(x) = 0. Hencebg(x) = f(x) yields bg(x) = f(x), so

b =f(x)

g(x)∈ kA(x),

as desired.Finally, it is clear from the above that ΓB = ΓA.

3.20. Lemma. Let v : K× → Γ be a valuation on a field K, and let L = K(x)be a field extension with x transcendental over K. Let δ, in some ordered abeliangroup extending Γ, satisfy nδ ∈ Γ for all n > 0. Then v extends uniquely to avaluation w : L× → Γ + Zδ such that w(x) = δ. Moreover, kv = kw.

3.21. Remark. If Γ is a divisible ordered abelian group, δ is in some orderedabelian group extension, and δ ∈ Γ, then nδ ∈ Γ for all n > 0.

Proof. Let w be as in the lemma. Then

w(1) = 0, w(x) = δ, w(x2) = 2δ, . . .

lie in different cosets of Γ, so by the proof of Proposition 3.17,

w(f0 + f1x+ · · ·+ fnxn) = min

i{v(fi) + iδ} (all fi ∈ K). (3)

Thus w is completely determined by v.(Existence.) Define w : K[x] \ {0} → Γ + Zδ by (3). Again it is clear that

(V1) is satisfied. To verify (V2) we mimic the proof of the previous lemma: takei0 minimal such that w(fi0) = mini{v(fi) + iδ}, and choose j0 minimal suchthat w(gj0) = minj{v(gj) + jδ}. Put n0 := i0 + j0. First note that

w((∑

i+j=n

figj)xn) ≥ min

i+j=n{v(fi) + v(gj) + nδ} ≥ w(f) + w(g).

Next,∑n≤n0

(∑

i+j=n

figj)xn = fi0gj0x

n0 + terms figjxn with i < i0 or j < j0︸︷︷︸

valuation > v(fi0) + v(gj0) + n0δ

.


Therefore:

w((∑

i+j=n0

figj)xn) = v(fi0) + v(gj0) + n0δ = w(f) + w(g).

So w is a valuation on K[x], hence yields a valuation on L extending v andsatisfying w(x) = δ.

It is an instructive exercise to prove kv = kw.

3.22. Let (K,A) be an algebraically closed valued field with A = K, andconsider it as a structure for the language of rings with a unary relation symbolU (interpreted as the set A). Then the binary relation “vA(x) ≤ vA(y)” on Kis definable in (K,A) by the formula ∃z

(U(z) ∧ xz = y

). However, one can

show that this relation is not qf-definable in (K,A), even if we allow names forthe elements of K to be used. Thus QE for algebraically closed valued fieldsneeds a different language. It turns out that the binary relation above is theonly obstruction to QE in the language above.

3.23 Definition. A valuation divisibility on a domain R is a binary relation |on R such that for all x, y, z ∈ R

(i) not 0 | 1;

(ii) x | y and y | z =⇒ x | z;

(iii) x | y and x | z =⇒ x | y + z;

(iv) x | y ⇐⇒ xz | yz for z = 0;

(v) x | y or y | x.

3.24. Lemma. Valuation divisibilities and valuations are related as follows:

(i) Each valuation ring A of a field K gives rise to a valuation divisiblity |Aon K by

x |A y ⇐⇒ vA(x) ≤ vA(y) ( ⇐⇒ ax = y for some a ∈ A).

(ii) Given a field K, the map

A 7→ |A : { valuation rings of K } −→ { valuation divisibilities on K }

is a bijection.

(iii) For each valuation divisibility | on a domain R there is a unique valuationdivisiblity |′ on K = Frac(R) such that (R, |) ⊆ (K, |′); it is given by

a1b|′ a2

b⇐⇒ a1 | a2 (a1, a2, b ∈ R, b = 0).


(iv) The valuation divisibilities on Z are exactly the |p (p a prime number) and|t, where |t is the trivial valuation divisibility:

a |p b ⇐⇒ vp(a) ≤ vp(b),

a |t b for all a, b with a = 0.

3.25 Definition. Let Lval := { 0, 1,+,−, ·, | } be the language of rings with abinary relation symbol |, and let ACFval be the theory in this language whosemodels are the structures (K, | ) such that K |= ACF , and | is a non-trivialvaluation divisibility on K. (Here a valuation divisibility on a field K is said tobe trivial if the corresponding valuation ring is K, equivalently, the associatedvaluation is trivial.)

We shall obtain quantifier elimination (QE) for ACFval using

QE-Test: Given a theory T , the following are equivalent:

1. T has QE;

2. given any modelsM,N of T where N is |M|+-saturated, and given anyembedding i : A → N of a proper substructure A of M into N , we canextend i to an embedding j : A′ → N of some substructure A′ ofM thatproperly extends A.

Thus the algebraic counterpart of T having QE is that isomorphisms betweensubstructures of models of T can be extended.

Any substructure of a model of ACFval is clearly a domain with a valuationdivisibility on it. Conversely, every domain with a valuation divisibility on it isa substructure of a model of ACFval: first extend the domain with its valuationdivisibility to its fraction field; then extend further to the algebraic closure ofthe fraction field; in case a valuation is trivial, adjoin a transcendental to makeit non-trivial.

Let (R, |) be a domain with valuation divisibility | on R, and let

i : (R, |)→ (F, |F )

be an embedding into an algebraically closed field F with valuation divisibility|F on F . By item (iii) of the lemma above, and using its notation, we can extend

i uniquely to an embedding i′ : (K, |′)→ (F, |F ). Let K be an algebraic closure

of K, and | a valuation divisibility on K such that (K, |′) ⊆ (K, | ). Then by

corollary 3.14 we can extend i′ to an embedding i : (K, | )→ (F, |F ).

3.26. Theorem. ACFval has QE.

Proof. By the remarks preceding the theorem, and the QE-test it suffices toprove the following:

Let (E,A), (F,B) be non-trivially valued algebraically closed fields such that(F,B) is |E|+-saturated. Let K be a proper algebraically closed subfield of E,


and i : (K,A∩K) −→ (F,B) a valued field embedding. Then there is x ∈ E \Ksuch that i extends to a valued field embedding j : (K(x), A∩K(x)) −→ (F,B).

E F

K(x)?�

OO

j //

K?�

OO

i // iK?�

OO

To find such x and j we shall distinguish three cases. To simplify notation,denote both vA and vB by v, and let z be the residue class of z in kA for z ∈ A,and also the residue class of z in kB for z ∈ B.

Case 1: kA∩K = kA. Then we take x ∈ A such that x ∈ kA∩K . SincekA∩K is algebraically closed, x is transcendental over kA∩K . Also x /∈ K, sox is transcendental over K. By the saturation assumption on (F,B) we canfind y ∈ B such that y ∈ kB∩iK . So y is transcendental over kB∩iK and y istranscendental over iK. Extend i to a field isomorphism j : K(x) −→ (iK)(y)by j(x) = y. By the uniqueness part of lemma 3.19, j is also an isomorphism(

K(x), A ∩K(x)) ∼−→

((iK)(y), B ∩ (iK)(y)

)⊆ (F,B).

Case 2: v(K×) = v(E×). Take any γ ∈ v(E×) \ v(K×). Note that v(K×)and v(iK×) are divisible. Since (F,B) is κ+-saturated where κ = |K|, so isv(F×) as an ordered abelian group. Also v(F×) = {0}. It follows that we cantake δ ∈ v(F×) \ v(iK×) such that for all a ∈ K×,

γ < v(a) ⇐⇒ δ < v(ia).

Thus the isomorphism of ordered abelian groups

va 7→ v(ia) : v(K×) −→ v(iK×), (a ∈ K×)

extends to an isomorphism of ordered abelian groups

va+ kγ 7→ v(ia) + kδ : v(K×) + Zγ −→ v(iK×) + Zδ.

Take x ∈ E× and y ∈ F× such that vx = γ, and vy = δ. Since x /∈ K, x istranscendental over K; likewise, y is transcendental over iK. By the uniquenesspart of lemma 3.20, the field isomorphism j : K(x) −→ (iK)(y) extending i andsending x to y is even a valued field isomorphism(

K(x), A ∩K(x)) ∼−→

((iK)(y), B ∩ (iK)(y)

)⊆ (F,B).

Case 3: kA∩K = kA and v(K×) = v(E×) =: Γ. Take any x ∈ E \K. Thevaluation v|K(x) : K(x) → Γ∞ is completely determined by v|K : K → Γ∞


and by the map a 7→ v(x − a) : K −→ Γ, since each f ∈ K[x] \ { 0 } factorsas f = c

∏i(x − ai) with c, ai ∈ K, so vf = vc +

∑i v(x − ai). To simplify

notation, let us identify (K,A ∩K) with (iK,B ∩ iK) via i (so both now havevalue group Γ). It is enough to find y ∈ F \ iK such that v(y − a) = v(x − a)for all a ∈ K, because for such y the field isomorphism j : K(x) −→ (iK)(y)extending i and sending x to y is even a valued field isomorphism

(K(x), A ∩K(x))∼−→ ((iK)(y), B ∩ (iK)(y)) ⊆ (F,B).

Such an element y exists by saturation and the following general lemma.

3.27. Lemma. Let (K,A) ⊆ (L,B) be a valued field extension such that kA =kB, and let v = vB. Let a1, . . . , an ∈ K, n ≥ 1, and let x ∈ L \ K be suchthat v(x − ai) ∈ v(K×) for i = 1, . . . , n. Then there exists a ∈ K such thatv(x− ai) = v(a− ai) for i = 1, . . . , n.

Proof. Any a ∈ K such that v(a − x) > v(a − ai) for i = 1, . . . , n has thedesired property. We may assume v(x− a1) ≥ v(x− ai) for i = 2, . . . , n. Sincev(x−a1) ∈ v(K×) we can take b ∈ K such that v(x−a1) = vb. So v(x−a1

b ) = 0,and since kA = kB ,

x−a1

b = c + ε with c ∈ K, v(c) = 0, v(ε) > 0. Thena = a1 + bc works because x− a = bε and v(bε) > v(x− ai).

3.3 The complete extensions of ACFval

For a valued field (K,A) with residue field k = kA the pair(char(K), char(k)

)can take the following values:

(0, 0) : “equicharacteristic 0”; an example is k((tΓ)) with char(k) = 0 andthe usual valuation.

(0, p), p a prime number: “mixed characteristic p”; an example is Qp withthe p-adic valuation, whose residue field is Fp.

(p, p), p a prime number: “equicharacteristic p”; an example is k((tΓ)) withchar(k) = p and the usual valuation.

It is easy to check that (p, 0) with p prime, and (p, q) with p, q distinct primes,are impossible as values of

(char(K), char(k)

). Let us call

(char(K), char(k)

)the characteristic of the valued field (K,A).

Let (a, b) be a possible characteristic of a valued field, and let ACFval(a, b)be obtained from ACFval by adding axioms specifying that the characteristic ofthe underlying field is a and that the characteristic of the residue field is b.

3.28. Corollary. ACFval(a, b) is complete.

Proof. Construing valued fields as fields with a valuation divisibility, those ofcharacteristic (0, 0) all have (an isomorphic copy of) (Z, |t) as substructure,those of charcteristic (0, p) have (Z, |p) and those of characteristic (p, p) have(Fp, |t) as substructure, where |t is the trivial valuation divisibility on Fp. Tosee this, use the classification of valuation divisibilities on Z, and the fact thatfinite fields only admit the trivial valuation divisibility.


Let (K,A) be a non-trivially valued algebraically closed field. The abovesuggests some natural model-theoretic questions.

1. Find natural models for the completions of ACFval.

2. What is the shape of the subsets of K definable in (K,A)?

3. What structure does (K,A) induce on kA and ΓA?

4. What is the definable closure (in (K,A)) of a subfield of K?

5. Does ACFval or some natural expansion of it by definable sorts admitelimination of imaginaries?

Later we shall prove the following answers to some of these questions:

1. For any algebraically closed field k and any divisible ordered abelian groupΓ, the power series field k((tΓ)) is also algebraically closed, in particular,C((tQ)) with its usual valuation divisibility is a model of ACFval(0, 0), and

Fp((tQ)) with its usual valuation divisibility is a model of ACFval(p, p).

(Here p is a prime number and Fp is the algebraic closure of Fp.)

2. Each definable subset of K is a disjoint union of finitely many swisscheeses, where a swiss cheese is a set D \ (E1 ∪ · · · ∪En) with D either adisc in K or D = K, and E1, . . . , En discs in K.

3. ToDo: reference

4. ToDo: ref to homework for case char 0

5. ToDo: reference

Exercise on Puiseux series fields. Let k be a field, and define the field ofPuiseux series over k to be the subfield of k((tQ)) given by

P (k) :=∞∪d=1

k((t1dZ)).

Note that P (k) contains the field k((t)) of Laurent series over k as a subfield.

Show that if d is a positive integer and α = t1d ∈ P (k), then

k((t1dZ)) = k((t)) + k((t))α+ · · ·+ k((t))αd−1.

Use this to show that k((t1dZ)) is an algebraic extension of k((t)) of degree d.

(Thus P (k) is an algebraic extension of k((t)).)Show that the closure of P (k) in k((tQ)) (with respect to the valuation

topology) is the subfield{f ∈ k((tQ)) | supp f ∩ (−∞, n) is finite for each n

}

4 IMMEDIATE EXTENSIONS 39

of k((tQ)). In particular, P (k) is not dense in k((tQ)), although it has the sameresidue field k and the same value group Q as k((tQ)).

Comment on exercise: It will be shown later that if k is algebraically closed ofcharacteristic 0, then P (k) is algebraically closed, and thus an algebraic closureof k((t)). On the other hand, if k is algebraically closed of characteristic p = 0,then P (k) is not algebraically closed, since the equation xp − x− t−1 = 0 overP (k) has a root

x = t−1/p + t−1/p2

+ t−1/p3

+ · · ·in k((tQ)) that does not lie in P (k).

4 Immediate Extensions

Consider a valued field extension (K, v,Γ) ⊆ (K ′, v′,Γ′). We identify kv witha subfield of kv′ in the usual way. This extension is said to be immediateif kv = kv′ and Γ = Γ′. The basic facts on immediate extensions, due toOstrowski, Krull, Kaplansky (first half of the 20th century), have their ownintrinsic interest, and are also crucial in the later work by Ax–Kochen andErsov.

These facts are related to the notion of pseudoconvergence. To explain this,consider for example the valued field P (k) of Puiseux series over a field k andits immediate valued field extension k((tQ)) (with valuation v). The series

a = 1 + t1/2 + t2/3 + t3/4 + · · · =∑n

tn/(n+1)

in k((tQ)) does not lie in P (k). While a is approximated in a certain sense bythe sequence {an} in P (k), where an :=

∑ni=0 t

i/(i+1), a is not the limit of thissequence in the valuation topology of k((tQ)): v(a − an) → ∞ as n → ∞, infact, v(a− an) < 1 for all n. On the other hand, v(a− an) is strictly increasingas a function of n, and this suggests a notion of (pseudo)limit that turns out tobe useful.

In this section (K, v,Γ) is a valued field, and O := Ov,m := mv,k := kv.

4.1 Pseudoconvergence

A well-indexed sequence in K is a sequence {aρ} in K whose terms aρ areindexed by the elements ρ of an infinite well-ordered set without a last element.Let {aρ} be a well-indexed sequence in K, and a ∈ K. Then {aρ} is said topseudoconverge to a (notation: aρ ⇝ a), if {v(a − aρ)} is eventually strictlyincreasing, that is, for some index ρ0 we have v(a − aσ) > v(a − aρ) wheneverσ > ρ > ρ0. We also say in that case that a is a pseudolimit of {aρ}. Note thatif aρ ⇝ a, then aρ + b ⇝ a+ b for each b ∈ K, and aρb ⇝ ab for each b ∈ K×.We also have the equivalence

aρ ⇝ 0 ⇐⇒ {v(aρ)} is eventually strictly increasing.


4.1. Lemma. Let {aρ} be a well-indexed sequence in K such that aρ ⇝ a wherea ∈ K. With γρ := v(a− aρ), we have:

(i) either va > vaρ eventually, or vaρ = va eventually.

(ii) {vaρ} is either eventually strictly increasing, or eventually constant.

(iii) For each b ∈ K: aρ ⇝ b ⇐⇒ v(a− b) > γρ eventually.

Proof. Let ρ0 be as in the definition of “aρ ⇝ a”. Suppose va ≤ vaρ for a certainρ > ρ0. Then we claim that va = vaσ for all σ > ρ. This is because for σ > ρ wehave v(a− aσ) > v(a− aρ) ≥ va, so va = vaσ. This proves (i). Now (ii) followsfrom (i) by noting that if va > vaρ eventually, then v(a− aρ) = vaρ eventually,so {vaρ} is eventually strictly increasing. We leave (iii) to the reader.

If {aρ} is a well-indexed sequence in K and a ∈ K ′ where (K ′, v′,Γ′) is a valuedfield extension of (K, v,Γ), then “aρ ⇝ a” is of course to be interpreted inthis valued field extension, that is, by considering the sequence {aρ} in K as asequence in K ′.

4.2. Lemma. Suppose (K ′, v′,Γ) is an immediate valued field extension of(K, v,Γ), and let a′ ∈ K ′ \K. Then there is well-indexed sequence {aρ} in Ksuch that aρ ⇝ a′ and {aρ} has no pseudolimit in K.

Proof. We claim that the subset {v′(a′−x) : x ∈ K} of Γ has no largest element.To see this, let x ∈ K; we shall find y ∈ K such that v′(a′ − y) > v′(a′ − x).

Take b ∈ K such that v′(a′ − x) = vb. Then v′(a′−xb ) = 0, so a′−x

b = c+ ϵ withc ∈ K, vc = 0, and v′ϵ > 0. Then a′ − x = bc + bϵ, and thus y = x + bc hasthe desired property. It follows that we can take a well-indexed sequence {aρ}in K such that the sequence {v′(a′ − aρ)} is strictly increasing and cofinal in{v′(a′−x) : x ∈ K}. Thus aρ ⇝ a′. If aρ ⇝ a ∈ K, then v′(a′− a) > v(a′− aρ)for all ρ by Lemma 4.1, part(iii), hence v′(a′ − a) > v(a′ − x) for all x ∈ K, acontradiction.

An illustration of this lemma comes from the beginning of this section: we havethe immediate valued field extension P (k) ⊆ k((tQ)), the sequence {an} in P (k)has no pseudolimit in P (k) but pseudoconverges to a ∈ k((tQ)). Note also thatthe pseudolimits of {an} in k((tQ)) are exactly the series a+ b with vb ≥ 1, bypart (iii) of Lemma 4.1.

To capture within (K, v,Γ) that a well-indexed sequence {aρ} in K has a pseu-dolimit in some valued field extension we make the following definition. Apseudo-cauchy sequence in K (more precisely, in (K, v,Γ)) is a well-indexedsequence {aρ} in K such that for some index ρ0 we have

τ > σ > ρ > ρ0 =⇒ v(aτ − aσ) > v(aσ − aρ).

We also write “pc-sequence” for “pseudo-cauchy sequence”.


4.3. Lemma. Let {aρ} be a well-indexed sequence in K. Then {aρ} is a pc-sequence in K if and only if {aρ} has a pseudolimit in some valued field extensionof (K, v,Γ). In that case, {aρ} has even a pseudolimit in some elementaryextension of (K, v,Γ).

Proof. Suppose {aρ} is a pc-sequence in K, and let ρ0 be as in the definitionof “pseudo-cauchy sequence”. We consider the partial type in the variable xconsisting of the formulas

v(x− aσ) > v(x− aρ), (σ > ρ > ρ0).

Every finite subset of this partial type is realized by some aτ . By compactness wecan realize this partial type by a suitable a ∈ K ′ for some elementary extension(K ′, v′,Γ′) of (K, v,Γ), and then aρ ⇝ a.

For the converse, suppose aρ ⇝ a where a ∈ K ′ and (K ′, v′,Γ′) is a valuedfield extension of (K, v,Γ). Let ρ0 be as in the definition of pseudolimit, and letσ > ρ > ρ0. Then aσ − aρ = (aσ − a)− (aρ − a), so v(aσ − aρ) = v(a− aρ). Soif in addition τ > σ, then v(aτ − aσ) = v(a− aσ) > v(a− aρ) = v(aσ − aρ).

This Lemma, and part (ii) of Lemma 4.1 yield:

4.4. Corollary. If {aρ} is a pc-sequence in K, then {vaρ} is either eventuallystrictly increasing, or eventually constant.

We are going to show that polynomials behave well on pc-sequences, and in thisconnection it is useful to fix some notation. For a polynomial f(x) ∈ K[x] ofdegree ≤ n we have a unique Taylor expansion for f(x+ y) in the ring K[x, y]of polynomials over K in the distinct indeterminates x, y:

f(x+ y) =

n∑i=0

f(i)(x) · yi,

where each f(i)(x) ∈ K[x]. For convenience we also set f(i) = 0 for i > n, so

f(0) = f and f(1) = f ′. (If char(K) = 0, then f(i)(x) = f (i)(x)/i! where f (i) is

the usual ith formal derivative of f .) It is convenient to record here the followingidentity, although is is needed only much later in this section:

4.5. Lemma. f(i)(j) =(i+ji

)f(i+j), (i, j ∈ N).

We also need an elementary fact on ordered abelian groups whose proof we leaveto the reader as an easy exercise.

4.6. Lemma For each i in a finite nonempty set I, let βi ∈ Γ, and ni ∈ N>0,and let λi : Γ → Γ be the linear function given by λi(γ) = βi + niγ. Assumethat ni = nj for distinct i, j ∈ I. Let ρ 7→ γρ be a strictly increasing functionfrom an infinite linearly ordered set without largest element into Γ. Then thereis an i0 ∈ I such that if i ∈ I and i = i0, then λi0(γρ) < λi(γρ), eventually.


We can now prove an important continuity property of polynomials:

4.7. Proposition. Let {aρ} be a well-indexed sequence in K such that aρ ⇝ a,where a ∈ K, and let f(x) ∈ K[x] be a nonconstant polynomial (that is, f /∈ K).Then f(aρ)⇝ f(a).

Proof. Let f be of degree ≤ n, so we have the identity

f(x+ y) = f(x) + f(1)(x)y + · · ·+ f(n)(x)yn

in the polynomial ring K[x, y]. Substituting a for x and aρ − a for y yields

f(aρ)− f(a) = f(1)(a)(aρ − a) + · · ·+ f(n)(a)(aρ − a)n =n∑

i=1

f(i)(a)(aρ − a)i.

Since f(x) = f(a) +∑n

i=1 f(i)(a)(x − a)i and f is not constant, there exists

i ∈ {1, . . . , n} such that f(i)(a) = 0. Now v(f(i)(a)(aρ − a)i

)= βi + iγρ where

βi := v(f(i)(a)

)and γρ := v(aρ−a). Since {γρ} is eventually strictly increasing,

there is by the previous lemma an i0 ∈ {1, . . . , n} such that for every i ∈{1, . . . , n} with i = i0 we have

βi0 + i0γρ < βi + iγρ, eventually.

Then v(f(aρ)− f(a)

)= βi0 + i0γρ eventually, in particular, {v

(f(aρ)− f(a)

)}

is eventually strictly increasing, that is, f(aρ)⇝ f(a).

4.8. Corollary. Suppose {aρ} is a pc-sequence in K, and f(x) ∈ K[x] isnonconstant. Then {f(aρ)} is a pc-sequence.

Proof. By Lemma 4.3 we have aρ ⇝ a with a in a valued field extension. Thenf(aρ)⇝ f(a) in this extension, and thus {f(aρ)} is a pc-sequence.

Let {aρ} be a pc-sequence in K, and let f(x) ∈ K[x] be nonconstant. Then bythe two corollaries above there are two possibilities: either

{v(f(aρ)

)} is eventually strictly increasing, (equivalently, f(aρ)⇝ 0),

or{v

(f(aρ)

)} is eventually constant, (equivalently, f(aρ) ⇝ 0).

We say that {aρ} is of algebraic type over K if the first possibility is realizedfor some nonconstant f ∈ K[x], and then such an f of least degree is called aminimal polynomial of {aρ} over K.

We say that {aρ} is of transcendental type over K if the second possibility isrealized for all nonconstant f ∈ K[x]. (Note that then {v

(f(aρ)

)} is eventually

constant for any f ∈ K[x], whether constant or not.) A pc-sequence in K oftranscendental type determines an essentially unique immediate extension:


4.9. Theorem Let {aρ} be a pc-sequence in K of transcendental type over K.Then {aρ} has no pseudolimit in K. The valuation v on K extends uniquely toa valuation v : K(x)× → Γ (x transcendental over K) such that

vf = eventual value of v(f(aρ))

for each f ∈ K[x]. With this valuation K(x) is an immediate valued fieldextension of (K, v,Γ) in which aρ ⇝ x.

Conversely, if aρ ⇝ a in a valued field extension of (K, v,Γ), then there isa valued field isomorphism K(x)→ K(a) over K that sends x to a.

Proof. If aρ ⇝ a with a ∈ K, then the polynomial x− a witnesses that {aρ} isof algebraic type over K. So {aρ} has no pseudolimit in K.

It is clear that by defining vf for f ∈ K[x] as in the above display, we havea valuation on K[x], and thus on K(x). The value group of this valuation isstill Γ, and one checks easily that aρ ⇝ x. To verify that the residue field ofK(x) is the residue field of K we first note that because the value groups areequal, each a ∈ K(x) with va = 0 has the form a = f/g where f, g ∈ K[x] withvf = vg = 0. So it is enough to consider a nonconstant f ∈ K[x] with vf = 0,and find b ∈ K with v(f − b) > 0. We have 0 = vf = v

(f(aρ)

)eventually, and

{v(f −f(aρ)

)} is eventually strictly increasing, so v

(f −f(aρ)

)> 0, eventually.

Thus b = f(aρ) for big enough ρ will do the job.Finally, suppose aρ ⇝ a with a in a valued field extension whose valuation

we also indicate by v for simplicity. For nonconstant f ∈ K[x] we have f(aρ)⇝f(a), and thus v

(f(a)

)= v

(f(aρ)

)eventually; in particular, f(a) = 0 and

v(f(a)) = vf ∈ Γ. Thus a is transcendental over K and the field isomorphismK(x)→ K(a) over K that sends x to a is even a valued field isomorphism.

Here is the analogue for pc-sequences of algebraic type over K:

4.10. Theorem Let {aρ} be a pc-sequence in K of algebraic type over K,without pseudolimit in K, and let µ(x) be a minimal polynomial of {aρ} overK. Then µ is irreducible in K[x] and deg µ ≥ 2. Let a be a zero of µ in anextension field of K. Then v extends uniquely to a valuation v : K(a)× → Γsuch that

v(f(a)

)= eventual value of v

(f(aρ)

)for each nonzero polynomial f ∈ K[x] of degree < deg µ. With this valuationK(a) is an immediate valued field extension of (K, v,Γ), and aρ ⇝ a.

Conversely, if µ(b) = 0 and aρ ⇝ b in a valued field extension of (K, v,Γ),then there is a valued field isomorphism K(a)→ K(b) over K sending a to b.

Proof. Much of the proof duplicates the proof for the case of transcendentaltype. A difference is in how we obtain the multiplicative law for v : K(a)× → Γas defined above. Let s, t ∈ K(a)×, and write s = f(a), t = g(a) with nonzerof, g ∈ K[x] of degree < degµ. Then fg = qµ + r with q, r ∈ K[x] and deg r <degµ, so st = r(a), and thus eventually

v(s) = v(f(aρ)

), v(t) = v

(g(aρ)

), v(st) = v

(r(aρ)

).


Also v(s) + v(t) = v(f(aρ)g(aρ)

)= v

(q(aρ)µ(aρ) + r(aρ)

), eventually. Since

{v(q(aρ)µ(aρ)

)} is either eventually strictly increasing, or eventually ∞, this

forces v(s) + v(t) = v(r(aρ)

), eventually, so v(s) + v(t) = v(st).

The minimal polynomial µ(x) in this theorem, even if we assume it to be monic,is in general highly non-unique. (For example, adding to its constant term anε ∈ K such that eventually vε > v(µa − µaρ), does not change its statusas minimal polynomial of {aρ}.) That is why we have to specify a particularminimal polynomial in this theorem to achieve the uniqueness up to isomorphismof K(a).

4.2 Maximal Valued Fields

A maximal valued field is by definition a valued field that has no immediateproper valued field extension. If k is algebraically closed and Γ is divisible, thenby Corollary 3.18 the algebraic closure of K equipped with a valuation thatextends v is necessarily an immediate extension of (K, v,Γ). Therefore:

4.11. Corollary. If (K, v,Γ) is maximal valued, k is algebraically closed andΓ is divisible, then K is algebraically closed.

Theorems 4.9 and 4.10 in combination with Lemma 4.2 yield:

4.12. Corollary. (K, v,Γ) is maximal valued if and only if each pc-sequencein K has a pseudolimit in K.

We use this equivalence to get the next result.

4.13. Corollary. The Hahn field k((tΓ)) is maximal valued. If k is algebraicallyclosed and Γ is divisible, then k((tΓ)) is algebraically closed.

Proof. The first assertion implies the second one, so it is enough to consideran arbitrary pc-sequence {aρ} in k((tΓ)) and show that it has a pseudolimit ink((tΓ)). Take an index ρ0 such that

τ > σ > ρ > ρ0 =⇒ v(aτ − aσ) > v(aσ − aρ).

Below we restrict ρ to be > ρ0. Note that we have γρ ∈ Γ with v(aσ − aρ) = γρfor all σ > ρ, and that {γρ} is strictly increasing. We split up aρ as follows:

aρ =∑γ<γρ

cγρtγ +

∑γ≥γρ

cγρtγ , (all cγρ ∈ k).

Then cγρ = cγσ for γ < γρ and σ > ρ. For any γ ∈ Γ, either γ < γρ for someρ, or γ > γρ for all ρ. Thus we can define the series a :=

∑γ cγt

γ ∈ k((tΓ)) bysetting cγ := cγρ whenever γ < γρ, and cγ := 0 if γ > γρ for all ρ. Then aρ ⇝ a,as is easily verified.

Under certain conditions on residue field and value group, a maximal valuedfield is isomorphic to a Hahn field, see Corollary 4.31. First a consequence ofKrull’s cardinality bound |K| ≤ |k||Γ| (Proposition 3.4) and Zorn:


4.14. Corollary. (K, v,Γ) has an immediate maximal valued field extension.

This raises of course the question whether such an extension is unique up toisomorphism. In the equicharacteristic 0 case the answer is yes (Corollary 4.28),but first we take care of a loose end in Theorem 4.10:

4.15. Lemma. Let a in an immediate valued field extension of (K, v,Γ) bealgebraic over K, and a /∈ K. Then there is a pc-sequence {aρ} in K of algebraictype over K that has no pseudolimit in K, such that aρ ⇝ a.

Proof. By Lemma 4.2 we have a pc-sequence {aρ} in K without pseudolimit inK, such that aρ ⇝ a. Let f(x) be the minimum polynomial of a over K. Thenby the Taylor identity preceding Lemma 4.5 we have

faρ = faρ − fa = (aρ − a) · gaρ, g(x) ∈ K(a)[x].

Thus v(faρ) = v(aρ − a) + v(gaρ). Since {v(aρ − a)} is eventually strictlyincreasing and {v(gaρ)} is eventually strictly increasing or eventually constant,{v(faρ)} is eventually strictly increasing, so {aρ} is of algebraic type overK.

A valued field is algebraically maximal if it has no immediate proper algebraicvalued field extension. (Thus maximal =⇒ algebraically maximal.) Obviously,every algebraically closed valued field is algebraically maximal.

Theorem 4.10 and Lemma 4.15 yield:

4.16. Corollary. (K, v,Γ) is algebraically maximal if and only if each pc-sequence in K of algebraic type over K has a pseudolimit in K.

Just from the definition of algebraically maximal (and Zorn) it follows that(K, v,Γ) has an immediate valued field extension that is algebraically maximaland algebraic over K. A natural question is whether such an extension is deter-mined up to isomorphism over (K, v,Γ). The answer is yes in equicharacteristic0 and in some other cases, see Theorem 4.27 and its consequences.

First we show that algebraic maximality implies henselianity. A valued fieldis said to be henselian if its valuation ring is henselian.

4.17. Lemma. Let f ∈ O[x] and a ∈ O be such that v(fa) > 0, v(f ′a) = 0,and f has no zero in a + m. Then there is a pc-sequence {aρ} in K such thatf(aρ)⇝ 0, and {aρ} has no pseudolimit in K.

Proof. We build such a sequence by the Newton approximation at the beginningof subsection 2.2, starting with a0 = a. Indeed, this method yields b ∈ O suchthat v(b−a) = v(fa) > 0 and v(fb) ≥ 2v(fa) > 0. In particular, a ≡ b mod m,hence f ′a ≡ f ′b mod m, so v(f ′b) = 0. We now take a1 = b, and continue withb as new input. More precisely, let λ be an ordinal > 0 and {aρ} a sequence ina+m indexed by the ordinals ρ < λ such that a0 = a, and v(aσ − aρ) = v(faρ)and v(faσ) ≥ 2v(faρ) whenever λ > σ > ρ. (For λ = 2 we have such a sequencewith a0 = a and a1 = b.) If λ = µ+ 1 is a successor ordinal, then we constructthe next term aλ ∈ a+m by Newton approximation so that v(aλ−aµ) = v(faµ)and v(faλ) ≥ 2v(faµ).


Suppose now that λ is a limit ordinal. Then {aρ} is clearly a pc-sequenceand f(aρ) ⇝ 0. If {aρ} has no pseudolimit in K, then we are done. If {aρ}has a pseudolimit in K, let aλ be such a pseudolimit. Then faρ ⇝ faλ; since{v(faρ)} is strictly increasing, this yields v(faλ) ≥ v(faρ+1) ≥ 2v(faρ) for eachindex ρ < λ. It is also clear that v(aλ − aρ) = v(aρ+1 − aρ) = v(faρ) for eachρ < λ, in particular, aλ ∈ a + m. Thus we have extended our sequence by onemore term. This building process must come to an end.

In combination with Theorem 4.10 the previous lemma yields:

4.18. Corollary. Each algebraically maximal valued field is henselian (andthus each maximal valued field is henselian).

For example, the Puiseux series field P (k) is henselian, since it is the directed

union of its maximal valued subfields k((t1dZ)) (d = 1, 2, 3, . . . ).

In the equicharacteristic 0 case, algebraically maximal is equivalent to henselian;see Corollary 4.22. For (K, v,Γ) to be henselian is a condition on polynomialsover its valuation ring O. It is convenient to have an equivalent condition forpolynomials over K. Let f(x) ∈ K[x] be of degree ≤ n, and let a ∈ K, so

f(a+ x) =n∑

i=0

f(i)(a)xi = fa+ f ′(a)x+

n∑i=2

f(i)(a)xi.

4.19 Definition. We say f, a is in hensel configuration if f ′a = 0, and eitherfa = 0 or there is γ ∈ Γ such that

v(fa) = v(f ′a) + γ < v(f(i)a

)+ i · γ

for 2 ≤ i ≤ n. (This γ is unique: γ = v(fa)− v(f ′a).)

Suppose f, a is in hensel configuration and fa = 0. Take γ as in the definition,take c ∈ K with vc = γ and put g(x) := f(cx)/f(a) and α := a/c. Theng(α) = 1, v

(g′(α)

)= 0, and v

(g(i)(α)

)> 0 for i > 1, as is easily verified. Thus

h(x) := g(α+ x) = 1 + g′(α)x+

n∑i=2

g(i)(α)xi

lies in O[x]. The set {u ∈ O : v(1 + g′(α)u

)> 0} is a congruence class modulo

m. For u in this set we have v(h(u)

)= v

(g(α+ u)

)> 0,

h′(u) = g′(α+ u) = g′(α) +

n−1∑j=1

g(1)(j)(α)uj ,

and v(g(1)(j)(α)

)≥ v

(g(1+j)(α)

)> 0 for j > 0, so v

(h′(u)

)= 0.

Suppose now that (K, v,Γ) is henselian; this allows us to pick u as abovesuch that g(α+ u) = 0. By Lemma 2.1 this u is unique in the sense that thereis no u′ ∈ O such that g(α + u′) = 0 and u′ = u. Now b := a + cu satisfiesf(b) = 0 and v(a− b) = γ. Summarizing:


4.20. Lemma. If (K, v,Γ) is henselian and f, a is in hensel configuration, thenthere is a unique b ∈ K such that f(b) = 0 and v(a− b) ≥ v(fa)− v(f ′a); thisb satisfies v(a− b) = v(fa)− v(f ′a).

Suppose (K, v,Γ) is of mixed characteristic (0, p). Then we say that (K, v,Γ) isfinitely ramified if the set {γ ∈ Γ : 0 ≤ γ < vp} is finite. For example, Qp withits p-adic valuation is finitely ramified.

4.21. Proposition. Suppose (K, v,Γ) is of equicharacteristic 0, or finitelyramified of mixed characteristic. Let {aρ} be a pc-sequence in K and let f(x) ∈K[x] be such that faρ ⇝ 0 and f(j)aρ ⇝ 0 for all j ≥ 1. Then

(i) f, aρ is in hensel configuration, eventually;

(ii) in any henselian valued field extension of (K, v,Γ) there is a unique b suchthat aρ ⇝ b and f(b) = 0.

Proof. Let a be a pseudolimit of {aρ} in some valued field extension of (K, v,Γ)whose valuation we also denote by v, and put γρ := v(a − aρ). The proof ofProposition 4.7 yields a unique j0 ≥ 1 such that for each k ≥ 1 with k = j0,

v(faρ − fa

)= v

(f(j0)a

)+ j0γρ < v

(f(k)a

)+ k · γρ, eventually.

Now f(aρ)⇝ 0, so for k ≥ 1, k = j0:

v(faρ

)= v

(f(j0)a

)+ j0 · γρ < v

(f(k)a

)+ k · γρ, eventually.

We claim that j0 = 1. Let k > 1; our claim will then follow by deriving

v(f ′a

)+ γρ < v

(f(k)a

)+ kγρ, eventually.

The proof of Proposition 4.7 applied to f ′ instead of f also yields

v(f ′aρ − f ′a

)≤ v

(f(1)(j)a

)+ j · γρ, eventually

for all j ≥ 1. Since v(f ′aρ

)= v

(f ′a

)eventually, this yields

v(f ′a

)≤ v

(f(1)(j)a


for all j ≥ 1. Using f(1)(j) = (1 + j)f(1+j) (Lemma 4.5), this gives for j ≥ 1:

v(f ′a

)≤ v(1 + j) + v

(f(1+j)(a)


For j = k − 1, this yields

v(f ′a) ≤ vk + v(f(k)(a)

)+ (k − 1) · γρ, eventually.

The assumption on (K, v,Γ) then yields:

v(f ′a) < v

(f(k)a

)+ (k − 1) · γρ, eventually, hence


v(f ′a

)+ γρ < v(f(k)a) + k · γρ, eventually.

Thus j0 = 1, as claimed. Since for each k ≥ 1 we have v(f(k)aρ) = v(f(k)a),eventually, the above equalities and inequalities also show that f, aρ is in henselconfiguration, eventually. This proves (i).

For (ii), let (K ′, v′,Γ′) be a henselian valued field extension of (K, v,Γ).After deleting an initial segment of the sequence {aρ} we can assume that {γρ}is strictly increasing, that v(aσ − aρ) = γρ whenever σ > ρ, and that f ′aρ = 0for all ρ. Likewise, by (i) and Lemma 4.20 we can assume that for every ρ thereis a unique bρ ∈ K ′ such that f(bρ) = 0 and v(aρ− bρ) = v(faρ)− v(f ′aρ). Theproof of (i) shows that we can also assume that v(faρ)− v(f ′aρ) = γρ for all ρ,so v(aρ − bρ) = γρ for all ρ. The uniqueness of bρ shows that bσ = bρ wheneverσ > ρ. Thus all bρ are equal to a single b, which has the desired properties.

4.22. Corollary. Let (K, v,Γ) be of equicharacteristic 0, or finitely ramified ofmixed characteristic. Then (K, v,Γ) is henselian if and only if it is algebraicallymaximal.

Proof. Assume (K, v,Γ) is henselian, and let {aρ} be a pc-sequence in K ofalgebraic type over K. Take a monic minimal polynomial f(x) of {aρ} over K.Then faρ ⇝ 0 and f(j)aρ ⇝ 0 for each j ≥ 1, so f, aρ is in hensel configuration,eventually, and thus by Lemma 4.20 we have b ∈ K such that f(b) = 0. Sincef(x) is irreducible, this gives f(x) = x− b, so aρ ⇝ b.

This leads to a partial converse to some earlier results:

4.23. Corollary. Let (K, v,Γ) be of equicharacteristic 0. Then K is alge-braically closed if and only if (K, v,Γ) is henselian, k is algebraically closed andΓ is divisible.

Puiseux series fields P (k) are henselian with value group Q, hence:

4.24. Application. Suppose k is algebraically closed of characteristic 0. Thenthe Puiseux series field P (k) is algebraically closed.

Exercise. Let k ⊆ k′ be a field extension of finite degree n, and let b1, . . . , bnbe a basis of k′ over k. Show that k′((tΓ)) is also of finite degree n over itssubfield k((tΓ)) with basis b1, . . . , bn.

Show that if k has characteristic 0 with algebraic closure k, then the algebraicclosure of k((t)) in the algebraically closed field extension k((tQ)) is the unionof the Puiseux series fields P (k′) where k′ ranges over the subfields of k thatcontain k and are of finite degree over k.

4.3 Henselization

The following notion is fundamental in valuation theory and beyond.


4.25 Definition. A henselization of (K, v,Γ) is a henselian valued field exten-sion (Kh, vh,Γh) of (K, v,Γ) such that any valued field embedding

(K, v,Γ)→ (K ′, v′,Γ′)

into a henselian valued field (K ′, v′,Γ′) extends uniquely to an embedding

(Kh, vh,Γh)→ (K ′, v′,Γ′).

As usual with such definitions, existence guarantees uniqueness: if (K1, v1,Γ1)and (K2, v2,Γ2) are henselizations of (K, v,Γ), then the unique embedding(K1, v1,Γ1) → (K2, v2,Γ2) over (K, v,Γ) is an isomorphism. Thus there isno harm in referring to the henselization of (K, v,Γ) if there is one. Of course,if (K, v,Γ) is henselian, it is its own henselization. Also, if (K, v,Γ) has ahenselization and (K ′, v′,Γ′) is any henselian valued field extension of (K, v,Γ),then the henselization of (K, v,Γ) in (K ′, v′,Γ′) is by definition the henselization(Kh, vh,Γh) of (K, v,Γ) such that

(K, v,Γ) ⊆ (Kh, vh,Γh) ⊆ (K ′, v′,Γ′).

4.26. Corollary. Any henselization of (K, v,Γ) is an immediate valued fieldextension of (K, v,Γ) and algebraic over K.

Proof. We already know that (K, v,Γ) has an immediate algebraically maximalvalued field extension that is algebraic over K. Any henselization of (K, v,Γ)must embed over (K, v,Γ) into such an extension.

Each valued field has a henselization, as we show later in the more generalsetting of local rings. For the Ax–Kochen–Ershov story it is enough to considerthe equicharacteristic 0 case and the finitely ramified mixed characteristic case.In those cases the henselization has a particularly nice characterization:

4.27. Theorem. Let (K, v,Γ) be of equicharacteristic 0, or finitely ramifiedof mixed characteristic. Let (K1, v1,Γ) be an immediate henselian valued fieldextension of (K, v,Γ) such that K1 is algebraic over K. Then (K1, v1,Γ) is ahenselization of (K, v,Γ).

Proof. Call a field L with K ⊆ L ⊆ K1 nice if each valued field embeddingfrom (K, v,Γ) into a henselian valued field (K ′, v′,Γ′) extends uniquely to anembedding from (L, vL,Γ) into (K ′, v′,Γ′), where vL := v1|L× . Consider a nicefield L such that L = K1. With Zorn in mind, it clearly suffices to show thatthen there is a nice field L′ with L ⊆ L′ ⊆ K1 and L = L′.

Lemma 4.15 gives a pc-sequence {aρ} in L of algebraic type over L withoutpseudolimit in L. Take a minimal polynomial µ(x) of {aρ} over L. Thenµ(aρ) ⇝ 0 and µ(j)(aρ) ⇝ 0 for all j ≥ 1. Item (ii) of Proposition 4.21 yieldsa ∈ K1 such that aρ ⇝ a and µ(a) = 0. Let a valued field embedding ifrom (K, v,Γ) into a henselian valued field (K ′, v′,Γ′) be given, and let iL bethe unique extension of i to an embedding from (L, vL,Γ) into (K ′, v′,Γ′). Tosimplify notation, identify (L, vL,Γ) with a valued subfield of (K ′, v′,Γ′) via iL.Item (ii) of Proposition 4.21 also gives a unique b ∈ K ′ such that aρ ⇝ b andµ(b) = 0. Now apply Theorem 4.10 to conclude that L(a) is nice.


Since (K, v,Γ) has an immediate algebraically maximal valued field extensionthat is algebraic over K, this theorem shows in particular:

If (K, v,Γ) is of equicharacteristic 0 or finitely ramified of mixed characteristic,then (K, v,Γ) has a henselization, and any immediate algebraically maximalvalued field extension of (K, v,Γ) that is algebraic over K is isomorphic over(K, v,Γ) to this henselization by a unique isomorphism.

4.4 Uniqueness of immediate maximal extensions

The results above have some very nice consequences:

4.28. Corollary. Let (K, v,Γ) be of equicharacteristic 0, or finitely ramified ofmixed characteristic. Then any two immediate maximal valued field extensionsof (K, v,Γ) are isomorphic over (K, v,Γ).

Proof. Let (K1, v1,Γ) and (K2, v2,Γ) be immediate maximal valued field ex-tensions of (K, v,Γ). Below, each subfield of Ki (i = 1, 2), is viewed as valuedfield by taking as valuation the restriction of vi to the subfield. Consider fieldsL1, L2 with K ⊆ L1 ⊆ K1 and K ⊆ L2 ⊆ K2, and suppose we have a valuedfield isomorphism L1

∼= L2. Note that L1 = K1 iff L2 = K2.Suppose L1 = K1 and L2 = K2. It suffices to show that then we can extend

the isomorphism L1∼= L2 to a valued field isomorphism L′

1∼= L′

2 where L′1 and

L′2 are fields with Li ⊆ L′

i ⊆ Ki and Li = L′i for i = 1, 2. If L1 is not henselian,

then we can take for L′i the henselization of Li in Ki, by Theorem 4.27. Suppose

L1 is henselian. Take b ∈ K1 \L1, and take a pc-sequence {aρ} in L1 such thataρ ⇝ b and {aρ} has no pseudolimit in L1. Since L1 is algebraically maximal byCorollary 4.22, {aρ} is of transcendental type over L1. The image of {aρ} underour isomorphism L1

∼= L2 has a pseudolimit c ∈ K2, and then Theorem 4.9allows us to extend this isomorphism to an isomorphism L1(b) ∼= L2(c).

The next variant is just what we need for the proof of the Ax-Kochen-Ershovtheorem in the next section.

4.29. Lemma. Let (K, v,Γ) be of equicharacteristic 0, or finitely ramified ofmixed characteristic, and let (K∗, v∗,Γ) be an immediate maximal extension of(K, v,Γ). Then we can embed (K∗, v∗,Γ) over (K, v,Γ) into any |Γ|+-saturatedhenselian valued field extension (K ′, v′,Γ′) of (K, v,Γ).

Proof. Exercise.

Are there useful conditions on k and Γ implying that if (K, v,Γ) is maximalvalued, then (K, v,Γ) is isomorphic to the Hahn field k((tΓ))? The answer isyes, and here liftings and cross-sections come into play. (A lifting of the residuefield of (K, v,Γ) is a field embedding i : k → K such that ia = a for all a ∈ k.A cross-section of the valuation v : K× → Γ is a group morphism s : Γ → K×

such that v(sγ) = γ for all γ ∈ Γ.) For example, the valuation of the Hahn fieldk((tΓ)) has the cross-section γ 7→ tγ . Also, if Γ = Z and v(π) = 1, π ∈ K, thenk 7→ πk : Z→ K× is a cross-section; in particular Qp has cross section k 7→ pk.


To see how a lifting of the residue field and a cross-section of the valuationcan help in answering the question above, we first single out the valued subfieldk(tΓ) of k((tΓ)): its underlying field is the subfield of k((tΓ)) generated over kby the multiplicative group tΓ. Note that k((tΓ)) is an immediate extension ofk(tΓ). Suppose now that we have both a lifting i : k → K of the residue fieldof (K, v,Γ) and a cross-section s : Γ→ K× of v. Jointly i and s yield a uniquevalued field embedding from k(tΓ) into (K, v,Γ) that extends i and maps tγ tosγ for each γ ∈ Γ. (Exercise.) It is clear that (K, v,Γ) is an immediate extensionof the image of this embedding. Making now the additional assumption that(K, v,Γ) is maximal valued and char(k) = 0, we conclude from Corollary 4.28that (K, v,Γ) is isomorphic to the Hahn field k((tΓ)) under an isomorphism thatmaps ia to a for each a ∈ k and sγ to tγ for each γ ∈ Γ.

4.30. Lemma. Suppose (K, v,Γ) is henselian, char(k) = 0 and the abeliangroups k× and Γ are divisible. Then K× is divisible, and the valuation v has across-section.

Proof. Let a ∈ K×, and n > 0; we shall find b ∈ K so that a = bn. First taked ∈ K with n · vd = va, so a = udn with vu = 0. Now the residue class u is annth power in k×, so u = cn with c ∈ K by an earlier exercise. Hence a = (cd)n.

The second statement follows from the first: From the divisibility of K× itfollows that U(O) is divisible. Thus the exact sequence

1→ U(O) ↪→ K× v−→ Γ→ 0

splits, so v has a cross-section.

The first two assumptions in this lemma imply also that there is a lifting of theresidue field of (K, v,Γ), by Theorem 2.9. Therefore:

4.31. Corollary. Suppose (K, v,Γ) is maximal valued, char(k) = 0 and theabelian groups k× and Γ are divisible. Then (K, v,Γ) is isomorphic to the Hahnfield k((tΓ)) via an isomorphism that induces the identity on k and on Γ.

4.32. Corollary. Suppose (K, v,Γ) is maximal valued, k is algebraically closed,char(k) = 0 and Γ is divisible. Then (K, v,Γ) is isomorphic to the Hahn fieldk((tΓ)) via an isomorphism that induces the identity on k and on Γ.

The exercise below gives another case where a maximal valued field is isomorphicto a Hahn field.

Exercise. Show that if Γ is free as an abelian group, then v has a cross-section.Show: if (K, v,Γ) is maximal valued, char(k) = 0, and Γ is free as an abeliangroup, then (K, v,Γ) is isomorphic to the Hahn field k((tΓ)) via an isomorphismthat induces the identity on k and on Γ.

5 THE THEOREM OF AX–KOCHEN AND ERSHOV 52

5 The Theorem of Ax–Kochen and Ershov

We prove here the original AKE-theorem, where AKE abbreviates Ax–Kochenand Ershov (really two groups who proved the theorem independently, one fromthe USA: Ax and Kochen, the other from Russia: Ershov). In later sections weshall prove more refined versions.

5.1. AKE-Theorem. Let (K,A) and (K ′, A′) be two henselian valued fieldswith residue fields k and k′, and value groups Γ and Γ′ (viewed as ordered abeliangroups). Suppose that char(k) = 0. Then

(K,A) ≡ (K ′, A′) ⇐⇒ k ≡ k′ and Γ ≡ Γ′.

The forward direction ⇒ should be clear (and does not need the henselian orresidue characteristic zero assumption). It will be convenient to prove firsta variant of theorem where we have also a lifting of the residue field and across-section of the valuation as part of the structure. So we begin with someadditional facts on cross-sections.

5.1 Existence of cross-sections in elementary extensions

In this subsection we fix a valued field (K, v,Γ). We shall write ×-section toabbreviate the word cross-section.

5.2 Definition. A partial ×-section of v is a group morphism s : ∆ −→ K×

from a subgroup ∆ of Γ back into K× such that v(sδ) = δ for all δ ∈ ∆.

A subgroup ∆ of Γ is said to be pure in Γ if Γ/∆ is torsion free, that is, whenevernγ ∈ ∆ with n > 0 and γ ∈ Γ, then γ ∈ ∆.

5.3. Lemma. Let s : ∆ −→ K× be a partial ×-section of v such that ∆ ispure in Γ. Then there is an elementary extension (K1, v1,Γ1) of (K, v,Γ) witha partial ×-section s1 : Γ −→ K×

1 of v1.

Proof. Let ∆′ be a subgroup of Γ that contains ∆ such that ∆′/∆ is finitelygenerated as an abelian group. Since ∆′/∆ is torsion-free, it is free as abeliangroup, so we can take nonzero γ1, . . . , γk ∈ ∆′ such that

∆′ = ∆⊕ Zγ1 ⊕ · · · ⊕ Zγk (internal direct sum of subgroups of Γ).

Take a1, . . . , ak ∈ K× such that v(a1) = γ1, . . . , v(ak) = γk. Extend s to agroup morphism s′ : ∆′ −→ K× by s′(γi) = ai, i = 1, . . . , k. It is easy to checkthat s′ is a partial ×-section of v.

Since ∆′ was arbitray, a compactness argument gives an elementary exten-sion (K1, v1,Γ1) of (K, v,Γ) with a partial ×-section s1 : Γ −→ K×

1 of v1.

Since Γ ≼ Γ1 as ordered abelian groups, Γ is pure in Γ1, so the purity assumptionon s is inherited by s1. Thus we can iterate the lemma:


5.4. Proposition. Let s : ∆ −→ K× be a partial ×-section of v such that ∆ ispure in Γ. Then there is an elementary extension (K ′, v′,Γ′) of (K, v,Γ) with a×-section s′ : Γ′ −→ K ′× of v′ that extends s.

Proof. Use the previous lemma to build an elementary chain

(K, v,Γ) = (K0, v0,Γ0) ≼ (K1, v1,Γ1) ≼ (K2, v2,Γ2) ≼ . . .

with a partial ×-section sn : ∆n −→ K×n of vn for each n, such that ∆0 = ∆,

∆n+1 = Γn and sn+1 extends sn, for each n. Then (K ′, v′,Γ′) =∪

n(Kn, vn,Γn)works with s′ =

∪n sn as ×-section.

By taking ∆ = {0} in this proposition we get:

5.5. Corollary. (K, v,Γ) has an elementary extension with a cross-section.

Exercise. Suppose K is algebraically closed and s : ∆ → K× is a partial×-section of v. Show that s extends to a ×-section of v.

5.2 Extending the value group

We need a few more basic results on how the value group can change when weextend a valued field. Let (K, v,Γ) be a valued field, and let

Γd = {γ/n : γ ∈ Γ, n > 0}

be the divisible hull of Γ.

5.6. Lemma. Let p be a prime number, and x an element in a field extensionof K such that xp = a ∈ K× but va /∈ pΓ. Then Xp − a is the minimumpolynomial of x over K, and v extends uniquely to a valuation w : K(x)× → ∆with ∆ ⊆ Γd (as ordered groups). Moreover, kw = kv, and [∆ : Γ] = p, with

∆ =

p−1∪i=0

Γ + iw(x) (disjoint union).

Proof. Let w : K(x)× → ∆ with ∆ ⊆ Γd be a valuation extending v. (By earlierresults we know that there is such an extension.) Since va /∈ pΓ, the elements

w(x0) = 0, w(x1) =va

p, . . . , w(xp−1) =

(p− 1)va

p

of ∆ are in distinct cosets of Γ, so 1, x, . . . , xp−1 are K-linearly independent,and thus Xp−a is the minimum polynomial of x over K. Also, for an arbitrarynonzero element b = b0 + b1x + · · · + bp−1x

p−1 of K(x) (b0, . . . , bp−1 ∈ K, notall zero), we have

w(b) = min{v(bi) +iva

p: 0 ≤ i ≤ p− 1},

showing the uniqueness of w. This also proves the claims made by the lemmaabout ∆. Since p = [K(x) : K] ≥ [∆ : Γ] · [kw : kv], this yields [kw : kv] = 1,that is, kw = kv.


5.7. Lemma. Let (K ′, v′,Γ′) be a valued field extension of (K, v,Γ), let x ∈ K ′,and put v′(x−K) := {v′(x− a) : a ∈ K}. If v′(x−K) has no largest element,then v′(x−K) ⊆ Γ. If v′(x−K) has a largest element v′(x− a), a ∈ K, thenv′(x−K) ⊆ Γ ∪ {v′(x− a)}.

Proof. For b, c ∈ K we have: v′(x−b) < v′(x−c) =⇒ v′(x−b) = v(b−c).

5.8. Corollary. Let (K, v,Γ) be algebraically maximal and have the valued fieldextension (K ′, v′,Γ′), and let x ∈ K ′ \K. Then one of the following holds:

1. x is a pseudolimit of a pc-sequence in K of transcendental type over K;

2. there exists a ∈ K such that v′(x− a) /∈ Γ;

3. there exist a, b ∈ K, b = 0, such that v′(x− a) = vb and the residue classof (x− a)/b does not lie in k.

Proof. Suppose v′(x−K) has no largest element. As at the end of the proof ofLemma 4.2, this yields a pc-sequence in K that pseudoconverges to x but hasno pseudolimit in K. By the algebraic maximality of (K, v,Γ) this pc-sequencemust be of transcendental type over K.

Suppose next that v′(x − K) has a largest element v′(x − a), a ∈ K, andthat v′(x− a) ∈ Γ. Then v′(x− a) = vb with b ∈ K, b = 0. If the residue classof (x− a)/b were in k, then (x− a)/b = u+ ϵ with u ∈ K, vu = 0, and v′ϵ > 0,so v′(x− (a+ bu)) > vb = v′(x− a), a contradiction. Thus the residue class of(x− a)/b cannot lie in k.

5.9. Corollary. Let (K ′, v′,Γ′) be a valued field extension of (K, v,Γ), and letx ∈ K ′, x /∈ K. If Γ = v(K×) is countable, so is v′(K(x)×).

Proof. By passing to an algebraic closure ofK ′ and extending the valuation v′ tothis algebraic closure we can assume that K ′ is algebraically closed. ReplacingK by its algebraic closure inside K ′ we further reduce to the case that K isalgebraically closed. Suppose first that v′(x − K) ⊆ Γ. Then, given any non-zero element f(x) ∈ K[x] we have a factorization f(x) = c(x− a1) · · · (x− am)with c, a1, . . . , am ∈ K, c = 0, so

v′(f(x)) = vc+ v′(x− a1) + · · ·+ v′(x− am) ∈ Γ,

hence v′(K(x)×) ⊆ Γ. Next, suppose that a ∈ K is such that v′(x − a) /∈ Γ.Then v′(x−K) ⊆ Γ∪{v′(x−a)} by Lemma 5.7, and thus by the same argumentv′(K(x)×) ⊆ Γ + Z · v′(x− a).

5.3 AKE-theorem with lifting and cross-section

Consider 3-sorted structures

K = (K,Γ,k; v, s, π, i)


where K and k are fields, Γ is an ordered abelian group, v : K× → Γ is avaluation with ×-section s, and π : Ov → k and i : k→ Ov are ring morphismssuch that π(ir) = r for all r ∈ k.

Note that then π is surjective, so π has kernel mv, and thus we have a fieldisomorphism k ∼= kv of k onto the residue field such that the diagram

............commutes. We call K the underlying field of K, Γ its value group, and k itsresidue field (even though the latter is only isomorphic to the residue field kv

of Ov).

Such a structure K will be called a cl-valued field , with “c” recalling the cross-section s and “l” the lifting i.

5.10. Remark. Each henselian valued field (K,A) of equicharacteristic 0 hasan elementary extension (K ′, A′) that can be expanded to a cl-valued field

(K ′,Γ′,k′; v′, s′, π′, i′)

where v′ := vA′ : K ′× → Γ′ := ΓA′ and π′ : A′ → k′ := kA′ is the residue map.

This follows from our lifting theorem for henselian local rings with residue fieldof characteristic 0, and Corollary 5.5.

In the rest of this section K and K′ are cl-valued fields,

K = (K,Γ,k; v, s, π, i), and K′ = (K ′,Γ′,k′; v′, s′, π′, i′).

An embedding K → K′ is a triple (f, fval, fres) consisting of a field embeddingf : K → K ′, an ordered group embedding fval : Γ → Γ′ and a field embeddingfres : k→ k′ such that

fval(va) = v′(fa) for a ∈ K×, f(sγ) = s′(fvalγ) for γ ∈ Γ,

fres(πa) = π′(fa) for a ∈ Ov, f(ir) = i′(fresr) for r ∈ k.

If K ⊆ K ′, Γ ⊆ Γ′, and k ⊆ k′ (as sets), and the corresponding inclusionmaps f : K → K ′, fval : Γ → Γ, and fres : k → k′ yield an embedding(f, fval, fres) : K → K′, then we call K an cl-valued subfield of K′; notation:K ⊆ K′.

Consider an embedding (f, fval, fres) : K → K′. Then fval and fres arecompletely determined by f . If f is a bijection, then fval and fres are bijec-tions, and we call (f, fval, fres) an isomorphism from K onto K′; note that then(f−1, f−1

val , f−1res ) is an isomorphism from K′ onto K.

In view of Remark 5.10 the AKE-theorem 5.1 is a consequence of the followingstronger result.

5.11. Theorem. Suppose K and K′ are henselian of equicharacteristic 0. Then

K ≡ K′ ⇐⇒ k ≡ k′ and Γ ≡ Γ′.


Proof. Since the forward direction ⇒ is clear, we assume below that k ≡k′ and Γ ≡ Γ′, and our task is to derive K ≡ K′. The case that Γ = {0} istrivial, so we assume from now on that Γ = {0} and thus Γ′ = {0}.

More drastically, we shall assume the Continuum Hypothesis CH, in order tomake our job a little easier. (Theorem 5.11 belongs to a class of mathematicalstatements with the property that any proof of a statement in that class fromZFC + CH can be converted in a, possibly longer, proof of the same statementfrom ZFC alone. See for example an exercise in the chapter on set theory inShoenfield’s Mathematical Logic.)

After replacing K and K′ by elementarily equivalent cl-valued fields, we may(and shall) assume thatK andK′ are saturated of cardinality ℵ1. (This is exactlywhere CH gets used.) In particular, Γ and Γ′ are saturated of cardinality ℵ1,so we have an isomorphism fval : Γ ∼= Γ′ of ordered abelian groups; likewise, kand k′ are saturated of cardinality ℵ1, so we have an isomorphism fres : k ∼= k′

of fields.To obtain Theorem 5.11 it suffices to establish:

Claim. There is a field isomorphism f : K ∼= K ′ which together with fval, andfres yields an isomorphism (f, fval, fres) : K ∼= K′.

We shall obtain such f by a back-and-forth construction. A good subfield of Kis by definition a subfield E of K such that ik ⊆ E, s(vE×) ⊆ E, and vE×

is countable. (Note that then E is the underlying field of a cl-valued subfieldof K.) Likewise, we define the notion good subfield of K ′. A good map is anisomorphism e : E ∼= E′ between good subfields E and E′ of K and K ′, suchthat e(E ∩ Ov) = E′ ∩ Ov′ and

fval(va) = v′(ea) for a ∈ E×, e(sγ) = s′(fvalγ) for γ ∈ vE×,

fres(πa) = π′(ea) for a ∈ E ∩ Ov, e(ir) = i′(fresr) for r ∈ k.

Note that ik and i′k′ are good subfields of K and K ′ and that

ir 7→ i′(fresr) : ik→ i′k′ (r ∈ k)

is a good map. Thus the claim above is a consequence of:

Back-and-Forth. The set of good maps has the back-and-forth property.

To prove Back-and-Forth, consider a good map e : E ∼= E′. We view E and E′

as valued subfields of (K, v,Γ) and (K ′, v′,Γ′) and we note that e : E → E′ isa valued field isomorphism. Before addressing Back-and-Forth we indicate twobasic ways in which e can be extended.

(1) Adjunction of an element of Γ that has prime-torsion modulo vE×. Letδ ∈ Γ be such that δ /∈ vE× but pδ ∈ vE× where p is a prime number. Takex = sδ, so xp = s(pδ) ∈ E×. Then by Lemma 5.6,

v(E(x)×) = vE× + Zδ =

p−1∪i=0

vE× + iδ (disjoint union).


Since s(vE×+Zδ) = s(vE×) ·xZ ⊆ E(x), it follows that E(x) is a good subfieldof K. Put x′ := s′(fvalδ) ∈ K ′. Then E′(x′) is likewise a good subfield of K ′,and Lemma 5.6 shows we have a good map E(x) → E′(x′) that extends e andmaps x to x′.

(2) Adjunction of an element of Γ that has no torsion modulo vE×. Let γ ∈ Γbe such that nγ /∈ vE× for all n > 0. Take x = sγ. By an earlier lemma x istranscendental over E and v(E(x)×) = vE× ⊕ Zγ. Since

s(vE× ⊕ Zγ) = s(vE×) · xZ ⊆ E(x),

E(x) is a good subfield of K. Put x′ := s′(fvalγ) ∈ K ′. Then E′(x′) is likewisea good subfield of K ′, and the same earlier lemma shows we have a good mapE(x)→ E′(x′) that extends e and maps x to x′.

Let ∆ be any countable subgroup of Γ that contains vE×. We now show howto extend our good map e to a good map whose domain has value group ∆.Clearly ∆ is the union

∪∞i=0 ∆i of an increasing sequence

vE× = ∆0 ⊆ ∆1 ⊆ ∆2 ⊆ . . .

of subgroups ∆i of ∆ such that for each i ∈ N, either ∆i+1 = ∆i, or ∆i+1 = ∆i+Zδi with δi ∈ ∆\∆i but pδi ∈ ∆i for some prime number p, or ∆i+1 = ∆i+Zδiand nδi /∈ ∆i for all n > 0. Then by repeated application of (1) and (2) andtaking a union we see that E∆ := E(sδi : i = 0, 1, 2, . . . ) is a good subfield of Kwith vE×

∆ = ∆, and that e extends (uniquely) to a good map with domain E∆.

In addition to (1) and (2), Lemma 4.29 provides a third way to extend goodmaps. We now combine these three ways of extending e to establish Back-and-Forth. Consider an element x ∈ K. We wish to find a good map extending ethat contains x in its domain. By Corollary 5.9 the group v(E(x)×) is countable,so by the construction above we can extend e to a good map e1 with domain E1

such that v(E×1 ) = v(E(x)×). In the same way we extend e1 to a good map e2

with domain E2 such that v(E×2 ) = v(E1(x)

×). By iterating this constructionand taking a union (and renaming this union as E) we reduce to the case thatE(x) is an immediate extension of E. The valued field E(x) has an immediatemaximal extension, which by Lemma 4.29 can be taken inside (K, v,Γ); let E•be the underlying field of such an a immediate maximal extension of E(x) inside(K, v,Γ). Since E• as valued field is also an immediate maximal extension ofE, we can use Lemma 4.29 to extend e to a valued field isomorphism e• fromE• onto a valued difference field E′

• inside (K ′, v′,Γ′). Then e• is a good mapwith x in its domain.

This takes care of the Forth part of Back-and-Forth. The Back part is donelikewise by interchanging the role of E and E′.

Note that Theorem 1.1 (announced in the Introduction) is a special case ofTheorem 5.1, since k[[t]] is the valuation ring of the Laurent series field k((t))(where k is a field).


Elementary classifications like Theorem 5.11 often imply a seemingly strongerresult to the effect that each sentence in a certain language is equivalent in asuitable theory to a sentence of a special form. We now turn our attention tothis aspect. It involves the following elementary fact on boolean algebras, whichis useful in a wide variety of model-theoretic situations.

5.12. Lemma. Let B be a boolean algebra and S(B) its Stone space of ultra-filters. Let Ψ be a subset of B and suppose the map F 7→ F ∩Ψ : S(B)→ P(Ψ)is injective. Then Ψ generates B: every element of B is a boolean combinationof elements of Ψ.

Proof. Let BΨ be the boolean subalgebra of B generated by Ψ. The inclusionBΨ ↪→ B induces the surjective map F 7→ F ∩ BΨ : S(B)↠ S(BΨ). This mapis also injective: if F1, F2 ∈ S(B) and F1∩BΨ = F2∩BΨ, then F1∩Ψ = F2∩Ψ,hence F1 = F2 by the hypothesis of the lemma. The bijectivity of F 7→ F ∩BΨ :S(B)→ S(BΨ) yields B = BΨ by the Stone representation theorem.

Let L be the 3-sorted language of cl-valued fields with 1-sorted sublanguagesLval for value groups and Lres for residue fields. Let T be the L-theory ofhenselian cl-valued fields with residue field of characteristic 0. Then:

5.13. Corollary. For each L-sentence σ there are sentences σ1val, . . . , σ

kval in

the language Lval and sentences σ1res, . . . , σ

kres in Lres such that

T ⊢ σ ←→ (σ1val ∧ σ1

res) ∨ · · · ∨ (σkval ∧ σk

res).

Proof. Let B be the boolean algebra of L-sentences modulo T -equivalence. LetΨ ⊆ B be the set of T -equivalence classes of sentences of the form σval ∧ σres

where σval is an Lval-sentence and σres is an Lres-sentence. Using the famil-iar bijective correspondence between ultrafilters of B and complete L-theoriesextending T , Theorem 5.11 shows that the map F 7→ F ∩ Ψ : S(B) → P(Ψ)from the Stone space S(B) of B into the power set of Ψ is injective. Thus byLemma 5.12 each element of B is a boolean combination of elements of Ψ. Weleave it as an exercise to show that this yields the desired result.

The equivalence displayed in Corollary 5.13 holds in all henselian cl-valued fieldsof equicharacteristic 0, and thus also in all henselian cl-valued fields with residuecharacteristic p > N where N ∈ N depends only on σ.

When we fix the value group to be Z, Corollary 5.13 becomes:

5.14. Corollary. For each L-sentence σ there is a sentence σres in Lres suchthat for all henselian K of equicharacteristic 0 with value group Z,

K |= σ ⇐⇒ k |= σres.

It is convenient to state some of the above also for (one-sorted) valued fields(K,A) without a cross-section or lifting as part of the structure.

6 UNRAMIFIED MIXED CHARACTERISTIC 59

5.15. Corollary. Let σ be a sentence in the language of rings with an extraunary relation symbol. Then there are sentences σ1

val, . . . , σkval in Lval and sen-

tences σ1res, . . . , σ

kres in Lres such that for all henselian valued fields (K,A) of

equicharacteristic 0:

(K,A) |= σ ⇐⇒ ΓA |= σival and kA |= σi

res, for some i ∈ {1, . . . , k}.

It should be clear how this follows from Corollary 5.13 in view of Remark 5.10.Note also that the equivalence holds for all henselian valued fields with residuecharacteristic p > N where N depends only on σ. Thus we have establishedTheorem 1.2 and the Ax-Kochen Principle 2.20.

There remain some claims in the Introduction—for example, Q[[t]] ⪯ C[[t]]—which belong to the circle of AKE-results, but where some extra work is needed.We also want to have equivalences like the above not just for sentences, but alsofor formulas (with applications to the structure of definable sets). These issueswill be addressed in a later version of the notes.

6 Unramified Mixed Characteristic

We now aim for results in the mixed characteristic case that are analogous tothe “equicharacteristic 0” theorems in the previous section. A key difference isthat in the mixed characteristic case we don’t have a lifting of the residue field.Fortunately, we do have a perfect substitute, namely the Teichmuller map.

After introducing the Teichmuller map we are naturally led to Witt vectorswhich allow us to construct, for each perfect field k of characteristic p > 0, adiscrete valuation ring W [k] with residue field k. This construction is functorialand has various other good properties. We shall determine the elementarytheory of the ring W [k] in terms of the elementary theory of the field k; thiscovers the elementary theories of the p-adic fields Qp as a special case.

Throughout we fix a prime number p. Recall that Fp = Z/pZ is the field ofp elements. For integers a and N ≥ 1 we let a mod N be the image of a in theresidue ring Z/NZ.

6.1 The Teichmuller Map

As we have seen, we cannot lift the residue fields of (henselian) local rings likeZ/p2Z and Zp for the simple reason that these rings do not contain any subfield.Fortunately, it turns out that there is a canonical system of representatives forthe residue field in these rings, but it only preserves the multiplicative structureof the residue field, not its additive structure. For Z/p2Z it is the map

τ : Fp = Z/pZ→ Z/p2Z, τ(a mod p) = ap mod p2, a ∈ Z.

To see that such a map τ exists, use the fact that for all a, b ∈ Z,

a ≡ b mod p =⇒ ap ≡ bp mod p2.


Because also a ≡ ap mod p for all a ∈ Z, this map τ is a system of represen-tatives for the residue field Fp, in the sense that for each x ∈ Fp the image ofτx in Fp is x. Note also that τ(xy) = τ(x)τ(y) for all x, y ∈ Fp. For the ringZ/p3Z, the canonical system of representatives for its residue field Fp is given

by a mod p 7→ ap2

mod p3, (a ∈ Z). Below we extend these observations tocomplete local rings with perfect residue field of characteristic p.

A field k of characteristic p is said to be perfect if its Frobenius endomorphismx 7→ xp is an automorphism. Given any local ring A with residue field k ofcharacteristic p, one shows easily by induction on n that for all x, y ∈ A,

(∗) x ≡ y mod m =⇒ xpn

≡ ypn

mod mn+1.

6.1. Theorem. Let A be a complete local ring with maximal ideal m and perfectresidue field k of characteristic p. Then there is a unique map τ : k → A suchthat τx = x and τ(xp) = (τx)p for all x ∈ k. This (Teichmuller) map τ has thefollowing properties:

(i) τ(k) = {a ∈ A : a is a pnth power in A, for each n};

(ii) τ(xy) = τ(x)τ(y) for all x, y ∈ k, τ(0) = 0, τ(1) = 1, and τ(k×) ⊆ U(A);

(iii) if p · 1 = 0 in A, then τ is a ring embedding.

Proof. Let x ∈ k, and choose for each n an an ∈ A such that

x = (an)pn

, (possible since k is perfect).

Then apn+1 ≡ an mod m, so apn+1

n+1 ≡ apn

n mod mn+1 by (*) above. Hence the

sequence {apn

n } converges in the m-adic norm to an element in A. This limitdoes not depend on the choice of {an}: if {bn} is another choice, then an ≡ bnmod m, so ap

n

n ≡ bpn

n mod mn+1 by (*) above, and thus {apn

n } and {bpn

n } havethe same limit. We define τx to be this limit. It is easy to check that τx = xand τ(xp) = (τx)p for all x ∈ k.

To show that these two identities uniquely determine τ , consider a mapι : k→ A such that ιx = x and ι(xp) = (ιx)p for all x ∈ k. For x ∈ k and {an}as above, put bn := ιan, so ιx = bp

n

n , so x = (bn)pn

(for all n), and thus

τx = lim bpn

n = ιx.

To prove (i), each element of τ(k) is a pnth power in A for each n, since eachelement of k is a pnth power in k for each n. Conversely, let a ∈ A be a pnthpower in A for each n. Choose for each n an an ∈ A such that a = ap

n

n . Thenthe construction of τ above shows that for x := a we have τx = a.

This construction also yields (ii), and (iii) follows from (i) by noting that ifp · 1 = 0 in A, then (a+ b)p

n

= apn

+ bpn

for all a, b ∈ A and all n.


The reader should verify that for A = Z/p2Z and A = Z/p3Z the Teichmullermap is the one given at the beginning of this subsection.

Exercise. With the same assumptions as in the Theorem, show that τ(k) isthe unique set S ⊆ A such that sp ∈ S for each s ∈ S, and S is mappedbijectively onto k by the residue class map. Show that for A = Zp one hasτ(Fp) = {x ∈ Zp : xp−1 = 1} ∪ {0}.

A local p-ring is a complete local ring A such that m = pA and k is perfect. Ifin addition pn = 0 in A for all n, then we say that A is strict. (Here p := p·1 ∈ A,and below we often commit the same abuse of notation by letting p denote theelement p · 1 in a ring.)

6.2. Corollary. Let A be a local p-ring. If pn = 0 and pn+1 = 0 in A, thenthere is for every a ∈ A a unique tuple (x0, . . . , xn) ∈ kn+1 such that

a =

n∑i=0

τ(xi)pn.

If A is strict, then there is for every a ∈ A a unique sequence {xn} ∈ kN suchthat

a =∞∑

n=0

τ(xn)pn.

Let A be a strict local p-ring. For a ∈ A, call {xn} as above the Teichmullervector of a. Note that the map

a 7→ Teichmuller vector of a : A→ kN

is a bijection. Given the Teichmuller vectors of a, b ∈ A, how do we obtain theTeichmuller vectors of a+ b and ab? It turns out that there is a good answer tothis question, but things become more transparent if we change from Teichmullervectors to Witt vectors: Let a ∈ A; then the unique sequence {xn} ∈ kN such

that a =∑∞

n=0 τ(xp−n

n )pn is called the Witt vector of a (and then {xp−n

n } is itsTeichmuller vector). Note that if k = Fp, then there is no difference betweenthe Teichmuller vector and the Witt vector of a.

We also want to show that, given any perfect field k, there is exactly onestrict local p-ring with residue field isomorphic to k, up to isomorphism. Toconstruct such a ring from k we shall take the set kN of Witt vectors over kand define a suitable addition and multiplication on it making it a strict localp-ring W [k]. This is done in the next subsection.

Exercises. For any ring R, let µ(R) := {x ∈ R : xn = 1 for some n > 0} bethe group of roots of unity in R, so µ(R) is a subgroup of U(R). Let A be astrict local p-ring. Show that if p = 2, then τ maps µ(k) bijectively onto µ(A).

6.2 Witt vectors

Under the familiar bijection F2p −→ Z/p2Z given by

(i mod p, j mod p) 7→ (i+ jp) mod p2, i, j ∈ {0, . . . , p− 1}


the addition and multiplication of Z/p2Z do not correspond to algebraicallynatural operations on F2

p. One motivation for Witt vectors is to find a bijectionF2p −→ Z/p2Z that makes the addition map of the ring Z/p2Z,

+ : Z/p2Z× Z/p2Z −→ Z/p2Z

correspond to an explicit polynomial map

F4p = F2

p × F2p −→ F2

p,

and likewise with multiplication on Z/p2Z. Corollary 6.2 applied to A = Z/p2Zturns out to provide just the right (Witt) bijection F2

p −→ Z/p2Z:

(a mod p, a′ mod p) 7→ (ap + pa′) mod p2, a, a′ ∈ Z.

To see which binary operations on F2p correspond under this bijection to addition

and multiplication on Z/p2Z we note that for integers a, a′, b, b′,

(ap + pa′) + (bp + pb′) = (a+ b)p + p(a′ + b′ −

p−1∑i=1

c(p, i)aibp−i),

(ap + pa′)× (bp + pb′) = (ab)p + p(apb′ + a′bp + pa′b′

),

where c(p, i) is the integer(pi

)/p for i = 1, . . . , p − 1. Thus addition on Z/p2Z

corresponds under the Witt bijection to the binary operation on F2p given by

((x0, x1), (y0, y1)

)7→

(x0 + y0, x1 + y1 −

p−1∑i=1

c(p, i)xi0y

p−i0

): F2

p × F2p → F2

p.

Likewise, multiplication on Z/p2Z corresponds under this bijection to((x0, x1), (y0, y1)

)7→ (x0y0, xp

0y1 + x1yp0 + px1y1) : F2

p × F2p → F2

p.

(Of course, in Fp we have the identity xp0y1+x1y

p0+px1y1 = x0y1+x1y0, but we

prefer to state it the way we did because later we shall replace Fp by an arbitraryring.) So our bijection turns the ring Z/p2Z into an algebraic-geometric objectliving in Fp so to say.

Next we try to extend all this to Z/p3Z, Z/p4Z, and so on, all the way upto Zp = lim←− Z/pnZ. Applying Corollary 6.2 to A = Z/p3Z yields the bijection

(a mod p, a′ mod p, a′′ mod p) 7→ (ap2

+ pa′p + p2a′′) mod p3, a, a′, a′′ ∈ Z.

from F3p onto Z/p3Z. Again, under this bijection the addition and multiplica-

tion of Z/p3Z correspond to binary operations on F3p given by easily specifiable

polynomials over Fp.Just as crucial is that this Witt bijection between F3

p and Z/p3Z is com-patible with the earlier one between F2

p and Z/p2Z in the sense that we have acommuting diagram


F3p −−−−→ F2

py yZ/p3Z −−−−→ Z/p2Z

where the vertical arrows are the Witt bijections, and horizontal arrows aregiven by

(x0, x1, x2) 7→ (x0, x1) (top),

a mod p3 7→ a mod p2, a ∈ Z (bottom).

It is also important that these constructions are functorial . Indeed, for any ringR we can define the ring W2[R] as follows: its underlying set is R2, its additionis given by

((x0, x1), (y0, y1)

)7→

(x0 + y0, x1 + y1 −

p−1∑i=1

c(p, i)xi0y

p−i0

): R2 ×R2 → R2

and its multiplication by((x0, x1), (y0, y1)

)7→ (x0y0, xp

0y1 + x1yp0 + px1y1) : R2 ×R2 → R2.

The zero element of W2[R] is (0, 0), and its multiplicative identity is (1, 0). Onecan view some of the above as providing an explicit isomorphism from the ringW2[Fp] onto the ring Z/p2Z. Likewise, we can define a ring W3[R], and so on,but at this point we may as well go the whole distance, and define W [R].

The Witt Polynomials. Fix distinct indeterminates

X0, X1, X2, . . . , Y0, Y1, Y2, . . . .

The polynomials W0(X0),W1(X0, X1), . . . ,Wn(X0, . . . , Xn), . . . are elements ofthe polynomial ring Z[X0, X1, X2, . . . ] defined as follows:

W0 = X0,

W1 = Xp0 + pX1,

W2 = Xp2

0 + pXp1 + p2X2,

· · · · · ·

Wn = Xpn

0 + pXpn−1

1 + · · ·+ pnXn =n∑

i=0

piXpn−i

i ,

· · · · · ·


Note that Wn+1(X0, . . . , Xn+1) = Wn(Xp0 , . . . , X

pn) + pn+1Xn+1. Let R be a

ring and consider the map θ : RN −→ RN defined by

θ(x0, x1, x2, . . . ) =(W0(x0),W1(x0, x1), . . . ,Wn(x0, . . . , xn), . . .

).

Exercise. Show that θ(1, 0, 0, 0, . . . ) = (1, 1, 1, 1, . . . ). Show also:

θ(−1, 0, 0, 0, . . . ) = (−1,−1,−1,−1, . . . ) if p is odd,

θ(−1,−1,−1,−1, . . . ) = (−1,−1,−1,−1, . . . ) if p = 2.

Below, x = (x0, x1, x2, . . . ) and y = (y0, y1, y2, . . . ) range over RN .

6.3. Lemma. Suppose p · 1 ∈ U(R). Then θ is a bijection.

Proof. We have θ(x) = y if and only if

x0 = y0, px1 = y1 − xp0, . . . , pnxn = yn − xpn

0 − · · · − pn−1xpn−1, . . . ,

so there is exactly one such x for each y.

Remark. By the same argument, θ is injective if pa = 0 for all a ∈ R \ {0}.

6.4. Lemma. Suppose pa = 0 for all nonzero a ∈ R, let m be given, and letx0, x1, . . . , xn, y0, y1, . . . , yn ∈ R. Then the following are equivalent

(1) xi ≡ yi mod pmR for i = 0, . . . , n;

(2) Wi(x0, . . . , xi) ≡Wi(y0, . . . , yi) mod pm+iR for i = 0, . . . , n.

Proof. That (1) implies (2) is because for all a, b ∈ R and i ∈ N,

a ≡ b mod pmR =⇒ api

≡ api

mod pm+iR.

The converse follows by induction on i, using the assumption on p.

6.5. Lemma. Let A := Z[X0, X1, . . . , Y0, Y1, . . . ]. There are polynomialsS0, S1, S2 . . . , P0, P1, P2, · · · ∈ A such that for all n,

Wn(X0, . . . , Xn) +Wn(Y0, . . . , Yn) = Wn(S0, . . . , Sn),

Wn(X0, . . . , Xn)×Wn(Y0, . . . , Yn) = Wn(P0, . . . , Pn).

These conditions determine the sequences {Sn} and {Pn} uniquely, and for eachn we have Sn, Pn ∈ Z[X0, . . . , Xn, Y0, . . . , Yn].

Proof. We make p ∈ A a unit by working in the larger domain

R := {a/pm : a ∈ A,m = 0, 1, 2 . . . } ⊆ Q[X0, X1, . . . , Y0, Y1, . . . ].

Then the bijectivity of the map θ of Lemma 6.3 shows that there are uniquesequences {Sn} and {Pn} in R such that the identities above hold. Note that


S0 = X0 + Y0. Assume for a certain n > 0 that Si ∈ Z[X0, . . . , Xi, Y0, . . . , Yi]for i = 0, . . . , n− 1. We have

Wn(S0, . . . , Sn) = Wn−1(Sp0 , . . . , S

pn−1) + pnSn

= Wn(X0, . . . , Xn) +Wn(Y0, . . . , Yn)

= Wn−1(Xp0 , . . . , X

pn−1) +Wn−1(Y

p0 , . . . , Y

pn−1) + pn(Xn + Yn)

= Wn−1

(S0(X

p0 , Y

p0 ), . . . , Sn−1(X

p0 , . . . , X

pn−1, Y

p0 , . . . , Y

pn−1)

)+ pn(Xn + Yn).

Also, Spi ≡ Si(X

p0 , . . . , X

pi , Y

p0 , . . . , Y

pi ) mod pA for i = 0, . . . , n − 1, hence by

Lemma 6.4 the polynomial Wn−1(Sp0 , . . . , S

pn−1) is congruent modulo pnA to

Wn−1

(S0(X

p0 , Y

p0 ), . . . , Sn−1(X

p0 , . . . , X

pn−1, Y

p0 , . . . , Y

pn−1)

).

It follows that all coefficients of Sn are in Z. The proof for the Pn is similar,starting with P0 = X0Y0.

Of course, the letters S and P were chosen to remind the reader of sum andproduct. The polynomials S0, S1, P0, P1 are easy to write down:

S0 = X0 + Y0, S1 = X1 + Y1 −p−1∑i=1

c(p, i)Xi0Y

p−i0 ,

P0 = X0Y0, P1 = Xp0Y1 +X1Y

p0 + pX1Y1.

It is also easy to check that the constant terms of Sn and Pn are zero.

Let W [R] be the set RN equipped with the binary operations

+ : W [R]×W [R]→W [R], · : W [R]×W [R]→W [R],

defined as follows:

x+ y :=(S0(x0, y0), S1(x0, x1, y0, y1), . . . , Sn(x0, . . . , xn, y0, . . . , yn), . . .

),

x · y :=(P0(x0, y0), P1(x0, x1, y0, y1), . . . , Pn(x0, . . . , xn, y0, . . . , yn), . . .

).

We also define the elements 0 and 1 of W [R] by 0 = (0, 0, 0, . . . ) and 1 =(1, 0, 0, . . . ). For any ring morphism ϕ : R→ R′ we define

W [ϕ] : W [R]→W [R′], W [ϕ](x) = (ϕx0, ϕx1, ϕx2, . . . ),

so W [ϕ] is a homomorphism for the addition and multiplication operations onW [R] and W [R′], and W [ϕ](0) = 0, W [ϕ](1) = 1.

6.6. Corollary. With these operations, W [R] is a ring with zero element 0 andmultiplicative identity 1.


Proof. By the previous lemma, the map θ : W [R]→ RN is a homomorphism foraddition and multiplication, where these operations are defined componentwiseon RN . Note also that θ(0) = 0 and θ(1) = (1, 1, 1, . . . ) are the zero andmultiplicative identity of the ringRN . Thus by the exercise preceding Lemma 6.3we have a subring θ(W [R]) of RN . In view of the remark following Lemma 6.3,we conclude that if pa = 0 for all nonzero a ∈ R, then W [R] is a ring with zeroelement 0 and multiplicative identity 1. In particular, W [A] is a ring with zeroelement 0 and multiplicative identity 1 for any polynomial ring A = Z[(Xi)i∈I ]in any family of distinct indeterminates (Xi)i∈I . The general case now followsby taking a surjective ring morphism ϕ : A→ R where A is such a polynomialring, and using the surjective map W [ϕ] : W [A]→W [R].

We call W [R] the ring of Witt vectors over R. The map θ is a ring morphismof W [R] into the product ring RN . Note also that if ϕ : R → R′ is a ringmorphism, so is W [ϕ] : W [R] → W [R′]. Below we consider x = (x0, x1, . . . )and y = (y0, y1, . . . ) as elements of the ring W [R] rather than of the productring RN ; we also put x(i) := Wi(x0, . . . , xi) ∈ R for i ∈ N, and x(i) := 0 ∈ R fornegative i ∈ Z.

Exercise. Show that if p is odd, then the additive inverse of x in W [R] is givenby −x = (−x0,−x1,−x2, . . . ).

We now define two further (unary) operations on W [R]:

V : W [R]→W [R], Vx = (0, x0, x1, x2, . . . ),

F : W [R]→W [R], Fx = (xp0, x

p1, x

p2, . . . ).

The map V is a shift operation (“Verschiebung” in German). The shift mapV can also be viewed as a unary operation on the ring RN , and as such it isan endomorphism of the additive group of RN ; this is also the case for V as anoperation on W [R], but this requires an argument:

6.7. Lemma. The following identities hold:

(Vx)(n) = px(n−1), V(x+ y) = Vx+Vy, (Vmx)(n) = pmx(n−m).

Proof. For n > 0 we have

Wn(0, x0 . . . , xn−1) = pxpn−1

0 + · · ·+ pnxn−1 = pWn−1(x0, . . . , xn−1),

which gives the first identity. In other words, θ(Vx) = pV(θx). Hence

θ(V(x+ y)) = pV(θ(x+ y)) = pV(θx+ θy)

= pV(θx) + pV(θy) = θ(Vx) + θ(Vy)

= θ(Vx+Vy).

Thus V(x+ y) = V(x) +V(y) if R is a polynomial ring over Z. The validity ofthis identity in this special case implies its validity in general since each ring isisomorphic to a quotient of a polynomial ring over Z. The third identity followsby induction on m from the first identity.


Next we put [a] := (a, 0, 0, . . . ) ∈W [R] for a ∈ R.

6.8. Lemma. The following identities hold, where a ∈ R:

(1) [a](n) = apn

;

(2) x =∑n−1

i=0 Vi[xi] +Vn(xn, xn+1, . . . );

(3) [a]x = (ax0, apx1, a

p2

x2, . . . ).

Proof. Identity (1) is immediate. It easily yields identity (2) for n = 1, initiallyin the case that R is a polynomial ring over Z, and then for any R by the usualargument. Induction gives identity (2) for all n. Identity (3) follows likewise bychecking that ([a]x)(n) = Wn(ax0, . . . , a

pn

xn).

Remark. The decomposition in (2) is clearly unique: if x =∑n−1

i=0 Vi[yi]+Vnz,then xi = yi for all i < n.

6.9. Lemma. Let y = px and z = V(Fx). Then yn ≡ zn mod pR.

Proof. We have y(n) = px(n) = p(F(x)(n−1) + pnxn

)by the recursion for the

Witt polynomials. Also p(Fx)(n−1) = (V(Fx)(n) = z(n) by the first identity ofLemma 6.7, so y(n) ≡ z(n) mod pn+1R. This holds for all n, so by Lemma 6.4we have yn ≡ zn mod pR for all n in case R is a polynomial ring over Z, andthus for all R by the usual argument.

By Lemma 6.3 the ring W [R] is just a disguised version of the product ring RN

if R is a field of characteristic 0. The situation is much more interesting if R isa perfect field of characteristic p, and from this point on we are just going toconsider this case. So in the remainder of this subsection k denotes a perfectfield of characteristic p, and x = (x0, x1, x2, . . . ), y = (y0, y1, y2, . . . ) ∈W [k].

Note that then F = W [Φ] where Φ : k→ k is the Frobenius automorphism.Thus F is an automorphism of the ring W [k]. The map V is related to F by

F ◦V = V ◦ F = p · Id,

as a consequence of Lemma 6.9. In other words,

px = p(x0, x1, x2, . . . ) = (0, xp0, x

p1, x

p2, . . . ),

and thus p2x = (0, 0, xp2

0 , xp2

1 , . . . ), and so on. Also

(∗) Vmx ·Vny = Vm+n(Fn(x)Fm(y)

).

To see this, note that upon setting x = Fmu and y = Fnz with u, z ∈ W [k],the identity becomes pmu · pnz = pm+nuz.

We define v : W [k]→ Z ∪ {∞} by

v(0) =∞, vx = n if xn = 0 and xi = 0 for all i < n.


6.10. Theorem. W [k] is a complete DVR with normalized valuation v, fractionfield of characteristic 0, and maximal ideal pW [k]. The map x 7→ x0 : W [k]→ kis a ring morphism with kernel pW [k], and thus induces an isomorphism of theresidue field of W [k] with the field k. Upon identifying k with the residue fieldvia this isomorphism, the Teichmuller map k → W [k] is given by a 7→ [a], andfor x ∈W [k] we have

x =∞∑

n=0

[xp−n

n ]pn.

Proof. Note that vx ≥ n iff x = Vnu for some u ∈ W [k] (and in that case,vx = n iff u0 = 0). Hence v(x + y) ≥ min(vx, vy) by the additivity of V. Wealso get v(xy) = vx + vy by the identity (∗), and the fact that this identityclearly holds when vx = vy = 0. Thus W [k] is a domain, and v is a valuationon this domain. Next, we claim that v(x − y) ≥ n iff xi = yi for all i < n. Tosee this, write x = y+z, so the claim becomes: v(z) ≥ n iff xi = yi for all i < n.Now decompose each of x, y, z according to identity (2) of Lemma 6.8, and usethe remark following this lemma to establish the claim. From this claim we getthat W [k] is complete with respect to the ultranorm |x| := p−vx. Identity (2)

of Lemma 6.8, together with [a] = Fn[ap−n

] for a ∈ k, now yields

x =

∞∑n=0

Vn[xn] =

∞∑n=0

[xp−n

n ]pn.

If |x| = 1, then x0 = 0, and thus by (2) and (3) of Lemma 6.8, [x−10 ]x = 1 + y

with |y| < 1, so x ∈ U(W [k]). We have now shown that W [k] is a local domainwith maximal ideal

{x : vx > 0} = V(W [k]

)= V

(F(W [k])

)= pW [k].

It also follows that vx = n iff x = pnu with u ∈ U(W [k]). Thus W [k] is acomplete DVR with normalized valuation v and residue field isomorphic to k.Since p · 1 = (0, 1, 0, 0, . . . ) = 0, the fraction field of W [k] has characteristic 0.The rest of the theorem now follows easily.

In particular, W [k] is a strict local p-ring with residue field isomorphic to k.This property determines the ring W [k] up to isomorphism, as a consequenceof the next theorem. (It is this consequence that is crucial in the proof of theAKE-results in the unramified mixed characteristic case.)

6.11. Theorem. Let A be a strict local p-ring and let ϕ : k → kA be a fieldembedding of k into the residue field kA of A. Then there is a unique ringmorphism f : W [k]→ A such that the diagram

W [k]f−−−−→ Ay y

k −−−−→ϕ

kA


commutes. The map f is injective. If ϕ is an isomorphism, so is f .

Proof. Let τ be the Teichmuller map of A. To simplify notation we identify kwith a subfield of kA via ϕ. We define f : W [k]→ A by

f(x) =

∞∑i=0

τ(xp−i

i )pi.

It is clear that f(0) = 0 and f(1) = 1. Let x + y = z in W [k]. To provef(x) + f(y) = f(z), put

F−nx :=(xp−n

0 , xp−n

1 , xp−n

2 , . . .)∈W [k],

τ(F−nx) :=(τ(xp−n

0 ), τ(xp−n

1 ), τ(xp−n

2 ), . . .)∈W [A].

Then (τ(F−nx)

)(n)=

n∑i=0

τ(xp−i

i )pi ∈ A,

solim

n→∞

(τ(F−nx)

)(n)= f(x).

Now F−nx+ F−ny = F−nz in W [k], so for all ν ∈ N:(τ(F−nx) + τ(F−ny)

)ν≡

(τ(F−nz)

)ν

mod pA,

and thus by Lemma 6.4:(τ(F−nx)

)(n)+(τ(F−ny)

)(n) ≡ (τ(F−nz)

)(n)mod pn+1A.

Taking the limit as n goes to infinity yields f(x) + f(y) = f(z). Likewise oneshows that f(x)f(y) = f(xy), so f is a ring morphism lifting ϕ. The otherclaims of the theorem are easy consequences of Corollary 6.2.

For A = Zp and residue field Fp this theorem yields a ring isomorphism

W [Fp] ∼= Zp, x = (x0, x1, . . . ) 7→∞∑i=0

τ(xn)pn,

where τ is the Teichmuller map of Zp (which assigns to each η ∈ F×p the unique

ζ ∈ Zp with ζp−1 = 1 that has residue class a.) It is usual to identify the ringsW [Fp] and Zp via this isomorphism.

Exercises. Show:

{x ∈W [k] : Fx = xp} = {[a] : a ∈ k}.

With Aut(R) denoting the group of automorphisms of a ring R, show that

ϕ 7→W [ϕ] : Aut(k)→ Aut(W [k])

is a group isomorphism.


6.3 Coarsening

To obtain AKE-type results for unramified henselian valued fields of mixedcharacteristic we proceed as follows: in a sufficiently saturated model we cancoarsen the valuation to have equicharacteristic zero and a residue field that iscomplete with respect to a discrete valuation. Using the uniqueness theorem atthe end of the previous subsection, this allows us to reduce the problem to anequicharacteristic zero situation. In this short subsection we just consider theoperation of coarsening in general.

Let Γ be an ordered abelian group. A subgroup ∆ of Γ is said to be convex (inΓ) if for all γ ∈ Γ and δ ∈ ∆,

|γ| < |δ| =⇒ γ ∈ ∆.

Given any γ ∈ Γ there is a smallest convex subgroup of Γ that contains γ,namely

{δ ∈ Γ : |δ| ≤ n|γ| for some n},

If γ ∈ Γ and γ = 0, there is also a largest convex subgroup of Γ that does notcontain γ, namely

{δ ∈ Γ : |δ| < n|γ| for all n}.

The subgroups {0} and Γ of Γ are convex, and Γ is archimedean iff there are noother convex subgroups of Γ. The set of all convex subgroups of Γ is linearlyordered by inclusion.

Let ∆ be a convex subgroup of Γ. Then we make the quotient group Γ/∆into an ordered abelian group, by defining γ+∆ > 0+∆ if γ > ∆. The naturalmap Γ→ Γ/∆ then preserves ≤: for γ1, γ2 ∈ Γ we have

γ1 ≤ γ2 =⇒ γ1 +∆ ≤ γ2 +∆.

Next, consider a valuation v : K× → Γ on a field K with valuation ring O,maximal ideal m of O, and residue field k = O/m. Let a convex subgroup ∆ ofΓ be given. Then we define the coarsening v of v as follows:

v : K× → Γ := Γ/∆, va := va+∆.

This coarsening is again a valuation on K, with valuation ring

O = {a ∈ K : va ≥ δ for some δ ∈ ∆}.

The maximal ideal of O is

m = {a ∈ K : va > δ for all δ ∈ ∆}.

The information that gets lost by this coarsening can be recovered in the residuefield k = O/m by means of the induced valuation

v : k→ ∆, v(a+ m) := va for a ∈ O \ m.


(To keep notations simple we use the same letter v for the initial valuation onK and this induced valuation on k.) Note that m is a prime ideal of O, and thatthe subring O/m of O/m = k is the valuation ring of the induced valuation onk, with maximal ideal m/m, and residue field k (after an obvious identification).

Thus in some sense the original valuation v : K× → Γ is decomposed into twosimpler valuations, namely v : K× → Γ with residue field k, and v : k× → ∆.

Exercise. Show that O is henselian if and only if O and O/m are henselian.

Let us next consider the situation where Γ has a smallest positive element, callit 1, and ∆ = Z1 is the smallest convex subgroup of Γ that contains 1. Picksome t ∈ O with vt = 1. It is easy to check that then

O = O[1/t], m = ∩∞n=0tnO.

For later use we note:

6.12. Lemma. If (K, v,Γ) is ℵ1-saturated, then the discrete valued field (k, v,Z1)is complete.

6.4 AKE for unramified mixed characteristic

For perfect k of characteristic p we let W (k) be the fraction field of W [k], andconsider it as the valued field that has W [k] as its valuation ring. A valuedfield (K, v,Γ) of mixed characteristic (0, p) is said to be unramified if vp is thesmallest positive element of Γ. (For example, the valued field W (k) with k asabove is unramified.)

6.13. Theorem. Let (K1, v1,Γ1) and (K2, v2,Γ2) be henselian unramified fieldsof mixed characteristic (0, p), with perfect residue fields k1 and k2. Supposethat k1 ≡ k2 (as fields), and Γ1 ≡ Γ2 (as ordered abelian groups). Then(K1, v1,Γ1) ≡ (K2, v2,Γ2).

Proof. As in the proof of Theorem 5.11 we shall take a short cut by assumingCH. Then we can reduce to the case that (Ki, vi,Γi) is saturated of size ℵ1 fori = 1, 2, so that we have isomorphisms

ϕ : k1∼= k2, h : Γ1

∼= Γ2.

For i = 1, 2 we identify Z with the convex subgroup Z · vip of Γi via k 7→ k · vip,and coarsen vi accordingly, obtaining the valuation

vi : K×i → Γi := Γi/Z.

Its residue field ki carries the distinguished discrete valuation vi : ki×→ Z.

Then h induces an isomorphism h : Γ1∼= Γ2. Because ki is complete with

respect to vi and has perfect residue field ki, we can use Theorem 6.11 to liftthe isomorphism ϕ : k1

∼= k2 to an isomorphism

(f, id) : (k1, v1,Z) ∼= (k2, v2,Z).


Since the valued field (K1, v1, Γ1) is henselian of equicharacteristic 0 we havea lifting i1 : k1 → K1 of the residue field of this valued field, and likewise wehave a lifting i2 : k2 → K2 of the residue field of the valued field (K2, v2, Γ2).It can be shown that we also have ×-sections si : Γi → K×

i of these valuedfields. (.....this needs some explanation...) Under these circumstances the proofof Theorem 5.11 constructs an isomorphism

(F, h, f) : (K1, Γ1, k1; v1, π1, s1, i1) ∼= (K2, Γ2, k2; v2, π2, s2, i2),

where πi : Oi → ki is the residue map of the valued field (Ki, vi, Γi). It followseasily that then we have an isomorphism

(F, h, ϕ) : (K1,Γ1,k1; v1, π1) ∼= (K2,Γ2,k2; v2, π2),

where πi : Oi → ki is the residue map of the valued field (Ki, vi,Γi).

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