Articulo 7.dvi


CUBO A Mathematical Journal
Vol.12, No¯ 02, (97–121). June 2010

The tree of primes in a field

Wolfgang Rump

Institute for Algebra and Number Theory,

University of Stuttgart,

Pfaffenwaldring 57, D-70550 Stuttgart, Germany

email: rump@mathematik.uni-stuttgart.de

Dedicated to B. V. M.

ABSTRACT

The product formula of algebraic number theory connects finite and infinite primes in a

stringent way, a fact, while not hard to be checked, that has never ceased to be tantalizing.

We propose a new concept of prime for any field and investigate some of its properties.

There are algebraic primes, corresponding to valuations, such that every prime contains a

largest algebraic one. For a number field, this algebraic part is zero just for the infinite

primes. It is shown that the primes of any field form a tree with a kind of self-similar

structure, and there is a binary operation on the primes, unexplored even for the rationals.

Every prime defines a topology on the field, and each compact prime gives rise to a unique

Haar measure, playing an essential part in the product formula.

RESUMEN

La fórmula producto de la Teoŕıa de Números Algebraicos conecta primos finitos e infinitos

de una formula estricta, un hecho no dif́ıcil de ser verificado, es que nunca cesa de ser

estudiado. Nosotros proponemos un nuevo concepto de primos para cualquier cuerpo

e investigamos algunas de sus propiedades. Hay primos algebraicos, correspondientes a

valuaciones, talque todo primo contiene un primo algebraico mayor. Para un número de

cuerpos, esta parte algebraica es cero solamente para primos infinitos. Es demostrado que

los primos de cualquier cuerpo forman un árbol con una clase de estructura auto-similar, y



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hay una operación binaria sobre los primos inexplorada incluso para los racionales. Todo

primo define una topoloǵıa sobre el cuerpo, y todo primo compacto da origen a una única

medida de Haar, jugando rol esencial en la fórmula producto.

Key words and phrases: prime, valuation, product formula.

AMS (MOS) Subj. Class.: Primary: 11S15, 11N80, 12J20, 13A18. Secondary: 28C10

Let K be a field, and let p 6= 0 be a polynomial in K[x]. Assume first that p is normed and

decomposes into powers of distinct linear factors:

p = (x − b1)
e1 · · · (x − br)

er . (1)

Then p is completely determined by the function e: K → Z with e(bi) := ei and e(b) := 0 for

all b /∈ {b1, . . . ,br}. Furthermore, e(b) > 0 for all b ∈ K. The same holds for rational functions

p = g
h
∈ K(x) if g,h ∈ K[x] are normed and decompose into linear factors. Here e may also attain

negative values in Z.

To complete the picture, replace K by the projective line P1(K) := K ∪ {∞}. If K = IR, we can

calculate the limit of p for x → ∞: In case that deg p := deg g − deg h < 0, we have lim
x→∞

p(x) = 0,

and −deg p can be regarded as the multiplicity of ∞ ∈ P1(K) as a zero of p. If deg p > 0, then p has

a pole in ∞, i. e. a zero with negative multiplicity −deg p. Therefore, we set

e(∞) := −deg p, (2)

and this definition makes sense for an arbitrary field K. The function e vanishes almost everywhere,

and ∑

b∈P1(K)

e(b) = 0. (3)

In general, instead of (1), a polynomial p is of the form

p = a · pe11 · · ·p
er
r (4)

with a ∈ K× and different irreducible polynomials p1, . . . ,pr. Then the pi take the rôle of the points

bi ∈ P
1(K). However, K can be extended such that every pi splits into deg pi linear factors. To take

this into account, a weight f(pi) := deg pi is attached to the point pi, i. e. pi counts f(pi)-fold. At

infinity ∞ we still have e(∞) = −deg p, and f(∞) := 1. So Eq. (3) turns into

r∑

i=1

e(pi)f(pi) + e(∞)f(∞) = 0. (5)

In other words, every point pi has two invariants: the multiplicity e(pi), that is, the ramification order,

and the weight f(pi), called the inertial degree.

Historically, the analogy between function fields and the field Q of rationals became more and

more apparent by the work of Gauß, Dedekind and Artin [11]. In this vein, it was natural to ask

whether a formula like (5) holds for the field Q.

In fact, every n ∈ Q× is of the form n = ±pe11 · · ·p
er
r with different primes p1, . . . ,pr and integers

ei. The unit ±1 corresponds to the unit a in formula (4). To get a multiplicative analogue of (5), we



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The tree of primes in a field 99

set e(pi) := ei and f(pi) := p
−1
i . For the infinite prime, we set e(∞) := 1 and f(∞) := |n|. So we get

the product formula for Q:
r∏

i=1

f(pi)
e(pi) · f(∞)e(∞) = 1. (6)

An analogous product formula holds for algebraic number fields (see, e. g., [8], III.1.3), or more

generally, for global fields [10]. Artin and Whaples [1, 2] have shown that the product formula

for algebraic number fields follows by formal arguments from its validity for Q, and that a similar

statement holds for function fields.

For a global field K, (finite or infinite) primes can be conceived as equivalence classes of absolute

values v : K → IR, normalized in a natural way using the modulus function of the locally compact

completion K̂v (see [4], VI.9, Proposition 1). In this way, the product formula admits a natural

formulation, but we feel that it remains to be surprising: it just happens to be true - still a mystery!

It deeply touches the relationship between the additive and the multiplicative structure of Q.

To reveal this mystery, attempts have been made to elaborate on the concept of “prime” in a

field [6, 7, 5, 13]. In the present paper, we give a new definition, partly anticipated by Krull [7],

which might also shed some more light upon the distinction between finite and infinite primes of an

algebraic number field. Namely, we define a prime in a field K to be a subset P ⊂ K with the following

properties:

(P1) 1 /∈ P .

(P2) a,b ∈ P ⇒ ab ∈ P .

(P3) ab ∈ P ⇒ a ∈ P or b ∈ P .

(P4) ∃d ∈ K× : P + P ⊂ dP .

As a consequence, the subsets aP of K with a ∈ K× form a chain. There are two types of primes

- we call them algebraic and transcendental - such that the algebraic primes correspond to valuation

rings in K. For every prime P , there is a largest algebraic prime Q ⊂ P (Proposition 7.2); we denote

it by P◦. If P◦ = 0, we call P purely transcendental. For an algebraic number field K, the purely

transcendental primes just correspond to the infinite primes.

There are unexpected properties of our concept of prime. As is well-known, the valuation rings

of a field form a dual tree under inclusion. More generally, we show that all primes of a field K form

a tree which is almost self-similar (section 2). Namely, we introduce a binary operation P ∗ Q on the

primes which satisfies P ⊂ P ∗Q (Proposition 2.4). Thus even for K = Q, there are a lot more primes

P ⊂ Q than rational prime numbers. A further investigation of this operation might be desirable.

Two primes define the same topology on K if and only if their intersection is non-zero (Propo-

sition 3.3). Up to a constant, we associate a unique pseudo-valuation v : K → [0,∞] to any prime of

K (Proposition 3.4). This means that v(a) = 0 may also occur for non-zero a ∈ K. For a minimal

(non-zero) prime P , the corresponding pseudo-valuation is of the form vt with t > 0 for some absolute

value v, and then P can be recovered from v. Every prime P makes K into a separated topological

field such that the completion K̂P is again a topological field. If P is open in its topology, then P is

induced by a prime of K̂P (Theorem 3.1).

We introduce compact primes, which are in a sense totally bounded (Definition 4.1). They always

contain a minimal prime which is again compact. For example, all the primes of Q - not only the

minimal ones - are compact. To any compact prime in any field, we associate a Haar integral and



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show that it admits a natural normalization (Theorem 6.1). For the minimal primes of Q or Fp[x], this

normalization corresponds to the standard normalization of absolute values. So the product formula

becomes a natural statement on the minimal primes of certain fields.

A special case of our concept of prime already occurs in a paper of W. Krull [7]. Such primes are

open in their topology, and we call them Krull primes. We show that every minimal prime is a Krull

prime, and that every prime P of a field K contains a largest Krull prime P◦ (Proposition 3.1). There

is a close relationship between algebraic primes and Krull primes. Namely, we show that P◦ ⊂ P
◦,

so that there is no other prime between P◦ and P
◦. Therefore, every compact Krull prime is either

algebraic or purely transcendental (Proposition 7.5). Furthermore, a non-zero purely transcendental

prime is a Krull prime if and only if it is minimal. The same holds for compact primes (Corollary of

Proposition 7.6).

1 Primes

The two examples of the introduction show that up to units, the elements of certain fields are given by

their values at minimal “primes”, where a relation between these values is given by a product formula

(6). Let us first give a general definition of a prime.

For subsets P,Q of a field K and c ∈ K, define P + Q := {a + b | a ∈ P,b ∈ Q} and PQ :=

{ab | a ∈ P,b ∈ Q}. If P = {a}, we simply write a + Q := {a} + Q and aQ := {a}Q. The unit group

of a ring R will be denoted by R×.

Definition 1.1. Let K be a field. We call a non-empty subset P ⊂ K a prime of K if the following

are satisfied (∀a,b ∈ K):

(P1) 1 /∈ P .

(P2) a,b ∈ P ⇒ ab ∈ P .

(P3) ab ∈ P ⇒ a ∈ P or b ∈ P .

(P4) ∃d ∈ K× : P + P ⊂ dP .

From (P1) and (P2), we get −1 /∈ P , since (−1)(−1) = 1. Therefore, (−1)(−a) = a gives

a ∈ P ⇔ −a ∈ P (7)

by virtue of (P3). By (P4), every a ∈ P satisfies 0 = a − a ∈ P + P ⊂ dP for some d ∈ K×. Hence

0 ∈ P. (8)

It is easily checked that P = 0 := {0} is a prime. We call it the trivial prime. By (P1) and (P2),

a finite field has only the trivial prime. For a prime P , we call

P̃ := {a ∈ K | aP ⊂ P} (9)

the hull and

∂P := P̃ r P (10)

the boundary of P . The trivial prime 0 satisfies 0̃ = K and ∂0 = K×.



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The tree of primes in a field 101

Proposition 1.1. Let K be a field with a prime P. The aP̃ with a ∈ K× form a chain.

Proof. For a,b ∈ K×, assume that aP̃ 6⊂ bP̃ . Then ac /∈ bP̃ for some c ∈ P̃ . For every p ∈ P ,

this gives acb−1 · ba−1p = cp ∈ P , hence ba−1p ∈ P by (P3). Therefore, ba−1 ∈ P̃ , and thus

bP̃ = a · ba−1P̃ ⊂ aP̃ . �

Proposition 1.2. Let K be a field with a prime P. Then

∂P = {a ∈ K× | aP = P} = {a ∈ K× | a,a−1 /∈ P}. (11)

Proof. Assume that a ∈ ∂P . Then a 6= 0 and aP ⊂ P . For b ∈ P , we have a · a−1b = b.

Thus (P3) gives a−1b ∈ P . Therefore, b ∈ aP , which yields P ⊂ aP ⊂ P . Now let a ∈ K× with

aP = P be given. Then aa−1 = 1 /∈ P implies a−1 /∈ P . Similarly, a−1P = P , hence a /∈ P . Thus

{a ∈ K |aP = P} ⊂ {a ∈ K× |a,a−1 /∈ P}. Finally, assume that a ∈ K× with a,a−1 /∈ P . For b ∈ P ,

this gives a−1 · ab = b, hence ab ∈ P by (P3). Consequently, a ∈ P̃ r P = ∂P . �

Note that (11) implies that ∂P is a subgroup of K×, and

a /∈ P̃ ⇔ a−1 ∈ P (12)

for all a ∈ K×, that is, P and P̃ determine each other. Furthermore, −1 ∈ K rP , and thus −1 ∈ ∂P .

Definition 1.2. We call a prime P of a field K algebraic if P + P ⊂ P , that is, (P4) holds for d = 1,

otherwise transcendental.

If P is algebraic, then (P̃ + P̃)P ⊂ P̃P + P̃P ⊂ P + P ⊂ P , which gives P̃ + P̃ ⊂ P̃. Thus P̃

is a subring of K, and by (12), it follows that K is the quotient field of P̃ . By Proposition 1.1, the

principal ideals of P̃ form a chain. Therefore, all ideals of P̃ form a chain. Namely, if I 6⊂ J are ideals

of P̃ , there is some a ∈ I r J. For all b ∈ J, this gives aP̃ 6⊂ bP̃, hence bP̃ ⊂ aP̃ ⊂ I. Furthermore,

P is an ideal of P̃ , and by (11) and (12), ∂P = (P̃)×. Thus P is the unique maximal ideal of P̃ . In

other words, P̃ is a valuation domain.

Example. For K = Q, every rational prime p gives rise to an algebraic prime

Pp = {
pa
b
∈ Q | a,b ∈ Z, p ∤ b}. (13)

By (12), we get P̃p = Zp, whence

P̃p/Pp = Zp/pZp ∼= Z/pZ. (14)

Moreover, Q has a transcendental prime

P∞ := (−1, 1) ∩ Q = {a ∈ Q | |a| < 1}. (15)

Here (P4) holds for d = 2.

2 The primes of Q

By (12), two primes P ⊃ Q of a field K satisfy

Q ⊂ P ⊂ P̃ ⊂ Q̃, (16)

hence cP ⊂ cQ̃ ⊂ Q for all c ∈ Q. More precisely:



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Proposition 2.1. Let K be a field with a prime P. For any submonoid U of K× with P̃ ⊂ U,

Q := {0} ∪ {a ∈ K× | a−1 /∈ U} (17)

is a prime with Q ⊂ P and Q̃ = U. Every prime Q ⊂ P is of this form.

Proof. Let U be a submonoid of K×, and P̃ ⊂ U. By (12), this implies that Q ⊂ P . Assume

that a,b ∈ Q ∩ K×. Then a−1,b−1 /∈ U. So b−1 /∈ P̃ , and therefore, b ∈ P ⊂ U. Now ab /∈ Q would

yield a−1b−1 ∈ U, whence a−1 = a−1b−1 · b ∈ U, which is impossible! Thus (P1) and (P2) hold for Q.

To prove (P3), we have to verify the implication a,b /∈ Q ⇒ ab /∈ Q, that is, a−1,b−1 ∈ U ⇒

a−1b−1 ∈ U. This follows since U is multiplicatively closed.

Finally, Q + Q ⊂ P + P ⊂ dP holds for some d ∈ K×. To prove (P4), we can assume that Q 6= 0.

For arbitrary a ∈ K r U and b ∈ P ∩ K×, we have a /∈ P̃ . Thus a−1 ∈ P . So b−1a ∈ U would

give a ∈ bU ⊂ U, a contradiction. Hence b−1a /∈ U, and thus ba−1 ∈ Q. So we get P ⊂ aQ, i. e.

Q + Q ⊂ daQ, which completes the proof of (P4).

Conversely, let Q ⊂ P be a prime. Then (12) shows that Q is of the form (17) with a submonoid

U := Q̃ of K×. �

Corollary. For each pair P,Q of primes of a field K, there is an infimum P ∧ Q.

Proof. By Proposition 2.1, the submonoid P̃Q̃ corresponds to a prime P ∧ Q with the desired

property. �

Definition 2.1. We say that two primes P and Q of a field K are dependent if P ∧ Q 6= 0, otherwise

independent. The minimal among the non-zero primes are called minimal primes.

Minimal primes can be characterized as follows.

Proposition 2.2. A non-trivial prime P of a field K is minimal if and only if for every pair a,b ∈

P ∩ K×, there exists some n ∈ IN with an ∈ bP.

Proof. Assume that a,b ∈ P ∩ K×. Then U :=
⋃

n∈IN a
−nP is a submonoid of K×, and P̃ ⊂

a−1P ( U. Suppose that an /∈ bP for all n ∈ IN. Then b−1 /∈ U. By Proposition 2.1, there is a

non-trivial prime Q ( P with Q̃ = U.

Conversely, let Q be a prime with 0 6= Q ( P . Choose a ∈ P r Q and b ∈ Q ∩ K×. Then a /∈ Q

implies that a−1 ∈ Q̃, hence a−n ∈ Q̃ for all n ∈ IN. Furthermore, bP ⊂ bP̃ ⊂ bQ̃ ⊂ Q, and thus

a−nbP ⊂ a−nQ ⊂ Q ⊂ P . Hence a−nb ∈ P̃ , that is, anb−1 /∈ P for all n. Thus an /∈ bP holds for

n ∈ IN. �

Now we are able to give an explicit description of the minimal primes of Q. To this end, we need

two auxiliary results.

Lemma 2.1. Let P be a prime of Q, such that P + P ⊂ dP for some d ∈ Q×. Then nP ⊂ dkP holds

for all k ∈ IN and n ∈ Z with |n| 6 2k.

Proof. For k = 0, this is trivial. If the assertion holds for some k ∈ IN, then every n ∈ Z with

|n| 6 2k+1 can be written as a sum n = r + s in Z such that |r|, |s| 6 2k. Hence nP ⊂ rP + sP ⊂

dkP + dkP ⊂ dk+1P . �



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The tree of primes in a field 103

Lemma 2.2. Let P be a prime of Q, and let a > b be relatively prime in IN, such that a(P +P) ⊂ bP.

Then

naaP ⊂ ba+k−1P (18)

holds for all k ∈ IN and n ∈ Z with |n| < ak.

Proof. We proceed by induction. For k = 0, the statement (18) is trivial. Suppose that (18) is

shown for some k ∈ IN. We prove (18) for k + 1. Assume that |n| < ak+1. Since a and b are relatively

prime, there are m,r ∈ Z with

n = ma + rbk.

If r is replaced by r + a, the equation holds for m − bk instead of m. So we can assume that |r| < a.

For r 6= 0, there are still two possibilities. We choose r such that the signs of n and r are not

different. Then |m| = 1
a
|n − rbk| < ak. So the inductive hypothesis gives maaP ⊂ ba+k−1P , hence

naaP ⊂ ma · aaP + rbk · aaP ⊂ ba+k−1aP + rbkaaP . Since |r| 6 a − 1 < 2a−1, Lemma 2.1 implies

that rP ⊂
(

b
a

)a−1
P . Therefore, we get naaP ⊂ ba+k−1aP + ba+k−1aP ⊂ ba+k−1bP = ba+kP . �

As a consequence of Lemma 2.2, we have

Lemma 2.3. Let P be a prime of Q, and assume that 1 < c ∈ P. Then there exists an integer e > 1

with Ze ⊂ P and e(P + P) ⊂ P.

Proof. By (P4), there is some d ∈ Q× with P + P ⊂ dP . Choose i ∈ IN such that ci > |d|. Then

ci · |d|−1 = a
b

with relatively prime a,b ∈ IN. Thus a > b and a(P + P) = bcid−1(P + P) ⊂ bciP ⊂ bP .

For every m ∈ Z, there exists some k ∈ IN with |m| · ba <
(

a
b

)k
. We set n := m · ba+k. Then

|n| = |m| · ba+k < ak. By Lemma 2.2, this implies that naaP ⊂ ba+k−1P , that is, maabP ⊂ P for all

m ∈ Z, and thus Zaa+1 = Zaab · a
b
⊂ P . Furthermore, aa+1(P + P) = aa · a(P + P) ⊂ aabP ⊂ P . �

Now we show that the list of minimal primes of Q given in (13) and (15) is complete. The first

part of the following theorem is essentially due to Ostrowski [9] (cf. [12]).

Theorem 2.1. Every minimal prime P of Q is of the form P = Pp with a rational prime p or p = ∞,

and every prime of Q contains a minimal prime.

Proof. By Proposition 2.2, the prime P∞ is minimal. Therefore, if |c| < 1 for all c ∈ P , then

P ⊂ P∞, and thus P = P∞. Otherwise, there exists some c ∈ P with c > 1. By Lemma 2.3,

there is an integer e > 1 with Ze ⊂ P and e(P + P) ⊂ P . Assume that e = p1 · · ·pr with rational

primes pi. Suppose that ep
−r
i ∈ P holds for all i. Then 1 = (ep

−r
1 ) · · · (ep

−r
r ) ∈ P , a contradiction.

Hence ep−r /∈ P for some prime p|e. This implies that e−1pr ∈ P̃ . For every n ∈ Z, we thus get

(np)r = nre · e−1pr ∈ PP̃ ⊂ P . Therefore, np ∈ P , and thus Zp ⊂ P .

Now let p ∤ n for some n ∈ Z. Then there are u,v ∈ Z with un + vpr+1 = 1. So the residue class

n + pr+1Z is invertible in Z/pr+1Z. Therefore, we find some k ∈ IN r 0 with nk ≡ 1 (pr+1). With

s := (r + 1)k, we also have ns ≡ 1 (pr+1), and thus 1 = ns + mpr+1 for some m ∈ Z. Suppose that

ns ∈ prP . Then 1 ∈ prP + prP = e−1pr · e(P + P) ⊂ P̃P ⊂ P , a contradiction. Hence nsp−r /∈ P ,

and thus n−spr ∈ P̃ . So we get (n−1p)s = n−sp(r+1)k ∈ n−sprP ⊂ P . This gives n−1p ∈ P , whence

Pp ⊂ P . �

Remark. There exist non-minimal primes in Q. For example,

P := 2Z2 ∪ (Z2 ∩ 3Z3)



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is a prime. In fact, this is a general phenomenon. We show first that the primes form a tree with the

trivial prime as a root:

Proposition 2.3. Let K be a field with a prime P. Then the primes Q ⊂ P form a chain.

Proof. Let Q,Q′ be primes with Q,Q′ ⊂ P and Q′ 6⊂ Q. Then there is an element a ∈ Q′ r Q.

For b ∈ Q, the inclusions (16) give bP̃ ⊂ bQ̃ ⊂ Q. Hence aP̃ 6⊂ bP̃. Therefore, Proposition 1.1 yields

b ∈ bP̃ ⊂ aP̃ ⊂ aQ̃′ ⊂ Q′, and thus Q ⊂ Q′. �

Moreover, the tree of primes has a strong symmetry:

Proposition 2.4. Let K be a field with two primes P and Q. Then

P ∗ Q := P ∪ (P̃ ∩ Q)

is a prime of K.

Proof. Obviously, 1 /∈ P ∗ Q. Assume that a,b ∈ P ∗ Q. If a ∈ P or b ∈ P , then ab ∈ P ⊂ P ∗ Q

since a,b ∈ P̃ . For a,b ∈ P̃ ∩ Q, however, we have ab ∈ P̃ ∩ Q. This proves (P2). To verify (P3),

let a,b ∈ K with ab ∈ P ∗ Q be given. If ab ∈ P , then a ∈ P or b ∈ P . Therefore, assume that

ab ∈ P̃ ∩Q and a,b /∈ P . Then a−1,b−1 ∈ P̃ . Hence a = ab ·b−1 ∈ P̃ , and similarly, b ∈ P̃ . Therefore,

a ∈ P̃ ∩Q or b ∈ P̃ ∩Q. Finally, assume that d ∈ K× with P +P ⊂ dP . If P = 0, we have P ∗Q = Q.

Otherwise, there exists some a ∈ P r 0. Now (P̃ + P̃)P ⊂ P + P ⊂ dP , hence (P̃ + P̃)d−1 ⊂ P̃ .

Consequently, (P ∗ Q) + (P ∗ Q) ⊂ P̃ + P̃ ⊂ dP̃ = da−1aP̃ ⊂ da−1(P ∗ Q). �

We call P ∗ Q the lexicographic product of P and Q. Thus for a fixed P , every prime Q yields

a prime P ∗ Q ⊃ P which may coincide with P . Apart from this exception, the tree of primes is

self-similar, i. e. the whole tree sits above each prime.

3 The topology of a prime

Let K be a field with a prime P . For a,b ∈ K, we define

a 6 b : ⇔ aP̃ ⊂ bP̃ (19)



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The tree of primes in a field 105

and a < b if a 6> b. Then
a 6 b 6 c ⇒ a 6 c

a 6 b ⇒ ac 6 bc
(20)

holds for a,b,c ∈ K, and Proposition 1.1 yields

a 6 b oder b 6 a. (21)

In particular, this implies the reflexity a 6 a for all a ∈ K. By (11), we have aP = bP ⇔ b−1aP =

P ⇔ b−1a ∈ ∂P for a,b ∈ K×. Therefore, (19) induces a linear order on K×/∂P . From (19) and

(12), we get

P̃ = {a ∈ K | a 6 1} ; P = {a ∈ K | a < 1}. (22)

For a ∈ K and c ∈ K×, we define the c-neighbourhood of a as follows:

Uc(a) = U
P
c (a) := a + cP. (23)

A subset U ⊂ K is open with respect to P if for every a ∈ U, there is an element c ∈ K× with

Uc(a) ⊂ U. The P-open sets define a topology on K, but it is not clear in advance that the c-

neighbourhoods (23) are neighbourhoods in this topology. The following proposition shows that this

is in fact the case.

Proposition 3.1. Let K be a field with a prime P. The set

P◦ := {a ∈ P | ∃n ∈ IN: an(P + P) ⊂ P} (24)

is open. If P is minimal, then P◦ = P.

Proof. Assume that a ∈ P◦, say, an(P + P) ⊂ P . Then a2n(P + P + P + P) ⊂ an(P + P) ⊂ P ,

and by induction, we get

amn(P + · · · + P) ⊂ P

for m ∈ IN, where P + · · · + P may have at most 2m summands. We set b := a2n(n+1). Now let p ∈ P

be arbitrary. Expanding (a + bp)2n gives 22n summands, namely, a2n plus 22n − 1 summands in bP .

So we get

(a + bp)2n(P + P) ⊂ (a2n + ba−2n
2

P)(P + P) ⊂ (a2n + a2nP)(P + P)

⊂ a2n(P + P) + a2n(P + P) ⊂ anP + anP ⊂ P.

This yields a + bP ⊂ P◦. Finally, let P be minimal and d ∈ K× with P + P ⊂ dP . Then

P ⊂ dP , hence d−1 ∈ P̃ . For a ∈ P , Proposition 2.2 yields an n ∈ IN with an ∈ ad−1P , whence

an(P + P) ⊂ ad−1(P + P) ⊂ aP ⊂ P . Thus a ∈ P◦. �

In section 7, we will show that P◦ is again a prime.

Corollary. Let K be a field with a minimal prime P. For every a ∈ P, there are b,c ∈ P ∩ K× such

that aP̃ + bP ⊂ cP̃.

Proof. Assume that P + P ⊂ dP with d ∈ K× and a ∈ P . By Proposition 3.1, there is some

b ∈ d−1P ∩ K× with a + bdP ⊂ P . We show first that aP̃ + bP ⊂ P . To this end, assume that e ∈ P̃ .



106 Wolfgang Rump CUBO
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Case I: e ∈ d−1P . Then ae + bP ⊂ d−1P + d−1P ⊂ P .

Case II: e /∈ d−1P . Then de /∈ P , hence d−1e−1 ∈ P̃ . This gives ae + bP = e(a + bdd−1e−1P) ⊂

e(a + bdP) ⊂ P .

If aP̃ = P , we can set c := a. Otherwise, there exists some c ∈ P r aP̃ . Then ca−1 /∈ P̃ , hence

c−1a ∈ P . By the above, we find an element b ∈ d−1P ∩ K× ⊂ P ∩ K× with c−1aP̃ + bP ⊂ P .

Whence aP̃ + cbP ⊂ cP ⊂ cP̃. �

Lemma 3.1. Let K be a field with a prime P. Assume that d ∈ K× with P + P ⊂ dP. Then every

a,b ∈ K× with b ∈ ad−1P satisfies

∀p ∈ P : (a + bp)−1 − a−1 ∈ da−2bP.

Proof. Assume that p ∈ P . Then bp ∈ aP implies that a+bp 6= 0. Suppose that (1+a−1bp)d ∈ P .

Then d = (1 + a−1bp)d − a−1bpd ∈ P + P ⊂ dP , a contradiction. Hence (1 + a−1bp)−1d−1 ∈ P̃ . From

(1 + a−1bp)−1(1 + a−1bp) = 1 we get

(1 + a−1bp)−1 + (1 + a−1bp)−1a−1bp = 1.

Multiplying by a−1 gives (a + bp)−1 + (1 + a−1bp)−1a−2bp = a−1, i. e. (a + bp)−1 − a−1 = −(1 +

a−1bp)−1a−2bp ∈ dP̃ · a−2bP ⊂ da−2bP . �

Theorem 3.1. Let K be a field with a prime P. The topology of P makes K into a separated

topological field, and the completion K̂P of K with respect to P is a topological field. If P is open,

then

P̂ :=
⋃

a∈P

aP̃ (25)

is a prime in K̂P with P̂ ∩ K = P, and P̂ defines the topology of K̂P .

Proof. By (P4), there is some d ∈ K with P +P ⊂ dP . This implies that K is a topological ring.

Hence K is a topological field by Lemma 3.1.

To show that the same holds for the completion K̂ of K, we have to verify that the image under

the mapping x 7→ x−1 of every Cauchy filter C in K which does not have a cluster point at 0, is a

Cauchy filter (see [3], III.6.8, Proposition 7). Thus let c ∈ K× and C ∈ C be such that C∩cP = ∅. We

can assume that a−b ∈ cd−1P for all a,b ∈ C. Choose any a ∈ C. Suppose that b ∈ C ∩cd−1P . Then

a = (a−b) +b ∈ cd−1P +cd−1P ⊂ cP , a contradiction. Hence C ∩cd−1P = ∅. This gives ac−1d /∈ P ,

and thus a−1cd−1 ∈ P̃ , i. e. a−1 ∈ c−1dP̃ . For an arbitrary e ∈ K×, we find some C′ ⊂ C in C with

a − b ∈ c2d−2eP for all a,b ∈ C′. This gives a−1 − b−1 = a−1b−1(b − a) ∈ c−2d2P̃ · c2d−2eP ⊂ eP for

a,b ∈ C′.

Now let P be open. We show first that for a ∈ P ,

aP̃ ⊂ P (26)

holds in K. For a = 0, this is trivial. So we assume that a ∈ P ∩ K×. Let b /∈ P be given. Then

b−1 ∈ P̃, which gives (a−1b)−1 = b−1a ∈ P . Since P is open, the continuity of x 7→ x−1 implies that

(a−1b + cP)−1 ⊂ P for some c ∈ K×. Hence (a−1b + cP) ∩ P̃ = ∅, and thus (b + acP) ∩ aP̃ = ∅.

This proves (26). Together with (25), we infer that P̂ ∩ K = P .



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The tree of primes in a field 107

Furthermore, (26) shows that K̂P satisfies (P1). If b,c ∈ P̂ , then b,c ∈ aP̃ for some a ∈ P . Hence

bc ∈ aP̃ · aP̃ ⊂ a2P̃ ⊂ P̂ , which gives (P2). Similarly, (P4) is easily verified.

To prove (P3), assume that b,c ∈ K̂P satisfy bc ∈ P̂ and b /∈ P̂ , say, bc ∈ aP̃ for some a ∈ P .

Then b /∈ pP̃ for all p ∈ P . For any p ∈ P , we find a neighbourhood U of 0 in K̂P such that

(b + U) ∩pP̃ = ∅. Hence (bc + Uc) ∩pcP̃ = ∅. On the other hand, (bc + Uc) ∩ aP̃ 6= ∅. Thus for any

p ∈ P , we get aP̃ 6⊂ pcP̃. If p 6= 0, this implies that p−1c−1a /∈ P̃ . Hence pca−1 ∈ P for all p ∈ P .

Consequently, ca−1 ∈ P̃ , and thus c ∈ aP̃ ⊂ P̂ .

It remains to verify that P̂ defines the topology of K̂P . For a ∈ P , we have aP ⊂ P̂ . Conversely,

(26) implies that P̂ ⊂ P . This completes the proof. �

In particular, Theorem 3.1 implies that the limit of a sequence in K with respect to a prime P

is unique. A minimal prime P is uniquely determined by its topology:

Proposition 3.2. Let K be a field with a minimal prime P. Then

P = {a ∈ K | lim
n→∞

an = 0}. (27)

Proof. Let a ∈ P . By Proposition 2.2, every c ∈ K× gives rise to some N ∈ IN with aN ∈ cP .

Therefore, an ∈ cP for all n > N, and thus lim
n→∞

an = 0. On the other hand, if a /∈ P , then a−1 ∈ P̃ ,

which gives a−n ∈ P̃ for all n ∈ IN. But this implies that an /∈ P for all n. �

If a prime contains a minimal prime P , then P can be determined topologically by means of

Proposition 3.2. The following proposition decides whether two primes define the same topology.

Proposition 3.3. For two primes P,Q 6= 0 of a field K, the following are equivalent.

(a) PQ 6= K.

(b) P̃Q̃ 6= K.

(c) P and Q are dependent (see Definition 2.1).

(d) There is some a ∈ K× with aP ⊂ Q.

(e) P and Q define the same topology.

Proof. (a) ⇒ (b): Suppose that P̃Q̃ = K. For a ∈ P ∩ K× and b ∈ Q ∩ K×, this implies that

K = abP̃Q̃ ⊂ PQ, which is impossible. The implication (b) ⇒ (c) follows by Proposition 2.1.

(c) ⇒ (d): For every a ∈ (P ∧ Q) ∩ K×, we have aP ⊂ P ∧ Q ⊂ Q.

(d) ⇒ (a): From aP ⊂ Q with a ∈ K×, we get aPQ ⊂ Q2 ⊂ Q, hence PQ ⊂ a−1Q 6= K.

Therefore, condition (d) is symmetric. It states that every Q-open set with respect to P is open.

Hence (d) ⇔ (e). �

Let K be a field with a prime P . A function f : K → [0,∞] is said to be monotonous (with

respect to P) if

a 6 b ⇒ f(a) 6 f(b) (28)

holds for all a,b ∈ K. We call f strictly monotonous if (28) can be replaced by an equivalence. For

the following definition, we set

∞ · a = a · ∞ := ∞ (29)

for 0 < a 6 ∞.



108 Wolfgang Rump CUBO
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Definition 3.1. Let K be a field with a prime P . We call v : K → [0,∞] a pseudo-valuation with

respect to P if v is monotonous with v(0) = 0 and v(1) = 1, such that

v(ab) = v(a)v(b) (30)

holds for all a,b ∈ K with v(a) /∈ {0,∞}. We call v finite if v(a) < ∞ for all a ∈ K. If v is finite and

strictly monotonous, we say that v is a P-valuation.

For a ∈ K×, the definition yields

v(a) = 0 ⇔ v(a−1) = ∞. (31)

Note that by (22), the prime P of a P-valuation v is given by P = {a ∈ K | v(a) < 1}. By [4], VI.6.1,

a P-valuation is of the form vt, where t > 0 and v is an absolute value on K.

Proposition 3.4. Let K be a field with a prime P. For every c ∈ K r P̃ and every real number

r > 1, there is exactly one pseudo-valuation v : K → [0,∞] with v(c) = r. If P is minimal, then v is

a P-valuation.

Proof. First, let v be a pseudo-valuation with v(c) = r, and let a ∈ K be given. If a 6 cn for

all n ∈ Z, then v(a) 6 rn for all n, and thus v(a) = 0. If, however, a > cn for all n ∈ Z, we get

v(a) = ∞. Otherwise, there exists some n ∈ IN with c−n 6 a 6 cn, and then r−n 6 v(a) 6 rn, hence

v(a) /∈ {0,∞}. For such an a and an integer n > 0, there exists some m ∈ Z with cm−1 6 an 6 cm,

which gives rm−1 6 v(a)n 6 rm, i. e.

r
m−1

n 6 v(a) 6 r
m
n .

Thus v is unique:

v(a) = inf{rm/n | m,n ∈ Z, n > 0, an 6 cm}. (32)

This formula extends to v(a) ∈ {0,∞}. Conversely, we show that (32) defines a pseudo-valuation.

Again, v(a) = 0 respectively v(a) = ∞ holds if a 6 cn resp. a > cn for all n ∈ Z. Therefore, assume

that v(a) /∈ {0,∞}. Then for every n > 0 in IN, there is an integer m with cm−1 6 an 6 cm. This

implies (30) for v(b) ∈ {0,∞}. Assume that ck−1 6 bn 6 ck. Then we get r
m−1

n 6 v(a) 6 rm/n and

r
k−1

n 6 v(b) 6 rk/n. Hence r
m+k−2

n 6 v(a)v(b) 6 r
m+k

n . On the other hand, ck+m−2 6 (ab)n 6 cm+k,

and thus r
m+k−2

n 6 v(ab) 6 r
m+k

n . For n → ∞, this gives (30). Furthermore, (32) implies that v is

monotonous, and v(0) = 0, v(1) = 1 holds together with v(c) = r. Hence v is a pseudo-valuation. If

P is minimal, Proposition 2.2 shows that for every a ∈ P , the inequality v(a) < 1 holds. Hence v is a

P-valuation. �

Examples. 1. For the transcendental prime P∞ = (−1, 1) ∩ Q of Q, the unique P∞-valuation

v : Q → [0,∞] with r = c > 1 (Proposition 3.4) is given by v(a) := |a|. So the completion is IR, and

P̂∞ = (−1, 1). For a prime p and P := pZp, every a ∈ Q
× admits a unique representation a = pn · b

c

with p ∤ b,c and n ∈ Z. Then v(a) = |a|p := p
−n with |0|p := 0 is a P-valuation v : Q → [0,∞], the

p-adic absolute value of Q. The c-neighbourhoods (23) of 0 are then of the form pnZp, n ∈ Z. As

usual, we write Q̂p for the completion of Q with respect to P = pZp. Here we have P̂ = pẐp, where

Ẑp denotes the ring of p-adic numbers.

2. For the field C of complex numbers, P := {a ∈ C | |a| < 1} is a prime, and the absolute value

is a P-valuation v : C → [0,∞]. Here we have ĈP = C.



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The tree of primes in a field 109

4 Compact primes and the Haar integral

In the sequel, we deal with primes like those occuring in algebraic number fields.

Definition 4.1. We call a prime P of a field K compact if for every b ∈ K× there exist a1, . . . ,an ∈ K

with

P ⊂ (a1 + bP) ∪ · · · ∪ (an + bP). (33)

Example. The primes of Q are compact: For the prime P∞ this is trivial; for a rational prime

p, we have pZp =
⋃pn−1

i=1 ip + p
nZp. So the minimal primes of Q are compact.

(34)

By the following result, the other primes are compact, too:

Proposition 4.1. Let K be a field. The primes Q which are dependent of a non-trivial compact

prime P are compact.

Proof. Assume that b ∈ K×. Since P ∧ Q 6= 0, there exists some a ∈ (P ∧ Q) ∩ K×. As P

is compact, there are a1, . . . ,an ∈ K with P ⊂ (a1 + a
2bP) ∪ · · · ∪ (an + a

2bP). Hence aQ ⊂ P ⊂⋃n
i=1(ai + a

2bP) ⊂
⋃n

i=1(ai + abQ), and thus Q ⊂
⋃n

i=1(a
−1ai + bQ). �

Every compact prime P of a field K leads to a concept of integral. Let CP (K) denote the set of

real functions f : K → IR such that f vanishes on KrcP for some c ∈ K×, and is uniformly continuous

with respect to P , i. e. for every ε > 0 there exists some b ∈ K× such that |f(a) −f(a′)| < ε holds for

all a,a′ ∈ K with a− a′ ∈ bP . Clearly, CP (K) is an IR-vector space. We write f 6 g for f,g ∈ CP (K)

if f(a) 6 g(a) holds for all a ∈ K. If, in addition, f 6= g, we write f < g. The subset of the f > 0 in

CP (K) will be denoted by C
+
P (K). We show first that C

+
P (K) 6= ∅.

Proposition 4.2. Let K be a field with a prime P. For every b ∈ K×, there exists a function

f ∈ C+P (K) which satisfies f(a) = 1 for all a ∈ bP.

Proof. Since a ∈ bP ⇔ ab−1 ∈ P , we can assume that b = 1. Furthermore, we can assume that

P 6= 0. By (P4), there is an element d ∈ K r P̃ with P + P ⊂ dP , and by Proposition 3.4, there

exists a pseudo-valuation v : K → [0,∞] with v(d) = 2. We show first that for every n ∈ IN there

exists some c ∈ K× such that

d−1 6 (1 + cp)n 6 d (35)

holds for all p ∈ P . From (P4) we get by induction P + · · ·+P ⊂ dnP , if the sum P + · · ·+P contains

at most 2n summands. For c ∈ P , we therefore have (1 + cp)n ∈ 1 + cP + · · · + cP ⊂ 1 + cdnP .

Replacing c by cd−n, the inequality (35) turns into d−1 6 1 + cp 6 d for some c ∈ K× and p ∈ P .

By Lemma 3.1 (with a = 1), this reduces to the single inequality 1 + cp 6 d. This is equivalent to

(1 + cp)d−1 ∈ P̃ , i. e. (1 + cp)d−1P ⊂ P . By (P4), the latter is satisfied for c = 1.



110 Wolfgang Rump CUBO
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Using v, we define f : K → IR by

f(a) :=





1 for a ∈ P

2 − v(a) for a ∈ dP r P

0 for a /∈ dP

and prove the uniform continuity of f. Let ε > 0 be given. We have to find c ∈ K× so that

|f(a) − f(a′)| < ε holds for all a,a′ ∈ K with a − a′ ∈ cP . By symmetry, we can assume that

a ∈ dP r P .

Case I: a′ ∈ P . For c ∈ d−1P , this implies that a′ /∈ d−1P , since otherwise, a = (a − a′) + a′ ∈

d−1P + d−1P ⊂ P . Hence if n ∈ IN, the inequality (35) yields an element c ∈ K× with (1 + c
a′
p)n 6 d

for all p ∈ P . For a− a′ = cp, we thus have a < a
a′

, hence d−1 < 1 6 an 6
(

a
a′

)n
= (1 + c

a′
p)n 6 d. So

we get 2−1 6 v(an) 6 2, and thus 2−
1
n 6 v(a) 6 2

1
n . Consequently, |f(a) − f(a′)| = |(2 −v(a)) − 1| =

|v(a) − 1| < ε holds for sufficiently large n.

Case II: a′ /∈ dP . For any n ∈ IN, the inequality (35) shows that there exists some c ∈ K× with

d−1 6 (1 + c
a′
p)n for all p ∈ P . For a − a′ = cp, this gives d−1 6 (1 + c

a′
p)n =

(
a
a′

)n
6

(
a
d

)n
6 1 < d.

Therefore, dn−1 6 an 6 dn+1, hence 2n−1 6 v(a)n 6 2n+1, and thus 2
n−1

n 6 v(a) 6 2
n+1

n . For

sufficiently large n, we get |f(a) − f(a′)| = |v(a) − 2| < ε.

Case III: a′ ∈ dP r P . Again by (35), for any n ∈ IN, there exists c ∈ K× with d−1 6

(1 + c
a′
p)n 6 d for all p ∈ P . For a − a′ = cp, we get d−1 6

(
a
a′

)n
6 d for all p ∈ P . This

gives 2−1 6 v(a)nv(a′)−n 6 2, hence 2−
1
n 6

v(a)
v(a′)

6 2
1
n . For sufficiently large n, it follows that

|f(a) − f(a′)| = |v(a) − v(a′)| < ε. �

Note that f ∈ CP (K) implies that |f| ∈ CP (K). Therefore, every f ∈ CP (K) admits a decompo-

sition

f = f+ − f− (36)

with f+ :=
1
2
(|f| + f) and f− :=

1
2
(|f| − f), so that |f| = f+ + f− and f+,f− > 0. For f ∈ CP (K)

and a ∈ K, we define fa ∈ CP (K) by

fa(b) := f(b − a). (37)

Definition 4.2. Let K be a field with a compact prime P . A linear form I : CP (K) → IR is said to

be a Haar integral if I(f) > 0 holds for f > 0, and I(fa) = I(f) for all a ∈ K.

So the Haar integral is linear (like any integral) and monotonous (f < g ⇒ I(f) < I(g)), and

invariant under translations. We will show first that a Haar integral exists and is unique up to a

constant. Our proof is similar to [14], but does not assume that the topology of K given by the

compact prime P is locally compact. In section 6, we will show that the constant can be chosen in a

canonical way.

For f,g ∈ C+P (K), let f : g be the infimum of all sums r1 + · · · + rn with r1, . . . ,rn > 0 in IR,

such that f 6 r1g
a1 + · · · + rng

an holds for suitable a1, . . . ,an ∈ K.

Proposition 4.3. Let K be a field with a compact prime P. Every function f ∈ CP (K) is bounded.

For f,g ∈ C+P (K), we have 0 < f : g < ∞.

Proof. First let f ∈ CP (K) and b ∈ K
× be given such that f(a) = 0 holds for a /∈ bP . For every

ε > 0, there exists some c ∈ K× such that |f(a)−f(a′)| < ε holds for a−a′ ∈ cP . Since P is compact,

there are a1, . . . ,an ∈ K with bP ⊂ (a1 + cP) ∪ · · · ∪ (an + cP). Hence f is bounded.



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The tree of primes in a field 111

Next let f,g ∈ C+P (K) be given. Choose a0 ∈ K with g(a0) > 0. Then we find ε > 0 in IR and

c ∈ K× with g(a) > ε for a − a0 ∈ cP . Assume that f(a) = 0 for a /∈ bP and a1, . . . ,an ∈ K with

bP ⊂ (a1 + cP) ∪ · · · ∪ (an + cP). Then there exists some m ∈ IN with f(a) 6 mε for all a ∈ bP .

Hence f 6 mga1−a0 + · · · + mgan−a0 , which gives f : g < ∞. Furthermore, g(a) 6 r for all a ∈ K and

a real r > 0. Now if f(a) > 0, we obtain f : g > f(a)r−1 > 0. �

Lemma 4.1. Let K be a field with a compact prime P. Let b ∈ K× and f,g ∈ C+P (K) satisfy f(a) = 0

for a /∈ bP and g(a) > ε for some ε > 0 in IR and a ∈ b(P + P). Then h := f
g
∈ C+P (K), where

h(a) := 0 for a /∈ bP.

Proof. For all a,a′ ∈ K, we have:

|h(a) − h(a′)| =
|f(a)g(a′) − g(a)f(a′)|

g(a)g(a′)

=

∣∣(f(a) − f(a′)
)
g(a′) + f(a′)

(
g(a′) − g(a)

)∣∣
g(a)g(a′)

6
|f(a) − f(a′)|

g(a)
+
f(a′)

g(a′)
·
|g(a) − g(a′)|

g(a)
.

For a,a′ /∈ bP , this expression vanishes. Otherwise, if a − a′ ∈ bP , then a,a′ ∈ bP + bP . Hence

|h(a) − h(a′)| 6 ε−1|f(a) − f(a′)| + ε−2f(a′) · |g(a) − g(a′)|

for all a,a′ ∈ K. Since f is bounded by Proposition 4.3, this gives h ∈ C+P (K). �

Proposition 4.4. Let K be a field with a compact prime P. For every f ∈ C+P (K) and c ∈ K
×,

there are f1, . . . ,fn ∈ C
+
P (K) and a1, . . . ,an ∈ K such that fi(a) = 0 holds for a − ai /∈ cP, so that

f = f1 + · · · + fn.

Proof. Assume that f(a) = 0 holds for a /∈ bP , where b ∈ K×. By Proposition 4.2, there

exists some f′ ∈ C+P (K) which satisfies f
′(a) = 1 for a ∈ b(P + P). Furthermore, there exists some

g ∈ C+P (K) that vanishes outside cP . By Proposition 4.3, we have (f + f
′) : g < ∞. Hence there are

real numbers r1, . . . ,rn > 0 and a1, . . . ,an ∈ K with f + f
′ 6 r1g

a1 + · · · + rng
an . By Lemma 4.1,

h := f
r1g

a1 +···+rngan
∈ C+P (K), whence f = r1hg

a1 + · · · + rnhg
an . �

5 Existence and Uniqueness

Now we are ready to prove existence and uniqueness of the Haar integral.

Proposition 5.1. For every compact prime P of a field K, there exists a Haar integral.

Proof. By (36), it is enough to find a map I : C+P (K) → (0,∞) which satisfies I(rf + sg) =

rI(f) + sI(g) for r,s > 0 and I(fa) = I(f) for a ∈ K. For r > 0 in IR and f,g ∈ C+P (K), we have

(rf : g) = r(f : g). Furthermore, for f,g,h ∈ C+P (K),

(f : g)(g : h) > f : h (38)



112 Wolfgang Rump CUBO
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In fact, assume that f 6 r1g
a1 + · · · + rng

an and g 6 s1h
b1 + · · · + smh

bm . Then

f 6
∑n

i=1 ri
(∑m

j=1 sjh
bj

)ai
=

∑n
i=1

∑m
j=1 risjh

bj +ai , and thus f : h 6
(∑n

i=1 ai
)(∑m

j=1 bj
)
.

To determine I(f), let f0 ∈ C
+
P (K) be fixed. Then f : f0 can be regarded as a first approximation

of I(f) : I(f0). An improvement is given by the function Ig : C
+
P (K) → (0,∞) with

Ig(f) :=
f : g

f0 : g
(39)

for small g ∈ C+P (K). By (38), we have

(f0 : f)
−1

6 Ig(f) 6 f : f0. (40)

For b ∈ K×, let Vb be the set of all Ig with g(a) = 0 for a /∈ bP . By Proposition 4.2, Vb 6= ∅ for all

b ∈ K×. By (40), we can regard Ig as a point of the compact space

X :=
∏

f∈C
+

P
(K)

[(f0 : f)
−1,f : f0]. (41)

As the closed sets Vb ⊂ X form a chain, their intersection is non-empty. Hence there exists some

I ∈
⋂

b∈K× Vb. Consequently, for every b ∈ K
× and arbitrary f1, . . . ,fn ∈ C

+
P (K) and ε > 0 in IR,

there exists some g ∈ C+P (K) with g(a) = 0 for a /∈ bP , so that for i ∈ {1, . . . ,n},

|I(fi) − Ig (fi)| < ε. (42)

We will prove that I is a Haar integral. Firstly,

(f1 + f2) : g 6 (f1 : g) + (f2 : g)

implies that Ig(f1 + f2) 6 Ig(f1) + Ig (f2) holds for f1,f2,g ∈ C
+
P (K). Hence (42) gives

I(f1 + f2) 6 I(f1) + I(f2) (43)

for all f1,f2 ∈ C
+
P (K). For 0 < r < ∞ and f,g ∈ C

+
P (K), we have (rf : g) = r(f : g), i. e.

Ig(rf) = rIg (f), and thus I(rf) = rI(f), again by (42). Similarly, Ig(f
a) = Ig(f) for all a ∈ K, hence

I(fa) = I(f). From (41) and (42), it follows that I is monotonous. Thus it remains to show that the

inequality (43) is an equation.

Let f1,f2,g ∈ C
+
P (K) be given. Then there exists some b ∈ K

× such that f1 +f2 vanishes outside

bP . By Proposition 4.2, we find f3 ∈ C
+
P (K) with f3(a) = 1 for a ∈ b(P + P). For an arbitrary ε > 0,

define f := f1 + f2 + εf3 and hj :=
fj
f

for j ∈ {1, 2}. By Lemma 4.1, this gives hj ∈ C
+
P (K). Hence

there exists some c ∈ K× such that |hj (a) − hj (a
′)| < ε holds for a − a′ ∈ cP and j ∈ {1, 2}. Choose

g ∈ C+P (K) with g(a) = 0 for a /∈ cP , and

f 6 r1g
a1 + · · · + rng

an.

Then fj(a) = f(a)hj(a) 6
∑n

i=1 rig
ai (a)(hj (ai) + ε), hence

fj : g 6

n∑

i=1

ri(hj (ai) + ε).

This implies (f1 : g) + (f2 : g) 6
∑n

i=1 ri(1 + 2ε) since h1 + h2 6 1. So we get (f1 : g) + (f2 : g) 6 (f :

g)(1 + 2ε), and thus Ig(f1) + Ig(f2) 6 Ig (f) · (1 + 2ε). By (42), this gives I(f1) + I(f2) 6 I(f)(1 + 2ε).

Using (43), we get I(f1)+I(f2) 6 (I(f1 +f2)+εI(f3))(1+2ε) for all ε, hence I(f1)+I(f2) 6 I(f1 +f2).

�



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The tree of primes in a field 113

Proposition 5.2. Let K be a field with a compact prime P. For every f0 ∈ C
+
P (K), there exists a

unique Haar integral I : CP (K) → IR with I(f0) = 1.

Proof. Existence follows by Proposition 5.1. Therefore, let I be a Haar integral, and f0 ∈ C
+
P (K).

Then rI is a Haar integral for any r > 0 in IR. Assume first that f,g ∈ C+P (K) and f 6
∑n

i=1 rig
ai

with ri > 0 and ai ∈ K. Then I(f) 6
∑n

i=1 riI(g), hence

I(f)

I(g)
6 f : g. (44)

For f ∈ C+P (K) and ε > 0, there exists some b ∈ K
× which satisfies |f(a) − f(a′)| < ε for a − a′ ∈ bP .

Choose g ∈ C+P (K) with g(a) = 0 for a /∈ bP . Passing from g(a) to g(a) + g(−a), we can assume that

g(a) = g(−a) for all a ∈ K. Therefore, if a,a′ ∈ K, we have f(a′)ga(a′) > (f(a) − ε)ga(a′), hence

I(fga) > (f(a) − ε)I(g), and thus

f(a) 6 ε +
I(fga)

I(g)
. (45)

For every δ > 0, there is some c ∈ K× with |g(a) − g(a′)| < δ for a − a′ ∈ cP . By Proposition 4.4,

there are elements f1, . . . ,fn ∈ C
+
P (K) and a1, . . . ,an ∈ K with f = f1 + · · · + fn and fi(a

′) = 0 for

a′ − ai /∈ cP . This gives fi(a
′)g(a′ − a) 6 fi(a

′)(g(ai − a) + δ) for all a,a
′ ∈ K, hence

I(fga) =

n∑

i=1

I(fig
a) 6

n∑

i=1

I(fi)(g
a(ai) + δ) =

n∑

i=1

I(fi)(g
ai (a) + δ). (46)

By Proposition 4.2, there exists some h ∈ C+P (K) such that h(a
′) = 1 holds for all a′ with f(a′) +

f0(a
′) > 0. Then (45) and (46) yield

f 6 εh +
n∑

i=1

I(fi)

I(g)
(gai + δh) = (ε +

I(f)

I(g)
δ)h +

n∑

i=1

I(fi)

I(g)
gai.

So we get f : g 6 ε(h : g) +
I(f )
I(g)

, and by virtue of (38) and (44),

f : g

f0 : g
6 ε ·

h : g

f0 : g
+

I(f)

I(g)(f0 : g)
6 ε(h : f0) +

I(f)

I(f0)
.

Similarly,
f0 : g

f : g
6 ε(h : f) +

I(f0)

I(f)
=
I(f0)

I(f)

(
1 + ε(h : f)

I(f )
I(f0)

)
,

and thus
I(f)

I(f0)

(
1 + ε(h : f)

I(f )
I(f0)

)−1
6

f : g

f0 : g
6 ε(h : f0) +

I(f)

I(f0)
.

This implies that the quotient f :g
f0:g

converges to
I(f )
I(f0)

for b → 0, which shows that I(f) is unique for

I(f0) = 1. �

6 The normalized Haar integral

Let K be a field with a compact prime P . By Proposition 5.2, there exists a Haar integral I : CP (K) →

IR, unique up to a real constant. Furthermore, the proof of Proposition 5.2 yields an explicit determi-

nation of the Haar integrals. In fact, for any b ∈ K×, let L(b) be the set of g ∈ C+P (K) which vanish



114 Wolfgang Rump CUBO
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outside bP . Moreover, we set

Ls(b) := {g ∈ L(b) | ∀a ∈ K : g(−a) = g(a)}. (47)

For f,f0 ∈ C
+
P (K) and every ε > 0 in IR, There exists some b ∈ K

× such that
∣∣∣∣
I(f)

I(f0)
−
f : g

f0 : g

∣∣∣∣ < ε (48)

for all g ∈ Ls(b). In this way, the quotient
I(f )
I(f0)

can be approximated by the expressions (39) with

g ∈ Ls(b).

For any c ∈ K×, consider the IR-linear map

ρc : CP (K) → CP (K) (49)

with ρc(f)(a) := f(c
−1a) for all f ∈ CP (K) and a ∈ K. Since f(K r bP) = 0 ⇔ ρc(f)(K r cbP) = 0,

we have

ρc(L(b)) = L(cb). (50)

Similarly, the uniform continuity is preserved under the map ρc. Furthermore, ρc is monotonous, i. e.

for f,g ∈ CP (K),

f 6 g ⇐⇒ ρc(f) 6 ρc(g). (51)

Since ρc is invertible, it induces a bijection

ρc : C
+
P (K) −→

∼ C+P (K). (52)

For a,b ∈ K and f ∈ CP (K), we finally have ρc(f
a)(b) = fa(c−1b) = f(c−1b − a) = f(c−1(b − ca)) =

ρc(f)(b − ca) = ρc(f)
ca(b), hence

ρc(f
a) = ρc(f)

ca. (53)

Consequently, for any Haar integral I : CP (K) → CP (K) and c ∈ K
×, it follows that Iρc is again a

Haar integral. So there is a real function vP : K → [0,∞] with vP (0) := 0 and

Iρc = vP (c)I (54)

for c 6= 0,which does not depend on the choice of the Haar integrals I.

Now we prove that vP is a pseudo-valuation. Using vP , the Haar integral can be normalized in

a natural way.

Proposition 6.1. Let K be a field with a compact prime P. Then the uniquely defined function

vP : K → [0,∞] in (54) is a pseudo-valuation with respect to P.

Proof. From (54) we infer that vP is multiplicative: vP (bc) = vP (b)vP (c) for all b,c ∈ K. In fact,

for b,c ∈ K× and a Haar integral I, vP (bc)I = Iρbc = Iρbρc = vP (b)Iρc = vP (b)vP (c)I. Similarly,

vP (1) = 1.

To prove that vP is monotonous, we use I to define a function vI : K → [0,∞) which is related

to vP . (In contrast to vP , however, vI depends on I.) For b ∈ K, we set U(b) := {f ∈ C
+
P (K) | ∀a ∈

bP̃ : f(a) = 1}, and

vI (b) := inf
f∈U(b)

I(f). (55)



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The tree of primes in a field 115

Although the function (55) is not multiplicative, in general, it is always monotonous. In fact, if

b,c ∈ K with b 6 c, then U(b) ⊃ U(c), and therefore, vI (b) 6 vI (c).

Before we compare vP with vI , we prove, in analogy to (50), that

ρc(U(b)) = U(cb) (56)

holds for b,c ∈ K×. This follows by the equivalence ρc(f) ∈ U(cb) ⇔ f ∈ U(b) for all f ∈ C
+
P (K). For

b,c ∈ K×, we have the equation vP (c) · inff∈U(b) I(f) = inff∈U(b) I(ρc(f)) = inff∈U(cb) I(f), which

gives

vI (cb) = vP (c) · vI (b). (57)

In particular, this gives vI (c) = vP (c) · vI (1), i. e.

vI = vI (1) · vP . (58)

By Proposition 4.2, we find a function g ∈ L(1) with g 6 1. For every f ∈ U(1), we get f > g, hence

I(f) > I(g), and thus vI (1) > I(g) > 0. By (58), the monotonous property of vI carries over to vP ,

i. e. vP is a pseudo-valuation. �

The fact that vI depends on I enables us to normalize the Haar Integral:

Theorem 6.1. Let K be a field with a compact prime P. There exists a unique Haar integral J with

vJ = vP .

Proof. Let I be a Haar integral, and J = rI with r > 0 in IR. Then vrI (1) = rvI (1). By Eq.

(58), the condition vJ = vP is equivalent to vJ (1) = 1, that is, r = vI (1)
−1. �

In the sequel, we simply call J the integral and vP = vJ the normed pseudo-valuation with

respect to P . Then (55) shows that vP (b) measures the content of bP̃ . Our normalization thus fixes

the content of P̃ to 1.

Next we determine the pseudo-valuations of a compact prime. First, let P be an arbitrary prime.

We call a pseudo-valuation v : K → [0,∞] with respect to P trivial if the values of v belong to

{0, 1,∞}.

Proposition 6.2. Let K be a field with a prime P. There is a natural bijection between the primes

Q ⊂ P and the trivial pseudo-valuations v : K → [0,∞] with respect to P.

Proof. First, let v : K → [0,∞] be a trivial pseudo-valuation with respect to P . By Proposi-

tion 2.1, the primes Q ⊂ P correspond to the multiplicatively closed subsets U ⊂ K with P̃ ⊂ U,

where U = Q̃. For such a Q, we define v by

v(a) :=





0 for a ∈ Q

1 for a ∈ ∂Q

∞ for a ∈ K r Q̃.

(59)

Then v is monotonous with v(0) = 0 and v(1) = 1. For a,b ∈ K with v(a) = 1, we have b ∈

Q ⇔ ab ∈ Q and b ∈ Q̃ ⇔ ab ∈ Q̃. Hence v is a pseudo-valuation by Definition 3.1. Conversely, let

v : K → [0,∞] be a trivial pseudo-valuation with respect to P . We set U := {a ∈ K |v(a) < ∞}. Then

U is multiplicatively closed with P̃ ⊂ U. The corresponding prime Q ⊂ P satisfies a ∈ Q̃ ⇔ v(a) 6 1,

and by (31), a ∈ Q ⇔ a−1 /∈ Q̃ ⇔ v(a−1) = ∞ ⇔ v(a) = 0 for a ∈ K×, in accordance with (59). �



116 Wolfgang Rump CUBO
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Proposition 6.3. Let K be a field with a finite, non-trivial pseudo-valuation v : K → [0,∞] with

respect to a prime P. Then Q := {a ∈ K | v(a) < 1} ⊂ P is a minimal prime.

Proof. For a,b ∈ Q ∩ K×, we have v(ab) = v(a)v(b) < 1, hence ab ∈ Q. If a,b /∈ Q, then

v(a),v(b) > 1, which gives v(ab) = v(a)v(b) > 1, i. e. ab /∈ Q. This proves (P1), (P2), and (P3). Thus

Q ⊂ P implies that Q is a prime. Now assume that a,b ∈ Q ∩ K×. Then there exists some n ∈ IN

with v(an) = v(a)n < v(b). Hence an < b, and thus an ∈ bP . So Q is minimal by Proposition 2.2. �

Corollary. Every compact prime P 6= 0 of a field K contains a minimal prime.

Proof. Proposition 6.1 implies that vP is a finite pseudo-valuation, and vP is non-trivial since

P 6= 0. Now Proposition 6.3 completes the proof. �

7 Krull primes

By Proposition 3.4, every minimal prime P of a field K gives rise to a family of P-valuations. Con-

versely,

Proposition 7.1. Let K be a field with a prime P 6= 0 and a P-valuation v : K → [0,∞]. Then P is

minimal.

Proof. Let a,b ∈ P ∩ K× be given. Then v(a−1) < ∞ implies that v(a) 6= 0, and a < 1 yields

0 < v(a) < 1. So there is an n ∈ IN with v(an) = v(a)n 6 v(b). Hence an 6 b. By Proposition 2.2, we

infer that P is minimal. �

In algebraic number theory, the distinction between finite and infinite primes is crucial. For an

arbitrary field, we have to distinguish between algebraic and transcendental primes. First, we prove

Lemma 7.1. A prime P of a field K is algebraic if and only if P̃ is a subring of K.

Proof. If P is algebraic and a,b ∈ P̃ , then (a + b)P ⊂ P + P ⊂ P , hence a + b ∈ P̃ . Therefore,

P̃ is a subring. Conversely, assume that P̃ + P̃ ⊂ P̃ and a,b ∈ P , say, a 6 b. Then ab−1 ∈ P̃ , which

implies that ab−1 + 1 ∈ P̃ . Hence a + b = (ab−1 + 1)b ∈ P . So we get P + P ⊂ P , and thus P is

algebraic. �

Proposition 7.2. For every prime P of a field K, there is a largest algebraic prime P◦ ⊂ P.

Proof. The subring of K generated by P̃ is

U := {a1 + · · · + an | ai ∈ P̃}. (60)

By Proposition 2.1, U corresponds to a prime P◦ ⊂ P with P̃◦ = U. By Lemma 7.1, this prime meets

the requirement. �

There is a close relationship between P◦ and the open subset P
◦ of P given by Proposition 3.1.

First, we prove

Proposition 7.3. If P is a prime of a field K, then P◦ is a prime, and P◦◦ = P◦.



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The tree of primes in a field 117

Proof. Clearly, P◦ satisfies (P1) and (P2). Let a,b /∈ P◦ be given. For any n ∈ IN, there are

elements pn,qn,rn,sn ∈ P with a
n(pn+qn) /∈ P and b

n(rn+sn) /∈ P . Hence a
nbn(pn+qn)(rn+sn) /∈ P ,

and thus anbn(P + P + P + P) 6⊂ P for all n ∈ IN. If ab ∈ P◦, there exists some n ∈ IN with

(ab)n(P + P) ⊂ P , whence (ab)2n(P + P + P + P) ⊂ (ab)n(P + P) ⊂ P , a contradiction. This proves

(P3) for P◦. To verify (P4), choose d ∈ K× with P + P ⊂ dP . We show that P◦ + P◦ ⊂ dP◦. To

this end, assume that a,b ∈ P◦, say, an(P + P) ⊂ P and bn(P + P) ⊂ P . Then (a + b)2n(P + P) =∑2n
i=0

(
2n
i

)
aib2n−i(P + P) ⊂

∑2n
i=0

(
2n
i

)
P ⊂

∑22n
i=1 P ⊂ d

2nP , which gives (a + b)d−1 ∈ P◦. Whence

P◦ + P◦ ⊂ dP◦.

Finally, assume that a ∈ P◦, say, an(P + P) ⊂ P . Then P + P ⊂ a−nP , and by the above, this

implies that P◦ + P◦ ⊂ a−nP◦. Hence an(P◦ + P◦) ⊂ P◦, and thus a ∈ P◦◦. �

Primes P with P = P◦ were first considered by Krull [7]. We call them Krull primes. By

Proposition 3.1, every minimal prime is a Krull prime. The connection to algebraic primes is given by

Proposition 7.4. Let K be a field with a prime P. Then every prime Q ( P◦ of K is algebraic.

Proof. Choose a ∈ P◦ r Q. So there is an n ∈ IN with an(P + P) ⊂ P . Let b ∈ Q be arbitrary.

Then a /∈ Q implies that a−1 ∈ Q̃, hence a−n−1b ∈ Q. This gives b(P̃ + P̃) ⊂ an+1Q(P̃ + P̃) ⊂

an(aP̃ + aP̃)Q ⊂ an(P + P)Q ⊂ PQ ⊂ Q. Assume that a,b ∈ Q, say aP̃ ⊂ bP̃ , and b 6= 0. Then

a + b = (ab−1 + 1)b ∈ (P̃ + P̃)b ⊂ Q. So we get Q + Q ⊂ Q. �

Corollary 1. Let K be a field with a prime P. Then P◦ ⊂ P
◦, and there is no other prime between

P◦ and P
◦.

Proof. Since P◦(P + P) ⊂ P◦ + P◦ = P◦ ⊂ P , we have P◦ ⊂ P
◦. For every prime Q ( P◦,

Proposition 7.4 gives Q ⊂ P◦. �

Corollary 2. Let K be a field with a prime P. A prime Q ⊂ P is algebraic if and only if Q ⊂ P◦.

Proof. Assume that Q ⊂ P◦. If Q 6= P◦, then Q ( P◦ ⊂ P
◦, hence Q is algebraic by Proposi-

tion 7.4. The converse is trivial. �

We call a prime P purely transcendental if P◦ = 0. If P 6= 0, then (P4) implies that P
◦ 6= 0. By

Corollary 1, this shows that P◦ is minimal if P 6= 0 is purely transcendental. Thus Proposition 3.1

gives

Corollary 3. A purely transcendental prime P 6= 0 is minimal if and only if P is a Krull prime.

In the paragraph that follows Definition 1.2, we have shown that algebraic primes P of a field K

correspond to the valuation rings P̃ with quotient field K. If P is principal, say, P = pP̃ 6= 0, and⋂∞
n=0 p

nP̃ = 0, then P̃ is a discrete valuation domain. Now we turn our attention to compact primes.

Proposition 7.5. A compact Krull prime P of a field K is either algebraic or purely transcendental.

Proof. Assume that P is not purely transcendental. Then P◦ 6= 0. Choose a ∈ P◦ r 0. Then

aP̃◦ ⊂ P◦. By Proposition 4.1, P◦ is compact. So there are a1, . . . ,an ∈ K with P◦ ⊂
⋃n

i=1(ai + aP◦),

hence P̃◦ ⊂ a
−1P◦ ⊂

⋃n
i=1(a

−1ai + P◦). Thus |P̃◦/P◦| < ∞. By (60), this gives P̃◦ ⊂ P̃ + · · · + P̃



118 Wolfgang Rump CUBO
12, 2 (2010)

with finitely many summands. Suppose that P 6= P◦. Then there exist b ∈ P r P◦ and n ∈ IN with

bn(P + P) ⊂ P . This gives bn+1(P̃ + P̃) ⊂ bn(bP̃ + bP̃) ⊂ P . Hence, for a suitably large m ∈ IN,

we get bmP̃◦ ⊂ b
m(P̃ + · · · + P̃) ⊂ P . If b−m ∈ P̃◦, then 1 = b

m · b−m ∈ bmP̃◦ ⊂ P , a contradiction.

Hence b−m /∈ P̃◦, and thus b
m ∈ P◦. So we obtain b ∈ P◦, contrary to our assumption. Thus P = P◦.

�

Proposition 7.6. For an algebraic prime P 6= 0 of a field K, the following are equivalent.

(a) P is compact and minimal.

(b) P is a compact Krull prime.

(c) P̃ is a discrete valuation domain with finite resudue field.

Proof. The implication (a) ⇒ (b) follows by Proposition 3.1.

(b) ⇒ (c): Assume that a ∈ P r 0. Then there exist a1, . . . ,an ∈ K with P ⊂
⋃n

i=1(ai +

a2P), hence P̃ ⊂ a−1P ⊂
⋃n

i=1(a
−1ai + aP). Thus |P̃/aP | < ∞. So we get |P̃/P | < ∞, and by

Proposition 1.1, P is a principal ideal, say P = pP̃ . Since |P̃/aP | < ∞ holds for all a ∈ P r 0, it

follows that
⋂

∞

n=0 P
n = 0.

(c) ⇒ (a): Assume that a,b ∈ P ∩K×. Then there is some n ∈ IN with b /∈ P n. Since an ∈ P n, we

have bP̃ 6⊂ anP̃ , and thus anP̃ ⊂ bP̃ by Proposition 1.1. Hence an+1 ∈ bP . Thus, by Proposition 2.2,

P is minimal. Assume that P = pP̃ . Then |P̃/P n| = |P̃/P |n, which implies that P is compact. �

Corollary. A compact prime P 6= 0 of a field K is a Krull prime if and only if P is minimal.

Proof. Assume that P is a compact Krull prime. If P is purely transcendental, Corollary 1 of

Proposition 7.4 implies that P is minimal. Otherwise, P is algebraic by Proposition 7.5, hence minimal

by Proposition 7.6. The converse follows by Proposition 3.1. �

As already shown in section 6, Proposition 4.1 implies that every non-trivial compact prime P

contains a minimal prime.

8 The product formula

In this section, we focus our attention upon the minimal compact primes P of a field K. If P is

algebraic, then Proposition 7.6 implies that P̃ is a discrete valuation domain with finite residue field.

So we have a chain

· · · ( P 2 ( P ( P 0 ( P−1 ( P−2 · · · , (61)

with P 0 = P̃ and P n := pnP̃. For a ∈ K×, let e(a) denote the greatest integer n with a ∈ P n. Thus

aP̃ = P e(a) (62)

holds for any a ∈ K×. To get the normalized P-valuation, we define a function χn : K → [0,∞) for

any n ∈ Z by

χn(a) :=

{
1 for a ∈ P n

0 for a /∈ P n.
(63)



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The tree of primes in a field 119

For a,b ∈ K with a − b ∈ P n, we have |χn(a) − χn(b)| = 0, which yields χn ∈ C
+
P (K).

We set rP := |P̃/P |. Then P
n/P n+1 = pnP̃/pn+1P̃ ∼= P̃/pP̃ , hence |P n/P n+1| = rP , and thus

|P n/P n+m| = rmP for m ∈ IN. This gives χn : χn+m = r
m
P , hence

χn : χn+m
χ0 : χn+m

=
rmP
rn+mP

= r−nP .

Therefore, by (48), there is a Haar integral I with I(χ0) = 1 and I(χn) = r
−n
P . For b ∈ K

×, Eq. (62)

shows that χe(b) is the smallest function in C
+
P (K), which takes the value 1 on bP̃ . Now (55) gives

vI (b) = r
−e(b)
P , hence vI (1) = 1. By (58), we get a normalized P-valuation

vP (a) = r
−e(a)
P . (64)

Let us return to the product formula (6) of the introduction. The normalized Pp-valuations

vp := vPp for the primes Pp of Q are obtained by (64):

vp(a) = p
−e(a), (65)

where e(a) is the exponent of p in a prime factorization of a. Thus vp coincides with the p-adic

absolute value (cf. section 3, Example 1).

The topology of the prime P∞ = (−1, 1) ∩ Q of Q is the usual one. Therefore, a function

f : Q → IR belongs to C∞(Q) := CP∞ (Q) if and only if it is uniformly continuous (in the usual

sense) and vanishes outside a bounded interval. The Haar integral with respect to P∞ is just the

Riemann-integral. Therefore, the normalized P∞-valuation v∞ can be obtained from (54). In fact,

for f ∈ C∞(Q) and c ∈ Q
×, we get by the substitution b = c−1a:

∫

Q

fρc(a)da =

∫

Q

f(c−1a)da = |c|

∫

Q

f(b)db.

So the normalized P∞-valuation is the absolute value:

v∞(a) = |a|. (66)

In the sequel, we denote the set of minimal primes of a field K by P(K). From (65) and (66) we thus

get for all a ∈ Q× the product formula:

∏

P ∈P(Q)

vP (a) = 1. (67)

Lemma 8.1. Let K be a field with a minimal prime P, and let a,b ∈ K× be such that an ⊂ bP̃ for

all n ∈ IN. Then a ∈ P̃.

Proof. Suppose that a /∈ P̃ . Then a−1 ∈ P . By Proposition 2.2, there is an n ∈ IN with

a−n ∈ b−1P . Hence 1 = ana−n ∈ bP̃b−1P ⊂ P , a contradiction! �

Proposition 8.1. Every minimal prime of a field K of characteristic p > 0 is algebraic.

Proof. Let F be the prime field of K. Suppose that a ∈ F rP̃ . Then a−1 ∈ P . Thus |F×| = p−1

implies that 1 = (a−1)p−1 ∈ P , a contradiction. Therefore, F ⊂ P̃ . Choose d ∈ K× with P̃ +P̃ ⊂ dP̃ .



120 Wolfgang Rump CUBO
12, 2 (2010)

For a,b ∈ P̃ and an arbitrary n ∈ IN we find some m ∈ IN with mn < 2m. Then (a + b)mn =∑mn
i=0

(
mn
i

)
aibmn−i ∈

∑2m
i=1 P̃ ⊂ d

mP̃ . Hence ((a + b)nd−1)m ∈ P̃ , and thus (a + b)nd−1 ∈ P̃ . By

Lemma 3.1, this gives a + b ∈ P̃ . So P̃ + P̃ ⊂ P̃ . Now Lemma 2.4 implies that P is algebraic. �

For a rational prime p, let Fp ∼= Z/pZ denote the field with p elements. Consider the quotient

field Fp(x) of Fp[x]. By Proposition 8.1, every minimal prime of Fp(x) is algebraic. For any normed

irreducible polynomial q ∈ Fp[x], consider the subring

Fp[x]q := {
f
g
∈ Fp(x) | f,g ∈ Fp[x],q ∤ g} (68)

of Fp(x). Then Fp[x]q is a discrete valuation ring with residue class field Fp. Hence Pq := qFp(x) is a

prime of Fp(x) with P̃q = Fp[x]q. For all these primes, P̃q ⊃ Fp[x]. Furthermore,

P∞ := x
−1Fp[x

−1]x−1 (69)

is a prime with x /∈ P̃∞, hence P̃∞ 6⊃ Fp[x]. In analogy with (15), we write P∞, which does not mean,

however, that P∞ is transcendental (which is impossible by Proposition 8.1.

The primes of the form Pq with q irreducible, or q = ∞, are compact Krull primes by Proposi-

tion 7.6. In fact, if m = deg q is the degree of an irreducible polynomial q ∈ Fp[x], then dimFp (P̃q/Pq) =

dimFp (Fp[x]/qFp[x]) = m, hence |P̃q/Pq| = p
m. For P∞, we have |P̃∞/P∞| = p. The normalized Pq-

valuation vq := vPq results from (64):

vq(f) = p
−eq (f )deg q, (70)

where eq(f) denotes the multiplicity of q in f. For example, ex−1(x
3 − 2x2 + x) = 2. Moreover, we

get

v∞(f) = p
deg f. (71)

Proposition 8.2. Let p be a rational prime. Every minimal compact prime of Fp(x) is of the form

Pq with q irreducible, or q = ∞.

Proof. Let P be a minimal compact prime of Fp(x). As in the proof of Proposition 8.1, we have

Fp ⊂ P̃ . Assume first that x ∈ P̃ . Then Fp[x] ⊂ P̃ . Since P is a prime ideal of P̃, it follows that

P ∩ Fp[x] is a prime ideal of Fp[x]. If P ∩ Fp[x] = 0, then a /∈ P for all a ∈ Fp[x] r 0. Hence a
−1 ∈ P̃ ,

and thus Fp(x) = P̃ , i. e. P = 0. Consequently, P ∩ Fp[x] is a maximal ideal of Fp[x]. As Fp[x] is

a principal ideal domain, there is an irreducible polynomial q ∈ Fp[x] with P ∩ Fp[x] = qFp[x]. For

every f ∈ Fp[x] with q ∤ f, we have f /∈ P , hence f
−1 ∈ P̃ . Therefore, Pq = qFp[x]q ⊂ P . As P is

minimal, we obtain P = Pq. Finally, assume that x /∈ P̃ . Then x
−1 ∈ P . Applying the preceding

argument to Fp[x
−1] instead of Fp[x], we get P∞ = x

−1Fp[x
−1]x−1 ⊂ P . Thus P∞ = P . �

From (70) and (71) we again obtain the product formula by Proposition 8.2. Note that in both

cases, the product formla arises in a natural way: The minimal primes P are all compact, and their

normalized P-valuations are unique. So the formula describes an intrinsic relation among the minimal

primes.

Received: November 2008. Revised: March 2009.



CUBO
12, 2 (2010)

The tree of primes in a field 121

References

[1] E. Artin and G. Whaples, Axiomatic characterization of fields by the product formula for

valuations, Bull. Amer. Math. Soc. 51 (1945), 469-492

[2] E. Artin and G. Whaples, A note on axiomatic characterization of fields, Bull. Amer. Math.

Soc. 52 (1946), 245-247

[3] N. Bourbaki, General topology, Chapters 1-4; Elements of Mathematics, Springer-Verlag, Berlin,

1998

[4] N. Bourbaki, Commutative algebra, Hermann, Paris 1972

[5] D. K. Harrison, Finite and infinite primes for rings and fields, Mem. Amer. Math. Soc. 68 (1966)

[6] W. Krull, Über geschlossene Bewertungssysteme, J. Reine Angew. Math. 190 (1952), 75-92

[7] W. Krull, Ordnungsfunktionen und Bewertungen von Körpern, Math. Z. 77 (1961) 135-148

[8] J. Neukirch, Algebraic number theory, Grundlehren der Mathematischen Wissenschaften, 322.

Springer-Verlag, Berlin, 1999

[9] A. Ostrowski, Über einige Lösungen der Funktionalgleichung ψ(x)·ψ(x) = ψ(xy), Acta Math. 41

(1916), 271-284

[10] I. Reiner, Maximal orders, London Mathematical Society Monographs, No. 5, Academic Press,

London-New York, 1975

[11] P. Roquette, Class field theory in characteristic p, its origin and development; Class field

theory-its centenary and prospect (Tokyo, 1998), 549-631, Adv. Stud. Pure Math. 30, Math. Soc.

Japan, Tokyo, 2001

[12] H. Weber, Ringtopologien auf Z und Q, Math. Z. 155 (1977), 287-298

[13] H. Weber, Charakterisierung der lokalbeschränkten Ringtopologien auf globalen Körpern,

Math. Ann, 239 (1979), 193-205

[14] A. Weil, L’Intégration dans les groupes topologiques et ses applications, Publ. Inst. Math. Univ.

Strasbourg, Hermann, Paris, 1965