Absolute Values on the Field Q

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2. Absolute Values on the Field Q

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but 0 < I p I < 1. I claim that for every integer ME Z prime top, we have Im I= 1.

Indeed, if m is prime to p, the Bdzout theorem asserts that there are integers u and v with up + vm = 1. Hence

1 = III = Iup+vmI <max(Iup1, Ivmp < 1.

Since by assumption I up I = I I I P I < Iu I < 1, the maximum must be I vmI = 1 and hence Imi = 1 (we know a priori that I v 1 < 1 and Im1 < 1). There is now a unique positive real number a such that

IPI = (I/P)a

(indeed, take a = (logipl)/(log(I/p)) - a quotient of two negative numbers - independent from the basis of logarithms chosen). Then if the rational number x is written in the form x = p"a/b c Q" with p prime to a and b (i.e., alb E Zop)), we shall have

IxI = IPIU = (1/0"a = IxI"

and the theorem is completely proved.

2.2.

Generalized Absolute Values

Observe that if I ^.I is an absolute value and a > 0, then is not an absolute value in general. For example if I I. I is the usual absolute value on Q and a = 2, then f (x) = IX 12 does not satisfy the triangle inequality

4= f(2)= f(I+1)> f(1)+f(1)=2.

But it satisfies

f(x +Y) = Ix + Y12 < (IxI + IYI)2 < (2max{IxI, IYI})2 = 4max(f(x). f(Y)).

This is one reason for considering generalized absolute values.

Definition. A generalized absolute value on afield K is a homomorphism f : K" -> R>0 extended by f (0) = O for which there exists a constant C > 0 such that

f (x + y) < C max(f(x), f (y)) (x, y E K).

Observations. (1) For any generalized absolute value f and any a > 0 (not only for 0 < a < 1), f is also a generalized absolute value: Replace C by C'.

(2) The ultrametric absolute values are those for which the above inequality holds with C = 1. Moreover, if f is a (usual) absolute value, then

f (x + y) < f (x) + f (Y) < 2 max(f (x), f (Y)),

and (usual) absolute values are generalized absolute values: The above inequality holds with C = 2. Let us prove a converse.

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2. Absolute Values on the Field Q 87

Theorem. Let f be a generalized absolute value on afield K for which f (x + y) < 2 max(f (x), f (y)) (x, y E K).

Then f is a usual absolute value: It satisfies the identity

f(x + y) f (x) + f(y)

(x, y E K).

PRooF. Iterating the defining inequality for generalized absolute values, we find that

Pal + a2 + a3 + a4) < C max(f (at + a2), f (a3 + a4))

< C2maxt<i<4 f(ai) More generally, by induction if n = 2', then

f(at

^{+. ..+an)}

^{< C'max}

^f(ai)-

Since we are assuming that the constant C = 2 can be taken in the preceding inequalities, we have

f(a, +---+a,,) <2'max f(ai)=nmax f(ai).

Now, if n is not a power of 2, say 2r-1 < n < 2', we can complete the sum by taking coefficients ai = 0 for n < i < 2' and still write

2nmaxf(ai).

We shall have to use two particular cases of this general inequality:

(1) f (n) < 2n (take ai = I for I < i < n),

(2) f

(E1 <i<n a,) < 2n max f (ai) < 2n Yt<i<n f (ai)

To estimate f (a + b), we shall estimate f ((a+b)') thanks to the binomial formula (the nth power of a + b is a sum of n + I monomials)

f((a+b)n) = f (1: (')a'b'1_')

< 2(n + 1)

f (()) f (a)' f (b)n-`

<2(n+1)2 (i

1f(a)`f(b)n-`

= 4(n +

1)(f(a)+/f(b))n.

Let us extract nth roots:

f(a+b)<41"n(n+l)t1n-(f(a)+f(b))-*

f(a)+f(b) (n-

^oo).

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2.3. Ultrametric Among Generalized Absolute Values

We can give a generalization of (1.6).

Theorem. Let f be a generalized absolute value on afield K. If f is bounded on the image of the natural numbers N in K, then it is an ultrametric absolute value.

PROOF. Let n = 2' be a power of 2 and consider a sum of n terms ai. As in (2.2), we see by induction that

.f (ai + ... + a,,) < C' max_{f (ai ).}

Take now x E K and consider the element

+x)n-t = F (n_I)1

0<i <n

Since this sum has n elements, we have

l1

(.f(1 ^+x))n-t

= f((1 +x)n-')

< C' max

[f ^((n_I))

^.

^f(x`)]

If f is bounded on the image of N in K, say f (k) < A for all k E N, we shall have

(.f(1 +x))n-t < C'A max (1, ^.f(x)n-t)

and

f(1 +x) <

Cr/(n-t)At/(n-u

max (1, f(x)).

Letting again n -+ oc, we obtain

f(1 +x) < max(1, f(x)).

If now a 94- 0 and b E K, then f (a) 0 and

f(a+b)= f(a)f(1+b/a)

< f (a) max (1, f (b/a)) = max (f (a), f (b)).

2.4. Generalized Absolute Values on the Rational Field

The ultrametric absolute values on the rational field Q have been determined in (2.1). Here, we treat the generalized absolute values.

Theorem. Any nontrivial generalized absolute value on the rational field Q is either a power of the usual absolute value or a power of the p-adic absolute value.

r°.

2. Absolute Values on the Field Q 89

Proof Take any nontrivial generalized absolute value f and assume that f (x + y) C - max (f (x), f (y)).

If C < 1, then f is ultrametric, and we conclude by (2.1). Assume now C > 1.

By induction - regardless of the size and number of addends - we can prove f(ao+...+ar) <Cr max f(ai).

Let us fix an integer n > 2 and put A = An = max (f (1), ..., f (n)) > 1. Now, any integer m > 2 can be expanded in base n, say

m = E min`

(0 < m, < n, m, :?4- 0).

0<i <r

Hence

f (m) < Cr ^{. Max}f (mi )f (n`)

< CrAn max f (n)` = CrAn max (1, f (n)r).

But mr 0 0, nr < m, and thus r < log m/ log n, so that we can write f (m) < An C'Ogm/'ogn - max (1, f (n)'ogm/loge)_, f(m)1/logm < An/logmCl/logn. max(1, f(n)'/logn)_

Let us replace m by mk (keeping n fixed), so that the left-hand side is unchanged, and let k --> oo, whence A,1,/(klogm)

1. We obtain

f (m)'/'ogm < Cl/'ogn. max (1, f(n)1/logn).

In other words, we have obtained an inequality in which the constant An does not appear. We can now replace n by nk, and since C11k -- 1, we have simply

f

(m)1/'09m <max (1, f (n)'/'09 n).

First case: There is an integer n > 2 with f (n) < 1.

We can use such an integer n in the inequality just found and deduce f (m) < 1 for every integer m > 2.

Hence f is an ultrametric absolute value by (2.3). Finally, Ostrowski's theorem (2.1) applies: f is a power of the p-adic absolute value

f(x) = IxI p (x E Q)

for some real a (determined by the condition f(p) = Ipl° = (1/p)a).

Second case: We have f (n) > I for every integer n > 2.

The general inequality

f(m)'/logm < max (1, f (n)l/logn)

is now simply

f(m)l/logm

< f(n)l/loge

Since we can permute the roles of n and m, we must even have

f(m)1/l09m

= f(n)l/loge Hence f (n)11 log' = ea is independent from n. This leads to

f(n) = ealogn = na for all integers n > 1, and with the usual absolute value

f(n) = I n l a (n E Z).

By the multiplicativity property, we also have f(x) = Ixlc' ^{(x E Q)-}

Since 0 < a < oo, the map f is a power of the usual absolute value, and the

theorem is completely proved. ^M

Comment. The preceding result shows that for a generalized absolute value f on the field Q, the only possibilities are

f is trivial,

Ip I < I for some prime p and f is a power of the p-adic absolute value, In I > I for all positive integers n and f is a power of the usual Archimedean absolute value.

Observe that the two nontrivial cases can also be classified according to the value of 121: If 121 < 1, f is a power of the 2-adic absolute value; if 121 = 1, f is a power of the p-adic absolute value for some odd prime p; if 121 > 1, f is a power of the usual Archimedean absolute value.