Norms and places on number fields

In this section, we will classify the norms on a number field. We start withQ. Let| · |∞

denote the usual real norm onQ. For any rational primep, let| · |pdenote thep-adic norm

discussed in Subsection 8.1.4. A fundamental fact for norms onQis the following

Theorem 10.1.1 (Ostrowski). — The norm| · |p is not equivalent to | · |q ifp6=q with

p, q≤ ∞. Every nontrivial norm | · | on Q is equivalent to | · |p for some prime p or for

p=∞.

Proof. — If one of p, q is ∞ (and the other one is finite), then it is clear that | · |p is

not equivalent to | · |_q. If both p and q are finite primes, then |p|_p = p−1 and |p|_q = 1. Therefore,| · |p and | · |q can not be equivalent.

Assume first that | · | is archimedean. By Proposition 8.2.5,| · | must be unbounded on

Z. Let n0 ≥1 be the first integer such that |n0|>1. Letc∈R>0 be such that |n0|=nc0. We have to prove that|n|=ncfor any positive integer n. Write

n=a0+a1n0+· · ·+asns0, with 0≤ai < n0, as6= 0.

Then one has

|n| ≤ |a₀|+|a₁||n₀|+· · ·+|a_s||n₀|s

=|a0|+|a1|nc0+· · · |as|ncs0 . By our choice ofn0 and since ai< n0, we have |ai| ≤1. Hence,

|n| ≤1 +nc₀+· · ·+ncs₀ ≤ncs₀ (1 +n−₀c+· · ·+n−₀cs)≤Anc,

where A is some constant independent of n. Now replacing n by nM and taking M-th radical, one gets

|n| ≤ N√Anc.

LettingN →+∞, one gets|n| ≤nc. We now deduce an inequality in the other direction as follows. If one writes n in terms of n0 as above, one has ns0+1 > n ≥ ns0. Thus, the

trigonometric inequality implies that |n| ≥ |n0|s+1−(|ns0+1−n|)≥n c(s+1) 0 −(ns0+1−n)c. Since (ns₀+1−n)c≤(n₀s+1−n0)c=nc0(s+1)(1− 1 n0 )c, we get |n| ≥nc₀(s+1)(1−(1− 1 n0 )c)≥A0nc

withA0 = 1−(1− _n01 )c_{. Replacingn byn}M _{and takingM-th root, we get}

|n| ≥ M√A0_nc_.

LettingM →+∞, we get|n| ≥nc. We conclude finally that|n|=ncif| · |is archimedean. Now assume that | · | is non-archimedean. Then we have |n| ≤ 1 for all n ∈ _Z. Let p ⊆ Z be the subset consisting of n ∈ Z with |n| < 1. Then one sees easily that p is a

prime ideal, and p6= 0 since | · | is non-trivial. Therefore, there exists a prime number p

such thatp= (p). By Proposition 8.2.13, | · | is equivalent to| · |p.

Definition 10.1.2. — Let K be a number field. A place v of K is an equivalence class of norms onK. If these norms are archimedean (resp. non-archimedean), we say the place

vis archimedean (resp. non-archimedean).

By Theorem 10.1.1 the set of places of Q is {rational primes } ∪ {∞}. This result can

be generalized to any number field.

10.1.3. Places of an arbitrary number field. — Let K be a number field. Let σi

with 1 ≤ i ≤ r1 denote the real embeddings of K, and σr1+j,¯σr1+j with 1 ≤ j ≤ r2 be the non-real complex embeddings of K. Then for each complex embedding σi with

1≤i≤r1+r2, we have a norm| · |σ on K given by

|x|σi =|σi(x)|C.

On the other hand, for each prime idealpof the integral ringO_K ofK, we have an additive valuation vp on K, which sends each x ∈ K to the exponent of p in the factorization of

(x). We define the normalized p-norm on K by

|x|_p= N(p)−vp(x)_.

Theorem 10.1.4. — (1) Any two of the absolute values of | · |v for v ∈ {σi |1 ≤i ≤

r1+r2} ∪ {prime ideals of OK} are not equivalent to each other.

(2) Every absolute value on K is equivalent to some | · |σi with 1 ≤i≤r1+r2 or | · |p

for a prime ideal p of O_K.

Proof. — It is obvious that a archimedean absolute value | · |σi is not equivalent to any

| · |p, which is non-archimedean. It is also clear that | · |p is not equivalent to | · |q ifp6=q,

since they induces different prime ideals in O_K. To prove that | · |_σi is not equivalent to

| · |σj fori6=j with 1≤i, j≤r1+r2, we recall that, under the map

10.1. NORMS AND PLACES ON NUMBER FIELDS 117

defined byλ(x) = (σi(x))1≤i≤r1+r2, the image ofOK is a (full) lattice in Rr1 ×Cr2. Then

λ(K) is dense in Rr1 ×Cr2. In particular, for any i with 1 ≤ i ≤ r1+r2, there exists

xi ∈ K such that |xi|σi < 1 and |xi|σj > 1 for any j 6= i. Hence, the sequence (x

n i)n≥1 converge to 0 for| · |σi but diverges for| · |σj withj6=i. Hence,| · |σi is not equivalent to

| · |σj.

Suppose now | · | is an absolute value on K, we have to prove that | · | is equivalent to some| · |σi or to some| · |p. Suppose first that| · |is non-archimedean. SinceOK is integral overZ, we have|x| ≤1 for allx∈ OK. Let p be the subset of x∈ OK such that |x|<1.

One verifies easily thatpis a prime ideal ofOK. ThenOK,pis the valuation ring of K for

| · |, which coincides that of K for| · |_p. By Proposition 8.2.13,| · |is equivalent to | · |_p. Assume now that | · | is archimedean. Let Kv denote the completion of K under | · |.

Then Kv is a finite extension of R, which is the completion of Q at its unique infinite

place. Thus,Kv is eitherRorC. In any case, the embeddingK ,→Kv is one ofσi or the

complex conjugate of someσi.

Theorem 10.1.4 says that the non-archimedean (or finite) places of K are in natural bijection with non-zero prime ideals of OK, and the archimedean (or infinite) places of

K are in natural bijection with the orbit of complex conjugate on the set of complex embeddings ofK.

Theorem 10.1.5. — Letv1,· · · , vrbe distinct places of K. Then the diagonal embedding

K ,→

r

Y

i=1

Kvi,

has dense image, where Kvi denotes the completion of K at vi.

Proof. — Step 1: SinceKis dense in eachKvi, it suffices to prove that, givenx1,· · ·, xr∈

K and >0, there exists ξ∈K such that

(10.1.5.1) |ξ−xi|vi < .

We claim that for eachiwith 1≤i≤r and anyδ >0, there exists a ξi ∈K such that

|ξi−1|vi < , |ξi|j ≤δ, forj6=i. Assuming the claim for a moment, thenξ =Pr

i=1ξixi satisfies |ξ−xi|vi =|xi(ξi−1) + X j6=i xjξj|vi ≤ |xi|viδ+ X j6=i |xj|viδ=δ r X j=1 |xj|vi.

Step 2. The proof of the claim can be reduced to showing that for anyiwith 1≤i≤r, there existsξ∈K such that|ξ|vj <1 for allj 6=i, and|ξ|vi >1. Since then, we have

| ξ n 1 +ξn −1|vi =| 1 1 +ξn|vi →0, | ξn 1 +ξn|vj →0 for all j6=i

as n → +∞. We may assume that i = 1, and proceed by induction on r. For r = 2, the existence of ξ follows from the non-equivalence of v1 and v2. Assume now r > 2. By induction hypothesis, there exists ξ ∈ K such that |ξ|_v1 > 1 and |ξ|_vj < 1 for j = 2,· · · , r−1. If |ξ|_vr < 1, then the assertion is proved. Consider the cases |ξ|_vr ≥ 1. As

v1 and vr are not equivalent, there exists α ∈ K such that |α|v1 > 1 and |α|vr < 1. If

|ξ|_vr = 1, then ξNα for sufficiently large N will answer the question. If|ξ|_vr >1, we can take ₁₊αξ_ξNN forN sufficiently.