Math 129: Algebraic Number Theory Lecture 15: Valuations

William Stein

Thursday, March 25, 2004

I don’t know about you, but Swinnerton-Dyer’s book is getting on my nerves (I will avoid more passionate words), so we’re switching to the venerable and famous book by Cassels-Frohlich. In particular, we’re going to systematically go through the article Global Fields by Cassels, which is chapter 2 of the book. The topics are similar to the ones in chapter 2 of Swinnerton-Dyer, but Cassels’s article is amazingly well written. Also, you are well prepared to read and appreciate it given what you’ve learned so far in this course.

A scan of the article is available on the web page for the course, and you can get a photocopy from me.

The notes for the rest of the course will be a rewrite of Global Fields meant to make it more accessible. I will copy Cassels’s article closely, except I will fix any typos found, reword things in a way consistent with the rest of these notes, and add exercises and comments you might have. I will also add the details of the implicit exercises and remarks that are left to the reader.

1 Valuations

Definition 1.1 (Valuation). A valuation | | on a field K is a function defined on K with values in R_{≥0} satisfying the following axioms:

(1) |a| = 0 if and only if a = 0,

(2) |ab| = |a| |b|, and

(3) there is a constant C ≥ 1 such that |1 + a| ≤ C whenever |a| ≤ 1.

The trivial valuation is the valuation for which |a| = 1 for all a ≠ 0. We will often tacitly exclude the trivial valuation from consideration.

From (2) we have

|1| = |1| · |1| ,

so |1| = 1 by (1). If w ∈ K and w^{n} = 1, then |w| = 1 by (2). In particular, the only valuation of a finite field is the trivial one. The same argument shows that | − 1| = |1|, so

| − a| = |a| all a ∈ K.

Definition 1.2 (Equivalent). Two valuations | |_{1} and | |_{2} on the same field are equivalent if there exists c > 0 such that

|a|_{2} = |a|_{1}^{c}

for all a ∈ K.

Note that if | |_{1} is a valuation, then | |_{2} = | |_{1}^{c} is also a valuation. Also, equivalence of valuations is an equivalence relation.

If | | is a valuation and C is the constant from Axiom (3), then there is a c > 0 such that C^{c} = 2 (i.e., c = log(C)/ log(2)). Then we can take 2 as constant for the equivalent valuation | |^{c}. Thus every valuation is equivalent to a valuation with C = 2. Note that if C = 1, e.g., if | | is the trivial valuation, then we could simply take C = 2 in Axiom (3).

Proposition 1.3. Suppose | | is a valuation with C = 2. Then for all a, b ∈ K we have

|a + b| ≤ |a| + |b|  (triangle inequality).  (1.1)

Proof. Suppose a_{1}, a_{2} ∈ K with |a_{1}| ≥ |a_{2}|. Then a = a_{2}/a_{1} satisfies |a| ≤ 1. By Axiom (3) we have |1 + a| ≤ 2, so multiplying by a_{1} we see that

|a_{1} + a_{2}| ≤ 2|a_{1}| = 2 · max{|a_{1}|, |a_{2}|}.

Also we have

|a_{1} + a_{2} + a_{3} + a_{4}| ≤ 2 · max{|a_{1} + a_{2}|, |a_{3} + a_{4}|} ≤ 4 · max{|a_{1}|, |a_{2}|, |a_{3}|, |a_{4}|},

and inductively we have for any r > 0 that

|a_{1} + a_{2} + ⋯ + a_{2^{r}}| ≤ 2^{r} · max |a_{j}|.

If n is any positive integer, let r be such that 2^{r−1} ≤ n ≤ 2^{r}. Thenn

|a_{1} + a_{2} + ⋯ + a_{n}| ≤ 2^{r} · max{|a_{j}|} ≤ 2n · max{|a_{j}|},

since 2^{r} ≤ 2n. In particular,

|n| ≤ 2n · |1| = 2n  (for n > 0).  (1.2)

Applying (1.2) to |\binom{n}{j}| and using the binomial expansion, we have for any a, b ∈ K that

|a + b|^{n} = |∑_{j=0}^{n} \binom{n}{j}a^{j}b^{n−j}|

≤ 2(n + 1) max_{j} \{|\binom{n}{j}| |a|^{j} |b|^{n−j}\}

≤ 2(n + 1) max_{j} \{2\binom{n}{j} |a|^{j} |b|^{n−j}\}

≤ 4(n + 1) max_{j} \{\binom{n}{j} |a|^{j} |b|^{n−j}\}

≤ 4(n + 1)(|a| + |b|)^{n}.

Now take nth roots of both sides to obtain

|a + b| ≤ \sqrt[n]{4(n + 1)} · (|a| + |b|).

We have by elementary calculus that

lim_{n→∞} \sqrt[n]{4(n + 1)} = 1,

so |a + b| ≤ |a| + |b|. (The “elementary calculus”: We instead prove that \sqrt[n]{n} → 1, since the argument is the same and the notation is simpler. First, for any n ≥ 1 we have \sqrt[n]{n} ≥ 1, since upon taking nth powers this is equivalent to n ≥ 1^{n}, which is true by hypothesis. Second, suppose there is an ε > 0 such that \sqrt[n]{n} ≥ 1 + ε for all n ≥ 1. Then taking logs of boths sides we see that \frac{1}{n} log(n) ≥ log(1 + ε) > 0. But log(n)/n → 0, so there is no such ε. Thus \sqrt[n]{n} → 1 as n → ∞.)

Note that Axioms (1), (2) and Equation (1.1) imply Axiom (3) with C = 2. We take Axiom (3) instead of Equation (1.1) for the technical reason that we will want to call the square of the absolute value of the complex numbers a valuation.

Lemma 1.4. Suppose a, b ∈ K, and | | is a valuation on K with C ≤ 2. Then

||a| − |b|| ≤ |a − b| .

(Here the big absolute value on the outside of the left-hand side of the inequality is the usual absolute value on real numbers, but the other absolute values are a valuation on an arbitrary field K.)

Proof. We have

|a| = |b + (a − b)| ≤ |b| + |a − b|,

so |a| − |b| ≤ |a − b|. The same argument with a and b swapped implies that |b| − |a| ≤ |a − b|, which proves the lemma.

2 Types of Valuations

We define two important properties of valuations, both of which apply to equivalence classes of valuations (i.e., the property holds for | | if and only if it holds for a valuation equivalent to | |).

Definition 2.1 (Discrete). A valuation | | is discrete if there is a δ > 0 such that for any a ∈ K

1 − δ < |a| < 1 + δ ⟹ |a| = 1.

Thus the absolute values are bounded away from 1.

To say that | | is discrete is the same as saying that the set

G = {log |a| : a ∈ K, a ≠ 0} ⊂ R

forms a discrete subgroup of the reals under addition (because the elements of the group G are bounded away from 0).

Proposition 2.2. A nonzero discrete subgroup G of R is free on one generator.

Proof. Since G is discrete there is a positive m ∈ G such that for any positive x ∈ G we have m ≤ x. Suppose x ∈ G is an arbitrary positive element. By subtracting off integer multiples of m, we find that there is a unique n such that

0 ≤ x − nm < m.

Since x − nm ∈ G and 0 < x − nm < m, it follows that x − nm = 0, so x is a multiple of m.

By Proposition 2.2, the set of log |a| for nonzero a ∈ K is free on one generator, so there is a c < 1 such that |a|, for a ≠ 0, runs precisely through the set

c^{Z} = {c^{m} : m ∈ Z}

(Note: we can replace c by c^{−1} to see that we can assume that c < 1).

Definition 2.3 (Order). If |a| = c^{m}, we call m = ord(a) the order of a.

Axiom (2) of valuations translates into

ord(ab) = ord(a) + ord(b).

Definition 2.4 (Non-archimedean). A valuation | | is non-archimedean if we can take C = 1 in Axiom (3), i.e., if

|a + b| ≤ max{|a|, |b|}.  (2.1)

If | | is not non-archimedean then it is archimedean.

Note that if we can take C = 1 for | | then we can take C = 1 for any valuation equivalent to | |. To see that (2.1) is equivalent to Axiom (3) with C = 1, suppose |b| ≤ |a|. Then |b/a| ≤ 1, so Axiom (3) asserts that |1 + b/a| ≤ 1, which implies that |a + b| ≤ |a| = max{|a|, |b|}, and conversely.

We note at once the following consequence:

Lemma 2.5. Suppose | | is a non-archimedean valuation. If a, b ∈ K with |b| < |a|, then |a + b| = |a|.

Proof. Note that |a + b| ≤ max{|a|, |b|} = |a|, which is true even if |b| = |a|. Also,

|a| = |(a + b) − b| ≤ max{|a + b|, |b|} = |a + b|,

where for the last equality we have used that |b| < |a| (if max{|a + b|, |b|} = |b|, then |a| ≤ |b|, a contradiction).

Definition 2.6 (Ring of Integers). Suppose | | is a non-archimedean absolute value on a field K. Then

O = {a ∈ K : |a| ≤ 1}

is a ring called the ring of integers of K with respect to | |.

Lemma 2.7. Two non-archimedean valuations | |_{1} and | |_{2} are equivalent if and only if they give the same O.

We will prove this modulo the claim (to be proved next time) that valuations are equivalent if (and only if) they induce the same topology.

Proof. Suppose suppose | |_{1} is equivalent to | |_{2}, so | |_{1} = | |_{2}^{c}, for some c > 0. Then |c|_{1} ≤ 1 if and only if |c|_{2}^{c} ≤ 1, i.e., if |c|_{2} ≤ 1^{1/c} = 1. Thus O_{1} = O_{2}.

Conversely, suppose O_{1} = O_{2}. Then |a|_{1} < |b|_{1} if and only if a/b ∈ O_{1} and b/a ∉ O_{1}, so

|a|_{1} < |b|_{1} ⟺ |a|_{2} < |b|_{2}.  (2.2)

The topology induced by | |_{1} has as basis of open neighborhoods the set of open balls

B_{1}(z, r) = {x ∈ K : |x − z|_{1} < r},

for r > 0, and likewise for | |_{2}. Since the absolute values |b|_{1} get arbitrarily close to 0, the set U of open balls B_{1}(z, |b|_{1}) also forms a basis of the topology induced by | |_{1} (and similarly for | |_{2}). By (2.2) we have

B_{1}(z, |b|_{1}) = B_{2}(z, |b|_{2}),

so the two topologies both have U as a basis, hence are equal. That equal topologies implies equivalence of the corresponding valuations will be proved later.

The set of a ∈ O with |a| < 1 forms an ideal 𝔭 in O. The ideal 𝔭 is maximal, since if a ∈ O and a ∉ 𝔭 then |a| = 1, so |1/a| = 1/|a| = 1, hence 1/a ∈ O, so a is a unit.

Lemma 2.8. A non-archimedean valuation | | is discrete if and only if 𝔭 is a principal ideal.

Proof. First suppose that | | is discrete. Choose π ∈ 𝔭 with |π| maximal, which we can do since

S = {log |a| : a ∈ 𝔭} ⊂ (−∞, 1],

so S is discrete and bounded above. Suppose a ∈ 𝔭. Then

|\frac{a}{π} | = \frac{|a|}{|π|} ≤ 1,

so a/π ∈ O. Thus

a = π · \frac{a}{π} ∈ πO.

Conversely, suppose 𝔭 = (π) is principal. For any a ∈ 𝔭 we have a = πb with b ∈ O. Thus

|a| = |π| · |b| ≤ |π| < 1.

Thus {|a| : |a| < 1} is bounded away from 1, which is exactly the definition of discrete.

Example 2.9. For any prime p, define the p-adic valuation | |_{p} : Q → R as follows. Write a nonzero α ∈ K as p^{n} · \frac{a}{b}, where gcd(a, p) = gcd(b, p) = 1. Then

|p^{n} · \frac{a}{b}|_{p} := p^{−n} = (\frac{1}{p})^{n} .

This valuation is both discrete and non-archimedean. The ring O is the local ring

Z_{(p)} = \{\frac{a}{b} ∈ Q : p ∤ b\} ,

which has maximal ideal generated by p. Note that ord(p^{n} · \frac{a}{b}) = p^{n}.

We will need the following lemma later.

Lemma 2.10. A valuation | | is non-archimedean if and only if |n| ≤ 1 for all n in the ring generated by 1 in K.

Note that we cannot identify the ring generated by 1 with Z in general, because K might have characteristic p > 0.

Proof. If | | is non-archimedean, then |1| ≤ 1, so by Axiom (3) with a = 1, we have |1 + 1| ≤ 1. By induction it follows that |n| ≤ 1.

Conversely, suppose |n| ≤ 1 for all integer multiples n of 1. This condition is also true if we replace | | by any equivalent valuation, so replace | | by one with C ≤ 2, so that the triangle inequality holds. Suppose a ∈ K with |a| ≤ 1. Then by the triangle inequality,

|1 + a|^{n} = |(1 + a)^{n}|
≤ ∑_{j=0}^{n} |\binom{n}{j}| |a|
≤1 + 1 + ⋯ + 1 = n.

Now take nth roots of both sides to get

|1 + a| ≤ \sqrt[n]{n},

and take the limit as n → ∞ to see that |1 + a| ≤ 1. This proves that one can take C = 1 in Axiom (3), hence that | | is non-archimedean.