LOCAL FIELDS 1. Absolute Values Let K Be a Field. an Absolute Value on K Is a Map X → |X| from K Into the Nonnegative Real

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LOCAL FIELDS 1. Absolute Values Let K Be a Field. an Absolute Value on K Is a Map X → |X| from K Into the Nonnegative Real CHAPTER II LOCAL FIELDS 1. Absolute Values Let k be a field. An absolute value on k is a map x → |x| from k into the nonnegative real numbers that satisfies the following three conditions: (i) |x| = 0 if and only if x = 0, (ii) |xy| = |x||y| for all x and y in k, (iii) |x + y| ≤ |x| + |y| for all x and y in k. The first results about a field with an absolute value are immediate consequences of the definition, and we organize them as a series of remarks. Every field k has at least one absolute value which we denote here by | |0. This is defined by 0 if x = 0, |x|0 = 1 if x 6= 0. The absolute value | |0 is called the trivial or improper absolute value. Any other absolute values on k will be called nontrivial or proper. When working in a field k the trivial absolute value is usually not considered. However, there is at least one situation in which the trivial absolute value may appear quite naturally. Suppose that K is an extension of k and | | is an absolute value on K. Then | | restricted to k is obviously an absolute value on k. It may happen that | | is a nontrivial absolute value on K, but the restriction | | to k is trivial. Let k× denote the multiplicative group of nonzero elements in k. If | | is an absolute value on k then the restriction of | | to k× is a homomorphism from the multiplicative × group k into the multiplicative group of positive real numbers. It follows that |1k| = 1 and more generally, if ζ is an nth root of unity in k× then |ζ| = 1. We also get | − x| = |x| for all x ∈ k, and |x−1| = |x|−1 for all x ∈ k×. If x is in k× and |x| 6= 1 then x must have infinite order in the multiplicative group k×. This shows that the only absolute value on a finite field is the trivial absolute value. When verifying that a function | | : k → [0, ∞) is an absolute value, it is sometimes useful to be able to check an inequality that is weaker than the triangle inequality. Typeset by AMS-TEX 1 2 II. LOCAL FIELDS Lemma 1.1. Assume that k is a field and | | : k → [0, ∞) satisfies the three conditions (i) |x| = 0 if and only if x = 0, (ii) |xy| = |x||y| for all x and y in k, (iv) |x + y| ≤ 2 max{|x|, |y|} for all x and y in k. Then | | satisfies the triangle inequality (iii) in the definition of an absolute value, and therefore | | is an absolute value on k. Proof. If M = 2m for some positive integer m, then it follows easily from (iv), by induction on m, that m (1.1) |x1 + x2 + ··· + xM | ≤ 2 max{|x1|, |x2|,..., |xM |} m−1 m for all x1, x2, . , xM in k. If N is a positive integer and 2 < N ≤ 2 , then by selecting xN+1 = xN+2 = ··· = xM = 0 in (1.1), we find that m |x1 + x2 + ··· + xN | ≤ 2 max{|x1|, |x2|,..., |xN |} (1.2) ≤ 2N max{|x1|, |x2|,..., |xN |} for all x1, x2, . , xN in k. In particular, if x1 = x2 = ··· = xN = 1, then it follows that (1.3) |N| ≤ 2N. Now let x and y be elements of k, and let L be a positive integer. Then using (1.2) and (1.3) we get |x + y|L = |(x + y)L| L X L = xlyL−l l l=0 L ≤ 2(L + 1) max |x|l|y|L−l : 0 ≤ l ≤ L l L ≤ 4(L + 1) max |x|l|y|L−l : 0 ≤ l ≤ L l L X L ≤ 4(L + 1) |x|l|y|L−l l l=0 L = 4(L + 1)|x| + |y| , and therefore 1/L (1.4) |x + y| ≤ 4(L + 1) |x| + |y|. 1. ABSOLUTE VALUES 3 The triangle inequality (iii) follows now by letting L → ∞ in (1.4). If | | is an absolute value on k then the map (x, y) → |x − y| from k × k into [0, ∞) is a metric, and therefore the absolute value induces a metric topology in k. The distance from x to y is |x − y|. Let Ak denote the set of absolute values on k. We say that two absolute values in Ak are equivalent if they induce the same metric topology. It is easy to verify that this is in fact an equivalence relation in Ak. The trivial absolute value | |0 clearly induces the discrete topology in k, and it is the unique absolute value in its equivalence class. We will show that an equivalence class determined by a nontrivial absolute value contains many distinct, nontrivial absolute values. An equivalence class determined by a nontrivial absolute value in Ak is called a place of k. Equivalent absolute values on k can be characterized in a simple way. Theorem 1.2. Let | |1 and | |2 be absolute values on k. Then the following assertions are equivalent: (1) | |1 and | |2 induce the same metric topology in k, (2) {x ∈ k : |x|1 < 1} = {x ∈ k : |x|2 < 1}, θ (3) there exists a positive number θ such that |x|1 = |x|2 for all x in k. Proof. Assume that (1) holds and let Uj = {x ∈ k : |x|j < 1}, for j = 1, 2. n ∞ If x belongs to U1 then {x }n=1 is a sequence that converges to 0 in k. As U2 is an N open neighborhood of 0 we must have x in U2 for some sufficiently large integer N. N N But |x |2 < 1 implies that |x|2 < 1, and therefore |x|2 < 1. It follows that U1 ⊆ U2, and by symmetry U2 ⊆ U1. Thus condition (2) is satisfied. Now assume that (2) holds. If either | |1 or | |2 is trivial then (3) follows immediately. Therefore we may assume that both | |1 and | |2 are nontrivial. Let y in k satisfy −1 |y|1 6= 0 and |y|1 6= 1. By replacing y with y if necessary, we may further assume that 0 < |y|1 < 1. From (2) we have 0 < |y|2 < 1. Hence log |y| θ = 2 log |y|1 θ is a positive real number. We claim that |x|1 = |x|2 for all x in k. If this claim is false then by replacing x with x−1 if necessary, there exists a point x in k with θ |x|1 < |x|2. Now select a rational number r/s such that θ θr/s r/s |x|1 < |y|1 = |y|2 < |x|2. 4 II. LOCAL FIELDS r/s This choice is possible because the image of the map r/s → |y|2 is dense in (0, ∞). Then we have s θ r θ r s |x |1 < |y |1 = |y |2 < |x |2, and it follows that s −r s −r |x y |1 < 1, and 1 < |x y |2. This contradicts (2) and so verifies our claim. Thus (3) is satisfied. Assume that (3) holds. Then the set of all open balls in the | |1-topology is equal to the set of all open balls in the | |2-topology. The assertion (1) follows. Let | | be an absolute value on k, and let 0 < θ < 1. It is obvious that x → |x|θ satisfies the conditions (i) and (ii) in the hypotheses of Lemma 1.1. Also, condition (iv) holds because θ |x + y|θ ≤ |x| + |y| ≤ 2θ max |x|θ, |y|θ . It follows then from Lemma 1.1 that | |θ is an absolute value on k. By Theorem 1.2, for each value of θ with 0 < θ < 1, the absolute value | |θ is equivalent to | |. If we further assume that | | is nontrivial, then |z| 6= 1 for some point z in k×, and we conclude that | |θ and | | are equivalent but not equal. In particular, this shows that the place determined by | | in Ak contains many distinct but equivalent absolute values. Let | | be an absolute value on k. Then define θ Θ(| |) = {θ > 0 : x → |x| is an absolute value in Ak}. It follows from the definition of an absolute value that Θ(| |) is a closed, nonempty subset of (0, ∞). By our previous remarks, if τ is in Θ(| |) then θτ is in Θ(| |) for all θ such that 0 < θ ≤ 1. This shows that either Θ(| |) = (0, ∞), or Θ(| |) = (0, τ] for some real number τ ≥ 1. We define a second function on absolute values in Ak by setting Φ(| |) = sup{|x + 1| : x ∈ k and |x| ≤ 1}. Obviously we have 1 ≤ Φ(| |) ≤ 2. Also, the inequality (1.5) |x + y| ≤ Φ(| |) max{|x|, |y|} holds for all x and y in k. To verify (1.5) assume that 0 < |x| ≤ |y|. Then we have |x + y| = |(xy−1 + 1)y| = |xy−1 + 1||y| ≤ Φ(| |)|y| = Φ(| |) max{|x|, |y|}. Note that if Φ(| |) = 1 then (1.5) is stronger than the triangle inequality (iii) in the definition of an absolute value. 1. ABSOLUTE VALUES 5 Theorem 1.3. Let | | be an absolute value on k. Then the following assertions are equivalent: (1) Θ(| |) = (0, ∞), (2) |x + y| ≤ max{|x|, |y|} for all x and y in k, (3) Φ(| |) = 1. Moreover, if 1 < Φ(| |) ≤ 2, then Θ(| |) = (0, τ] for some real number τ ≥ 1, and (1.6) Φ(| |)τ = 2. Proof. Assume that Θ(| |) = (0, ∞). Then the map x → |x|n is an absolute value for all positive integers n. It follows that 1/n |x + y| ≤ |x|n + |y|n for all x and y in k.
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