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1 Absolute Values and Valuations Algebra in positive Characteristic Fr¨uhjahrssemester 08 Lecturer: Prof. R. Pink Organizer: P. Hubschmid Absolute values, valuations and completion F. Crivelli ([email protected]) April 21, 2008 Introduction During this talk I’ll introduce the basic definitions and some results about valuations, absolute values of fields and completions. These notions will give two basic examples: the rational function field Fq(T ) and the field Qp of the p-adic numbers. 1 Absolute values and valuations All along this section, K denote a field. 1.1 Absolute values We begin with a well-known Definition 1. An absolute value of K is a function | | : K → R satisfying these properties, ∀ x, y ∈ K: 1. |x| = 0 ⇔ x = 0, 2. |x| ≥ 0, 3. |xy| = |x| |y|, 4. |x + y| ≤ |x| + |y|. (Triangle inequality) Note that if we set |x| = 1 for all 0 6= x ∈ K and |0| = 0, we have an absolute value on K, called the trivial absolute value. From now on, when we speak about an absolute value | |, we assume that | | is non-trivial. Moreover, if we define d : K × K → R by d(x, y) = |x − y|, x, y ∈ K, d is a metric on K and we have a topological structure on K. We have also some basic properties that we can deduce directly from the definition of an absolute value. 1 Lemma 1. Let | | be an absolute value on K. We have 1. |1| = 1, 2. |ζ| = 1, for all ζ ∈ K with ζd = 1 for some 0 6= d ∈ N (ζ is a root of unity), 3. |x−1| = |x|−1, 4. ||x| − |y|| ≤ |x − y|. Proof: 1. |1|2 = |12| = |1| ⇒ |1| = 1. d 2. |ζ| = ζd = |1| = 1 ⇒ |ζ| = 1. 3. 1 = |xx−1| = |x| |x−1| ⇒ |x−1| = |x|−1, 4. Set a := x − y, b := y. We have |a + b| ≤ |a| + |b|, therefore |x| − |y| = |a + b| − |b| ≤ |a| = |x − y|. Another definition about absolute values. Definition 2. Two absolute values on K are equivalent if they define the same topology on K. The next theorem and its corollary give us other conditions to verify if two absolute values are equivalent. Theorem 1. Let | |1 and | |2 be two absolute values on K. They are equivalent if, and only if, there exists s ≥ 0 real such that s |x|1 = |x|2 , ∀x ∈ K. s Proof: If we have |x|1 = |x|2, s > 0, clearly the two absolute values are equivalent since they define the same open sets. If | |1 and | |2 are equivalent, a series converging to 0 with respect to | |1 will converge to 0 with respect to | |2. Moreover, for all K field and | | on K, the n inequality |x| < 1 is equivalent to saying that the sequence {x }n∈N converges to 0. Therefore if | |1 and | |2 are equivalent, we have |x|1 < 1 ⇔ |x|2 < 1. 2 Let y ∈ K be an element such that |y| > 1 and let x ∈ K, x 6= 0. Then there exists n o1 α ∈ such that |x| = |y|α. Let mi be a sequence of rational numbers R 1 1 ni i∈N mi ∗ α ni (mi ∈ Z, ni ∈ N ) converging to α from above. Then |x|1 = |y|1 < |y|1 and hence xni xni < 1 ⇒ < 1. mi mi y 1 y 2 mi n o ni α mi This gives |x|2 < |y|2 and consequently |x|2 ≤ |y|2 . Taking a sequence n i i∈ α α N converging to α from below will give |x|2 ≥ |y|2 and therefore |x|2 = |y|2 . So, for all 0 6= x ∈ K, we have log |x| log |y| 1 = 1 =: s log |x|2 log |y|2 s and hence |x|1 = |x|2. Finally, |y|1 > 1 implies |y|2 > 1 and so s > 0. Corollary. Two absolute values | |1 and | |2 on K are equivalent if, and only if, |x|1 < 1 ⇐⇒ |x|2 < 1. We continue with this important Definition 3. An absolute value is called non-Archimedean if we have |x + y| ≤ max{|x| , |y|}, ∀x, y ∈ K. Otherwise the absolute value is called Archimedean. Note that if | | is non-Archimedean, for x, y ∈ K we have |x|= 6 |y| =⇒ |x + y| = max{|x| , |y|}. Indeed: w.l.o.g we can assume |x| > |y|, then obviously |x + y| ≤ max{|x| , |y|} = |x|. On the other side, |x| = |x − y + y| ≤ max{|x + y| , |y|}. Assume that max{|x + y| , |y|} = |y|, then |x| ≤ |y| < |x|. This contradicts the hypothesis |x| > |y|, hence max{|x + y| , |y|} = |x + y| and so |x| ≤ |x + y|. 1.2 Valuations We introduce the symbol ∞ with the convention that for all a ∈ R we have a < ∞, a + ∞ = ∞ and ∞ + ∞ = ∞. As for absolute values, we start with a basic Definition 4. A valuation on K is a function v : K → R ∪ {∞} satisfying these properties, for all x, y ∈ K: 3 1. v(x) = ∞ ⇔ x = 0, 2. v(xy) = v(x) + v(y), 3. v(x + y) ≥ min{v(x), v(y)}. Note that if we set v(x) = 0 for all 0 6= x ∈ K and v(0) = ∞, we have a valuation on K, called the trivial valuation. From now on, when we speak about a valuation v, we assume that v is non-trivial. Now, a lemma with some basic properties induced by the definition of valuation Lemma 2. Let v be a valuation on K. We have 1. v(1) = 0, 2. v(ζ) = 0, for all ζ ∈ K root of unity, 3. v(x−1) = −v(x), ∀ x ∈ K∗, 4. if x, y ∈ K and v(x) 6= v(y), v(x + y) = min{v(x), v(y)}. Proof: 1. v(1) = v(12) = v(1) + v(1) ⇒ v(1) = 0. 2. Let ζ ∈ K with ζd = 1 for some 0 6= d ∈ N. We have dv(ζ) = v(ζd) = v(1) = 0, therefore v(ζ) = 0. 3. 0 = v(1) = v(xx−1) = v(x) + v(x−1) ⇒ v(x−1) = −v(x). 4. W.l.o.g. we can assume that v(x) > v(y) and so y 6= 0. If x = 0, obvious. Assume x 6= 0. Then v(x + y) ≥ min{v(x), v(y)} = v(y). We have also v(y) = v(y + x − x) ≥ min{v(x + y), v(x)}. Assume that min{v(x + y), v(x)} = v(x), then v(x) > v(y) ≥ v(x) and we get a contradiction. So, min{v(x + y), v(x)} = v(x + y) and finally v(y) ≥ v(x + y). Some further terminology regarding valuations. Definition 5. A valuation v on K is called discrete if v(K∗) = sZ, for a real s > 0. Moreover, v is normalized if s = 1. We introduce now the equivalence between valuations. Definition 6. Two valuations v1 and v2 on K are equivalent if there exists a real s > 0 such that v1 = sv2. Note that if we have a discrete valuation on K with v(K∗) = sZ, dividing it by s we obtain an equivalent normalized valuation. 4 1.3 Relations between non-Archimedean absolute values and valuations The following theorem provides a relation between the non-Archimedean absolute values and the valuations on K. Theorem 2. Let | | be a non-Archimedean absolute value on K and s ∈ R, s > 0, then the function vs : K −→ R ∪ {∞} −s log |x| if x 6= 0 x 7→ ∞ if x = 0 0 0 0 is a valuation on K. Furthermore, if s, s ∈ R, s, s > 0 and s 6= s , vs is equivalent to vs0 . Conversely, if v is a valuation on K and q ∈ R, q > 1, the function | |q : K −→ R q−v(x) if x 6= 0 x 7→ 0 if x = 0 0 0 0 is an absolute value on K. Besides, if q, q ∈ R, q, q > 1 and q 6= q , | |q is equivalent to | |q0 . Proof: We just need the check the definitions of an absolute value and a valuation. We start with vs. Clearly we have that vs(x) = ∞ if, and only if, x = 0. Let x, y ∈ K, if x = 0 or y = 0, xy = 0 and ∞ = vs(xy) = v(x) + v(y) = ∞. Assume x, y 6= 0, then vs(xy) = −s log |xy| = −s log (|x| |y|) = −s log |x| − s log |y| = vs(x) + vs(y). Let again be x, y ∈ K. If x = y = 0, then ∞ = vs(x + y) = min {vs(x), vs(y)} = ∞. If x = 0, y 6= 0 (or y = 0, x 6= 0), vs(x + y) = vs(y) = min {vs(x), vs(y)} (or vs(x + y) = vs(x) = min {vs(x), vs(y)}). Assume now x, y 6= 0. We have vs(x + y) = −s log |x + y| ≥ −s log (max {|x| , |y|}) = min {−s log |x| , −s log |y|} = min {vs(x), vs(y)} . 0 0 Therefore vs is a valuation on K. Assume now s, s > 0, s 6= s . For all 0 6= x ∈ K, we have s 0 s v (x) = −s log |x| = (−s log |x|) = v 0 (x). s s0 s0 s This means that vs and vs0 are equivalent. We continue with | |q. Clearly we have that |x| = 0 if, and only if, x = 0 and since q > 1 > 0 that |x|q ≥ 0 for all x ∈ K. Let x, y ∈ K, if x = 0 or y = 0, xy = 0 and 0 = |xy|q = |x|q |y|q = 0. Assume x, y 6= 0, then −v(xy) −v(x)−v(y) −v(x) −v(y) |xy|q = q = q = q q = |x|q |y|q .
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