ABSOLUTE VALUES AND VALUATIONS YIFAN WU, wuyifan@umich.edu Abstract. We introduce the basis notions, properties and results of absolute values, valuations, discrete valuation rings and higher unit groups. Contents 1. Absolute Values 1 2. Valuations 4 3. Unit Groups 8 References 9 1. Absolute Values Definition 1.1 (Absolute Value). Let K be a field. An absolute value on K is a function : K R 0 satisfying the following three conditions simultaneously: (1) x = 0 x = 0. (2) (Multiplicative) xy = x y. (3) (Triangel Inequality) x + y x + y. [1] uses the term multiplicative valuationstead of absolute value. Example 1.2 (Trivial Absolute Value). Let K be a field. The absolute value { 0, x = 0 : K R, x 1, x 0 is known as the trivial absolute. If F q is a finite field, we claim that the only absolute value defined for F q is the trivial absolute value. This is because the multiplicative group F q is cyclic and is generated by some g F q. Consider g R 0. being multiplicative implies g q 1 = g q 1 = 1 = 1 g = 1. Date: First Version: June 30, 2017. Last Update: June 30, 2017.
Absolute Values and Valuations It follows that every 0 x F q has x = 1. Definition 1.3 (Metric Topology Induced by the Absolute Value). Let K be a field and let be an absolute value defined on K. The distance on K by d : K K R 0, d(x, y) = x y will be metric on K. This makes K into a metric topological space. Definition 1.4 (Equivalence of Absolute Values). Let K be a field and let 1, 2 be two absolute values on K. 1, 2 are said to be equivalent provided they induce the same topology on K. While this definitios seemingly non-straightforward, we do have a more straightforward condition for two absolute values to be equivalent: Proposition 1.5 (Alternative Definitions of Equivalence of Absolute Values). Let 1, 2 be two absolute values on a field K. The following three statements are equivalent: (1) 1, 2 are equivalent. (2) There exists a real number s > 0 such that 1 = s 2. (3) For all x K, x 1 < 1 implies x 2 < 1. Proof. (2) (1) : This is clear. (1) (3) : Suppose x 1 < 1. Then 1 being multiplicative implies x n 1 approaches 0 R as n approaches infinity, so x n approaches 0 K. Now since the topology induced by 1 and 2 are same, it better be case x 2 < 1, otherwise x n 2 would not approach 0 K. (3) (2): Fix an element y K such that y 1 > 1. Let 0 x K, then x 1 = y α 1 for some α R. Let m i Q, Z >0 be a sequence of rational numbers such that m i converges to α. We have m i m i > α x 1 = y α n 1 < y i 1 x y m 1 < 1. i By (3), it follows that x y m i 2 < 1. We then have x y m i 2 < 1 x 2 < y m i 2 x 2 y α 2. 2 > α and m i
Yifan Wu, wuyifan@umich.edu Similarly, let m i and m i Q, Z >0 be a sequence of rational numbers such that m i converges to α. We have m i m i < α x 1 = y α n 1 > y i 1 x y m 1 > 1. i By (3), it follows that x y m i 2 > 1. We then have x y m i 2 > 1 x 2 > y m i 2 x 2 y α 2. The conclusion we draw, from choosing two sequence approaching α, is that Therefore, for any 0 x K, we have x 2 = y α 2. < α log x 1 log x 2 = α log y 1 α log y 2 = log y 1 log y 2. Define s = log y 1 log y 2. This number s will be positive since y 1 > 1 and y 2 > 1. Then for all 0 x K. x 1 = x s 2 Definition 1.6 (Non-Archimedean Absolute Value). Let be an absolute value on a field K. The absolute value is said to be non-archimedeaf the set { n : n = 1 + + 1 n times} R 0 is bounded. The absolute value is said to be Archimedean otherwise. Proposition 1.7 (Equivalent Definitions of Non-Archimedean Absolute Value). Let be an absolute value on a field K. is non-archimedeaf and only if satisfies the strong triangle inequality: for all x, y K. x + y max{ x, y } Proof. ( ) Suppose there exists a N R 0 such that { n : n = 1 + + 1 n times} R 0 is bounded above by N. Let x, y K and assume x y. Then we have x i y n i x n Triangle Inequality n ( ) n n x + y x i y n i i i=0 3 n N x n = nn x n i=0
Absolute Values and Valuations x + y n nn x. Letting n allow us to conclude x + y max{ x, y }. ( ) Suppose strong inequality holds then 1 + 1 max{ 1, 1 } = 1. Use an easy induction to finish. Proposition 1.8. Let be a non-archimedean absolute value on a field K. Then Proof. Suppose x > y. On one hand, On the other hand, x y x + y = max{ x, y }. x + y Non-Archimedean y < x = x + y y Non-Archimedean x. max{ x + y, y }. Then clearly max{ x + y, y } = x + y, and x x + y. so x = x + y. Definition 1.9 (Extension of Non-Archimedean Absolute Value to Function Field). Let be a non-archimedean absolute value on a field K. can be extended to the polynomial ring K(t) via a 0 + a 1 t + + a n t n = max{ a 0, a 1,, a n }. After this, extends multiplicatively to the function field K(t). The axioms of 1.1 can be verified so is an absolute value on K(t). 2. Valuations Definition 2.1 (Valuation). Let K be a field. A function v : K R { } is called a valuation on K if it satisfies the following three conditions simultaneously: (1) v(x) = x = 0. (2) v(xy) = v(x) + v(y). (3) v(x + y) min{v(x), v(y)}. [1] uses the term exponential valuationstead of valuation. 4
Yifan Wu, wuyifan@umich.edu Remark 2.2 (Relation between Non-Archimedean Absolute Values and Valuations). Let be a non-archimedean absolute value on a field K. Set { log x, x 0 v(x) =., x = 0 Then v : K R { } will be a valuation on K. Conversely, given a valuation we can set v : K R { }, : K R 0, x { q v(x), x 0 0, x = 0 for any q R >1. This will be a non-archimedean absolute value on K. Example 2.3 (Trivial Valuation). In 1.2, we have seen trivial absolute value. Translating the trivial absolute value using the prescribed method 2.2, we get trivial valuation: { 0, x 0 v(x) =, x = 0. Definition 2.4 (Equivalence of Valuations). We can again translate (2) of 1.5 into the language of valuations: we say two valuations v 1, v 2 : K R { } are equivalent if v 1 = sv 2 for some real number s > 0. Similar as in 1.8 we have the following result: Proposition 2.5. Let v be a valuation on a field K. Then v(x) v(y) v(x + y) = min{v(x), v(y)}. Proof. One can prove this using the translation 2.2. Alternatively, suppose v(x) < v(y). On one hand, v(x + y) > v(x). On the other hand, v(y) > v(x) = v(x + y y) min{v(x + y), v(y)}. Then min{v(x + y), v(y)} = v(x + y), v(x) v(x + y). So v(x) = v(x + y). Proposition 2.6 (Valuation Ring). Let v : K R { } be a valuation on a field K. Let be defined as in 2.2. (1) The subset O = {x K : v(x) 0} = {x K : x 1} 5,
is a ring with group of units Absolute Values and Valuations O = {x K : v(x) = 0} = {x K : x = 1} and a unique maximal ideal p = {x K : v(x) > 0} = {x K : x < 1}. (2) O is antegral domain with fraction field K. Furthermore, O is valuation ring, meaning for every x K, either x O or x 1 O. (3) O is integrally closed. Proof. (1) This follows from definition. (2) Clear. (3) In general, every valuation ring is integrally closed. The reasos that: suppose x K is integral over O. Let x n + a 1 x n 1 + + a n = 0, a 1,, a n O. Suppose x / O then x 1 O. Multiply both sides of the equation by x (n 1) then x + a 1 + + a n x (n 1) = 0 x O, which is a contradiction. Definition 2.7 (Residue Class Field). The field O/p is called the residue class field of O. Definition 2.8 (Discrete Valuation). Let v be a valuation on a field K. v is said to be discrete valuation provided v(k) admits a smallest positive value s. If this happens it must be the case that v(k ) = sz. A discrete valuation v is said be normalized provided this s = 1. For a non-normalized v, if we consider the equivalent valuation 1 v, then we obtain s a normalized discrete valuation. O, O, p as defined in 2.6 won t change if we move from v to 1v. s Definition 2.9 (Uniformizer). Let v be a normalized discrete valuation on a field K. An element π O with v(π) = 1 is called an uniformizer. [1] uses the term prime element instead of uniformizer. Lemma 2.10. Let v be a normalized discrete valuation on a field K. Let π be a uniformizer. Every element x K has a unique representation with m Z and u O. x = uπ m 6
Yifan Wu, wuyifan@umich.edu Proof. Let m = v(x). Then v(xπ m ) = v(x) v(π m ) = m m = 0. So xπ m O is a unit. If we are handed a discrete valuation v on a field K, then O defined as in 2.6 will be a discrete valuation ring. This is the content of next result: Proposition 2.11 (Discrete Valuation Ring). Let v be a discrete valuation on a field K. We might as well assume it is normalized. Let π be a uniformizer. Then the following holds: (1) O = {x K : v(x) 0} is a principal ideal domain with unique maximal ideal p = {x K : v(x) 1} = (π)o. (2) Every non-zero ideal of O looks like p n = {x K : v(x) n} = (π n ) for some n Z 1. (3) Let O/p be the residue class field. We have asomorphism of O/p vector spaces p n /p n+1 = O/p. Proof. (1) Let a O be adeal. Let a a be such that v(a) = m is smallest among all of a. We know from 2.10 that a = uπ m, m Z 0. Then clearly π m a. So (π m ) = p m a. By our choice of a there won t be any left over. This shows every ideal is principal of form (π m ) = p m. (2) Done in (1). (3) Use the map p n /p n+1 O/p, uπ n + p n+1 u + p. 7
Absolute Values and Valuations 3. Unit Groups In this section, let v : K Z be a normalized discrete valuation ring. Set = q v( ), q > 1 as in 2.2. Then 2.11 gives a chain of ideals O p p 2 p 3 {x : v(x) 0} {x : v(x) 1} {x : v(x) 2} {x : v(x) 3} {x : x 1} {x : x 1 q 1 } {x : x } {x : x 1 } q 2 q 3. {x : x < q} {x : x < 1} {x : x < 1 1 } {x : x < } q q 2 The last row immediately tells us that the collection {O, p, p 2, p 3, } forms a basis of neighborhoods of 0 K. We want a similar basis of neighborhoods of 1 K. This is given by translating the fundamental basis of 0 to 1 to get a fundamental basis {1 + p, 1 + p 2, 1 + p 3, } of neighborhood of 1 K. One way to see O 1 + p is to use 2.5. O 1 + p 1 + p 2 1 + p 3 {x : x = 1} {1 + x : x < 1} {1 + x : x < 1 1 } {1 + x : x < } q q 2. {x : x < q} {y : 1 y < 1} {y : 1 y < 1 1 } {y : 1 y < } q q 2 Definition 3.1 (Higher Unit Groups). Define U (0) = O. Define U (n) = 1 + p n. U (n) is called n th higher unit group, and U (1) is called the group of principal units. Lemma 3.2. (1) U (n) = 1 + p s closed under multiplication. (2) If x U (n) then so is x 1. So higher unit groups are indeed groups. Proof. (1) Let y 1, y 2 U (n) then x 1 = y 1 1, x 2 = y 2 1 p n. y 1 y 2 = (1 + x 1 )(1 + x 2 ) = 1 + x 1 + x 2 + x 1 x 2 1 + p n. 8
(2) We have Yifan Wu, wuyifan@umich.edu U (n) = 1 + p n = {y : 1 y < 1 }. qn 1 If y U (n) then 1 y 1 = y 1 y 1 = y 1 < 1 q n 1 y 1 U (n). Proposition 3.3 (Successive Quotients of Higher Unit Groups). Let n 1. (1) As groups, O /U (n) = (O/p n ). (2) As groups, U (n) /U (n+1) = O/p. Proof. (1) We have a group homomorphism: O (O/p), u u + p n. This homomorphism is surjective because: Suppose x O is such that x + p n (O/p). Then there exists y O such that xy + p n = 1 + p n u O with xy = 1 + uπ n. Then v(xy) = v(x) + v(y) = v(1 + uπ n ) 2.5 = v(1) = 0. However v(x) 0 and v(y) 0. Hence it must be the case that v(x) = 0 and x O. It follows that this homomorphism is surjective. The kernel of this homomorphism is precisely 1 + p n. Now use first isomorphism theorem. (2) We have a group homomorphism U (n) = 1 + p n O/p, 1 + uπ n u + p. It is obvious that this group homomorphism is surjective. The kernel of this map is precisely U (n+1) = 1 + p n+1. References [1] J. Neukirch, Algebraic Number Theory, Springer-Verlag, Berlin Heidelberg, 1999. 1, 4, 6 9