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1 OSTROWSKI S THEOREM GEUNHO GIM Contents. Ostrowski s theorem for Q 2. Ostrowski s theorem for number fields 2 3. Ostrowski s theorem for F T ) 4 References 5. Ostrowski s theorem for Q Definition.. A valuation of a field K is a function : K R such that i) x 0, x = 0 iff x = 0 ii) xy = x y iii) x + y x + y iii) x + y max{ x, y } If satisfies iii), it is called nonarchimedean. Otherwise, it is called archimedean. Definition.2. Two valuations, 2 of a field K are equivalent if = s 2 for some s > 0 They define the same topology on K.) Example.3. Let p be a prime number. For any 0 a Q, we can write a = p m b where c m, b, c Z, bc, p) =. Define a p = and 0 p m p = 0, then p is a nonarchimedean valuation on Q. Note that for different primes p and q, p and q are not equivalent. For z C, define z = z the usual absolute value). Then is an archimedean valuation on Cthus is not equivalent to p for any p). Theorem.4. Ostrowski for Q) Every nontrivial valuation on Q is equivalent to p for a unique prime p or. Proof. Suppose is a nonarchimedean valuation on Q. Let a = {a Z a < }. Note that for any n N, n = + + +, thus a is an ideal of Z. If p = for every prime p, then is trivial. Take a prime p a, and pz a Z shows a = pz. Let s = log p log p. For a = pm b where m, b, c Z and bc, p) =, c a = p m b c = pm = p ms = a s p

2 Therefore is equivalent to p. Suppose is an archimedean valuation on Q. For integers m, n >, write m = a 0 + a n + + a r n r, 0 a i n, n r m We get r log m and a i } + {{ + } n. If n <, then we have m a i elements ai n i n for all m. This contradicts the assumption that is archimedean. n So we have n and r m a i n i + log m ) n n log m i=0 Substitute m k for m, then m k + k log m ) n n k log m m log m n + k log m )) k n By letting k, we get m log m n. We can interchange m and n, so m log m is a constant for m Z. Define s = log m log m ). For any 0 n Z, n = n = e s log n = n s Therefore is equivalent to. Theorem.5. Product Formula) For any a 0 in Q, a p = where the product runs over all the prime numbers p and. p 2. Ostrowski s theorem for number fields Example 2.. Let K be a number field and O K be the ring of integers of K. For a prime ideal p and α K, write α) = p m b c where m Z and b, c are ideals of O K relatively prime to p. Define v p α) = m and α p = Np) vpα). This gives a nonarchimedean valuation on K. For two different primes p and q, the Chinese remainder theorem lets us find β K such that β p and β mod q. So we have β p < and β q =, which shows that p is not equivalent to q. For any embedding σ : K C, define α σ = σα) t where t = for real embeddings and t = 2 for complex embeddings. This gives an archimedean valuation on K. Note that complex conjugate pair σ, σ of embeddings have α σ = α σ. Suppose we have σ = τ 2

3 for two different embeddings σ, τ : K C. Choose θ such that K = Qθ). Consider the isomorphism στ : Qτθ)) Qσθ)). Since τy) τx) = y x τ = y x σ = σy) σx), στ is continuous. Also στ fixes Q, so we can extend it to an isomorphism στ : Rτθ)) Rσθ)). This shows τθ) = σθ), thus τx) = σx) for all x K. Let r resp. r 2 ) be the number of real embeddingsresp. pairs of complex embeddings) of K, then we get r + r 2 inequivalent valuations of this type. Lemma 2.2. Let K be a number field and p be a prime ideal in O K. If α K and v p α) 0, then we can find x, y O K such that α = x y and v py) = 0. Proof. Let α) = a b where a, b are relatively prime ideals in O K. Note that b, p) =. By CRT, we can find an integral ideal c such that c a and c, p) =. Since a b, both ac and bc are integral. Let ac = x) and bc = y). With suitable choice of x, we can write α = x with v y py) = 0. Theorem 2.3. Ostrowski for number fields) Let K be a number field. Every nontrivial valuation on K is equilvalent to p for a unique prime ideal p of O K or σ for a real or a complex embedding σ : K C. Proof. Suppose is a nonarchimedean valuation on K. For α O K, we have for a i Z. Since a i for all i, α n + a n α n + + a α + a 0 = 0 α n = α n = a n α n a 0 max a iα i 0 i n max 0 i n α i This shows that α for all α O K. Define a = {α O K α < }. We can easily see that a is a nonzero ideal of O K if a = 0, then the valuation is trivial). Suppose αβ a for α, β O K. Since αβ <, α < or β <. This proves that a = p for some prime ideal p. Suppose v p α) = 0. By the previous lemma, we can write α = x with x, y O y K and v p y) = 0. Also we have v p x) = v p α) + v p y) = 0. Thus x, y / p and α = x ) =. y α Fix γ p p 2. Since v p = 0 v γ vpα) p γ) = ), we have α γ vpα) =. Therefore we have α = γ vpα) which shows is equivalent to p. The proof of archimedean part comes from the following theorem. 3

4 Theorem 2.4. Let K be a field complete with an archimedean valuation. Then there is an isomorphism σ : K R or C satisfying for a K and some fixed s 0, ]. a = σa) s Proof. Because the valuation is archimedean, char K = 0. Without loss of generality, we can assume that R K and the restriction of to Q is equal to by replacing to /s if necessary. Fix ξ K and define f : C R by fz) = ξ 2 z + z)ξ + zz. We can see that f is continuous and lim fz) =. Thus f has a minimum m. Suppose m > 0. S = f m) z is nonempty and compact, so we can choose z 0 S with maximal absolute value. For 0 < ɛ < m, define gx) = x 2 z 0 + z 0 )x + z 0 z 0 + ɛ R[x] Gx) = gx) ɛ) n ɛ) n = x α i ) = x α i ), α i C Suppose gz ) = 0, then we have z z = z 0 z 0 + ɛ and z > z 0. By assumption, we have fz ) > m. Since Gz ) = 0, we can assume z = α. Gξ) 2 = ξ 2 α i + α i )ξ + α i α i = fα i ) fz )m 2n Gξ) fz 0 ) n + ɛ n = m n + ɛ n From these, we have fz ) ɛ ) n ) 2 m + m as n. This contradicts the fact that fz ) > m. Therefore m = 0 and ξ is a root of quadratic polynomial over R, which shows K = R or C. Theorem 2.5. Product Formula) For any α 0 in K, α v = where the product runs over all the inequivalent valuations on K. v 3. Ostrowski s theorem for F T ) Example 3.. Let F be a field and F T ) be the field of rational functions. Since F T ) = F racf [T ]) and F [T ] is a PID, we can define v π f) for a monic irreducible polynomial πt ) and ft ) F [T ] ft ) = πt ) vπf) gt ) where gt ), ht ) F [T ] are relatively ht ) prime to πt ).) Choose a number c 0, ) and define ft ) π = c deg π ) vπf), then π is a nonarchimedean valuation on F T ). Also define ft ) = c deg f, then is a nonarchimedian valuation. We can see that these are not equivalent to each other. 4

5 Example 3.2. Choose a real transcendental number γ. Since γ is transcendental, the evaluation map e γ : QT ) R is injective. Define γ on QT ) so that ft ) γ = fγ), then we can check that this gives an archimedean valuation on QT ). Theorem 3.3. Ostrowski for F T )) Every nontrivial valuation on F T ) which is trivial on F is equivalent to π for some monic irreducible π or. Proof. Let ft ) = a n T n + + a T + a 0 with a n 0. Let be a nontrivial valuation on F T ). Note that has to be nonarchimedean since it is trivial on F. Firstly, suppose T >. Since a n = and a i T i a i T i < T n for 0 i n, we have ft ) = a n T n = T deg f, which is equivalent to. Now suppose T. For any ft ) = a n T n + + a 0 F [T ], we have ft ) max a i T i. Since is nontrivial on F T ) and trivial on F, the set a = {gt ) F [T ] gt ) < } contains at least one nonconstant polynomial. You can see that a is a prime ideal of F [T ] by similar argument to the previous theorems, thus is generated by some monic irreducible polynomial πt ) F [T ]. This shows that for any gt ) relatively prime to πt ) has gt ) =. For general ft ) F [T ], we have ft ) = πt ) vπf) ht ) for some ht ) relatively prime to πt ) = πt ) vπf) which is equivalent to π. Note 3.4. A finite field F can only have trivial valuation. So this theorem works for any F T ) with finite field F. Theorem 3.5. Product Formula) For any ft ) 0 in F T ), ft ) π = where the product runs over monic irreducible polynomials in F [T ] and. π References [] K. Conrad, Ostrowski for FT), at kconrad/blurbs/. [2] K. Conrad, Ostrowski for number fields, at kconrad/ blurbs/. [3] J. Neukirch, Algebraic Number Theory, Springer, 999. [4] P. Ribenboim, The Theory of Classical Valuations, Springer,

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