CHAPTER 3. The Norm Map

Transcription

1 CHAPTER 3 The Norm Map In this chapter we study the norm map acting on Henselian discrete valuation fields. Section 1 studies the behaviour of the norm map on the factor filtration introduced in section 5 Chapter I for cyclic extensions of prime degree. Section 2 demonstrates that almost all cyclic extensions of degree p can be described by explicit equations of Artin Schreier type. Section 3 associates to the norm map a real function called the Hasse Herbrand function; properties of this function and applications to ramification groups are studied in sections 3 and 4. The long section 5 presents a relatively recent theory of a class of infinite Galois extensions of local fields: arithmetically profinite extensions and their fields of norms. The latter establishes a relation between the fields of characteristic 0 and characteristic p. We will work with complete discrete valuation fields leaving the Henselian case to Exercises. 1. Cyclic Extensions of Prime Degree In this section we describe the norm map on the factor filtration of the multiplicative group in a cyclic extension of prime degree. The most difficult and interesting case is of totally ramified p-extensions which is treated in subsections 1.4) and 1.5). Using these results we will be able to simplify expositions of theories presented in several other sections of this book. et be a complete discrete valuation field and its Galois extension of prime degree n. Then there are four possible cases: / is unramified; / is tamely and totally ramified; / is totally ramified of degree p = char ) > 0; / is inseparable of degree p = char ) > 0. Since the fourth case is outside the subject of this book, we restrict our attention to the first three cases still, see Exercise 2). The following results are classical and essentially due to H. Hasse. 67

2 68 III. The Norm Map 1.1). emma. et / be a separable extension of prime degree n, γ M. Then N / 1 + γ) = 1 + N / γ) + Tr / γ) + Tr / δ) with some δ O such that v δ) 2v γ) N / and Tr / are the norm and the trace maps, respectively). Proof. Recall that for distinct embeddings σ i of over into the algebraic closure of, 1 i n, one has see [a1, Ch. VIII]) n n N / α = σ i α), Tr / α = σ i α), α. Hence N / 1 + γ) = i=1 n 1 + σ i γ)) i=1 = 1 + n i=1 i=1 n ) σ i γ) + σ i i=1 1 j n or δ = 1 j n γσ jγ) + we get v δ) 2v γ). ) γσ j γ) + + n σ i γ). i=1 Our nearest purpose is to describe the action of the norm map N / to the filtration discussed in section 5 Ch. I. with respect 1.2). Proposition. et / be an unramified extension of degree n. Then a prime element π in is a prime element in. et U i, = 1 + π i O, U i, = 1 + π i O and let λ i,, λ i, i 0), be identical to those of Proposition 5.4) Ch. I, for π = π. Then the following diagrams are commutative: v Z N / n N / U λ 0, N / U i, N / λ i, Tr / v Z λ 0, U U i, λ i, Proof. The first commutativity follows from 2.3) Ch. II. Proposition 3.3) Ch. II implies that N / α) = N / α) for α O, i.e., the second diagram is commutative. The preceding emma shows that N / 1 + θπ i ) = 1 + Tr / θ)π i + N / θ)π ni + Tr / δ) with v δ) 2i and, consequently, v Tr / δ) 2i. Thus, we get N / 1 + θπ i ) 1 + Tr / θ)π i mod π i+1 and the commutativity of the third diagram.

3 1. Cyclic Extensions of Prime Degree 69 Corollary. In the case under consideration N / U 1, = U 1,. 1.3). Proposition. et / be a totally and tamely ramified cyclic extension of degree n. Then for some prime element π in, the element π = π n is prime in Proposition 3.5) Ch. II) and =. et U i, = 1 + π i O, U i, = 1 + π i O, and let λ i,, λ i, be identical to those of Proposition 5.4) Ch. I, for π = π and π = π. Then the following diagrams N / v Z id N / U λ 0, n v Z λ 0, U U ni, N / λ ni, = n U i, λ i, are commutative, where id is the identity map, n takes an element to its n th power, n is the multiplication by n, i 1. Moreover, N / U i, = N / U i+1, if n i. Proof. Since π n = π and / is Galois, then Gal/ ) is cyclic of order n and σπ ) = ζπ for a generator σ of Gal/ ), where ζ is a primitive n th root of unity, ζ. The first diagram is commutative in view of Theorem 2.5) Ch. II. Proposition 4.1) Ch. II shows that σα) = α for σ Gal/ ), α O, and we get the commutativity of the second diagram. If j = ni, then 1 + θπ j for θ O, and N / 1 + θπ j ) = 1 + θπi ) n 1 + nθπ i mod π i+1 by Proposition 5.4) Ch. I. Applying Corollary 5.5) Ch. I, we deduce U i, = U n i, = N / U ni,. inally, X n 1 = n 1 j=0 X ζj ), therefore for n i and for θ O one has Thus N / U i, = N / U i+1,. n 1 N / 1 + θπ) i = 1 + ζ j θπ) i = 1 θ) n π i. j=0 Corollary. In the case under consideration N / U 1, = U 1,. If is algebraically closed then N / =.

4 70 III. The Norm Map 1.4). Now we treat the most complicated case when / is a totally ramified Galois extension of degree p = char ) > 0. Then Corollary 2 of 2.9) Ch. II shows that O = O [π ], = π ) for a prime element π in, and =. et σ be a generator of Gal/ ), then σπ )/π U. One can write σπ )/π = θε with θ U, ε 1 + M. Then and σ 2 π )/π = σθε) θε = θ 2 ε σε), 1 = σ p π )/π = θ p ε σε) σ p 1 ε). This shows that θ p 1 + M and θ 1 + M, because raising to the p th power is an injective homomorphism of. Thus, we obtain σπ )/π 1 + M. Put σπ ) π = 1 + ηπ s with η U, s = s ) 1. ) Note that s does not depend on the choice of the prime element π and of the generator σ of G = Gal/ ). Indeed, we have σ i π ) 1 + iηπ s mod π s+1 π and σρ) ρ 1 mod π s+1 for an element ρ U. We also deduce that σα) α U s, for every element α. This means that G = G s, G s+1 = {1} see 4.3) Ch. II). emma. et fx) = X p + a p 1 X p a 0 be the irreducible polynomial of π over. Then ) π j { 0 if 0 j p 2, Tr / f = π ) 1 if j = p 1. Proof. get Since σ i π ) for 0 i p 1 are all the roots of the polynomial fx), we p 1 1 fx) = 1 f σ i π ) ) X σ i π ) ). i=0

5 1. Cyclic Extensions of Prime Degree 71 Putting Y = X 1 and performing the calculations in the field Y )), we consequently deduce fx) = Y p 1 + a p 1 Y + + a 0 Y p ), 1 fx) = 1 X σ i π ) = Y p 1 + a p 1 Y + + a 0 Y p Y p mod Y p+1, Y 1 σ i π )Y = j 0 because 1/1 Y ) = i 0 Y i in Y )) ). Hence or as desired. Tr / π j f π ) p 1 j 0 i=0 ) = σ i π j j+1 )Y σ i π j )Y j+1 f σ i π ) ) Y p mod Y p+1, p 1 i=0 σ i π j ) { 0 if 0 j p 2, f σ i π ) ) = 1 if j = p 1, Proposition. et [a] denote the maximal integer a. or an integer i 0 put ji) = s [ i 1 s)/p ]. Then Tr / πo i ) = π ji) O. Proof. One has f π ) = p 1 i=1 π σ i π ) ) and σ i π )/π 1 + iηπ s mod π s+1. Then f π ) = p 1)! η) p 1 π p 1)s+1) ε with some ε 1 + M p 1)s+1)+1. Since =, for a prime element π in one has the representation π = π p ε with ε U. The previous emma implies ) { 0 if 0 j < p 1, Tr / π j+s+1 ε j+s+1 = π s+1 if j = p 1 for ε j+s+1 = ε ) s+1 / p 1)! η) p 1 ε ). Taking into consideration the evident equality Tr / π i α) = πi Tr / α) we can choose the units ε j+s+1, for every integer j, such that Tr / π j+s+1 ε j+s+1 ) = 0 if p j +1) and = π s+j+1)/p if p j +1). Thus, since the O -module π i O is generated by π j ε j, j i, we conclude that Tr / π i O ) = π ji) O.

6 72 III. The Norm Map 1.5). Proposition. et / be a totally ramified Galois extension of degree p = char ) > 0. et π be a prime element in. Then π = N / π is a prime element in. et U i, = 1 + π i O, U i, = 1 + π i O and let λ i,, λ i, be identical to those in Proposition 5.4) Ch. I, for π = π and π = π. Then the following diagrams are commutative: N / v Z id N / U λ 0, p v Z λ 0, U U i, N / λ i, = p if 1 i < s, U i, λ i, U s, N / λ s, = θ θ p η p 1 θ U s, λ s, U s+pi, λ s+pi, = N / η p 1 ) if i > 0. U s+i, λ s+i, Moreover, N / U s+i, ) = N / U s+i+1, ) for i > 0, p i. Proof. The commutativity of the first and the second diagrams can be verified similarly to the proof of Proposition 1.3). In order to look at the remaining diagrams, put ε = 1 + θπ i with θ U. Then, by emma 1.1), we get N / ε = 1 + N / θ)π i + Tr / θπ i ) + Tr / θδ) with v δ) 2i. The previous Proposition implies that v Tr/ π) i ) [ ] i 1 s s + 1 +, v Tr/ δ) ) [ ] 2i 1 s s p p and for i < s v Tr/ π i ) ) i + 1, v Tr/ δ) ) i + 1.

7 1. Cyclic Extensions of Prime Degree 73 Therefore, the third diagram is commutative. urther, using ) of 1.4), one can write ) σπ ) 1 = N / 1 + N / η)π s + Tr / ηπ π ) s mod π s+1. We deduce that Tr / ηπ s ) N / η)π s mod π in view of U U U 1,, we conclude that mod πs+1 N / 1 + θηπ) s 1 η p π s θ p θ) π ps+1 θo. Since N / η) η p for θ O. This implies the commutativity of the fourth putting θ O ) and the fifth when θ π i O ) diagrams. inally, if p i, θ O, then σ1 + θπ i ) 1 + θπ i 1 + iθηπ i+s mod π i+s+1. This means that N / 1 + iθηπ i+s) N / U s+i+1, and N / U s+i, ) = N / U s+i+1, ). Remark. Compare the behaviour of the norm map with the behaviour of raising to the p th power in Proposition 5.7) in Ch. I. Corollary. U s+1, = N / U s+1,. If is algebraically closed then N / =. Proof. It follows immediately from the last diagram of the Proposition, since the multiplication by η) p 1 is an isomorphism of the additive group. Exercises. 1. a) et be a Henselian discrete valuation field, and / a cyclic extension of prime degree. Show that U i, N / U for sufficiently large i. b) Show that all assertions of this section hold for a Henselian discrete valuation field. 2. et / be a Galois extension of degree p = char ), and let / be an inseparable extension of degree p. et θ U be such that = θ). et σ be a generator of Gal/ ). a) Show that v σθ) θ) > 0. Put σθ) = 1 + ηπ s θ for a prime element π in and some η U, s 1. b) Show that Tr / π i O ) = π ji) O with ji) = p 1)s + i. c) Show that N / 1 + ηπ i ) 1 + N / η)π pi d) Show that mod π ps+1 N / 1 + cθ i ηπ s ) mod πpi+1 if i < s. { 1 + c p π ps N / θ i η), 0 < i p 1, 1 + c p c)π ps N / η), i = 0,

8 74 III. The Norm Map where c O. e) Show that U ps+1, N / U s+1,. 3. et / be a Galois extension of degree p = char ) > 0, that is not unramified. Show that ) γ v = max {v Tr / γ) α) : Tr / α) = 1}, α O where γ = α 1 σα) 1 for a generator σ of Gal/ ) and an element α O, such that p v α) when e ) = p and ᾱ when e ) is equal to Artin Schreier Extensions A theorem of E. Artin and O. Schreier asserts that every cyclic extension of degree p over a field K of characteristic p is generated by a root of the polynomial X p X α, α K see Exercise 6 section 5 Ch. V or [a1, Ch. VIII]). In this section we show in Proposition 2.4) and 2.5), following R. MacKenzie and G. Whaples [MW]), how to extend this result to complete discrete valuation fields of characteristic 0. An alternative proof of the main results of this section can be obtained by using formal groups, see for example [VZ]. 2.1). irst we treat the case of unramified extensions. The polynomial X p X is denoted by X) see 6.3) Ch. I). emma. et / be an unramified Galois extension of degree p = char ). Then = λ), where λ is a root of the polynomial X p X α = 0 for some α U with α / ). Proof. et = θ), where θ is a root of the polynomial X p X η = 0 for some η / ). Then the polynomial X p X α = 0, with α U, such that α = η, has a root λ in, by Hensel emma 1.2) Ch. II. Thus, = λ). 2.2). Now we study the case of totally ramified extensions. et / be a totally ramified Galois extension of degree p = char ). et σ be a generator of Gal/ ), π a prime element in and s = v π 1 σπ ) 1). emma. or β there exists an element b such that v σβ β) = v β b)+s. Proof. Ch. II). Then et β = a 0 + a 1 π + + a p 1 π p 1 with a i see Proposition 3.6) σβ) β = a 1 π γ + + a p 1 π p γ) p 1 1 ),

9 2. Artin Schreier Extensions 75 where γ = π 1 σπ ) 1. Since v γ) = s > 0, we get 1 + γ) i 1 iγ mod π s+1 for i 0. Hence, v a i π i 1 + γ) i 1 )) are distinct for 1 i p 1. Put b = a 0. Then v σβ) β) = v β b)γ) = v β b) + s, as desired. 2.3). Proposition. et be a complete discrete valuation field with residue field of characteristic p > 0. et be a totally ramified Galois extension of degree p. If char ) = p then p s. If char ) = 0, then s pe/p 1), where e = e ) is the absolute index of ramification of. In this case, if p s, then a primitive p th root of unity belongs to, and s = pe/p 1), = p α) with some α, α / U p. Proof. irst assume that char ) = p and s = pi. Then 1 + θπ i )p = 1 + θ p π pi for θ U. One can take π = N / π for a prime element π in. Then it follows from 1.4) that π π p mod πp+1. Since N / U pi+1, U pi+1,, we get the congruence 1 + θ p π pi N / 1 + θπ pi ) mod πpi+1, which contradicts the fourth diagram of Proposition 1.5). Hence, p s. Assume now that char ) = 0 and s > pe/p 1). et ε = 1 + θπ s U s, with θ U. Corollary 2 of 5.8) Ch. I shows that ε = ε p 1 for some ε 1 = 1 + θ 1 π s e U with θ 1 U. Then N / U ps e), U s+1,, but ps e) s + 1, which is impossible because of Corollary 1.5). Hence, s pe/p 1). By the same reasons as in the case of char ) = p, it is easy to verify that if s = pi < pe/p 1), then 1+θ p π pi N / 1+θπ pi ) mod πpi+1, which is impossible. Therefore, in this case we get s = pe/p 1). One can write σπ )π 1 1+θπe/p 1) mod π pe/p 1)+1. Then, acting by N /, we get θπ e/p 1) ) p mod π pe/p 1)+1. But U pe/p 1)+1, U p e/p 1)+1, see Corollary 2 of 5.8) Ch. I), that permits us to find an element ζ 1 + θπ e/p 1) mod π e/p 1)+1, such that ζ p = 1; ζ is a primitive p th root of unity in, hence = p α) for some α, by the Kummer theory. Writing α = π a ε 1 with ε 1 U and assuming p a, we can replace α with ε 1. Since = we obtain ε 1 p otherwise / would not be totally ramified) and ε 1 ε p 2 mod π for some ε 2 U. Replacing ε 1 with ε 3 = ε 1 ε p 2, we get ε 3 U 1,, = p ε 3 ). Note that σ1 + ρπ i ) 1 + ρπ i 1 + ρiηπ i+pe/p 1) mod π 1+i+pe/p 1) for ρ U. Hence ε 1 3 σε 3) 1 mod π 1+pe/p 1), but ε 1 3 σε 3) is a primitive p th root of unity. This contradiction proves that α / U p. 2.4). Proposition. et be a complete discrete valuation field with residue field of characteristic p > 0. et be a Galois totally ramified extension of degree p. et

10 76 III. The Norm Map s pe/p 1) if char ) = 0, e = e ). Then = λ), where λ is a root of some polynomial X p X α with α, v α) = s. Proof. The previous Proposition shows that p s. irst consider the case of char ) = p. Then, by the Artin Schreier theory, = λ), where λ is a root of a suitable polynomial X p X α with α. et σ be a generator of Gal/ ). Then σλ) λ) p = σλ λ. Since λ /, we get σλ) λ = a with an integer a, p a. Then λ 1 σλ) = 1 + aλ 1, and hence Proposition 1.5) implies 1 + aλ 1 U s,. This shows v λ) s and v α) s. Put t = v α). If t = pt, then we can write λ π t θ mod πt+1 with θ U and a prime element π in. Therefore, α π pt θp π pt θp mod π pt+1, where π = N / π π p mod πp+1 is a prime element in. Replacing λ by λ = λ π t θ and α by α = α π pt θp + π t θ, we get λ p λ = α and = λ ), v α ) > v α). Proceeding in this way we can assume p t because v α ) s. Then it follows from 1.4) that v λ 1 σλ) 1) = s and v α) = s. Now we consider the case of char ) = 0. irst we will show that there is an element λ 1, such that v λ 1 ) = s and v σλ 1 ) λ 1 1) > 0. Indeed, put β = π s ρs 1 with ρ U. Then σβ) β = π s ρs ηπ s ) s 1 ) ρη mod π. Hence, if we choose ρ = η 1, then v σβ) β 1) > 0. Put λ 1 = β b. Since s < pe/p 1) = e)/p 1), we get v σλ p 1 ) λp 1 1) > 0 and v σ λ 1 ) λ 1 )) > 0. Second we will construct a sequence {λ n } of elements in satisfying the conditions v λ n ) = s, v λ n+1 λ n ) v λ n λ n 1 ) + 1, v σ λ n+1 ) λ n+1 )) v σ λn ) λ n ) ) + 1. Then for λ = lim λ n we obtain σ λ) = λ), or in other words λ p λ = α and v α) = s. Put λ 0 = 0. et δ n = σ λ n ) λ n ). Then v δ n ) > 0. If δ n = 0, then put λ m = λ n for m > n. Otherwise, by emma 2.2), there exists an element c n such that v σ λ n ) λ n )) = v λ n ) c n ) + s. Put µ n = λ n ) c n, λ n+1 = λ n +µ n. Then σµ n = µ n +δ n, v σλ n+1 ) λ n+1 1) > 0 and v µ n ) > s, v λ n+1 ) = s. So v λ n+1 λ n ) = v µ n ) = s + v σ λ n ) λ n )) s v σ λ n 1 ) λn 1 ) ) = v λ n λ n 1 ) + 1

11 2. Artin Schreier Extensions 77 for n > 1, and v λ 2 λ 1 ) = s + v σ λ 1 ) λ 1 )) v λ 1 λ 0 ) + 1. urthermore, p ) p σ µ n ) µ n ) = µ n + δ n ) µ n ) = δ n + µ n p i δ i i n. We also get Moreover, and i=1 v σ µ n ) µ n ) + δ n ) v δ n ) + 1. σ λ n+1 ) λ n+1 ) = σ λ n ) λ n ) σλ p i n p 1 + σ µ n ) µ n ) + µ i n) λ p i n µ i n = λ p i n µ i n i=1 ) p σλ p i n i ε p i n 1 + δ n µ 1 n ) i 1 ), µ i n) λ p i n µ i ) n where λ 1 n σλ n = ε n U s,, v δ n µ n 1 ) = v δ n ) + s v δ n ) = s. Hence, for 1 i p 1 we get v σλ p i n µ i n) λ p i n µ i n) p 1)s + v δ n ) pe + v δ n ) + 1. As a result we obtain the following inequality which completes the proof. v σ λn+1 ) λ n+1 ) ) v δ n ) + 1, 2.5). The assertions converse to Propositions 2.1) and 2.4) can be formulated as follows. Proposition. et be a complete discrete valuation field with a residue field of characteristic p > 0. Then every polynomial X p X α with α, v α) > pe/p 1) if char ) = 0 and e = e ), either splits completely or has a root λ which generates a cyclic extension = λ) over of degree p. In the last case v σλ) λ 1) > 0 for some generator σ of Gal/ ). If α U, α / ), then / is unramified; if α M, then λ ; if α / O and p v α), then / is totally ramified with s = v α). Proof. et α M, fx) = X p X α. Then f0) M, f 0) / M, and, by Hensel emma 1.2) Ch. II, for every integer a there is λ M with fλ) = 0, λ a M. This means that fx) splits completely in. If α U, α / ), then Proposition 3.2) Ch. II shows that λ)/ is an unramified extension and Proposition 3.3) Ch. II shows that λ)/ is Galois of degree p. The generator σ Gal/ ), for which σᾱ = α + 1, is the required one.

12 78 III. The Norm Map If α / O, then let λ be a root of the polynomial X p X α in alg and = λ). Put ) ) p p gy ) = λ + Y ) p λ + Y ) α = Y p + λy p λ p 1 Y Y. 1 p 1 If char ) = p, then / is evidently cyclic of degree p when α / ). If char ) = 0, then v p i) ) λ i > e )e ei/p 1)) 0 for i p 1 and gy ) = Y p Y over. Hence by Hensel emma gy ) splits completely in. Therefore, / is cyclic of degree p if fx) does not split over. et σ be a generator of Gal/ ), such that σλ) λ is a root of gy ) and is congruent to 1 mod π. Then v σλ) λ 1) > 0. If p v α), then the equality pv λ) = v α) implies e ) = p, and / is totally ramified. It follows from the definition of s in 1.4) that s = v σλ) λ 1 1), and consequently s = v σλ) λ) v λ) = v λ) = v α). Corollary. et λ be a root of the polynomial X p X + θ p α with θ U, v α) = s > pe/p 1), p s. et = λ). Then α N / and 1 + θ p O ) α 1 + π s+1 O N /, where O ) = { β) : β O }. Proof. The preceding Proposition shows that / is a totally ramified extension of degree p and that v σπ )π 1 1) = s for a generator σ of Gal/ ) and a prime element π in. Put fx) = X p X + θ p α. Then we get N / λ) = f0) = θ p α and α = N / λθ 1 ). or β O put Then gy ) = fβ Y ) = β Y ) p β Y ) + θ p α. N / β λ) = g0) = β) + θ p α. Therefore, 1 + β) θ p α 1 N /. It remains to use Corollary 1.5). 2.6). Remark. Another description of totally ramified extensions of degree p can be found in [Am]. or a treatment of Artin Schreier extensions by using ubin Tate formal groups and a generalization to n-dimensional local fields see [VZ]. Exercises. 1. et / be a Galois extension of degree p = char ), and let / be an inseparable extension of degree p. et θ, σ, s be as in Exercise 2 section 1. et char ) = 0 and e = e ) the absolute index of ramification of. a) Prove an analog of emma 2.2) with θ instead of π ). b) Show that s e/p 1). c) Show that s < e/p 1) if and only if there exists an element λ 1, such that v σλ 1 ) λ 1 1) > 0, v λ 1 ) = ps.

13 3. The Hasse Herbrand unction 79 d) Show that if s < e/p 1), then = λ), where the element λ is a root of the polynomial X p X α with α, v α) = ps, v σλ) λ 1) > 0. e) Maintaining the conditions in d) show that α = β 1 β p 2 v β 2 ) = s. with β 1 U, β 1 / p, f) Show that if = λ), where λ p λ = α and α is as in e), then / is Galois of degree p and / is inseparable of degree p. 2. R. MacKenzie and G. Whaples [MW]) a) et = Q, and let be the unique cyclic subextension of prime degree p in ζ p 2)/ ζ p 2 is a primitive p 2 th root of unity). Show that the equation X p X α = 0 for α can have at most three real roots. However, for p 3 any defining equation of over splits into real linear factors in C. Hence, / is not generated by a root of any Artin Schreier equation for p > 3. b) et = Q, and let be the splitting field of the polynomial X p X 1. Show that / is not a cyclic extension when p et = γ), γ p γ = α, be a cyclic extension of degree p over. Assume that itself is a cyclic extension of K with a generator σ. Describe what condition should satisfy σα so that /K is a Galois abelian) extension? 4. V.A. Abrashkin [Ab1]) et be a complete discrete valuation field of characteristic 0 with residue field of characteristic p. et p n for some integer n 1. et λ be a root of the polynomial X pn X α with α, v α) > p n e )/p n 1). Then the extension λ)/ is said to be elementary. a) Show that λ)/ is Galois. b) Show that if p v α), v α) < 0, then λ)/ is a totally ramified Galois extension of degree p n, and if G = Gal λ)/ ), then G = G 0 = = G s, G s+1 = = {1} for the ramification groups of G, where s = v α). c) Show that if α 1 α 2 M, then λ 1 ) = λ 2 ). d) Show that if α 3 α 1 α 2 M, then λ 3 ) is contained in the compositum of λ 1 ) and λ 2 ). e) Show that if is algebraically closed, then M can be replaced with O in c), d). or additional properties of elementary extensions see [Ab]. The theory of such extensions is used to show that there are no abelian schemes over Z. 5. Generalize the results of this section to Henselian discrete valuation fields. 3. The Hasse Herbrand unction In this section we associate to a finite separable extension / a certain real function h / which partially describes the behaviour of the norm map from arithmetical point of view. In subsections 3.1), 3.2) we study the case of Galois extensions and in subsection 3.3) the case of separable extensions. In 3.4) we derive first applications. Then we relate the function h / which was originally introduced in a different way by H. Hasse and J. Herbrand to properties of ramification subgroups and prove in section 3.5) a theorem of J. Herbrand on the behaviour of ramification groups in extensions; further properties of ramification subgroups are studied in 3.6) and 3.7).

14 80 III. The Norm Map We maintain the hypothesis of the preceding sections concerning, and assume in addition that all residue field extensions are separable. 3.1). Proposition. et the residue field be infinite. et / be a finite Galois extension, N = N /. Then there exists a unique function such that h0) = 0 and h = h / : N N NU hi), U i,, NU hi), U i+1,, NU hi)+1, U i+1,. Proof. The uniqueness of h follows immediately. Indeed, for j > hi) NU j, U i+1,, hence if h is another function with the required properties, then hi) hi), hi) hi), i.e., h = h. As for the existence of h, we first consider the case of an unramified extension /. Then Proposition 1.2) shows that in this case hi) = i because N / ) 1 and Tr / = ). The next case to consider is a totally ramified cyclic extension / of prime degree. In this case Proposition 1.3) and Proposition 1.5) describe the behavior of N /. By means of the homomorphisms λ i,, the map N / is determined by some nonzero polynomials over. The image of under the action of such a polynomial is not zero since is infinite. Hence, we obtain hi) = : i, if / is totally tamely ramified, and { i, i s, hi) = s1 p) + pi, i s, if / is totally ramified of degree p = char ) > 0. Now we consider the general case. Note that if we have the functions h /M and h M/ for the Galois extensions /M, M/, then for the extension / one can put h / = h /M h M/. Indeed, N / U h/ i), N M/ U hm/ i),m U i,. urthermore, the behavior of N / is determined by some nonzero polynomials the composition of the polynomials for N /M and N M/, the existence of which can be assumed by induction). Hence Since N / U h/ i), U i+1,. N / U h/ i)+1, N M/ U hm/ i)+1,m U i+1,m, we deduce that h = h / is the desired function.

15 3. The Hasse Herbrand unction 81 In the general case we put h / = h /0 for 0 = ur and determine h /0 by induction using Corollary 3 of 4.4) Ch. II, which shows that / 0 is solvable. 3.2). To treat the case of finite residue fields we need emma. et / be a finite separable totally ramified extension. Then for an element α we get N / α) = N ur / urα) where ur is the completion of ur, ur = ur. Proof. et = π ) with a prime element π in, and let α. et n 1 απ i = c ij π j with c ij, 0 i n 1, n = :. j=0 Then N / α) = detc ij ) see [a1, Ch. VIII]). Since ur = ur π ) and we get ur : ur = e ur ur ) = e ur ) = e ) = :, N ur / urα) = detc ij) = N / α). inally, let E/ ur be a finite totally ramified Galois extension with E ur. et G = GalE/ ur ), H = GalE/ ur ), and let G be the disjoint union of σ i H with σ i G, 1 i ur : ur. Then N ur / urα) = σ i α) = N ur / urα), because G and H are isomorphic to GalÊ/ ur ) and GalÊ/ ur ) by 4) in Theorem 2.8) Ch. II. This emma shows that for a finite totally ramified Galois extension / the functions h / and h ur / ur coincide. Now, if / is a finite Galois extension, we put h / = h /0 = h ur / ur. In particular, if is finite we put h / = the residue field of h ur is infinite ur / ur as the separable closure of a finite field). It is useful to extend this function to real numbers. or unramified extension, or tamely totally ramified extension of prime degree, or totally ramified extension of degree p = char ) > 0 put { x, x s, h / x) = x, h / x) = : x, h / x) = s1 p) + px, x s

16 82 III. The Norm Map for real x 0 respectively. Using the solvability of / 0 Corollary 3 of 4.4) Ch. II) and the equality h / = h /M h M/ define now h / x) as the composite of the functions for a tower of cyclic subextensions in / 0. Proposition. Thus defined function h / : [0, + ) [0, + ) is independent on the choice of a tower of subfields. The function h / is called the Hasse Herbrand function of /. It is piecewise linear, continuous and increasing. Proof. It suffices to show that if M 1 /M, M 2 /M are cyclic extensions of prime degree, then *) h E/M1 h M1 /M = h E/M2 h M2 /M where E = M 1 M 2. Note that each of h M1 /M x), h M2 /M x) has at most one point at which its derivate is not continuous. Therefore there are at most two points at which the function of the left resp. right) hand side of ) has discontinuous derivative. By looking at graphs of the functions it is obvious that at such points the derivative strictly increases and there is at most one such noninteger point for at most one of the composed functions of the left hand side and the right hand side of ). At this point if it exists) the derivative jumps from p to p 2. rom the uniqueness in the preceding Proposition we deduce that the left and right hand sides of ) are equal at all nonnegative integers. Thus, elementary calculus shows that the left and right hand sides of ) are equal at all nonnegative real numbers. 3.3). et the residue field of be perfect. or a finite separable extension / put h / = h 1 E/ h E/, where E/ is a finite Galois extension with E. Then h / is well defined, since if E / is a Galois extension with E and E = E E, then h 1 E / h E / = h E /E h 1 E /) he /E h ) E / = h 1 E / h E / and, similarly, h 1 E / h E / = h 1 E/ h E/. We can easily deduce from this that the equality holds for separable extensions. h / = h /M h M/ Proposition. et / be a finite separable extension, and let be perfect. Then h / N) N and the left and right derivatives of h / at any point are positive integers. )

17 3. The Hasse Herbrand unction 83 Proof. et E/ be a finite Galois extension with E. Then from emma 3.2) we get h / = h 1 E/ h E/ = h 1 h = Ê ur / ur Ê ur / ur h ur / ur. Put G = GalÊur / ur ), H = GalÊur / ur ). Since G is a solvable group, there exists a chain of normal subgroups G G 1) G m) = {1}, such that G i) /G i+1) is a cyclic group of prime order. Then we obtain the chain of subgroups G G 1) H... G m) H = H, for which G i+1) H is of prime index or index 1 in G i) H. This shows the existence of a tower of fields ur M 1 M n 1 M n = ur, such that M i+1 /M i is a separable extension of prime degree. Therefore, it suffices to prove the statements of the Proposition for such an extension. If M i+1 /M i is a totally tamely ramified extension of degree l, then π = π1 l is a prime element in M i for some prime element π 1 in M i+1. Since l is relatively prime with char ), we obtain, using the Henselian property of M i and the equality M i = sep, that a primitive l th root of unity belongs to M i. This means that M i+1 /M i is a Galois extension and h Mi+1 /M i x) = lx. If M i+1 /M i is an extension of degree p = char ) > 0, then let K/M i be the smallest Galois extension, for which K M i+1. et K 1 be the maximal tamely ramified extension of M i in K; then l = ek 1 M i ) = ek M i+1 ) is relatively prime to p. Choose prime elements π and π 1 in M i+1 and K such that π = π l 1. et fx) M i [X] be the monic irreducible polynomial of π over M i. Then p 1 f π) = π σ i π) ) = π l 1 σ i π1 l )), i=1 where σ GalK/M i ) is such that σ Mi+1 is a generator of GalM i+1 /M i ). et s be defined for K/K 1 as in 1.4). Then v K π l 1 σ i π1 l )) = l + s for 1 i p 1, and p 1)l + s) = v K f π) ) is divisible by l. We deduce that l p 1)s and p 1 h Mi+1 /M i x) = 1 { x, x sl 1 l h, K/K 1 lx) = s1 p)l 1 + px, x sl 1. These considerations complete the proof. i=1

18 84 III. The Norm Map Corollary. The function h / is piecewise linear, continuous and increasing. Remark. h / possesses the properties of Proposition 3.1) in the general case of a separable extension / see Exercise 1). 3.4). The following assertion clarifies relation between the Hasse Herbrand function and the norm map. Proposition. et / be a finite separable extension. Then for ε O and if v α β) > 0 for α, β, then h / v N/ ε) 1 )) v ε 1), h / v N/ α) N / β) )) v α β). Proof. irst we show that the second inequality is a consequence of the first one. Indeed, if v β) v α β), then v α) v α β), and applying Theorem 2.5) Ch. II we get v N/ α) N / β) ) v α β). Since h / x) x, we obtain the second inequality. If v β) < v α β), then v 1 αβ 1 ) v α β) v β) > 0, and putting ε = αβ 1, we deduce h / v N/ α) N / β) )) v β) + v 1 αβ 1 ) v α β). We now verify the first inequality of the Proposition. By the proof of the previous Proposition, we may assume that / is totally ramified and is algebraically closed. It is easy to show that if the first inequality holds for /M and M/, then it holds for /. The arguments from the proof of the previous Proposition imply now that it suffices to verify the first inequality for a separable extension / of prime degree. If / is tamely ramified, then / is Galois, and the inequality follows from Proposition 1.3). If : = p = char ) > 0, then Proposition 1.5) implies the required inequality for the Galois case. In general, assume that E/ is the minimal Galois extension such that E, and let E 1 is the maximal tamely ramified subextension of in E. et l = E : = E 1 :. Then N / U i, ) = N E/ U li,e ) N E1 / U j,e1 ) with j h 1 E/E li). Hence, N 1 / U i, ) U k, with lk h 1 E/E li), i.e., k h 1 1 / i), as desired.

19 3. The Hasse Herbrand unction ). We will relate the Hasse Herbrand function to ramification groups which are defined in 4.3) Ch. II. If H is a subgroup of the Galois group G, then H x = H G x. As for the quotients, the description is provided by the following Theorem Herbrand). et / be a finite Galois extension and let M/ be a Galois subextension. et x, y be nonnegative real numbers related by y = h /M x). Then the image of Gal/ ) y in GalM/ ) coincides with GalM/ ) x. Proof. The cases x 1 or e M) = 1 are easy and left to the reader. Due to solvability of Galois groups of totally ramified extensions it is sufficient to prove the assertion in the case of a ramified cyclic extension /M of prime degree l. If l p, then using Proposition 3.5) Ch. II choose a prime element π of such that π M = π l is a prime element of M. Then for every τ Gal/ ) 1 we have π 1 M τπ M = π 1 τπ) l and therefore v π 1 τπ 1) = v π 1 τπ) l 1 ) = lv M π 1 M τπ M 1). Consider now the most interesting case l = p, x 1. et π be a prime element of. Put s = s M), see 1.4). The element π M = N /M π is a prime element of M. et τ Gal/ ) y. We have π 1 M τπ M = N /M π 1 τπ ). rom Proposition 3.4) we get h /M v M π 1 M τπ M ) 1) = h /M v M N /M π 1 τπ ) 1)) y, so τ M belongs to GalM/ ) x. Conversely, if τ M GalM/ ) x, then i = v M π 1 M τπ M 1) x. If i s then applying 1.5) we deduce that τ Gal/ ) i = Gal/ ) y. If i > s then Proposition 4.5) Ch. II and 1.5) show that j = v π 1 τπ 1) = s + pr for some nonnegative integer r. If r > 0 then Proposition 1.5) implies that i = s + r and τ Gal/ ) j = Gal/ ) y. If r = 0 then since i > s from the same Proposition we deduce that τπ π σπ mod M s+1 π for an appropriate generator σ of Gal/M). Then τσ 1 belongs to Gal/ ) k for k > s. Due to the previous discussions view k as j > s above) k = h /M i) and τ belongs to Gal/ ) y Gal/M), as required. Corollary. Define the upper ramification filtration of G = Gal/ ) as Gx) = Gal/ ) h/ x).

20 86 III. The Norm Map Then for a normal subgroup H of G the previous theorem shows that G/H)x) = Gx)H/H. Definition. or an infinite Galois extension / define upper ramification subgroups of G = Gal/ ) as Gx) = lim GalM/ )x) where M/ runs through all finite Galois subextensions of /. Real numbers x such that Gx) Gx + δ) for every δ > 0 are called upper ramification jumps of /. 3.6). The following Proposition is a generalization of results of section 1. Suppose that / is a finite totally ramified Galois extension and that : is a power of p = char ). Put G = Gal/ ). or the chain of normal ramification groups G = G 1 G 2... G n > G n+1 = {1} let m be the fixed field of G m ; then we get the tower of fields = 1 2 n n+1 =. Proposition. et 1 m n. Then Gal m+1 / m ) coincides with the ramification group Gal m+1 / m ) m, Gal m+1 / m ) m+1 = {1}, and h m+1 / m m) = m. Moreover, if i < m, then h m+1 / m i) = i and the homomorphism induced by N m+1 / m is injective; if i > m, then the homomorphism U i,m+1 /U i+1,m+1 U i,m /U i+1,m U hi),m+1 /U hi)+1,m+1 U i,m /U i+1,m induced by N m+1 / m for h = h m+1 / m is bijective. urthermore, the homomorphism U hi), /U hi)+1, U i, /U i+1, induced by N / for h = h /, is bijective if hi) > n. Proof. Induction on m. Base of induction m = n. Since Gal/ n ) x is equal to the group Gal/ ) x Gal/ n ), we deduce that Gal/ n ) n = Gal/ n ) and Gal/ n ) n+1 = {1}, and h /n x) = x for x n. All the other assertions for m = n follow from Proposition 1.5). Induction step m + 1 m. The transitivity property of the Hasse Herbrand function implies that h /m+1 x) = x for x m + 1. Now from the previous Theorem Gal m+1 / m ) x = Gal/ m ) h/m+1 x) Gal m+1 / m )/ Gal m+1 / m ).

21 3. The Hasse Herbrand unction 87 We deduce that Gal m+1 / m ) m = Gal m+1 / m ) and Gal m+1 / m ) m+1 = {1}. The rest follows from Proposition 1.5). To deduce the last assertion note that k = h / i) > n implies j = h m / i) > m. Corollary. The word injective in the Proposition can be replaced by bijective if is perfect. 3.7). Proposition. et / be a finite Galois extension, and let G = Gal/ ), h = h /. et h l and h r be the left and right derivatives of h. Then h l x) = G 0 : G hx), and { G0 h : G hx) if hx) is not integer, rx) = G 0 : G hx)+1 if hx) is integer. Therefore h / x) = x 0 G 0 : G ht) dt. Proof. Using the equality ) of 3.3), we may assume that / is a totally ramified extension the degree of which is a power of p = char ) > 0. Then G = G 0 = G 1. We proceed by induction on the degree :. et n be identical to that of 3.6); then n : < :. Since G/G n ) m = G m /G n for m n due to 3.6), we deduce the following series of claims. If h n / x) n, then, by Proposition 3.6), h / x) = h n / x) and h lx) = G/G n ) : G/G n ) hx) = G : G hx). If h n / x) < n and h / x) = h n / x) is not integer, then h rx) = G : G hx). If h n / x) is an integer < n, then h rx) = G/G n ) : G/G n ) hx)+1 = G : G hx)+1. Since the derivative right derivative) of h /n x) for x > n resp. equal to G n : G n ) n+1 = G n, we deduce that if h n / x) > n, then h lx) = G n G : G n = G = G : G hx). x n ) is So if h n / x) n, then h rx) = G n G : G n = G. This completes the proof. Remarks. 1. The function h / often appears under the notation ψ / ; in which case it is defined in quite a different way by using ramification groups, not the norm map. This function is inverse to the function ϕ / = x 0 dt G 0 :G t. 2. Information encoded in the Hasse Herbrand function can be extended using some additional ramification invariants introduced by V. Heiermann [Hei]. These arise when

22 88 III. The Norm Map one investigates more closely Eisenstein polynomials corresponding to prime elements see also Exercise 6 in section 4). Exercises. 1. Show that the three properties of the Hasse Herbrand function obtained in Proposition 3.1) hold for a finite separable extension / with a separable residue extension. 2. In terms of the proof of Proposition 3.2) show that h E/M1 h M1 /M = h M1 M 2 /M 2 h M2 /M by calculating the functions in accordance with the steps below. a) Suppose that M 1 : M = l is prime to p and M 2 : M = p. Choose a prime element π of E such that π l is a prime element of M 2 and calculate all the functions. b) Suppose that M 1 /M and M 2 /M are totally ramified extensions of prime degree p and M 1 M 2 = M. Using Proposition 4.5) Ch. II deduce that s 1 = se M 2 ) is congruent to s 2 = se M 1 ) modulo p. Show that if sm 2 M) > sm 1 M), then sm 1 M) = s 1 and sm 2 M) = s 1 + r. Show that if s = sm 2 M) = sm 1 M), then s 1 = s 2 s. 3. Y. Kawada [Kaw1]) et be an infinite Galois extension of a local field. a) et M 1 /, M 2 / be finite Galois subextensions of /. Show that the set of upper ramification jumps of M 1 / is a subset of upper ramification jumps of M 2 /. Denote by B/ ) the union of all upper ramification jumps of finite Galois subextensions of /. b) or a real x define x) = M Mx) where M runs over all finite Galois extensions of in and Mx) is the fixed field of GalM/ )x) inside M. Show that if x 1 < x 2, then x 1 ) x 2 ) if and only if [x 1, x 2 ) B/ ). c) Show that if x is the limit of a monotone increasing sequence x n, then x) = x n ). d) Show that if x is the limit of a monotone decreasing sequence x n and x B/ ), then x) = x n ). e) et x be the limit of a strictly monotone decreasing sequence x n. Define [x] = M n Mx n )) where M runs over all finite Galois extensions of in. Show that [x] = n x n ). Show that [x] = x) is and only if x B/ ). f) A subfield E of, E is called a ramification subfield if for every finite Galois subextension M/ of / there is y such that E M = My). Show that every ramifications subfield of over coincides either with some x) or with some [x]. g) Deduce that the set of all upper ramification jumps of / is the union of B/ ) and the limits of strictly monotone decreasing sequences of elements of B/ ). 4. The Norm and Ramification Groups We continue the study of ramification groups and the norm map. After recalling Satz 90 in 4.1) we further generalize results of section 1 as Theorem 4.2). In subsection 4.3) we study ramification numbers of abelian extensions. In this section is a complete discrete valuation field.

23 4. The Norm and Ramification Groups ). The following assertion is of general interest. Proposition Satz 90 ). et / be a cyclic Galois extension, and let N / α) = 1 for some α. Then there exists an element β such that α = β σ 1, where σ is a generator of Gal/ ). Proof. et βγ) denote γ + α 1 σγ) + α 1 σα 1 )σ 2 γ) + + α 1 σα 1 )... σ n 2 α 1 )σ n 1 γ) for γ, n = :. If βγ) were equal to 0 for all γ, then we would have a nontrivial solution 1, α 1, α 1 σα 1 ),... for the n n system of linear equations with the matrix σ i γ j ) ) 0 i,j n 1, where γ j) 0 j n 1 is a basis of over. This is impossible because / is separable see [a1, sect. 5 Ch. VIII]). Hence βγ) 0 for some γ. Then β = βγ) is the desired element. Corollary. If is a cyclic unramified extension of and N / α) = 1 for α, then α = γ σ 1 for some element γ U. Proof. In this case a prime element π in is also a prime one in. By the Proposition, α = β 1 σβ) with β = π i ε, ε U. Then α = ε 1 σε). Recall that in section 4 Ch. II we employed the homomorphisms ψ i : G i U i, /U i+1, we put U 0, = U ), where G = Gal/ ), π is a prime element in, i 0. Obviously these homomorphisms do not depend on the choice of π if / is totally ramified. The induced homomorphisms G i /G i+1 U i, /U i+1, will be also denoted by ψ i. 4.2). Theorem. et / be a finite totally ramified Galois extension with group G. et h = h /. Then for every integer i 0 the sequence 1 G hi) /G hi)+1 ψ hi) U hi), /U hi)+1, N i Ui, /U i+1, is exact the right homomorphism N i is induced by the norm map). Proof. The injectivity of ψ hi) follows from the definitions. It remains to show that if N / α U i+1, for α U hi),, then α σπ ) mod U hi)+1, π for some σ G hi). If / is a tamely ramified extension of degree l, then the fourth commutative diagram of Proposition 1.3) shows that N i is injective for i 1, and the kernel of N 0

24 90 III. The Norm Map coincides with the group of l th roots of unity which is contained in. Since π = l π is a prime element in for some prime element π in, we get kern 0 ) imψ 0 ), and in this case the sequence of the Theorem is commutative. If / is a cyclic extension of degree p = char ) > 0, then the fourth commutative diagram of Proposition 1.5) shows that kern s ) imψ s ) for s = v π 1 σπ )) and a generator σ of Gal/ ). Other diagrams of Proposition 1.5) show that N i is injective for i s. We proceed by induction on the degree :. Since we have already considered the tamely ramified case, we may assume that the maximal tamely ramified extension 1 of in does not coincide with. Since : 1 is a power of p, the homomorphism induced by N /1 U 0, /U 1, U 0,1 /U 1,1 is the raising to this power of p, and kern 0 ) is equal to the preimage under this homomorphism of the kernel of U 0,1 /U 1,1 U 0, /U 1,. In other words kern 0 ) coincides with the group of all l th roots of unity for l = 1 : which is contained in. Hence the kernel of N 0 is contained in the image of ψ 0, since ψ 0 is injective and G 0 : G 1 = l. Now suppose i 1. In this case we may assume 1 = because the homomorphism N i induced by N 1 / is injective for i 1. et n be as in Proposition 3.6). Then one can express N i as the composition U hi), /U hi)+1, N U h1 i), n /U h1 i)+1, n N U i, /U i+1,, where N and N are induced by N /n and N n / respectively, and h 1 i) = h n / i). If h 1 i) n, then by Proposition 3.6) Gal n / ) h1 i) = {1}, and we may assume that N is injective. Then by the induction assumption ker N i = ker N coincides with the set of elements π 1 σπ ) mod U hi)+1,, where σ runs over Gal/ n ) n = G n. If h 1 i) < n and N i α) U i+1, for some α U hi),, then hi) = h 1 i), and by the induction assumption, N α) σπ n ) π n mod U h1 i)+1, n for a prime element π n in n and some σ Gal/ ). We can take π n = N /n π. Hence ) N α) N σπ ) π mod U h1 i)+1, n. The homomorphisms U j, /U j+1, U j,n /U j+1,n

25 4. The Norm and Ramification Groups 91 induced by N /n, are injective for j < n by Proposition 3.6). Therefore, the element π 1 σπ ) belongs to U hi), and so σ G hi), α σπ ) π mod U hi)+1,. 4.3). Now we study ramification numbers of abelian extensions. We shall see that these satisfy much stronger congruences than that of Proposition 4.5) Ch. II. Theorem Hasse Arf). et / be a finite abelian extension, and let the residue extension / be separable. et G = Gal/ ). Then G j G j+1 for an integer j 0 implies j = h / j ) for an integer j 0. In other words, upper ramification jumps of abelian extensions are integers. Proof. We may assume that j > 0 and that / is totally ramified. et E/ be the maximal p-subextension in /, and m = : E. et π be a suitable prime element in such that π m E. or σ G j, σ G j+1 we get π m σπm = 1+mθπj for some θ U ; therefore j = mj 1, and σ E GalE/ ) j1, σ GalE/ ) j1 +1. If we verify that j 1 = h E/ j ) for some integer j, then j = h / j ). Thus, we may also assume G = G 1. If / is cyclic of degree p = char ), then the required assertion follows from Proposition 1.5). In the general case we proceed by induction on the degree of /. In terms of Proposition 3.6) it suffices to show that n h n / N) where G n {1} = G n+1. et σ G n, σ 1. Assume that there is a cyclic subgroup H of order p such that σ / H. Then denote the fixed field of H by M. or a prime element π in the element π M = N /M π ) is prime in M, and M = π M ) by Corollary 2 of 2.9) Ch. II. Then ε = N /M π 1 σπ )) = N /M π 1 )σn /M π )) 1, since σπ M ) π M. Put n = v M ε 1); then σ M G/H) n, σ M / G/H) n +1. By the induction hypothesis, n = h M/ n ) for some n N. Proposition 1.5) implies n h /M n ), and we obtain n h / n ). If n < h / n ), then, by Proposition 3.7) the left derivative of h / at n is equal to :, and the left derivative of h /M at n is equal to : M. Therefore, the left derivative of h M/ at n, which is equal to G/H) : G/H) n by Proposition 3.7), coincides with M :. This contradiction shows that n = h / n ). It remains to consider the case when there are no cyclic subgroups H of order p, such that σ / H. This means that G is itself cyclic. et τ be a generator of G. The choice of n and Theorem 4.2) imply that σ = τ ipm 1, where p i, p m = G. We can assume m 2 because the case of m = 1 has been considered above. et n 1 = v π 1 τ pm 2 π ) 1). Since G : G n = p m 1, Proposition 3.7) shows now that it suffices to prove that p m 1 n n 1 ). This is, in fact, a part of the third statement of the following Proposition.

26 92 III. The Norm Map Proposition. et / be a totally ramified cyclic extension of degree p m. et π be a prime element in. or σ Gal/ ) and integer k put σ k ) π ) c k = c k σ) = v 1. π Then 1) c k depends only on v p k), where v p is the p-adic valuation see section 1 Ch. I); 2) there exists an element α k such that v α k ) = k, v σαk ) α k 1 3) if v p k 1 k 2 ) a, then v p c k1 c k2 ) a + 1. ) = c k ; Proof. After Sh. Sen [Sen1]) 1) Note that c k does not depend on the choice of a prime element in by the same reasons as s in 1.4). et k = ip j with p i, j 0. Then σ k 1 = ρ 1)µ for ρ = σ pj, µ = ρ i 1 + ρ i As c k does not depend on the choice of a prime element in and v µπ )) = 1, then c k = c p j. 2) Put α k = k 1 i=0 σi π ) for k 0 and α k = α 1 k for k < 0. 3) Assume, by induction, that if v p k 1 k 2 ) a for a n 2, then v p c k1 σ) c k2 σ)) a + 1 for σ Gal/ ). irst we show that all the integers c p n 1, k+c k for v p k) n 1 are distinct. Indeed, let k 1 + c k1 = k 2 + c k2, v p k 1 ) v p k 2 ). Then v p k 1 k 2 ) = v p c k2 c k1 ) v p k 1 k 2 )+1, and thus k 1 = k 2. We also obtain that v p c p n 1 c k ) v p p n 1 k)+1 > v p k) and c p n 1 c k + k. Assume that v p c p n 1τ) c p nτ)) < n for a generator τ of Gal/ ). Our purpose is to show that this leads to a contradiction. Then, obviously, v p c k1 σ) c k2 σ)) a + 1 for v p k 1 k 2 ) a, a n 1. Put d = c p n 1τ) c p nτ). Since v p d) = v p c p n 2τ p ) c p n 1τ p )) n 1, we get v p d) = n 1. By 2), there exists an element α such that v α) = d, v τ p α) α) = d + c d τ p ) = d + c p nτ) = c p n 1τ). Put β = τ p 1 + τ p )α. Since v τ p α) α) = c p n 1τ) > 0, we get v τα) α) > v α) and v β) > d. We also obtain v τβ) β) = v τ p α) α) = c p n 1τ). Recalling that O = O [π ], we deduce that β can be expanded as β = β k, k v β) with β k possessing the same properties with respect to τ as α k of 2). Then τβ) β = τβ k ) β k ) + τβ k ) β k ). k v β) v p k)<n k v β) v p k) n

27 4. The Norm and Ramification Groups 93 The elements of the first sum in the right-hand expression do not cancel among themselves because v τβ k ) β k ) = k + c k τ) are all distinct and none of them coincides with c p n 1τ) = v τβ) β). Therefore, c p n 1τ) = v τβ k ) β k )). In this sum k v β) v p k) n v τβ k ) β k ) = k + c k τ) v β) + c p nτ) > d + c p nτ) = c p n 1τ), a contradiction. Remarks. 1. This Theorem can be naturally proved using local class field theory see 3.5) Ch. IV and 4.7) Ch. V). In addition, one can show that a finite Galois totally ramified extension / is abelian if and only if for every finite abelian totally ramified extension M/ the extension M/ has integer upper ramification jumps [e8]. or several other proofs of the Hasse Arf Theorem see [Se3], [N2]. 2. The arguments of the previous Proposition are valid for the more general situation of so called wildly ramified automorphisms, see Remark 3 in 5.7) and [Sen1]. 3. In the study of properties of ramification subgroup of finite Galois extensions of local fields one can use a theorem of. aubie [au1] which claims that for every finite Galois totally ramified extension of a local field there exists a Galois totally ramified extension of a local field with finite residue field such that the Galois groups are isomorphic and the ramification groups of the extensions are mapped to each other under this isomorphism. Exercises. 1. Prove Proposition 4.1) for a complete discrete valuation field and a cyclic extension / of prime degree using explicit calculations in section Show that if / is a finite totally ramified Galois extension, then ker N i G. 3. In terms of 4.3) show that { ) } σα) c k = max v α 1 : v α) = k, k = max {v α) : v σα) α) = k + c k }. i 0 4. ) Sh. Sen) et / be a cyclic totally ramified extension of degree p n, p = char ) > 0. et σ be a generator of Gal/ ), and let π be a prime element in. et c k be identical to those of Proposition 4.3). et A = {α O : Tr / α) = 0}, B = 1 σ)o.

28 94 III. The Norm Map a) Show that A/B is isomorphic k=pn 1 O /π g k O, where g k = [p n k + p n c k ]. k=1 b) Show that pa B. This assertion can be generalized to the case of arbitrary Galois extensions. It implies J. Tate s Theorem on invariants : let char ) = 0, char ) = p, and let = sep be the completion of sep. The Galois group G = Gal sep / ) operates on by continuity. Then G = [T2], [Sen1], [Ax]). 5. ) B.. Wyman [Wy]) et / be a cyclic totally ramified extension of complete discrete valuation fields, : = p n. et char ) = 0, char ) = p, and let be perfect. a) Show that / has n ramification numbers x 1 < x 2 < < x n. b) Show that if x i are divisible by p, then x i = x 1 + i 1)e for 1 i n, where e = e ). c) or the rest of this Exercise assume that a primitive p th root of unity ζ belongs to. et N / α) = ζ and v α 1) = i. Show that if x 1 < e/p 1), then x 1 i h / e/p 1)) and if x 1 e/p 1), then i = e/p 1). d) Assume that M/ is cyclic of degree p n 1 and = M p α) with α M. et α 1 σα) = β p for a generator σ of Gal/ ). Show that N M/ β) is a primitive p th root of unity. e) Show that if x 1 e/p 1), then x i = x 1 + i 1)e for 1 i n. f) et n 2. Show that if x n 1 p n 2 e/p 1), then x n = x n 1 + p n 1 e, and if x n 1 p n 2 e/p 1), then 1 + pp 1))x n 1 x n p n e/p 1) p 1)x n ) et / be a Galois totally ramified p-extension. et π be a prime element of and put π = N / π. Investigating the Eisenstein polynomial of π over show that a) or every i > 0 there exists j = ji) and g i O [X] such that g i 0 and N / 1 απ i ) = 1 + g i α)π j for every α O. Show that jh / k)) = k for every integer k > 0. b) Show that the sequence 1 Gal/ ) i / Gal/ ) i+1 U i, /U i+1, U j, /U j+1, is exact where the left arrow is induced by the norm map and is described by the c) polynomial g i. Put a i = deg g i. Show that i a i = :. d) et i 1 < < i m be the indices of all a i1,... a im which are > 1. Show that ji 1 ) < < ji m ). e) Assume in addition that char ) = p. Put b k = log p a ik+1 ) + + log p a im ) for 0 k m 1 and b m = 0. Prove that for all α O N / 1 απ ) = 1 + α pn π + f 1 α)π j f mα)π j m