A constructive theory ofr extensions of p-adic fields

8/10/2019 A constructive theory ofr extensions of p-adic fields

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6.3. P-GROUPS AS SPLITTING FIELD

automorphism of order p of the normal closure of L over F and 1 i < p , andL/F is Galois if and only if g(X ) = X p 1 + + g1X + g0 splits in linear factorsin L. If we can write () = +1 + . . . with U K , then () can beapproximated in L better than by any other conjugate, and consequently L/F is Galois by Krasner lemma. On the other hand if L/K is Galois we certainlyhave such an expression for some . Since

g0 = p 1

i =1

( i ) (6.6)

p 1

i =1

i +1 p 1 ( p 1)( +1) (6.7)

we have that L/F is Galois if and only if g0 is a ( p 1)-th power.

The monomial in X p

of f (X + ) is p2 p

pf p p

2 2 pX p = h0X p

where h0 is the constant term of h(X ), while the monomial in X is

Xf () = ( p2 r )f r p2 r 1X = g0h0X

where r should be p2 ( p 1) + p and v p(f r ) = 2, considering that f () is thedifferent and has valuation ( p2 p) 2 + ( p 1) ( + 1).

Since p2 p p 1 (mod p), by the denition of r we have taking the ratio

of the coefficients of the monomials above that

g0 ( p 1)( +1)

= rf r ( p 1) p 1

f p p2 2 p ( p 1)( +1) + . . .

= rf r / f p p2

+ = rf r / f p f p 2 + . . . ,

being p2

= f 0 + . . . .Since r (mod p) we obtained that p 1 is equal to f r / f p f p 2 , and it is

contained in p 1K if and only if g0 is a ( p 1)-th power. Put again F p = f p/ p,F p2 = F p 2/ p and G i = f i/ p2 for i = p, p2 .

Condition 6.3.6. L/F is Galois if and only if Gr / F p F p 2 is in p 1K , where r is

equal to p2 ( p 1) + p.

Lets recall that from 3.2.1, Chap. 3 we have that

N L/F (1 + ) = 1 + ( p p 1)F + . . . ,

while ( 1)F = 1 k F + . . .

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Maurizio Monge - A constructive theory for p-adic fields

for p 1 = F p/ F p 2 and some integer k prime with p, by Prop. 6.3.1. Fromwhat observed at the beginning, we obtain that the length of Gal( L/K ) is when X p Gr / F p F p 2 X = has solution in K , and + 1 if this is not the case.Replacing X by X and dividing by this is equivalent to testing if

( F p/ F p 2 ) X p Gr / F p F p 2 X = 1

has solution in K .Consequently we obtain

Theorem 6.3.7. Let 2 p 1 an let r = p2 ( p 1) + p, and assume that f (X ) is such that

1. v p(f p) = 1 , and v p(f pi ) 2 for i 2, p 1 ,

2. v p(f i ) 2 for i 1, r 1 , v(f r ) = 2 and v p(f i ) 3 for i r +1 , p2 1 ,

putting F p = f p/ p, F p2 = f p 2/ p, and Gi = f i/ p2 for all i = p, p2 we have

3. F p/ F p 2 = p 1 for some K ,

4. Gr / F p F p 2 = p 1 for some K .

Let L be the extension determined by f (X ), L the normal closure over K , and F the unique subextension of degree p contained in L. Then Gal( L/K ) is a split extension of G = Gal( F/K ) by the indecomposable F p[G]-module M =Gal( L/F ) and furthermore dening

U (X ) = ( F p/ F p 2 ) X p Gr / F p F p 2 X 1

we have that if

U (X ) has no root in K , then M has length + 1 and L/F is formed by an unramied extension followed by a totally ramied with upper ramication breaks 1, 2, . . . , ,

U (X ) has some root in K , then M has length and L/F is totally ramied with upper ramication breaks 1, 2, . . . , .

What is left is the easy case for = 1, which is considered separately. Inthis case L/K has 1 as unique ramication break, v p(f 1) = 1 while v p(f i ) 2for i 2, p2 1 , and consequently put F i = f i/ p for i = 1 ,p ,p2 . The mapU 1 ,L / U 2 ,L U 1 ,K / U 2 ,K induced by N L/K is described by the additive polynomial

A(Y ) = F p2 Y p2

+ F pY p + F 1Y , and L/K is Galois precisely when N L/K (U 1,L ) =1 + pW for a subspace W of codimension 2 in K , that is when A(Y ) splitscompletely in K . On the other hand the normal closure L/K is a p-extension if and only if L becomes abelian elementary over the unique unramied extensionof degree p of K , or equivalently if A(Y ) splits completely over the uniqueextension of degree p of K .

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6.4. CYCLIC EXTENSIONS OF DEGREE P3

Theorem 6.3.8. Assume that f (X ) is such that

1. v p(f p) 1, and v p(f pi ) 2 for i 2, p 1 ,

2. v p(f 1) = 1 , and v(f i ) 2 for i 2, p2 1 ,

and putting F i = f i/ p for i = 1,p ,p2

3. the polynomial F p2 Y p2

+ F pY p + F 1Y has a root in K , and F 1/ F p 2 (K )

p 1 .

Let L be the extension determined by f (X ), and L be the normal closure over K . Then

if F p2 Y p2

+ F pY p + F 1Y does not split in K then L/K has a unique subex-

tension F , Gal(L/F ) has length 2,

L/F is formed by an unramied extension followed by a totally ramied extension with upper ramication break 1, and

Gal( L/K ) is a split extension of Gal(F/K ) by Gal( L/F ),

if F p2 Y p2

+ F pY p + F 1Y has all roots in K then L/K is an abelian elemen-tary p-extension.

Theorems 6.3.5, 6.3.7 and 6.3.8 cover all possible ramication breaks of theextension L/F , so they completely describe the Galois groups of polynomials of degree p2 whose splitting eld is a p-extension.

6.4 Polynomials of degree p3 generating a cyclic

extensionWe proceed with the same strategy used for the polynomials of degree p2 , start-ing from the conditions on the valuations of the coefficients.

Let f (X ) = X p3

+ + f p3 1X + f p3 , since the different has now valuation4 p3 p2 p 2 it will be determined by the monomial f p2 + p+1 X p

3 p2 p 1 ,v p(f p2 + p+1 ) = 3, v p(f i ) 3 if (i, p) = 1 and v p(f i ) 4 if furthermore i > p 2 + p+1. Let be a root, the coefficients of the term of degree p of the ramicationpolynomial f (X + ) will have valuation ( p3 p2) 2 + ( p2 p) ( p + 1) =3 p3 p2 2 p and has to come from a monomial f p3 i (X + ) i contributingthe term i p f p3 i X

p i p, and we deduce the i has to be i = p3 p2 p, thatv p(f p2 + p) = 2, that v p(f pi ) 2 for (i, p) = 1 and v p(f pi ) 3 if furthermorei p + 2. Similarly considering the coefficient of the term of degree p2 of the ramication polygon, which shall have valuation 2 p3 2 p2 , we obtain thatv p(f p2 ) = 1 and v p(f p2 i ) 2 for all indices such that ( i, p) = 1.

Condition 6.4.1. We must have

1. v p(f p2 ) = 1 and v p(f p2 i ) 2 for i 2, p 1 ,

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2. v p(f pi ) 2 for all i 1, p 1 , v p(f p2 + p) = 2 , and v p(f pi ) 3 for all i p + 1 , p2 1 ,

3. v p(f i ) 3 for all i 1, p2 + p 1 , v p(f p2 + p+1 ) = 3 and v p(f i ) 4 for all i p2 + p + 2 , p3 1 .

Again working like in degree p2 , we shall require N L/K (U 1,L ) pi 1

U i +1 ,Lto be contained in 1 + pi V for 1 i 3 and some F p-vector space V , and afterdetermining V we will have to verify the condition on the combinations of thenorms of elements of the form 1 + for a unit , and 1 p2 + p + 1 and( , p) = 1.

Lets expand again i=0 f ( i T ) modulo p4 , taking into account the valua-tions of the f i and evaluating directly the ( ) via Prop. 6.1.4 it can be writtenwith the terms in increasing valuation as

p K + ,1f p2 T p2

+ ,1f p3 T p3

(6.8)

p 2K

+12

[2],2f 2

p2 T 2 p2 + [2],p+1 f p2 f p3 T

p3 + p2 + 12

[2],2f 2

p3 T 2 p3

+j 2,p 1

,j f p2 j T p2 j +

k 1,p +1

,k f pk T pk (6.9)

p 3K

+13

[3],3f 3

p2 T 3 p2 + [3],p+2 f p3 f

2 p2 T

p3 +2 p2 + [3],2 p+1 f 2

p3 f p2 T 2 p3 + p2 +

13

[3],3f 3

p3 T 3 p3

+j 2,p 2

[2],j +1 f p2 f p2 j T p2 + p2 j + [2],1f p2 f p3 p2 T

p3 +k 1,p +1

[2],k+ pf p2 f pk T p2 + pk

+j 2,p 1

[2],j + p

f p

3 f p

2

jT p

3 + p2 j +k 1,p +1

[2],k+ p

2 f p

3 f pk

T p3 + pk

+j p+2 ,p 2 1

,j f pj T pj +k 1,p 2 + p+1

,k f k T k

(6.10)

While this expansion looks scary we can start noticing that since raising to a p-thpower induces an automorphism on the set of multiplicative representatives wehave considering the expansion modulo p3 that the conditions stated in Theorem6.2.5 must be satised with f pi in place of f i . Consequently put F i = f i/ p fori = p2 , p3 , Gi = gi/ p2 for i p 1, p + 1 or i p2 2, p 1 , let A(Y ) =F p3 Y p + F p2 Y and put V = A(K ). Such conditions are satised if and onlyif V has codimension 1 in K and the norms contained in U 1,K or U 2,K arerespectively in 1 + pV and 1 + p2V .

Similarly to the case in degree p2 , for 2 this sum can be written asD () = |k dk (k ) where the d (T ) are the polynomial obtained if every [m ],i is interpreted as a Kroneckers delta and 1+ d () N (E ( )) (mod p4).For = 1 there are exceptions because [m ],i = 0 for < m.

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We require the norms in U 3,K to be in 1+ p3V , and lets concentrate rst onthe case of p + 2 , p2 + p + 1 so that the norms N L/K (1 + ) already livein U 3,K , and the rst few terms of the expansion disappear. For such indices, d () shall be in p3V for each representative , and dividing by p3 we can

consider the additive polynomials A (Y ) = d (Y )/ p3 , which, depending on , are

G p( p2 ) F p3 Y p + H Y p2 + 1 , p2 + p + 1 ,H p Y p + H Y 2 p + 2 , p2 1 ,

F 2 p3 F p2 Y p2 + ( H p F p2 G p( p+1) )Y p + H Y = 2 p + 1 ,

F p3 F p2 ( p) Y p2

+ ( H p F p2 G p( p) )Y p + H Y p + 3 , 2 p 1 ,

(F p3 F 2 p2 F p3 F 2 p2 )Y p2 + ( H p F p2 G2 p)Y p + H Y = p + 2 ,

where we have put H k = f k/ p3 for k 1, p2 + p + 1 and k p p + 2 , p2 1 .

Condition 6.4.2. For each p+2 , p2 + p+1 we shall have A (K ) A(K ).For p + 1 the question is a bit more complicated because in general

the norms of 1 will not be contained in U 3,K , but a proper combinationof norms of elements of this form may be, and we should require it to be in1+ p3V . However for varying the elements N L/K (1 ) p have norms coveringall classes in U 2,K /U 3,K , and consequently each N L/K (1 ) can be reducedinto U 3,K by multiplication by an suitable N L/K (1 ) p for some , and weshould verify that all such reduction are actually in 1 + p3V . The condition formore complicated combinations will certainly also be ensured.

Since the map p 2K / p 4K U 2 ,K / U 4 ,K induced by x 1+ x is still an isomorphismwe have that a proper combination of the of 1 + (e.g. via the Artin-Hasseexponential) has norm of the form 1 + d (). In other words depending on

2 p + 1 we have that the remaining term, which we call g (Y ), is f p3 f p2 Y p

2+ f p2 + pY p f p2 f pY p + f p+1 Y [ p + 1] ,

f p2 Y p2

+ f p Y p f p2 f p2 ( 1) Y p2

+ f Y [4, p 1],

f 3 p2 Y p2

+ f 3 pY p + 13 f 3

p3 Y p3 + 13 f

3 p2 Y

p2 f p2 f 2 p2 Y p2

+ f 3Y [3],

12 f 2

p3 Y p3 + f 2 p2 12 f

2 p2 Y

p2 + f 2 pY p + f 2Y [2],

where under braces are the terms that are not identically in p3K . On the otherhand

N (1 + ) = 1 + f p3 p3

+ f p2 p2

+ f p p mod p2V and consequently

N (1 + ) p = 1 + pf p3 p3 + pf p2 p

2 + pf p p mod p3V,

with again under braces are the terms that are not identically in p3K . Conse-quently lets consider the polynomial

h(Z ) = { pf p3 Z p2

+ pf p2 Z p} + pf pZ,

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we are looking for values of Z = (Y ), that will be the lifting of some additivepolynomials in Y , such that g (Y ) h( (Y )) p3K , to impose the conditionthat it shall be in p3V as well.

The additive polynomials g (Y )/ p2 , which we denote by B (Y p) replacing Y pby Y , are forced to have image contained V , that is the image of h (Y )/ p2 = A(Y p),and the condition is that B (Y ) = A(D (Y )) for some other additive polynomialD (Y ) whose coefficients can be deduced easily.

In particular, being A(Y ) = F p3 Y p + F p2 Y and B (Y ) the polynomials

F p2 F p3 Y p + G p2 + pY [ p + 1] ,G p2 Y p + G p Y [3, p 1],

12 F 2

p3 Y p2 + G2 p2 12 F

2 p2 Y

p + G2 pY [2],(6.11)

in view of Prop. 6.1.2 we can take as D (Y ) respectively the polynomials

G p / F p 2 Y [3, p + 1] , 12 F

1/ p p3 Y

p2 + G 2 p/ F p 2 Y [2]. (6.12)

Now, B (Y p) = A((D1/ p (Y )) p) where D

1/ p (Y ) is D (Y ) with the map x x1/ p applied to the coefficients. Given the denitions of A(Y ) and B (Y ) interms of the h(Y ) and g (Y ), we have that we can take as (Y ) any lifting of D

1/ p (Y ) to OK [Y ].For 3 p + 1 lets take a OK such that p = G p / F p 2 = f p / pf p 2 , then

D (Y ) = pY and we can take (Y ) = Y , and the polynomials 1 p3 (g (Y ) h( (Y )))should take values in V . Considering that

h( (Y )) = pf p3 p2Y p

2+ pf p2 pY p + pf pY,

depending on they are

f p 3 f p 2/ p3 f p 3 p2/ p2 Y p

2

+ (f p 2 + p/ p3 f p 2 p/ p2 ) F p2 G p Y p + ( H p+1 G p)Y

for = p + 1,

f p 2 / p3 f p 3 p2/ p2 F p2 G p2 ( 1) Y p

2

+( f p / p3 f p 2 p/ p2 )Y p + ( H G p)Y,

for 4 = p 1, and

13

F 3 p3 Y p3 + f 3 p 2/ p3 f p 3 p

2/ p2 +

13

F 3 p2 F p2 G2 p2 Y p2

+( f 3 p/ p3 f p 2 p/ p2 )Y p + ( H 3 G p)Y

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for = 3.For = 2 lets take , OK such that p = Gp 2/ F p 2 = f p 2/ pf p 2 and

p2 = 12 F p3 = 12

f p 3/ p3 . Then D (Y ) = pY p + pY so that we can take2(Y ) = Y p + Y , and we have

h(2(Y )) = pf p3 (Y p + Y ) p2

+ pf p2 (Y p + Y ) p + pf p(Y p + Y )

= pf p3 p3Y p

3+ pf p3 p

2Y p

2+ pf p3

p 1

i =1

p2

ip ip ( p i ) pY ip

2 +( p 1) p + O( p4)

+ pf p2 p2Y p

2+ pf p2 pY p + pf p2

p 1

i =1

pi

i ( p i ) Y ip +( p 1)

+ pf pY p + pf pY.

Considering that 1 p pi 1 p p2

ip (mod p) and the terms in the sums can bepaired in elements that are pf p3 p

2

ip Z p + pf p2 pi Z for Z =

i ( p i ) Y ip +( p 1)

and hence in p3V for each Z , we have that up to some element in p3V we canwrite h(2(Y )) as

pf p3 p3Y p

3+ ( pf p3 p

2+ pf p2 p

2)Y p

2+ ( pf p2 p + pf p )Y p + pf pY.

Consequently up to some element of V the polynomial 1 p3 (g2(Y ) h(2(Y ))) isthe

12

f 2p 3/ p3 f p 3 p3/ p2 Y p

3+ f 2 p 2/ p3

12

f 2p 2/ p3 f p 3 p2/ p2 f p 2 p

2/ p2 Y p

2

+ (f 2 p/ p3 f p 2 p/ p2 ) G p Y p + ( H 2 G p) Y,

which is required to take values in V .One last effort is required: for = 1 in the case that 1 has norm in

U 2,K (and hence in 1 + p2V ), that is when is such that A( p2) = 0, we should

also have that taking such that (1 )(1 ) p has norm in U 3,K , thanthat norm is required to be actually in 1 + p3V .

Let = T be as required, the terms that disappear because = 1 are

12

f 2 p2 T 2 p2 + f p3 f p2 T p

3 + p2 + 12

f 2 p3 T 2 p3 =

12

f p2 T p2

+ f p3 T p3 2

,

then

13

f 3 p2 T 3 p2 + f p3 f 2 p2 T

p3 +2 p2 + f 2 p3 f p2 T 2 p3 + p2 + 1

3f 3 p3 T

3 p3 = 13

f p2 T p2

+ f p3 T p3 3 ,

and the sums can be decomposed as sums of ( f p2 T p2

+ f p3 T p3)f p2 j T p

2 j and of (f p2 T p

2+ f p3 T p

3)f pk T pk , and in particular all such terms are in p4K considering

the hypotheses on T .

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Consequently such terms can be assumed to be present, and removing theextra terms we already studied (or considering the norm of E ()) the remainingterms are

w(T ) = f p3 T p3

+ f p2 T p2

f p2 f p3 p2 T p2

+ f pT p + f 1T.

Assume 1 p2 f p3 p3 + f p3 p

2 + f p p can be written as F p3 p2

+ F p2 p for some , then taking any lift of we can consider w() h(), which comes froma norm of the required type, and should be in p3V .

At last, we can state the

Theorem 6.4.3. The Eisenstein polynomial f (X ) = X p3

+ f 1X p3 1 + +

f p3 1X + f p3 determines a Galois extension of degree p3 over K if and only if

1. v p(f p2 ) = 1 and v p(f p2 i ) 2 for i 2, p 1 ,

2. v p(f pi ) 2 for all i 1, p 1 , v p(f p2 + p) = 2 , and v p(f pi ) 3 for all i p + 1 , p2 1 ,

3. v p(f i ) 3 for all i 1, p2 + p 1 , v p(f p2 + p+1 ) = 3 and v p(f i ) 4 for all i p2 + p + 2 , p3 1 ,

putting F p2 = f p 2/ p, F p3 = f p 3/ p, and Gi = f i/ p2 for all i in p2 2, p 1 or in p 1, p + 1 we have

4. F p 2/ F p 3 p 1K ,

5. G p p( p+1) = F p+1 p2 ,

6. G p

2 = F p

3 G p/ F p

2 p for 3, p 1 ,

7. G2 p2 = F p3 G 2 p/ F p 2 p + 12 F p2 F p2 F

1/ p p3 ,

if is such that p( p 1) = F p 2/ F p 3 we have (independently of )

8. 1 p2 f p3 p2 + f p2 p + f p = F p3 p + F p2 for some K ,

putting H i = f i/ p3 for i in 1, p2 + p + 1 or in p p + 2 , p2 1 we have

9. G p( p2 ) F p3 = F p3 (H / F p 2 ) p for p2 + 1 , p2 + p + 1 ,

10. H p = F p3 (H / F p 2 ) p for 2 p + 2 , p2 1 ,

11. H p(2 p+1) F p2

G p( p+1) = F p3

(H 2 p +1

/ F p 2 ) p

+ F p2

(F p3

F p2

)1/ p

,12. H p F p2 G p( p) = F p3 (H / F p 2 ) p F p2 (F p2 ( p) )

1/ p for p + 3 , 2 p 1 ,

13. H p( p+2) F p2 G2 p = F p3 (H p +2 / F p 2 ) p + F p2 (F 2 p2 F 2 p2 )1/ p ,

for each 3, p + 1 , let be such that p = G p / F p 2 . Then

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6.5. SUMS OF ROOTS OF UNITY

14. putting P p+1 = H p+1 G p p+1 ,

Q p+1 = ( f p 2 + p/ p3 f p 2 pp +1 / p2 ) F p2 G p, R p+1 = f p 3 f p 2/ p3 f p 3

p 2p +1 / p2 ,

we have Q p+1 = F p3 (P p +1 / F p 2 ) p + F p2 (R p +1 / F p 3 )1/ p ,

15. for each 4 p 1 putting P = H G p ,

Q = ( f p / p3 f p 2 p/ p2 ) F p2 G p, R = f p 2 / p3 f p 3 p 2 / p2 F p2 G p2 ( 1) ,

we have Q = F p3 (P / F p 2 ) p + F p2 (R / F p 3 )1/ p ,

16. putting putting P 3 = H 3 G p3 ,

Q3 = ( f 3 p/ p3 f p 2 p3/ p2 ), R3 = f 3 p 2/ p3 f p 3 p 23 / p2 + 13F 3 p2 F p2 G2 p2

we have 13 F p2 (F 2

p3 )1/ p + F p3 (Q 3/ F p 2 ) p = R3 + F p3 (F p 3/ F p 2 ) p(P 3/ F p 2 ) p

2,

let 2 , 2 OK such that p2 = G p 2/ F p 2 and p22 = 12 F p3 . Then

17. putting

P 2 = H 2 G p, Q2 = ( f 2 p/ p3 f p 2 p/ p2 ) G p ,

R2 = f 2 p 2/ p3 12

f 2p 2/ p3 f p 3 p2/ p2 f p 2 p

2/ p2 , S 2 =

12

f 2p 3/ p3 f p 3 p3/ p2

we have F p2 (S 2/ F p 3 )1/ p + F p3 (Q 2/ F p 2 ) p = R2 + F p3 (F p 3/ F p 2 ) p(P 2/ F p 2 ) p

2

,

if , are such that p2 ( p 1) = F p 2 / F p 3 and

1 p2

f p3 p3 + f p2 p

2 + f p p = F p3 p2

+ F p2 p,

18. we have that

1 p3

f p3 ( p3 p2 ) + f p2 ( p2 p) + f p( p ) f p2 f p3 p2 p

2 + f 1

is also of the form F p3 p + F p2 for some K .

6.5 Sums of roots of unityWe nally prove the lemma about the ( ), it is actually much more thanneeded but nevertheless is has a nice statement, which could still be useful insimilar circumstances:

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Lemma 6.5.1. Let = ( 1 , 2 , . . . , r ) be a partition, then

( ) = = j J ( j )

# J jJ

( 1)# ( j ) 1(# ( j ) 1)!

where the sum is over all the partitions = jJ ( j ) (as set) such that for each j J the sum | ( j ) | of the elements in ( j ) is multiple of and # ( j ) is the cardinality of the subset ( j ) .

Proof. Let A( i,j ) be the sets (indexed by the pairs ( i, j )) of indices ( 1 , . . . , r )such that i = j , let A0 be the set of all possible indices, and for A A0 denoteby ( A) the sum over all the indices in A. By inclusion-exclusion we have that

( ) = ( A0) (( i,j ) A( i,j ) )

= ( A0) ( i,j )

( A( i,j ) ) +( i,j )=( i ,j )

( A( i,j ) A( i ,j ) ) . . .

Now let A be the intersection of all the sets A( i k ,j k ) for a collection of pairsP = {(i1 , j 1), . . . , (is , j s )}sS , if we consider the graph with R = {1, . . . , r }as vertices and the ( ik , j k ) as edges we have that if we split R in connectedcomponents R = tT R t then the allowed indices are those constant on eachR t , and calling t the value taken on R t the sum ( A) becomes

( A) =tT

1

t =0

t ( r R t r ) ,

and this sum is # T

when all the rR t r are multiple of and 0 if not. Note

that ( A) appears with sign equal to ( 1)# S in the inclusion-exclusion, so foreach partition of R in sets R t such that the sum of r for r R t is multiple of

we have that to consider the all graphs with set of vertices R and such thateach Rt is a connected component, and count the number of graphs with aneven number of edges minus those with a odd number of edges. Now the totaldifference is the product of the differences over all the connected components,so we have

( ) = = j J ( j )

# J K # ( j )

where for each i we denote by K i the difference of the number of connectedgraphs on i vertices having an even and odd number of edges.

The difference of the number of connected graphs K i on i vertices with aneven or odd number of vertices can be computed xing an edge, and consideringthe graphs obtained adding or removing that edge. Those such that with orwithout it are connected come in pairs with an even and odd number of edges,the other graphs are obtained connecting two other connected graphs on j andi j vertices. In particular choosing j 1 vertices to make one component with

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6.5. SUMS OF ROOTS OF UNITY

the rst vertex of our distinguished edges we obtain

K i +2 = i

j

i j

K i j +1 K j +1

for i 0, and K 1 = 1. Calling G(X ) the exponential generating functioni =0

K i +1i ! X

i we obtain that

ddX

G(X ) = G(X )2

with the additional condition that K 1 = 1, and this equation is clearly satisedby 1/ (1+ X ) , which can be the only solution. Consequently K i+1 = ( 1)i i! andthe lemma is proved.

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Chapter 7

Special Eisensteinpolynomials

In this chapter we show how it is possible to dene a normal form for Eisensteinpolynomials, which can be used for quickly enumerating totally ramied exten-sions of a local eld, for selecting a special dening polynomial to representextensions, and for identication of the extensions. Unluckily it doesnt seempossible to produce easily exactly one special polynomial for each isomorphismclass of extensions, but we show how to obtain a very restricted set of polyno-mials generating each extension. The number of special polynomials generatinga xed extension L/K is not greater than the number of conjugates of L overK , so that each Galois extensions is generated by exactly one polynomial. Infact, the problem of selecting exactly one generating polynomial for each iso-morphism class appears to be as hard as that of determining the cardinality of the group of automorphisms of the extension generated by a polynomial.

As shown in [PR01] , it is possible to enumerate and identify the extensionsgenerated by Eisenstein polynomials selecting one polynomial for each neigh-borhood with respect to a suitable distance, and applying Panayi root ndingalgorithm to collect the polynomials generating the same extension. The searchspace can be drastically reduced by just taking into account Eisenstein polyno-mials in normal form.

Furthermore, for each Eisenstein polynomial generating an extension L/K of degree n there exists a quick way to recover all the special polynomials attachedto the extension. The procedure does not require an exhaustive search over thespace of all extensions of degree n of K , not even a search within the set of

polynomials generating extensions with xed ramication data.Indeed, any Eisenstein polynomial can be put into normal form by applyinggreedily a reduction algorithm, which however allows some free choices duringthe reduction. The full set of special polynomials is obtained as the set of allpossible outputs of the reduction algorithm, over all possible choices.

A family of unique representatives for Eisenstein polynomials was already

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dened long time ago by Krasner [Kra37], and it turns out that Krasner specialrepresentative is one of the special polynomials we consider, when the sets of representatives in the denition are taken to force as many terms to be 0 in the p-adic expansions of the coefficients. The unique Krasner representative is noteasy to characterize in terms of the coefficients, while from our point of view wehave a very restricted set of polynomials that can be easily described. CuriouslyKrasner work appears to be unknown to many people working in the eld, andis not taken into account in many recent works, such as in [ PR01, JR06 ].

We exhibit a criterion for establishing a priori that an Eisenstein polynomialf (T ) may not be converted to another polynomial g(T ) via such a reductionapplied greedily, and when f (T ) and g(T ) are any two Eisenstein polynomialssuch that one of them is known to generate a Galois extension then the criterioncan be used to show that f (T ) and g(T ) generate non-isomorphic extensions.The criterion takes into account the higher order terms appearing in the p-adic

expansion of the coefficients, not just the valuation (or rst-order expression)of f () g() for a uniformizer of an extension L/K of degree n. Thecriterion established in [Yos12] for totally ramied Galois extensions over Q p isalso recovered in a more general context.

In the last section we describe an algorithm that allows to construct theunique special Eisenstein polynomial generating a totally ramied class eld,given a suitable description of a norm subgroup. In particular, we show thatthere exists an ordering of the terms appearing in the p-adic expansions of the coefficients allowing to recover all the terms of the special polynomial, bysolving inductively linear equations over the residue eld. An algorithm for theconstruction of polynomials generating class eld was described in [ Pau06] forcyclic extensions, where an extension of degree pm is constructed inductivelyby steps of degree p. In our construction an Eisenstein polynomial generating

an arbitrary totally ramied class eld is constructed directly. For each closedsubgroup we obtain exactly one extension of degree equal to the index, whichhas additionally to be Galois, so we also obtain an alternate and constructiveproof of the Existence Theorem of local class eld theory.

Notation

We will assume that K has nite residue eld, and set eK = vK ( p) as usual. If L/K is any totally ramied extension of degree n , with k distinct ramicationbreaks say, we will usually denote with t1 < t 2 < < t k the ramicationbreaks. We will denote as 0 > 1 > > k the cardinalities of the corre-sponding ramication subsets (see 3.3, Chap. 3), so that 0 = # = n and

i = # t+i for 1 i k. The i are all powers of , except possibly for 0 = n.We will denote by L t i and L t +i the elds xed by t i and t +i respectively.

If ps is the biggest power of p dividing n, for each 0 s it will also beconvenient putting to be equal to the smallest real t such that n L/K (x) hasslope p for x t , it will be equal to either 0, or some ramication break t i .The are weakly decreasing and exhaust all the lower ramication breaks ti ,

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7.1. REDUCED POLYNOMIALS

and one break ti > 0 is repeated r times if ( t i : t +i ) is equal to pr , so each

ramication break is taken with its multiplicity in a suitable sense. It willalso be convenient dening

= nL/K ( ) p , = nL/K ( ),

for each 0 s. Up to a factor n the are the upper ramication breaks of the extension.

7.1 Reduction algorithm and the family of re-

duced polynomialsLet f (T ) = T n + f n 1T n 1 . . . f 1T + f 0 be a monic Eisenstein polynomial of de-gree n, let be a root in a xed algebraic closure K alg and put L = K (). Thenclearly f (T ) is the minimal polynomial of , which is a uniformizing elementof the extension determined by f (T ), and we are interested in understandinghow the coefficients of the minimal polynomial of a uniformizer change when is replaced by another uniformizer = + m +1 + . . . , for some unit U K and integer m 1. Since the following computation only depends on at therst order, may be taken to be a multiplicative representative.

Let us consider the ramication polynomial ( T ) = n f (T + ), its New-ton polygon is fully described by the lower ramication breaks. For OK we can compute a lower bound for the valuation of ( ) as function of vL ()starting from the Newton polygon of ( T ). The construction produces natu-rally the Newton copolygon , which is essentially the dual convex body of theNewton polygon, and is connected to the Hasse-Herbrand transition function asalready observed in [Lub81, Li97]; in such references f (T + ) was used insteadso the function obtained was slightly different from the classical Hasse-Herbranddened in [FV02, Ser79].

Indeed, the Newton polygon of the polynomial ( m T ) resulting by thesubstitution T m T can obtained from the polygon of ( T ) moving thepoints with abscissa x up by mn x. In other words, if N : [1, n ] R is the realfunction describing the polygon of ( T ), the polygon of ( m T ) is describedby N (x) + mn x.

The function N (x) is convex and piecewise linear, and by the well knownproperties of Newton polygons the slopes are t k/ n , . . . , t 1/ n where t 1 < t 2 < < t k are the lower ramication breaks of the extension generated by a root,and it has slope t i/ n in the interval [ i , i 1] where 0 > 1 > > k arethe cardinalities of the corresponding ramication subsets. We put t0 = + ,tk+1 = for convenience.

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0 1 2 3

t 1/ n

t 2/ n

t 3/ n

Lets consider the minimum achieved by the function N (x) + mn x in the

interval [1 , n ], as a function of the real parameter m. It is again a piecewise linearfunction with slope i/ n for ti m ti +1 , and we obtain that this minimumvalue function is exactly the Hasse-Herbrand function L/K (m). Hence this isthe smallest valuation (with respect to K ) of the coefficients of (m T ), and n L/K (m ) is the exact power of such that n L/K (m ) (m T ) is in OL [T ]and non trivial modulo pL .

Lets dene the valuation of a polynomial to be the smallest valuation of thecoefficients, we can resume what proved in the following

Proposition 7.1.1. Let f (T ) be an Eisenstein polynomial, a root, L = K ()and (T ) = n f (T + ) its ramication polynomial. Then

vL ((m T )) = nL/K (m).

It will also be convenient to deduce an expression for the values N ( p ) foreach 0 such that p | n. Starting from p the function N (x) has slope / n ,so N (x) + n x has inmum equal to L/K ( ), which is achieved for x = p andis also equal to N ( p ) + n p , so we obtain

N ( p ) = L/K ( ) n

p = n

.

Lemma 7.1.2. For each 0 such that p n we have N ( p ) = / n .

We will also prove another Lemma, which will be needed require later. If p is the abscissa of a vertex of the Newton polygon we have that the terms

contributing to the coefficient of T p in the ramication polynomial give to thecoefficient of T p

jcontributions having K -valuation at most eK ( j ) bigger, for

j < . In other words we have that N ( pj ) < e K ( j ) + N ( p ), for each j .Considering the last vertex of one side of the Netwon polygon, and since for each the slope is equal to / n in the interval [ p , p +1 ] and N ( p ) < e K + N ( p +1 ),

we have that then has to be at most n eK( p +1 p ) = eL / ( p+1 p ) . Hence we have

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Lemma 7.1.3. We have j eL ( j ) +

for each j < , and furthermore

e L / ( p +1 p )

for each .

We now study the points ( j, vK (j )), coming from a monomial j T j , thatmay lie on the boundary of the Newton polygon of ( T ) = ni =0 i T i . Weclaim that either their ordinate j is a power of p, either 1 | j , and the latter isonly possible when 0 = n is not a power of p, so that 1 is the biggest powerof p dividing n , and the polygon of ( T ) has slope 0 in the interval [ 1 , 0].

Indeed, for each r we have

r =n

i= r

ir

f i i n ,

and since the summands have different valuations modulo n the valuation of rhas to be equal to the minimal valuation of such terms. Consider vK ( ir f i

i n )as a function of r : then its minimum is obtained when r is the biggest power of pdividing i, and if p |i then vK ( ir f i

i n ) is at least vK ( i p f i i n ) when p +1

r , and strictly bigger if p r . So when p r we have that vL (r ) vL ( p ),and ( r, v K (r )) cannot be on the boundary of the polygon unless possibly whenthe segment containing p has horizontal slope, p = 1 and p | r .

For integer m 0 lets consider the polynomials

S m (T ) = n L/K (m ) (m T ).

If m 1, or n is a power of p, then S m (T ) is of the form

S m (T ) =b

i = a

ci T pi

for some coefficients ci , where pa = pb = i when m is not a ramication breakand m (t i +1 , t i ) say, while i = pa and i 1 = pb if m = ti for some i. Inparticular they are additive polynomials.

On the other hand if m = 0 and n is not a power of p (and hence L has anon-trivial tamely ramied subextension) the terms appearing in S 0(T ) = ( T )are all coming from the leading monomial T n , so that putting n = n/ 1 we have

S 0(T ) =n

j =1

n 1 j

T 1 j

=n

j =1

n j

T 1 j

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= (1 + T 1 )n 1.

We collect these facts in the following proposition.

Proposition 7.1.4. If m 1 the polynomial S m (T ) is an additive polynomial,which is composed by more than one monomial if and only if m is a lower ramication break. For m = 0 we have S 0(T ) = (1 + T p

s)n 1, where n = ps n

and ( p, n) = 1 .

When the context is clear, we will abuse of notation and also denote by S mthe induced map S m () over the residue eld or an extension thereof.

7.1.1 Change induced on the coefficients by a substitutionWe study now the effect of replacing the minimal polynomial f (T ) of with

the minimal monic polynomial g(T ) of a different uniformizer .Lets take = + m +1 + . . . , we will identify the term ( f i gi )i thathas minimal valuation for general , which gives information about the mostsignicant change induced on the coefficients f i gi as consequence of thesubstitution .

The non-zero terms ( f i gi )i have valuations with different remaindersmodulo n , and furthermore we have

n 1

i =0(f i gi )i = f () g()

= f () = n ( m + . . . ),

considering the denition of . If m 1, being (mod p2K

) we obtain thefollowing Lemma, after dividing by n ( L/K (m )+1) and reducing the expressionmodulo pK .

Lemma 7.1.5. If m 1 and g(T ) is the minimal monic polynomial of an element of the form = + m +1 + . . . we have

(f () g()) n (L/K (m )+1) = S m ().

Since n | vL (f i gi ) for each i, the unique term ( f i gi ) i of f () g()that may contribute to the left hand side is for i satisfying

i n(L/K (m) + 1) (mod n),

so i is uniquely determined being 0 i < n . We observe that if m 1 isnot a lower ramication break then S m is surjective being K nite and henceperfect, while if m = ti for some i then it may not be surjective, when theadditive polynomial S t i (T ) has a root over the residue eld K .

Assume that t i is an integer, we will later show that the polynomial S t i (T )only depends on the eld extension L/K and on the class of mod p2 , as a

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consequence of a stronger result, Theorem 7.2.5, which is proved independently.For the moment we can give a denition of reduced polynomial without assum-ing this invariance, even though the denition will be less manageable from apractical point of view.

Let I m be the image of S m , and also its preimage in OL when the context isclear. Lemma 7.1.5 says that passing to the minimal polynomial of an elementof the form + m +1 + . . . , if n(L/K (m)+ 1) = jn + i with 0 i n, we canchange the corresponding term f i by an element of nj I m while all other termsf r r are unchanged modulo jn + i+1 . Since the polynomials S r for r m arecertainly unchanged too, this observation motivates the following denition.

Denition 7.1.6. Let f (x) be an Eisenstein polynomial, and assume that eachcoefficient f i has an expansion

f i =

j 1 f i,j jK

with f i,j R for a xed set of residue representatives R, and where K is a xeduniformizer of K , and let f = f 0/ K . Assume the choice of a set A0 K of representatives of K / (

K )

n , and for each additive polynomial S m (T ) form 1 a set of elements Am K that are representatives of the cokernel of the map jf S (), where j = [L/K (m) + 1].

We say that f (x) is reduced (with respect to the choice of the A i ) when wehave

1. f = f 0/ K is in A0 ,

2. for each m 1, if n (L/K (m) + 1) = jn + i for positive integers i, j withi < n , then we have f i,j Am

We say that f (x) is reduced up to the level r when condition 1 is satised, andcondition 2 holds for all m r .

If f (T ) is any Eisenstein polynomial, its easy to see that the polynomialn f ( 1T ) satises condition 1 for some suitable . A polynomial reduced upto level 0 can be obtained by the following algorithm.

Algorithm 1 Reduction (step 0)

f 0/ K , Representative (, (K )n ), Solve(T n = / ), Lift (),return n f ( 1T ).

If i, j are as above, we have shown above f i,j jK can be changed by anyelement in nj I m modulo nj +1 , so f i,j can be changed by any element of

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( n / K )j I m . Since n = f 0 + . . . we have that ( n / K ) = f , so f i,j is changedby an element of j

f I m , while f is unchanged when passing to the minimal

polynomial of an uniformizer of the form + m +1 + . . . .In particular, if f (x) is reduced up to the level m 1 we can obtain a

polynomial reduced up to the level m via the following reduction step.

Algorithm 2 Reduction (step m) j L/K (m) + 1 ,i n {L/K (m) + 1 }, f i,j , Representative (, image(jf S m )), Solve(jf S m (T ) = ), Lift (),

F (T ) T + T m +1

+ {any terms of degree m + 2 },return Resultant U (f (U ), T F (U )).

Indeed, if g(T ) is the returned polynomial we have

(f i,j gi,j ) jK i n (L/K (m )+1) = ( f i,j gi,j ) jf = S m (),

and consequently gi,j = Am . Since we allow any higher order term in thechoice of F (T ) = T + T m +1 + . . . , we anticipate that for a suitable F (T ) itwill not be necessary to compute the resultant appearing in the algorithm asthe determinant of a big matrix with coefficients in K [T ], see Remark 7.2.6.

Remark 7.1.7. If m is bigger than the biggest lower ramication break tk , then S

m(x) is surjective, and the function n(

L/K (m)+1) assumes as possible values

all integers > n (L/K (tk ) + 1) . Consequently we can arbitrarily change all the representatives f i,j whenever

vL ( jK i ) = nj + i > n (L/K (tk ) + 1) , (7.1)

without affecting the generated extension, turning them all to 0 for instance.In this way we recover the well known quantitative criterion on the distance of two Eisenstein polynomials ensuring that they generate the same extension, as considered in [ Kra62 , PR01 , Yos12] .

7.1.2 Characterizing reduced polynomialsWe start with a few remarks about Denition 7.1.6. Since we allow a different

choice of the representing sets Am for each m, where the 0 element of the imageof the map is not even requested to be represented by 0, we have that eachEisenstein polynomial is reduced for a suitable choice of the Am . This choice isvery far from what would be recommended in a computer algebra system, butit will be useful to be able to consider each Eisenstein polynomial as alreadyreduced.

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On the other hand on a computer algebra system we can expect to havea more or less canonical way for selecting representing elements of a quotient,and selecting 0 as representative of the zero element in the quotient. Under thishypothesis we clarify here how a reduced polynomial looks like. In particularwe will see that, for each 0 such that p divides n, the possible valuations of the terms f i,j jK

i such p i belong to one xed interval, with some exception .Fix and let us consider the terms f i,j jK T

i with p i, we deduce a lowerbound for the value of nj + i from the shape of the Newton polygon of theramication polynomial. Indeed, the contribution to the coefficient of T p in(T ) is

n f i,j jK i i

p,

and since the contributions coming from different monomials of f (T ) have dif-ferent valuations modulo n then their smallest valuation should be at leastnN ( p ) = . In the same way we obtain that any term f i,j jK T

i with p i andnj n + i is compatible with the ramication data, and when p and p is the abscissa of a vertex of the ramication polygon then there should bea term f i,j jK T

i such that the valuation nj n + i of the contributed term isexactly , this case corresponds to a vertex of the Netwon polygon and hencethe minimum is reached.

We will show now that all the terms f i,j jK T i with p i and such that nj + i

is big enough are turned to 0 by the reduction algorithm, with a few exceptions.Indeed, we claim that the integers that are multiple of p and > n ( ) = areall of the form n L/K (m) for some m > t (note that may not be a multipleof p itself, we are considering non-Galois extensions and the t r and (t r ) may

not be integers).To show the claim we work by induction on the number of ramication

breaks. If p < k 1 then = tk , and nN (1) is certainly an integer beingequal to vL (D L/K ), and n L/K (m) for integer m assumes as values all integersthat are > n L/K (tk ), being n L/K (x) equal to nN (1) + x for integer m > t k .Assume instead p k 1 , then by induction n k 1 L t +k 1 /K

(m) takes as values

any multiple of p / k 1 bigger than n k 1 L t +k 1 /K ( ) for integer m > . So

nLt +k 1

/K (x) satises the required property with respect to p , and so does

nL/K (x), which is obtained as the minimum of n Lt +k 1

/K (x) and nN (1) + x.

Consequently we have from the claim that all the terms f i,j jK with p i and

nj n + i can be forced to satisfy f i,j = 0, except possibly when nj n + iis itself equal to r for some r , in this case we can only force f i,j to be asuitable representative depending on the image of the polynomial S r (T ), whichmay not be surjective as a function over K .

In the case of three breaks we have the following gure representing thevalues nj n + i of the terms of a reduced polynomial.

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t3( 0)t2( 1)t1( 2)

1

2

3

p i

f i,j jK T i

p i

p2 i

p3

p2

p1

3

1

2

We state the above results in the following proposition.Proposition 7.1.8. Let f (x) be a reduced Eisenstein polynomial, and assume that each coefficient f i has an expansion

f i =

j 1f i,j jK .

Assume p i, then f i,j is non-zero only when

nj n + i < ,

or when nj n + i is equal to some r and the corresponding additive polynomial S r (T ) has a root in K .

In other words we have that starting from a certain points all terms f i,j jK for p i can all be simplied to 0, except at upper ramication breaks. We willlater see how this phenomenon can be interpreted in terms of local class eldtheory for abelian extensions, or in connection with Serre mass formula [Ser78]in some simple case.

7.1.3 Representation of automorphism as power seriesApplying such substitutions for increasing m we are taking into account alltransformations F () of by a power series without constant coefficient F (T ) =1T + 2T 2 + . . . that may provide an element whose minimal polynomial isreduced, because any such power series can be written as a composition of

polynomials of the form T and T (1 + T m ). Applying the above reductionstep for increasing m, when m is not equal to a ramication break t i we have aunique possible choice for the class of in the substitution (1 + m ).When m = t i for some i , the choice for is dened up to an element that is aroot of S t i (T ), and taking into account representatives for all possible choicesfor we can track all possible outputs. We can run this algorithm starting

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from the set {f (T )} and replacing each polynomial with the set of all possibleoutputs, which may not be unique at the ramication breaks ti , and do so upto the level tk . After this last step we obtain reduced polynomials turning to 0all the f i,j for i, j such that nj + i > n (L/K (tk ) + 1).

Algorithm 3 All reduced polynomials{t1 , . . . , t k } LowerRamificationBreaks (f (T ))A {f (T )}for m = 0 tk do

B for g(T ) A do

B B AllReductions (g(T ), m)end forA B

end fora [L/K (tk ) + 1]b n {L/K (tk ) + 1 }return A mod ( a +1 , a T b)

Since some outputs may be repeated we end with a multiset of reducedpolynomials. Clearly different power series F (T ), G(T ) may give the samevalue F () = G() when evaluated in , but we will show that we took intoaccount all the different values F () L such that the minimal polynomial of F () is reduced.

Indeed, in step 0 we considered all possible values for F () modulo p2L , andassume by induction that all the F () taken into account up to step m 1 coverall possible values modulo pm +1

L . The values F () + F ()m +1 + . . . covered

in step m, for all admissible representatives , will provide all possible valuesmodulo pm +2L .

Let i (K ) be the cardinality of K / (K )

n if ti = 0, and let it be thenumber of roots of S t i (x) contained in K if ti is an integer and > 0. Thecardinality of the multiset of polynomials obtained as output of the algorithmcan be computed counting for each m the number of possible choices, which isindeed equal to i (K ) for ti = m when there is no unique choice. The totalcardinality is equal to the product of the i (K ) over all i such that ti is aninteger, that is

BL/K =1 i k

t i Z

i (K ).

We give now an interpretation of the i (K ) as the number of automorphismof intermediate extensions. Indeed, if t i = 0 then i (K ) counts the number of n-th roots of the unity in K , or equivalently of n -th roots if n = ps n with(n , p) = 1, which is also the number of automorphisms of a tame extension of degree n of K , like L0+ /K is.

For ti > 0 lets consider the intermediate extension Lt +i /L t i : if g(T ) is the

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minimal polynomial of over L t i (which is a factor of f (T )) then

n ( i 1 (t i )+1) g( t i +1 T + ) = S t i (T ),

and consequently representatives of the roots of S t i (T ) are exactly those suchthat

()/ = 1 + t i + . . .

for some Lt i -automorphism Aut( Lsep /L t i ). Now after extending the ele-ments of t +i to the normal closure we have t +i (|L ) = t +i , this is immediateconsidering t +i as the image of elements of a ramication (normal) subgroupof a bigger Galois extension containing L. Consequently averaging over t +i weobtain

(Lt +i

)/ Lt +i

= 1 + i t iLt +i

+ . . . ,

where L t +i = N L/L t +i (). The equality holds because ti is smaller than theall ramication numbers of the extension L/L t +i , by keeping into account theproperties of the norm map N L/L

t +i(see [FV02, Chap. 3, 1, Prop. 1.5]).

If (Lt +i

) Lt +i then is in K , and on the other hand if K then

(Lt +i

) can be approximated better than any other conjugate of Lt +i

havingL t +i /L t i only one ramication break, and consequently (L t +i

) Lt +i by Kras-ner Lemma. In other words we have one root of S t i (T ) in K for each conjugateof L

t +icontained in L t +i , and i (K ) = # Aut( L t +i /L t i ).

So we have that BL/K is an invariant of the extension L/K . Consideringthe subgroups Aut( L/L t i ) of Aut( L/K ) and the corresponding quotients as

subgroups of Aut( L t +i /L t i ), we observe that BL/K provides a naive upperbound to the cardinality of Aut( L/K ), but which is in general tighter than thefull degree [L : K ].

Let f 1(x), . . . , f r (x) be all the reduced polynomials obtained applying theabove algorithm. The number of times we obtain the same polynomial f i (x) isequal to the number of distinct F j () such that f 1(F j ()) = 0, and is conse-quently equal to the number of roots of f i (x) contained in L, in another wordsto the cardinality of Aut( L/K ).

Theorem 7.1.9. Each extension L/K is generated by a reduced polynomial,and the number of reduced polynomials generating a xed extension L/K is

BL/K / # Aut( L/K ).

If f (x) is an Eisenstein polynomial such that a root generates an extension isomorphic to L, then the reduction algorithm outputs a multiset of cardinality BL/K formed by the reduced polynomials, each having multiplicity # Aut( L/K ).

We remark that if F (T ) is a power series such that F () is a conjugate of , the algorithm giving the set of special polynomials can collect all the used

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in the substitutions + m +1 to produce an expression of

F (T ) mod ( f (T ), T t k +1 ),

which can be used to realize the group Aut( L/K ) as group of truncated powerseries under composition, we omit the details of the construction.

Note that there is a unique reduced representative for Eisenstein polynomialsgenerating Galois extensions, while in general we have a set of polynomials withcardinality equal to the ratio of the naive bound on the number of automor-phisms to the real number of automorphisms. We remark that extinguishingthe redundancy from the above family of reduced polynomials seems to be atleast as hard as computing the cardinality of the automorphism group. This canprobably be done in a few particular cases, possibly for polynomials of degree p2over an unramifed extension of Q p, but a criterion to determine the cardinalityof the group of automorphisms is required.

When L/K has only one ramication break, or when L/K is Galois, thenwe have a unique representative, so the unique Krasner representative is easilydescribed in terms of the coefficients. This was already remarked in the originalKrasner paper [ Kra37 , page 167, after the proof of Theorem V].

7.1.4 Amano polynomials and Serre mass formulaWe provide here some qualitative observation, without being completely rig-orous. First, if the degree n is prime with p its easy to say what are reducedpolynomials, and they are all of the form T n + K for some representative Rsuch that A0 , where A0 is the chosen set of representatives of K / (K )n .

When n = p, Amano dened in [Ama71] a set of special generating poly-

nomials composed by trinomials. The equations considered here turn out tolook much more complicated visually because they are no longer trinomials,but the number of parameters is clearly the same, and nevertheless Amanopolynomials do not seem to be easily generalizable to higher degree.

Reduced polynomials of degree p are of the form

T p + p

i =1 pj + i ( p 1) t + p pj + i


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We give one last interpretation of this fact, under the light of the proof of Serre mass formula. Considering the map

uniformizers of ex-tensions of degree p

minimal polynomial Eisenstein polyno-mials of degree p

we have a p-to-1 correspondence between measure spaces, whose scaling factorturns out to be determined by the discriminant of the extensions as proven in[Ser78]. Lets restrict the map to the uniformizers of a xed extension L/K inthe algebraic closure, then either the extension is Galois and the map is still p-to-1, either the extension is not Galois and the map becomes 1-to-1, but inthis case the image has bigger measure.

In other words, for xed degree and restricting to extensions with xeddiscriminant, the smaller is the space of polynomials generating one xed iso-morphism class of extensions, the bigger will be the automorphism group of

these extensions. When applying the reduction algorithm to a polynomial of degree p generating L/K , we have that when the unique ramication break tis an integer and the additive polynomial S t (T ) is not surjective we can do less simplications to the coefficients of the Eisenstein polynomial. Since any Eisen-stein polynomial generating L is a possible output of the reduction algorithm(for a suitable choice of the Ai ) we have that the set of possible polynomialsgenerating L turns out to be smaller, and that L/K is Galois having somenon trivial automorphism and degree p.

For higher degree, and in particular when there are more ramication breaks,it becomes difficult to generalize this observation, because a modication thatappears to be trivial at the rst order may actually provoke some change tothe higher order terms in the expansions. This fact also justies the claim thatreducing the family to have exactly one polynomial for each isomorphism classappears to be at least as hard as the computation of the number of isomorphismsfor the extension determined by one Eisenstein polynomial.

7.2 A criterion to rule out possible reductionsTo complement the above reduction algorithm we give a synthetic criterion toexclude an Eisenstein polynomial from generating an extension of which weknow the set of all the reduced polynomials. In particular given two polynomi-als f (T ) and g(T ) we can often rule out early the possibility that a sequence of substitutions + m +1 + . . . , starting from level m = r say, may trans-form the minimal polynomial f (T ) of into the new minimal polynomial g(T ),without having to compute the complete reduction.

Lets consider the monomial ( f i gi ) i having smallest valuation, whichdetermines the valuation of f () g(), and assume that its valuation is equalto v = n(L/K (r )+1) for some real number r . Lets select sets of representativesAm that make g(T ) reduced, then we say that f (T ) can be reduced to g(T )greedily if g(T ) is a possible output of the reduction algorithm applied to f (T )starting from step m = r . The proof of the following proposition is clear.

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Proposition 7.2.1. If r is not an integer than f (T ) cannot be reduced greedily to g(T ).

We also have the following proposition, whose proof is immediate as well.

Proposition 7.2.2. If r is an integer equal to a lower ramication break and (f i gi ) i v is not in the image of S t i (T ), then f (T ) cannot be reduced greedily to g(T ).

These observations are well complemented by the following proposition,which makes them particularly effective in the case of Galois extensions.

Proposition 7.2.3. Assume that one of f (T ) or g(T ) is known to generate a Galois extension, then f (T ) and g(T ) generate the same extensions if and only if one polynomial can be greedily reduced to the other.

Proof. For a suitable choice of representatives g(T ) is already reduced, and forGalois extensions there is only one reduced polynomial in view of Theorem 7.1.9,so applying greedily the reduction algorithm to f (T ) we obtain g(T ) as uniquepossible output. The other implication is clear.

In other words for Galois extensions if g(T ) can be obtained in some wayfrom f (T ), then it can also be obtained in the greedy way.

When considering wildly ramied Galois extensions over Q p the S t i (T ) arethe zero map over the residue eld F p, so if r is a ramication break thenf (T ) and g(T ) certainly generate non-isomorphic extensions, and we essentiallyrecovered the main result of [Yos12].

However, it is possible to give a deeper criterion, which is more selectivethan what can be obtained via an inspection of f () g() at the rst order.

Consider the range of monomials f i,j j

K T i

corresponding to one ramica-tion break as described in Prop. 7.1.8, then the intuitive idea is that if we canobtain f (T ) from g(T ) applying reductions of parameter m r then the rst rterms in each such interval must be equal, because such terms are not going tobe changed by any reduction of order r . Such ranges can be independentlybrought up to the front (with respect to the p-adic valuation) computing for-mally a ramication polynomial of f (T ) g(T ), and considering the coefficientsof T , T p, T p

2, . . . , as we can see observing the contributions to the coefficient of

T p in the ramication polynomial. Consequently taking into account a rami-cation polynomial for f (T ) g(T ) provides a synthetic and effective formalismto describe how some sets of coefficients must be equal in order to be able topass from f (T ) to g(T ) via reduction step.

What we are going to prove is closely related to what was done in [Hei96]

and Theorem 4.6 in particular, and shares the philosophy that the sets of mono-mials f i T i with a xed valuation of i live an independent life from the othermonomials, up to a certain extent, and that when a uniformizer is changed + m +1 + . . . the change induced on minimal polynomial satises acertain continuity (in [Hei96] a different kind of dening equation formed by apower series with coefficients in a set of representatives was used rather than

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Eisenstein polynomials, but the underlying principle is the same). From a moreeffective point of view, such a continuity provides an easily veriable criterionto exclude a polynomial from generating one xed extension, which is particu-larly effective in the case of Galois extensions thanks to Prop. 7.2.3. What weneed seems not to follow directly from the results of [Hei96] and additional stepswould be needed to switch to power series and back to Eisenstein polynomials,so we will avoid using the slightly cumbersome notation of [Hei96] and proveour result directly.

For integers a 0 and w lets dene P pa , resp. P pa (w), as the modulegenerated over OK by the monomials cT i such that pa | i, resp. those monomialssuch that additionally vL (c) + i w. If cT i P pa (w) for some w, than we have

c T + T m +1 + . . .i cT i

a

j =0P pj (w + eL (a j ) + pj m) (7.2)

as we can verify at once expanding the left hand side. Furthermore if g P pa (w)and h P pb (z) than clearly we have gh P pmin { a,b } (w + z).

Lets consider the ramication polynomial ( T ) = n f (T + ), then thecoefficient of T p

ais has valuation at least a . Assume pa i, from a monomial

f i T i we have a contribution i pa f i i n T p

ato the coefficient of T p

ain (T ),

so vL (f i ) + i n should be at least a , and consequently the monomial f i T i iscontained in P pa ( a + n), being vL (f i ) + i a + n.

Consequently we have obtained that

f (T ) s

j =0

P pj ( j + n), (7.3)

where s is the biggest integer such that ps | n (recall that s = 0).What observed above we obtain the following.

Proposition 7.2.4. Let F (T ) = T + m +1 T m +1 + m +2 T m +2 + . . . , then

f (T ) f (F (T )) mods

j =0

P pj ( j + n + pj m). (7.4)

Proof. Lets consider f (T ) f (F (T )), we will show that a monomial f i T i , whichis contained in P pa ( a + n) by (7.3) say, yields various terms each having valu-ation at least j + n + pj m and in P pj , for some j < a . But we obtain terms inP pj ( a + n + eL (a j )+ pj m) by (7.2), and a + eL (a j ) j by Lemma 7.1.3.

Assume = F () for a root of g(T ), we have now obtained a congruenceproperty for the power series f (F (T )), which clearly satises f (F ()) = 0. Theminimal polynomial of is clearly a factor of f (F (T )) of degree n , and observethat the valuation the coefficient of T n is 0 while the constant term has valuation1, so its Newton polygon has exactly one side of length n and slope 1/ n . Inparticular g(T ) is obtained by the factorization along the Newton polygon, or

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equivalently collecting the roots with positive valuation, which is exactly whatis provided by p-adic Weierstrass Preparation Theorem.

We will however show a reduction that allows to approximate the minimalmonic polynomial of starting from f (F (T )), and keeping the congruence ( 7.4).Lets start putting h1(T ) = f (F (T )), and consider the polynomial H 1(T ) ob-tained taking the monomials of degree n of h1(T ) T n . If H (T ) = 0 thenthere is no such monomial, and g(T ) is a monic polynomial of degree n , whichis Eisenstein being F () a root.

Let cT r a monomial of H 1(T ) that minimizes the quantity vL (c) + r , andtake the monomial with r as big as possible among those achieving the minimumof vL (c) + r , which are in a nite number. In other words, consider the highervaluation on OL [[T ]] dened as

(cT r ) = ( 1(cT r ), 2(cT r )) = ( vL (c) + r, r ) Z2

where Z2

is ordered lexicographically. We take cT r

to be the monomial of H 1(T )minimizing (cT r ).Lets replace now h1(T ) with the new polynomial

h2 = h1(T ) h1(T ) cT r n = h1(T ) 1 cT r n .

Apply iteratively such step. At the i-th step say, either the minimum of thequantity vL (c) + r for the monomials of degree n of H i (T ) is increased, eitheris decreased the biggest degree of the monomials achieving the minimum. Sincethe whole computation is done in OK [[T ]], the latter can only happen a nitenumber of times, and such minimum is increased after a nite number of steps.

Algorithm 4 Lifting stepH i (T ) {sum of monomials of degree n of h i (T ) T n }

cT r

(monomial of H i (T ) minimizing )return hi (T ) (1 cT r n )

After a sufficient number of iterations we can replace h i (T ) with the polyno-mial h(T ) formed by T n plus the monomials of degree < n of h i (T ). We obtainan Eisenstein polynomial such that h(F ()) is arbitrarily small, so h(T ) is itself an arbitrarily good approximation of the minimal polynomial of F ().

We need to show that while the above procedure approximating g(T ) iscarried on the congruence satised by f (F (T )) is preserved. Indeed, assumethat the congruence is satised by hi (T ) and let hi +1 (T ) = h i (T )(1 cT r n ).Then cT r P p ( + n + p m) for some , being cT r a monomial of H i (T ) andin view of the congruence that we assume to be satised by f (T ) and hi (T ).Let bT s be a monomial of h i (T ), then bT s P pk ( k + n) for some k by equation(7.3) and by the congruence satised by h i (T ). If k < we have

bT s cT r n P pk ( k + n + + n + p m n) P pk ( k + n + pk m),

while when k we have

bT s cT r n P p ( k + n + + n + p m n) P p ( + n + p m).

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We obtained that subtracting hi (T ) cT r n from hi (T ) preserves the congruence.Considering also an analogue of a ramication polynomial for f (T ) g(T ) wehave the following theorem.

Theorem 7.2.5. Let f (T ) be an Eisenstein polynomial of degree n and a root, if g(T ) is another Eisenstein polynomial of degree n having K () as root, and = + m +1 + . . . then we have that

f (T ) g(T ) mods

j =0

P pj ( j + n + pj m),

and the polynomial

f ( + T ) f () g( + T ) + g()

has its Newton polygon contained in the Newton polygon of f ( + m +1 T ).

Proof. We only need to prove the second assertion, but if cT r is in P pj ( j + n + pj m) then for each k j the contribution of c( + T )r to the coefficient of T p

k

has valuation at least j + n + pj m + ( j k)eL , which is at least k + n + pk mas shown above.

Remark 7.2.6. We point out that the algorithm used during the proof to recover (an approximation of) the minimal polynomial of = F 1() can be used toproduce the minimal polynomial of a uniformizing element obtained deforming in a much quicker way than by computing a resultant Res U (g(U ), T (U +U m +1 )) as the determinant of a (n + m) (n + m) matrix with coefficients in OK (T ). Consequently taking F (T ) = T T m +1 and computing via the above approximation the minimal polynomial of the uniformizer such that =

m +1

, we obtain the minimal polynomial of a uniformizer = + m +1

+ . . . ,and this observation allows to exploit the free choice of F (T ) in Algorithm 2 toavoid the computation of the resultant.

7.3 Construction of totally ramied class eldsIn this section we show how it is possible to convert a norm subgroup, repre-senting a totally ramied abelian extension via local class eld theory, into theunique reduced Eisenstein polynomial generating the extension.

We suppose given a nite index closed subgroup N K such that NU 0,K =K , so that the corresponding extension by local class eld theory is totallyramied. Being closed we have N U u for u sufficiently big, this hypothesisis automatically satised when K is a nite extension of Q p and N has niteindex.

We assume that N is described by a set of linear maps, one for each upperramication break. That is for all u 0 such that U u,K NU u +1 ,K we assumegiven a surjective homomorphism

u : N U u,K V u

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having kernel exactly equal to NU u +1 ,K , for some abstract group V u , which isnaturally an F p-vector space for u 1. Take u to be the trivial map to thetrivial group 1 when u is not an upper break, that is N U u,K = N U u +1 ,K . Notethat the knowledge of all the maps u determines uniquely the group N .

The map 0 , when non-trivial, gives a condition on the representative f 0,1 ,or equivalently on the residue class f 0 / K , this correspond to the well knownexplicit description of local class eld theory for tamely ramied extensions. Onthe other hand the terms appearing in a reduced polynomial in connection tothe cokernels of the polynomials S t i (T ) attached to the lower breaks t i are all of the form f 0,j jK , because the upper breaks are integers by Hasse-Arf Theorem.

Consequently the choice of such representatives f 0,j is determined by thecondition that f 0 should be a norm from the extension determined by N , anda suitable f 0 can be selected changing appropriately K .

The ramication data is described by the upper breaks u 1 and the dimen-

sions of the corresponding V u . After selecting f 0 we have a well dened skeletonfor the reduced Eisenstein polynomial, formed by a set of terms f i,j jK T i with

i = 0, where the f i,j will be considered as unknowns in the set of representativesR. We will describe how it is possible to recover the f i,j from the maps u .

The terms f i,j in a xed range as in Prop. 7.1.8 can be evaluated at thelevel when p i say, making use the map /n 1 as we will now show. If m issuch that nj + i + p m = n + n it will be possible to describe the dependenceof N K ( ) /K (1 m ) on the coefficient f i,j at the rst order, obtaining a linearsystem from /n 1 .

Denition 7.3.1. If p +1 |n, we dene R to be the set of pairs ( i, j ) such that j 1, 0 i < n , p i and

+ n nj + i < + n.

We assume that R is ordered depending on the value of nj + i. We deneM (i, j ) to be the number m such that

nj + i + p m = + n.

We remark that if the extension is abelian then n | by Hasse-Arf theorem,so the m dened above is always an integer and prime with p.

7.3.1 Dependence of norms on a f i,jWe will now track the dependence of a norm N K ( ) /K (1 m ) on a representa-tive f i,j appearing in the expansion of a coefficient. To do so, lets treat f i,j as

an indeterminate, and apply a sufficient number of steps of Algorithm 4 to passfrom f (T + T m +1 ) to the minimal polynomial g(T ) of , where = + m +1 .

Clearly = m +1 + . . . , and g0 /f 0 will be the norm of an element of the form 1 m + . . . . For some r > m , the changes induced changing f i,j onall N K ( ) /K (1 r + . . . ) turn out to be even smaller p-adically, so it will bepossible to ignore any term that is O(m +1 ).

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In the expansion of f (T + T m +1 ) a term f i,j jK (T + T m +1 )i appears,

and it has f i,j jK

T i as main term. In the algorithm we start with h0(T ) =f (T + T m +1 ), and at the i-th step we subtract h i (T ) cT r n from h i (T ), wherecT r n P pk ( k + pk m) for some k s. From the monomial f i,j

jK T

i the otherterms in f i,j that may appear in the algorithm have coefficient with valuationat least

nj + i + min0 k s

{ k + pk m} = nj + i + n L/K (m),

with respect to the mixed valuation 1(aK T b) = na + b on OL [[T ]].Note that the minimum of k + pk m is obtained as the minimum of the

piecewise linear function nN (x) + mx , which is nL/K (m) in view of whatproved before proposition 7.1.1. We will denote for convenience this quantity as

Ai,j (m) = nj + i + nL/K (m).

The term f i,j jK T i is the main term coming from f i,j jK (T + T m +1 ) i , andthe second contribution can be found considering the expansion

(1 + T m )i = 1 + i p

(T m ) p + i

p 1(T m ) p

1+ . . . .

Putting as usual p i, we denote the valuation of the second term as

B i,j (m) = nj + i + min0 k

eL ( k) + pk m .

Such term is equal to i p

jK T i (T m ) p

as long as mp < e L + mp 1 , that is m < eL / ( p p 1 ) , and if m M (i, j ) this condition is certainly satised because

m e L / ( p +1 p )

by Lemma 7.1.3.By Proposition 7.1.8 we can assume < s, and if m we always have

B i,j (m) = nj + i + p m < A i,j (m).

So the main contribution to N K ( ) /K (1 m ) = g0 /f 0 originated from f i,j jK T i

is only coming from f i,j i p jK T

i (T m ) p .When m = M (i, j ) we obtain a condition on f i,j to have N K ( ) /K (1 m )

in the norm group for each , which can be used to determine the representativef i,j .

This can be made to work when only one representative f i,j is unknown, buta more rened study is needed if we have to determine them all. In particular,we will see that there exists an ordering of such unknowns that allows to de-termine them all inductively. What complicates this idea is that it will to be

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necessary to interleave in a suitable way the ranges of representatives consideredin Prop. 7.1.8.

It will be convenient to write down a comfortable lower bound for the func-tions A i,j and B i,j , obtaining a function describing the biggest quotient OK / p uK where we can ignore the value of f i,j while computing N K ( ) /K ( m +1 ). Inparticular we can take

C i,j (m) = nj + i + min0 k

k + pk m ,

where = v p(i) as usual. We resume the properties proved in the followingLemma.

Lemma 7.3.2. Denote with a root of f (T ), and let (i, j ) R . Then, for each U K , the value of

N K ( )/K (1 m

) mod puK

does not depend on f i,j , whenever u is C i,j (m)/n 1. If m = M (i, j ) then C i,j (m) = + n , and for u = C i,j (m)/n 1 = L/K ( ) we have

N K ( ) /K (1 m ) = N K ( 0 ) /K (1 m0 ) +

uK f i,j i,j

p + . . . , (7.5)

where i,j is a xed unit dened as

i,j = i p

( f 0/ K )( i + p m )/n 1 ,

and 0 is a root of the polynomial obtained from f (T ) setting f i,j to 0.

Proof. We just have to prove the ( 7.5). For m = M (i, j ) the variation of theconstant term comes from the monomial f i,j i p

jK T

i (T m ) p in the expansionof f (T + T m +1 ), and during the reduction each T n is transformed into f 0 .Dividing by f 0 we obtain that the variation for N K ( ) /K (1 m ), which modulou +1K is

f i,j jK i p

( f 0)(i + p m )/n 1 p = uK f i,j i,j p .

We will now assume that some f i,j have been determined and some not yet,and will show that it is possible to determined some of the unknown ones. For such that p +1 n, consider the range of terms R like in Prop. 7.1.8, and let

(i , j ) be the smallest pair ( i, j ) R (i.e. the pair in R with nj + i as smallas possible) such that the corresponding f i,j has not been identied yet.

We rst prove a couple of technical lemmas about the functions C i,j (x).

Lemma 7.3.3. The functions C i,j (x) are strictly increasing, and for (i, j ) =(i , j ) then the functions C i,j (x) and C i ,j (x) are always different except p

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