Maximal tame extensions over Hopf orders in rings of integers of p-adic number fields

We recall the definition of a Larson order L in brief. Let L be a subalgebra o






 i σ p − 1 /π ti 0
i
Then by [5, (17.1) Theorem and (18.1) Theorem], we have L is a Hopf order iff pti
ti+1

(0
i < n − 1) and

ti
valk (p)/pn−i−1 (p − 1) = pi e0

(0
i < n).

(3)

In this paper, we call L a Larson order if integers ti satisfy (3). Then a set 



σ

pi

li

− 1 /π

ti li



| 0
li < p (0
i < n)

0
i
is clearly a basis of L over o. Next we define elements a0 , a1 , . . . , an−1 by 

ai =

i

ζ p j en−i−1 (j )

(4)


 i
ζ p j en j
+ pn−i−1 j

.

(5)

0
j

(cf. [9, p. 719]). Then we see by (2) 



0
j

0
j

ai =

For brevity, we denote en (j ) by e(j ), then by (5) i

σ p − ai =

 0
j

=



j
,j



i

ζ p j e(j ) −



 i

ζ p j e j
+ pn−i−1 j



j
,j



 
i

n−1

ζ p j ζ p j − 1 e j
+ pn−i−1 j

.

(6)

Y. Miyata / Journal of Algebra 276 (2004) 794–825

Let G be the factor group G/σ p G → G. Then we have by (4) ai =



ζ

pi j

0
j



i+1

797

 and σ be the image of σ under the canonical map



ζ

−p i+1 j l p n−(n−i−1) l

σ

 /pn−i−1 = 1.

(7)

0
l

Define ri by ri = pn−1 e0 − tn−i−1 , i

and let H = o[{(σ p − ai )/π ri }0
i
 σ i , σ j = ζ ij .

Then clearly 

 e(i), σ j = δij .

(8)

Let L∗ be a dual lattice of L, so L∗ is a Hopf order (cf. [5, (1.4)]). Proposition 1. H ⊆ L∗ . Proof. Write j in the form j = j0 + j1 p + · · · + jn−1 pn−1 with 0
jh < p. Put  i h x = (σ p − ai )/π ri , 0
h
Since (σ p − 1)jh =




jh  jh −lh σ p h lh , 0
lh
jh lh (−1)

by (8)

     
  j0 jn−1 x = 1/π ri + th jh (−1)j0 −l0 · · · (−1)jn−1 −ln−1 l0 ln−1 lh ,h

×ζ where

 lh ,h




x = 1/π

p i (l0 +pl1 +···+p n−i−2 ln−i−2 )

denotes the summation   

  ri + th jh

lh ,h



0
l0
j0

 0
h


0
l1
j1


pn−1 l  n−i−1 ζ −1 ,

···



0
ln−1
jn−1 .

   jh jh −lh p i p h lh (−1) ζ lh

   
n−1 jn−i−1 × (−1)jn−i−1 −ln−i−1 ζ p ln−i−1 − 1 ln−i−1

Then we have

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Y. Miyata / Journal of Algebra 276 (2004) 794–825

 × 




= 1/π

  ri + th jh

    jh (−1)jh −lh lh n−i−1
0
h
    
 jn−i−1 jn−i−1 −ln−i−1 p n−1 ln−i−1 × ζ −1 (−1) ln−i−1 ln−i−1



 (1 − 1) , jh

n−i−1




where li denotes the summation 0
li
ji . To investigate x, we separate three cases. In case jn−i pn−i + · · · + jn−1 pn−1 = 0, then for some h
, jh
= 0, so (1 − 1)jh
= 0. Thus 

(1 − 1)jh = 0,

n−i−1
so x = 0 ∈ o. In case jn−i pn−i + · · · + jn−1 pn−1 = 0 and jn−i−1 = 0,   
pn−1 ·0  0 ζ −1 (1 − 1) = 0, n−i−1
so x = 0 ∈ o again. In case jn−i pn−i + · · · + jn−1 pn−1 = 0 and jn−i−1 = 0, we have    

pi+h jh 
pn−1 j ri + th jh ζ x = 1/π ζ −1 − 1 n−i−1 − (1 − 1)jn−i−1 . 0
h
By (1) and the definition of ri ,  valk (x) = − pn−1 e0 − tn−i−1 + =

 0
h


 th jh +



jh pi+h e0 + jn−i−1 pn−1 e0

0
h

 
jh pi+h e0 − th + (jn−i−1 − 1) pn−1 e0 − tn−i−1 .

0
h
By (3), valk (x)  0, so (σ p − ai )/π ri ∈ L∗ , which completes the proof. 2 i

Applying similar arguments as in the proof of Proposition 1, we can prove Proposition 2. ai ∈ L∗ for 0
i < n. A subalgebra L
of kσ p  is defined by 


pn−2 

L
= o σ p − 1 /π t0 , . . . , σ p − 1 /π tn−2 , and (L
)∗ is defined to be the dual of L
in kσ p  similarly as for L in kσ .

(9)

Y. Miyata / Journal of Algebra 276 (2004) 794–825

799

Lemma 1. (i) L
is a Larson order. (ii) a0 ∈ (L
)∗ . Proof. (i) As ti
valk (p)/(pn−i−1 (p − 1)) = pi e0 ,
 valk (p)/ p(n−1)−i−1 (p − 1) − ti = pi+1 e0 − ti > 0, then t0 , . . . , tn−1 satisfy the condition (3) for L
and hence L
isa Larson order. (ii) We remark en−1 (j ), σ pi  = δij in kσ p . By (5), a0 = ζ j en−1 (j ), so    
h j σ p − 1 h /π h th−1 jh a0 , 1
h
=

      jn−1 j1 n−2 (−1)j1 −l1 · · · (−1)jn−1 −ln−1 ζ l1 +···+p ln−1 /π th−1 jh l1 ln−1 lh ,h


n−2 j = (ζ − 1)j1 · · · ζ p − 1 n−1 /π th−1 jh .

By (1) and (3),  valk

a0 ,




h

σp − 1

jh

   

h−1   valk ζ p − 1 − ph−1 e0 jh = 0, /π h th−1 jh 

h

so a0 ∈ (L
)∗ .

h

2

Let H
= H ∩ kσ p  and H the image of H under the canonical map G → G/σ p . Then by (7), a0 = 1 and so H = o[(σ − 1)/π r0 ]. Then we have a Hopf short exact sequence 0 → H
→ H → H → 0 (cf. [5, p. 32]), and H
= o


pi 

σ − ai /π ri 1
i
(10)

Let [L1 : L2 ] denote the module index of lattices L1 and L2 over o (for definition, cf. [3, p. 10]). We have

 [L1 : L2 ] = L∗2 : L∗1  (cf. [3, p. 11, Proposition 3]). For the maximal order M = 0
j
800

Y. Miyata / Journal of Algebra 276 (2004) 794–825

Lemma 2. [L∗ : oG] = pnp

n /2

π −(p−1)p

n(



0
h
.

We are ready to prove Theorem 1. Theorem 1. Let H, L be as above and L∗ the dual of L in kG. Then H = L∗ . Proof. We use induction on the exponent n. For n = 1, by [5, p. 110, Corollary 21.2], H = L∗ . Assume the result holds for the smaller exponent than n. By (9) and the inductive assumption, we have

∗  p   r





pn−2




L = o σ p − a0
/π r0 , σ p − a1
/π r1 , . . . , σ p /π n−2 , − an−2  i where ri
= pn−1 e0 − t(n−1)−i−1 and ai
= 0
j
ai
=

ζp

i+1 j

en−(i+1)−1 (j ) = ai+1 .

0
j

Then by (10), it follows

∗ L = H
.

(11) p

Put y = (σ − a0 )/π r0 , so σ = a0 + π r0 y. We see a0 = yp +

 0
j

ζ pj en−1 (j ) = σ p and

 p  p−h a π r0 (h−p) y h = 0. h 0

1
h
Then by (1) and the definition of r0 , for 1
h < p, valk (p) + (h − p)r0 = (h − 1)pn−1 e0 + (p − h)tn−1  0, so


p r (p−h) 0 ∈ o. By Lemma 1(ii), a0 ∈ H
, so h π H = H
+ H
y + · · · + H
y p−1 .

Let H
[σ ] = H
+ H
σ + · · · + H
σ p−1 , so H
[σ ] = H
+ H
(σ − a0 ) + · · · + H
(σ − a0 )p−1 . Then we see [H : H
[σ ]] = π ((p−1)pr0/2)p σ p , 

n−1

. By (11) and Lemma 2 with replaced G by




∗ 

n−1  n−1 H
: o σ p = L
: o σ p = p(n−1)p /2 π −(p−1)p ( 0
h
Y. Miyata / Journal of Algebra 276 (2004) 794–825

801

Thus 





 

 H : o[σ ] = H : H
[σ ] H
[σ ] : o[σ ] = H : H
[σ ] H
[σ ] : o σ p [σ ] = p(n−1)p = pnp

n /2

n /2

π (p−1)p

π −(p−1)p

n(

n (p n−1 e −t n 0 n−1 )/2−(p−1)p (



h


h
.

By Lemma 2, we have [L∗ : oG] = [H : o[σ ]], and L∗ = H because L∗ ⊇ H by Proposition 1. The proof of Theorem 1 is completed. 2

i

2. Actions of elements (σ p − ai ) As in the introduction, let K/k be a cyclic Kummer extension of degree pn with the ith ramification numbers ci (1
i
n) and assume there is an one-unit α satisfying σ (α) = ζ α and (valK (α − 1), p) = 1. We see, by [8, Lemma 2], valK (α − 1) + c1 = pn e0 and so
i

n−1   valK α p − 1 = pi pn e0 − c1 < pn−1 pn e0 = valK ζ p − 1 .

(12)

By [8, Proposition 3], we have
 ci = c1 + pe + · · · + pi−1 e = c1 + pn e0 pi−1 − 1 .

(13)

Define integers dj by  j
n−1 j 



dj = valK (α − 1)j0 α p − 1 1 · · · α p − 1 n−1 /pn , where [x] denotes an integer n with n
x < n + 1. Then from (valK (α − 1), p) = 1, it  h follows {( 0
h
   
h j o α p − 1 h /π dj . 0
h
We may assume σ α = ζ α and know e(j )α l = δj l α l for 0
j, l < pn . Then by (6), for j = j
+ pn−i−1 j

,


  i i

n−1

σ p − ai α j = ζ p j ζ p j − 1 α j .

(14)

802

Y. Miyata / Journal of Algebra 276 (2004) 794–825

Lemma 3.




j
n−1 j  i σ p − ai (α − 1)j0 α p − 1 1 · · · α p − 1 n−1 j
n−2 n−i−2 j
i − 1 n−i−2 = ζ p α − 1 0 · · · ζ p αp 
n−1 n−i−1 j
n−i−1 j × ζ p αp − 1 n−i−1 − α p − 1 n−i−1
n−i j
n−1 j × α p − 1 n−i · · · α p − 1 n−1 .

Proof. By (14), we have


i

σ p − ai



  
ph j α −1 h 0
h
 

 
i    jh n−1 = σ p − ai (−1)jh −lh α l0 +pl1 +···+p ln−1 lh lh ,h 0
h
  jh    h
n−1 l i n−i−2 l n−i−2 ) ζ p n−i−1 − 1 α h p lh (−1)jh −lh ζ p (l0 +pl1 +···+p = lh lh ,h h     jh  i h h = (−1)jh −lh ζ p p lh α p lh lh h


−

0
ln−i−1
jn−i−1



×

n−i−1
 

jn−i−1 n−i−1 (−1)jn−i−1 −ln−i−1 α ln−i−1 p ln−i−1

   jh jh −lh p h lh α (−1) , lh lh

2

from which we can conclude Lemma 3. Let m =




0
h
σ p − ai

li

h

=

with 0
mh < p, so by (6), for 0
li < p, 

i

ζp (



h
h )l

i




ζp

n−1 m n−i−1

l − 1 i e(m)

0
m

and   
i li  
pi (
li  p mh p h )li p n−1 mn−i−1 h
mh ,h

0
i
Y. Miyata / Journal of Algebra 276 (2004) 794–825

803

Then we have


i

σ p − ai =

li

  
ph j α −1 h 0
h


i

ζp (



h
h )l

i




ζp

n−1 m n−i−1

  l − 1 ie p h mh

mh ,h

h

         jh h (−1)jh −mh α h p mh × mh m =



ζ

0
h
ζ

−1

li

mh ,h

   jh h (−1)jh −mh α h p mh mh h     jh  jh −mh p h mh (−1) = α mh mh ,h h>n−i−1     
li pn−i−1 m jn−i−1 jn−i−1 −mn−i−1 p n−1 mn−i−1 n−i−1 ζ −1 α × (−1) mn−i−1 mn−i−1    i h  jh  p li p mh jh −mh p h mh × α ζ (−1) mh hn−i−1 mh     
n−1 l n−i−1 m jn−i−1 n−i−1 × (−1)jn−i−1 −mn−i−1 ζ p mn−i−1 − 1 i α p mn−i−1 mn−h−1     i h  jh  h × ζ p li p mh (−1)jh −mh α p mh . (15) mh m ×



h
h

 i − 1)jh by the action of 0
i
i0 (σ p − ai )li  i for some fixed i0 . Using induction on i0 , we verify that for 0
i


0
h
ph

    
pi li
ph jh σ − ai α −1 0
i
=





n−i0
h
mh

0
h
  
ln−h−1 m ph jh jh −mh p h mh ( 0
i
804

Y. Miyata / Journal of Algebra 276 (2004) 794–825



×






(

ζ

 0
i
p i li )

α

p h

−1

jh  .

(16)

0
h
For i0 = 1, we obtain the result by (15). Assuming (16) (for i0 ) holds, we shall verify (16) for i0 + 1. Let h0 = n − i0 − 1. Then by (15) and (16) we can deduce  


i

σ p − ai

0
i
i0

li

 


h

αp − 1

jh



0
h

i0 = σp

     j l
pi l
ph − ai0 i0 σ − ai i α −1 h 0
i
0
h
      jh  l h i
n−1 h = (−1)jh −mh ζ p mh ( 0
i
×



   h ln−h −1 mh ph0 jh0 jh0 −mh0 p 0 mh0 ( 0
i
mh0



×

h



0
h
n−i0 −1
h



×

mh





=

   h i jh jh −mh p mh ( 0
i
i0 p li ) mh p h (−1) ζ α mh

mh

   l h i
jh n−1 h (−1)jh −mh ζ p mh ( 0
i





ζ

(

 0
i
i0

p i li )

α

p h

−1

jh 

.

(17)

0
h
Moreover we have by (17)     
pi l
ph j σ − ai i α −1 h 0
i
0
h
    jh  l h i
n−1 h = (−1)jh −mh ζ p mh ( 0
i
h

We have the following lemma. Lemma 4. Let αh = ζ (



i 0
i
α for 0
h < n. Then


ph
h   valK αh − 1 = valK α p − 1 and

(18)

Y. Miyata / Journal of Algebra 276 (2004) 794–825

valK

 


i

σ p − ai

li

  
ph j α −1 h

0
i


=

805

0
h


n−h−1 l
ph j  valK σ p − an−h−1 n−h−1 αh − 1 h .

0
h
h

Proof. By (12), valK (αh − 1) = valK (α p − 1). Let i = n − h − 1, then by (15),


σp

l
ph j − an−h−1 n−h−1 αh − 1 h 
p h
0 
pi ph
ph
0 αh − 1 αh − 1 ζ

ln−h−1

=

h
>n−i−1





×

h



0
mh
jh

  l
n−1 jh ph m (−1)jh −mh ζ p mh − 1 n−h−1 αh h , mh

because jh
= 0 for h
= n − i − 1 in (15). Therefore by (18) valK

 


i

σ p − ai

li

  
ph j α −1 h

0
i
n

=



0
h


n−h−1 l
ph j  valK σ p − an−h−1 n−h−1 αh − 1 h .

0
h
In the following, for brevity, for h = n − i − 1, put j = jh ,
 p h i
ph β = ζ ( 0
i

and θ = ζ p

n−1

.

h

Then we note valK (β − 1) = valK (α p − 1). By Lemma 3,


 i σ p − ai (β − 1)j


j = (θβ − 1)j − (β − 1)j = β − 1 + β(θ − 1) − (β − 1)j  j  = (θ − 1)l β l (β − 1)j −l l 1
l
j

=

      j  l (θ − 1)l (β − 1)m (β − 1)j −l l m l

=

 0
j −l+m
j



0
m
l

 0
m
l
j 0
    j l l (θ − 1) (β − 1)j −l+m . l m

(19)

806

Y. Miyata / Journal of Algebra 276 (2004) 794–825 i

Thus we can write (σ p − ai )(β − 1)j in the form


  i σ p − ai (β − 1)j = j (θ − 1)(β − 1)j −1 + aj,1,h (β − 1)h 0
h
j

with aj,1,h ∈ o.

2

Lemma 5. (i) aj,1,j ∈ o(θ − 1), aj,1,j −1 ∈ o(θ − 1)1+1 , aj,1,h ∈ o(θ − 1)j −h for h < j − 1. i (ii) valK ((σ p − ai )(β − 1)j ) = j valK (β − 1) + (valK (θ − 1) − valK (β − 1)). Proof. (i) Take the exponent j − l + m in (19) to be j , then l = m, so  j l  (θ − 1)l ∈ o(θ − 1). aj,1,j = l l 1
l
j

Next we consider the case with the exponent h = j − 1. Then j − l + m (in (19)) = j − 1, so m = l − 1, hence m = 0 for l = 1. Thus  j  l   j  l  (θ − 1)l = j (θ − 1) + (θ − 1)l , l l −1 l l−1 1
l
j

2
l
j

so aj,1,j −1 ∈ o(θ − 1)1+1 . For h < j − 1, h = j − l + m (in (19)), so l = j − h + m  j − h and aj,1,h ∈ o(θ − 1)j −h . (ii) For h = j − l + m,

 valK (β − 1)h (θ − 1)l − valK (β − 1)j −1 (θ − 1)
 = (l − 1) valK (θ − 1) − valK (β − 1) + m valK (β − 1)  0. The equality holds iff l = 1 and m = 0, which completes the proof. 2 i

Proposition 3. For 1
l
j , (σ p − ai )l (β − 1)j is written in the form


l i σ p − ai (β − 1)j = j (j − 1) · · · (j − l + 1)(θ − 1)l (β − 1)j −l  + aj,l,h (β − 1)h with aj,l,h ∈ o. 0
h
j

Then (i) and (ii) hold (i) For h > j − l, aj,l,h ∈ o(θ − 1)l and for h < j − l, aj,l,h ∈ o(θ − 1)j −h . Moreover for h = j − l, aj,l,h ∈ o(θ − 1)l+1 . i (ii) valK ((σ p − 1)l (β − 1)j ) = j valK (β − 1) + l(valK (θ − 1) − valK (β − 1)).

Y. Miyata / Journal of Algebra 276 (2004) 794–825

807

Proof. (i) We use induction on the exponent l. For l = 1, the assertion follows immediately from Lemma 5. Applying the inductive assumption, we shall prove the assertions for the exponent l + 1: for h > j − l − 1,

aj,l+1,h ∈ o(θ − 1)l+1 ,

for h < j − l − 1,

aj,l+1,h ∈ o(θ − 1)j −h ,

for h = j − l − 1,

aj,l+1,h ∈ o(θ − 1)l+1+1 .

First we investigate (σ p − ai )(j (j − 1) · · · (j − l + 1)(θ − 1)l (β − 1)j −l ). Then we have i





 i σ p − ai j (j − 1) · · · (j − l + 1)(θ − 1)l (β − 1)j −l = j (j − 1) · · · (j − l + 1)(θ − 1)l  × (j − l)(θ − 1)(β − 1)j −l−1 +



aj −l,1,h
(β − 1)

h




0
h

j −l

= j (j − 1) · · · (j − l)(θ − 1)l+1 (β − 1)j −(l+1) 
+ j (j − 1) · · · (j − l + 1)(θ − 1)l aj −l,1,h
(β − 1)h . h


For h
= j − l, by Lemma 5, aj −l,1,j −l ∈ o(θ − 1), and so (θ − 1)l aj −l,1,j −l ∈ o(θ − 1)l+1 . For h
= j − l − 1, (θ − 1)l aj −l,1,j −l−1 ∈ o(θ − 1)(l+1)+1. For h
< j − l − 1,





(θ − 1)l aj −l,1,h
∈ o(θ − 1)l+j −l−h = o(θ − 1)j −h . i

Next we investigate other terms (σ p − ai )(aj,l,h (β − 1)h ) and have   
pi 

σ − ai aj,l,h (β − 1)h = aj,l,h h(θ − 1)(β − 1)h−1 + ah,1,h
(β − 1)h . 0
h

h

We separate three cases h > j − l, h = j − l and h < j − l, and first consider the case h > j − l. By the inductive assumption, aj,l,h ∈ o(θ − 1)l . Then for h
= h, aj,l,h ah,1,h ∈ o(θ − 1)l+1 . For h
= h − 1 > j − l − 1, aj,l,h ah,1,h−1 ∈ o(θ − 1)l+1

and haj,l,h (θ − 1) ∈ o(θ − 1)l+1 .

808

Y. Miyata / Journal of Algebra 276 (2004) 794–825


For h
< h − 1, ah,1,h
∈ o(θ − 1)h−h by Lemma 5, and


aj,l,h ah,1,h
∈ o(θ − 1)l+h−h . Then if h
 j − l − 1, l + h − h
> l + 1 by h
< h − 1 and hence aj,l,h ah,1,h
∈ o(θ − 1)l+2 . If h
< j − l − 1 (< h − 1), l + h − h
> j − h
by h > j − l and hence
aj,l,h ah,1,h
∈ o(θ − 1)j −h . Therefore in case h > j − l, the assertions for l + 1 are verified. Similarly for the rest cases we can verify the assertions. From the above arguments, we see i (σ p − ai )l+1 (β − 1)j −l−1 is of course written in the desired form. (ii) We have 
pi l σ − ai (β − 1)j = j (j − 1) · · · (j − l + 1)(θ − 1)l (β − 1)j −l + aj,l,h (β − 1)h . h

Then for h > j − l, by (i)
 valK aj,l,h (β − 1)h  valK (θ − 1)l + h valK (β − 1) > l valK (θ − 1) + (j − l) valK (β − 1). For h = j − l,
 valK aj,l,j −l (β − 1)j −l  valK (θ − 1)l+1 + (j − l) valK (β − 1), and for h < j − l, then j − h > l. Then noting valK (θ − 1) > valK (β − 1) by (12), we have
 valK aj,l,h (β − 1)h  valK (θ − 1)j −h + h valK (β − 1) > valK (θ − 1)l + (j − h − l) valK (β − 1) + h valK (β − 1)
 = j valK (β − 1) + l valK (θ − 1) − valK (β − 1) , which implies (ii). The proof is completed. 2 i

Proposition 4. For l > j , (σ p − ai )l (β − 1)j is written in the form  l
pi σ − ai (β − 1)j = aj,l,h (β − 1)h

with aj,l,h ∈ o.

0
h
j

Then (i) aj,l,h ∈ o(θ − 1)l . i (ii) valK ((σ p − ai )l (β − 1)j ) > j valK (β − 1) + l(valK (θ − 1) − valK (β − 1)).

Y. Miyata / Journal of Algebra 276 (2004) 794–825

809

Proof. (i) By Proposition 3, we have for l = j ,


 j i aj,j,h (β − 1)h σ p − ai (β − 1)j = j !(θ − 1)j +

with aj,j,0 ∈ o(θ − 1)j +1 and aj,j,h ∈ o(θ − 1)j for h > 0. By (6)



 i σ p − ai j !(θ − 1)j = 0

and by Lemma 5
pi 
 σ − ai aj,j,h (β − 1)h = haj,j,h (θ − 1)(β − 1)h−1   
+ aj,j,h ah,1,h
(β − 1)h 0
h

h

with aj,j,h (θ − 1) ∈ o(θ − 1)j +1 and aj,j,h ah,1,h
∈ o(θ − 1)j +1 . Therefore the assertion (i) is proved for l = j + 1. Similarly we can prove the assertion (i) for l > j + 1. Moreover i we see (σ p − ai )l (β − 1)j is written in the desired form. (ii) For l > j , by (i), we verify easily

i l 
 valK σ p − ai (β − 1)j  valK (θ − 1)l > l valK (θ − 1) − (l − j ) valK (β − 1), which completes the proof of Proposition 4. 2 Let EndoG (O) = {g ∈ kG | g(O) ⊆ O}. Let l = before. Then we remark for an integer fl
, 


σ

pi

i

iff valK

− ai

li



i 0
i
and j =



0
h
hj

h

as


/π fl ∈ EndoG (O)

    
pi l
ph j
σ − ai i α − 1 h /π fl +dj  0 i

for 0
j < pn .

h

Define fl by  

 
pi l
fl = max fl
 σ − ai i /π fl ∈ EndoG (O) . i

Immediately we have       
pi li jh dj n fl = min valK /p (α − 1) /π σ − ai i

|0


p h jh < p n .

h

(20)

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Y. Miyata / Journal of Algebra 276 (2004) 794–825

By Lemma 4 and Proposition 3 with (12), for j satisfying jh  ln−h−1 for 0
h < n, valK

 

=



  
ph jh dj /π α −1

i

h



n−1
h    ln−h−1 valK ζ p − 1 − valK α p − 1 + jh ph valK (α − 1) − pn dj

h

=


pi l σ − ai i



h

  
n−1  h n h ln−h−1 p c1 + p e0 ln−h−1 p −p

h

h

  
ph j + res valK α −1 h ,

(21.1)

h

where res(x) denotes the residue of x to modulus pn . By Lemma 4 and Proposition 4 with (12), for any j with jh < ln−h−1 for some h, we have valK

  i

>




pi l σ − ai i

  
ph jh dj α −1 /π h

  
 ln−h−1 pn−1 − ph ln−h−1 ph c1 + pn e0 h

  
ph jh + res valK α −1 .

(21.2)

h

Then we have Theorem 2. Let fl be as above for l = fl = e0

 



i 0
i

 ln−h−1 pn−1 − ph

Then



0
h
     
ph j α −1 h + min ln−h−1 ph c1 + res valK /pn h



  jh  ln−h−1 for 0
h < n . 

h

Proof. For j with jh  ln−h−1 for 0
h < n, by (21.1), we have pn fl




   
ph j α −1 h ln−h−1 ph c1 + res valK

h

+ p e0 n

  0
h
h


n−1  h . ln−h−1 p −p

Y. Miyata / Journal of Algebra 276 (2004) 794–825

811

For j such that jh < ln−h−1 for some  h, we take an integer m as follows: mh = max{ln−h−1 , jh } for 0
h < n and m = h mh ph . Then j
m, so dj
dm . By Lemma 4, Proposition 4 and the definition of m,     
pi li 
ph jh dj σ − ai α −1 valK /π i

h







h

n−1
h    jh valK α p − 1 + ln−h−1 valK ζ p − 1 − valK α p − 1

jh ln−h−1



+


n−1  ln−h−1 valK ζ p − 1 − pn dj

jh




  

h

n−1
h mh valK α p − 1 + ln−h−1 valK ζ p − 1 − valK α p − 1

jh ln−h−1



+





h

n−1
h    mh valK α p − 1 + ln−h−1 valK ζ p − 1 − valK α p − 1 − pn dm

jh
= valK

 


i

σ p − ai

   li 
p h m α − 1 h /π dm .

i

h

Noting mh  ln−h−1 for 0
h < n, we can conclude Theorem 2.

2

Here we recall the results of Bertrandias and Ferton [1]. Let K/k be a cyclic extension of degree p and a be the residue res(c1 ) of the first ramification number c1 . For 0
l < p, let nl be nl = l[c1 /p] +

min

0
j
p−1−l






 (l + j + 1)a/p − (j + 1)a/p .

Then they showed EndoG (O) =



o(σ − 1)l /π nl .

0
l
Corollary 1 [1, Proposition 3]. Let K/k be a totally ramified cyclic Kummer extension of degree p. Let nl and fl be as above. Then nl = fl . Proof. From n = 1, it follows that a0 = 1 and (σ − a0 )l = (σ − 1)l . Then for 1
j
p − 1, 



valK (α − 1)j − p valK (α − 1)j /p = j (pe0 − c1 ) − p j (pe0 − c1 )/p = p − res(j a). By Theorem 2, for l = l0

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Y. Miyata / Journal of Algebra 276 (2004) 794–825

fl = min

p>j l







lc1 + res valK (α − 1)j /p + e0 p0 − p0




= l[c1 /p] + [la/p] + min res(la) + p − res(j a) /p . As [(l + j + 1)a/p] = [la/p] + [(j + 1)a/p] + [(res(la) + res((j + 1)a))/p], we have nl = l[c1 /p] + [la/p] +

min

1
j +1
p−l





res(la) + res (j + 1)a /p .

Clearly res((p − (j + 1))a) = p − res((j + 1)a) for 1
j + 1
p. We note p > p − (j + 1)  l from which we have nl = fl .

iff p − l  j + 1 > 0,

2

3. Maximal dual Larson order in EndoG (O) In this section, we seek a maximal dual Larson order in EndoG (O). Let si be si = fpi for l = pi with 0
i < n. Then by (21.1), valK




i

σ p − ai

  

  h (α p − 1)jh /π dj

h

=p

n−i−1




c1 + p e0 p n




n−1

−p

n−i−1



 n−i−1

  
ph jh + res valK α −1 h


 


= pn−i−1 c1 + pn e0 pn−1 − p + j pn e0 − c1 − pn j pn e0 − c1 /pn
 

= pn−i−1 c1 + pn e0 pn−1 − pn−i−1 − j c1 − pn −j c1 /pn . As (c1 , p) = 1, [−j c1 /pn ] = −1 − [j c1 /pn ]. By (20), we have  si =

min

j with 1
jn−i−1

−

   
 j c1 j c1 c1 n−i−1 i p − −1 − p − 1 . + + e 0 pn pn pi+1

Further by (13), 

  
 j c1 c1 ci+1 j c1 n−i−1 i p + + e p − 1 − − 0 pn pn pi+1 pi+1     j c1 j c1 c1 c1 =1+ −  0. + − pn pn pi+1 pi+1

1+

Y. Miyata / Journal of Algebra 276 (2004) 794–825

813

Put zj = 1 + l[j c1 /pn ] − (j c1 /pn ) + (c1 /pi+1 ) − [c1 /pi+1 ], then 0
zj < 2, hence [zj ] = 0 or 1. Thus we know     c  i+1  si = min [zj ]  j = ph jh with jn−i−1  1 + i+1 . p

(22) i

Let Gci be the ci th ramification group and Ki be the subfield of K corresponding to σ p . i i+1 Then Gci+1 = σ p . Denote the factor group G/σ p  by G, then (G)ci+1 +1 = 1. By (7), ai = 1 and
i  i (σ p − ai )/π si = σ p − 1 /π si . i

i

As (σ p − ai )/π si ∈ EndoG (O) by the definition of si , (σ p − 1)/π si ∈ EndoG (OKi+1 ) and hence   ci+1 + 1 , (23) si
pi+1 i+1

because (G/σ p )ci+1 +1 = 1. Next we define an integer i0 as follows: 


c1 = pn c1 /pn + cn−1 pn−1 + · · · + ci
0 pi0 + pi0 − 1

(24)

with 0
ci
< p for i > i0 and 0
ci
0 < p − 1. Then clearly 0
i0
n. Lemma 6. If i  i0 , [(ci+1 + 1)/pi+1 ] = [ci+1 /pi+1 ] and if i < i0 , [(ci+1 + 1)/pi+1 ] = [ci+1 /pi+1 ] + 1. Proof. In case i  i0 , by (13) 

   
c1 + 1 ci+1 + 1 = + e0 pn−i−1 pi − 1 i+1 i+1 p p     
 ci+1 c1 ph−i−1 ch
+ e0 pn−i−1 pi − 1 = = pn−i−1 n + , p pi+1 i+1
h
since ci
0 + 1 < p by (24). In case i < i0 , i + 1
i0 , so 

     i0
p (ci0 + 1) − 1 
ci+1 ci+1 + 1 i0 −i−1
− = p = 1. + 1 − c i0 pi+1 pi+1 pi+1

2

Lemma 7. Let i < i0 . Then, if i0
n − i − 1, si = [ci+1 /pi+1 ] and if i0 > n − i − 1, si = [ci+1 /pi+1 ] + 1 = [(ci+1 + 1)/pi+1 ].

814

Y. Miyata / Journal of Algebra 276 (2004) 794–825

Proof. Write j = j
+ pn−i−1 jn−i−1 + pn−i j

with 0
j
< pn−i−1 , 0
j

< pi
and 0 < jn−i−1 < p. Let c
be defined by pi0 c
= pn−1 cn−1 + · · · + pi0 (ci
0 + 1), so
n−i 0 . By the assumption i < i0 and (24), 1
c

j
1 pn−i−1 − j
+ 1 + 1 − = 1 − < 1, pn pn pi+1

because j
< pn−i0
pn−i−1 . Then [zj ] = 0 and by (22), si = [ci+1 /pi+1 ] if i0
n − i − 1. Next we consider the case i0 > n − i − 1 (and i < i0 ). Let pi0 b = res(pi0 j
c
) with 0
some b < pn−i0 . First we assume pi0 b − j  0. Then we have res(pi0 j
c
− j ) = pi0 b − j and  zj = 1 + =1−

 j c1 pn−i−1 −pi0 b + j pn−i−1 j c1 =1+ +1− − n +1− n n n p p p p pn

pi0 b j − pn−i−1 + + 1  1, pn pn

because jn−i−1  1. Next we consider the case pi0 b − j < 0 and then we have
 res pi0 j
c
− j = pn + pi0 b − j. By the assumption, i0 > n − i − 1, so i0  n − i. Noting j − pi0 b  0, we have j
+ pn−i−1 jn−i−1 + pn−i j

− pi0 b  0 and pn−i j

− pi0 b  0. Further by jn−i−1 > 0, j − pi0 b − pn−i−1  0. Suppose zj < 1, then 1 > zj = 1 +

−pn − pi0 b + j pn−i−1 j − pi0 b − pn−i−1 +1− =1+ , n n p p pn

Y. Miyata / Journal of Algebra 276 (2004) 794–825

815

so j − pi0 b − pn−i−1 < 0, which is a contradiction. Therefore we have zj  1 for 0
j < pn with jn−i−1 > 0 and si = [ci+1 /pi+1 ] + 1. The proof of Lemma 7 is completed. 2 We are ready to prove the following theorem. i

Theorem 3. Let si be si = max{s | (σ p − ai )/π s ∈ EndoG (O)} for 0
i < n. Then (i) If i  i0 ,      
ci+1 + 1 ci+1 c1 n−i−1 i = = e si = p − 1 + i+1 . 0p pi+1 pi+1 p 

(ii) If i < i0
n − i − 1,  si =

   
ci+1 c1 n−i−1 i = e . p − 1 + p 0 pi+1 pi+1

(iii) If i < i0 and n − i − 1 < i0 ,      
ci+1 ci+1 + 1 c1 n−i−1 i = + 1 = e0 p si = p − 1 + i+1 + 1. pi+1 pi+1 p 

Proof. For i < i0 , Theorem 3 follows from Lemma 7. Thus we consider the remaining case i  i0 . Then by Lemma 6, [(ci+1 + 1)/pi+1 ] = [ci+1 /pi+1 ]. By (22) and (23), we have si = [ci+1 /pi+1 ]. The proof is completed. 2 Let B be a subalgebra of kG. We call B a dual Larson order if there is a Larson order L in kG with B = L∗ . i

Theorem 4. Let A = o[{(σ p − ai )/π si }0
i
   ci+1 c1 n−1 n−i−1 si = e0 + i+1 + 1. + 1 = p e0 − p pi+1 p Let ti = pn−1 e0 − sn−i−1 , then ti = pi e0 − [c1 /pn−i ] − 1 < pi e0 . Thus we have 

 c1 ti+1 − pti = − n−i−1 − 1 + p n−i + p p p

   c1   c1 p n−i−1 − n−i−1 + p − 1  0. =p p p c1





816

Y. Miyata / Journal of Algebra 276 (2004) 794–825

By (3), L = o[{(σ p − 1)/π ti }i ] is a Larson order and by Theorem 1, A = L∗ , hence A is a dual Larson order. Next we treat the case i0 < n. For n = 1, a0 = 1 and A = o[(σ − 1)/π e0 −t0 ], hence A is a dual Larson order (cf. [5, Chapter 5, §21]). The condition res(c1 ) < p is satisfied for n = 1. Therefore in the following we may assume n  2. Let A be a dual Larson order. For i = 0, n − i − 1 = n − 1  i0 , so s0 = [c1 /p] by Theorem 3. Similarly for i = n − 1, i = n − 1  i0 by i0 < n, so sn−1 = pn−1 e0 − e0 + [c1 /pn ]. Let ti = pn−1 e0 − sn−i−1 as before. Then tn−1 = pn−1 e0 − [c1 /p] and t0 = e0 − [c1 /pn ]. Let i L = o[{(σ p − 1)/π ti }i ], so L = A∗ is a Larson order by the assumption, whence by (3), n−1 tn−1 − p t0 = −[c1 /p] + pn−1 l[c1 /pn ]  0. Therefore [c1 /p] = pn−1 [[c1 /p]/pn−1 ], so immediately res(c1 ) < p. Conversely assume res(c1 ) < p. Then as in the above proof, we have i

 ti = p

n−1

e0 − sn−i−1 = p e0 − p i

i

 
    c0 c1 c1 i − n−i = p e0 − n . pn p p

Then t0 , t1 , . . . , tn−1 satisfy (3), so L is a Larson order and A = L∗ by Theorem 1. Hence A is a dual Larson order. The proof of Theorem 4 is completed. 2 i

Theorem 5. Assume n  2 and let A = o[{(σ p − ai )/π si }0
i
n−1

e0 − un−i−1 = p

n−1

e0 − p

n−1

e0 + p

n−(n−i−1)−1

e0 − p

n−(n−i−1)−1

c1 + 1 pn



   c1 + 1 . = p e0 − pn i

Therefore L = o[{(σ p − 1)/π p e0 −un−i−1 }i ] is a Larson order, and hence H is a dual Larson order. Clearly ui
si by Theorem 3, so H ⊆ A. Here we assume A is a dual Larson order. In case i0 = n, as in the proof of Theorem 4, si = pn−1 e0 − pn−i−1 e0 + [c1 /pi+1 ] + 1. By i0 = n, c1 = pn [c1 /pn ] + pn − 1, so i



n−1

     c1 n−i−1 c1 n−i−1 n−i−1 c1 + 1 + 1 = p + p . = p pn pn pi+1

Y. Miyata / Journal of Algebra 276 (2004) 794–825

817

Therefore we have si = ui and H = A. In case i0 < n, by Theorem 4, res(c1 ) < p and so i0
1. Then by Theorem 3,  si = pn−1 e0 − pn−i−1 e0 +

   c1 n−1 n−i−1 n−i−1 c1 = p . e − p e + p 0 0 pn pi+1

As i0
1 and n  2, [(c1 + 1)/pn ] = [c1 /pn ], hence si = ui and H = A. (ii) As H
is a dual Larson order, (H
)∗ is a Larson order in kG. By [5, (19.5) Theorem],

i i (H
)∗ = [{(σ p − 1)/π ti }i ] for the given generator σ of G. Then H
= o[{(σ p − ai )/π ri }i ]

n−1
with ri = p e0 − tn−i−1 . From the assumption H ⊆ EndoG (O) and Theorem 3, we see ri

si and H
⊆ A . Then we have H ⊆ H
⊆ A and A∗ ⊆ L
= (H
)∗ ⊆ L = H ∗ . As L

i is a Larson order, L
is generated by (σ p − 1)/π ti (0
i < n). Let ti = pn−1 e0 − sn−i−1 , then ti
ti

pn−1 e0 − un−i−1 . If A is a dual Larson order, by the above (i), H = H
= A. Thus in the following we may assume A is not a dual Larson order. Then i0 < n, so by (24), 

    n−1
   p cn−1 + · · · + pi0 (ci
0 + 1) c1 + 1 c1 c1 = + = . n n n p p p pn

Then     c1 c1 + 1
n−1 n−1 t0 = e0 − n
t0
p e0 − p e0 − e0 + , p pn 

so t0 = t0
= e0 − [c1 /pn ]. Thus pi t0
pi t0

ti

pn−1 e0 − un−i−1 = pi e0 − pi



 c1 = pi t0 , pn

hence ti
= pn−1 e0 − un−i−1 and H
= H . The proof is completed. 2

4. Maximal tame extension In this section, we construct a maximal tame extension in the ring O. Let S be

i a subalgebra of O generated by elements (α p − 1)/π ti for 0
i < n, where ti

= i [(valK (α p − 1))/pn ]. For brevity, we denote ti

by ti in the following. Then we have  ti =

     pi (pn e0 − c1 ) −c1 c1 i i = p = p . e + e − 1 − 0 0 pn pn−i pn−i

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Y. Miyata / Journal of Algebra 276 (2004) 794–825

Let ri = pn−1 e0 − tn−i−1 , then  c1 e0 + i+1 + 1. p 

ri = p

n−1

e0 − p

n−i−1

(25)

i

Let H (S) = o[{(σ p − ai )/π ri }0
i
Proof. Let L = o[{(σ p − 1)/π ti }i ]. Then     c1 c1 i+1 i+1 ti+1 − pti = p e0 − 1 − n−i−1 − p e0 + p + p n−i  0. p p Clearly ti
pi e0 , hence L is a Larson order by (3). Thus by Theorem 1, we have H (S) = L∗ and H (S) is a dual Larson order. 2 By Theorem 3 with (25), we remark that if i  i0 or i < i0
n − i − 1,  c1 ri − si = p +1 pi+1   
 c1 n−i−1 i − p p − 1 e0 + i+1 =1 p n−i−1


i  p − 1 e0 +



(26.1)

and if i < i0 and n − i − 1 < i0 ,  c1 +1 ri − si = p pi+1     
c1 n−i−1 i − p p − 1 e0 + i+1 + 1 = 0. p n−i−1




 p − 1 e0 +



i

(26.2)

Lemma 8.  (i) fl
0
i
li ri − fl

i



  c1  
  li i+1 + li pn−1 e0 − pn−i−1 e0 + li p i i i  
 − li pn−1 e0 − pn−i−1 e0 i

Y. Miyata / Journal of Algebra 276 (2004) 794–825

−

 

819

     
ph j α −1 h li pn−i−1 c1 + res valK /pn

i

h

  n−i−1    c1   )c1 ( li p = li i+1 + li − n p p         
ph jh n−i−1 n − res α −1 li p c1 + res valK /p i

h

i

h

     c1    c1   ( li res(pn−i−1 c1 )) = li i+1 − li i+1 + li − pn p p          jh
ph n−i−1 n − res c1 + res valK /p . li p α −1 As (res(pn−i−1 c1 )/pn ) < 1, [( 

li ri − fl  1 −



li res(pn−i−1 c1 ))/pn ]




li − 1. Therefore

        
ph j α −1 h res li pn−i−1 c1 + res valK /pn i

h

 0. If [(res((



i li p

n−i−1 )c

1 ) + res(valK (





h (α

ph

− 1)jh )))/pn ] = 0, then we have

li ri − fl > 0.

The proof of Lemma 8 is completed. 2 We are ready to prove the next theorem. Theorem 6.   i (i) EndoG (O) = 0
l
i

h

  i h Let xl,j = valK (bl ( i (σ p − ai )li )( h (α p − 1)jh /π dj )). By (21.1), we have that if j satisfies jh  ln−h−1 for 0
h < n, then

820

Y. Miyata / Journal of Algebra 276 (2004) 794–825

xl,j = valK (bl ) +




 ln−h−1 pn e0 pn−1 − ph + pn

 

h

   ln−h−1 ph c1 /pn

h

      
ph j h h + res α −1 ln−h−1 p c1 + res valK . h

h

  h h Further xl,j ≡ res(( h ln−h−1 ph )c1 ) + res(valK ( h (α p − 1)j )) (mod pn ), so xl
,j =
n
xl,j for 0
l = l < p for j such that jh  ln−h−1 (and jh  ln−h−1 ). We consider first the case that there is some l such that both valK (bl ) = 0 and         
ph jh h n α −1 ln−h−1 p c1 + res valK /p  1 res h

h

always for j satisfying jh  ln−h−1 for 0
h < n. Then by Theorem 2, fl =

 

   
 n ln−h−1 p c1 /p + 1 + ln−h−1 e0 pn−1 − ph , h

h

h

so 0
xl,j − pn fl < pn . Let j = pn − 1, then jh = p − 1  ln−h−1 , whence xl
,j = xl,j   i for l
= l. Therefore u = 0 and so ( bl ( i (σ p − ai )li )/π fl )/π u ∈ E. We consider next the case that if valK (bl ) = 0, then for some j with jh  ln−h−1 ,         
ph j /pn = 0. α −1 h res ln−h−1 ph c1 + res valK h

h

Then p fl = n






n

ln−h−1 p e0 p

n−1

−p

h



+p

n

 

h

   n ln−h−1 p c1 /p , h

(27)

h

  ph so xl,j − pn fl = res(( ln−h−1 ph )c 1)jh )) < pn . We can choose 1 ) + res(valK ( h (α −

h l with valK (bl
) = 0 such that res(( ln−h−1 p )c1 ) < res(( ln−h−1 ph )c1 ) for 0
l =  lh ph < pn with valK (bl ) = 0, because (c1 , p) = 1. By valK (bl
) = 0 and the assumption
in the considered case, for some j
with jh
 ln−h−1 ,         
ph j

h n h α −1 ln−i−1 p c1 + res valK res /p = 0. h

h

For l satisfying jh
 ln−h−1 for 0
h < n and valK (bl ) = 0, we have by (27),        j

ph xl,j
− pn fl = res ln−h−1 ph c1 + res valK α −1 h . h

h

Y. Miyata / Journal of Algebra 276 (2004) 794–825

821

Then xl,j
− pn fl > xl
,j
− pn fl
 
because res(( ln−h−1 ph )c1 ) > res(( ln−h−1 ph )c1 ). For l satisfying jh
< ln−h−1 for some h and valK (bl ) = 0, by (21.2) xl,j
>




 ln−h−1 p e0 pn−1 − ph + pn n

 

h

   n ln−h−1 p c1 /p h

h

     
ph j
α −1 h , + res ln−h−1 ph c1 + res h

h

 
h so xl,j
− pn fl > res(( h ln−h−1 ph )c1 ) + res(valK ( h (α p − 1)jh )) by (27). Hence by virtue of the choice of l
, we have again xl,j
− pn fl > xl
,j
− pn fl
. Noting xl,j
 pn for l with valK (bl ) = 0 and xl
,j
− pn fl
< pn , we know valK

        
pi l
ph j
σ − ai i /π fl α − 1 h /π dj bl l

i

h

      
ph j

h h α −1 = res ln−h−1 p c1 + res valK < pn . h

h

  
i h Therefore, since ( l bl ( i (σ p − ai )li )/π fl )( h (α p − 1)jh )/π dj ) ∈ π u O, we have u = 0 and EndoG (O) ⊆ E. Hence EndoG (O) = E. (ii) By Lemma 8(i), we have H (S) ⊇ E = EndoG (O) and by the definition of si , i E = EndoG (O) ⊇ A(= o[{(σ p − ai )/π si }i ]). The proof is completed. 2 Here we remember the definitions of a tame H -extension and a Galois H -extension for a Hopf order H (cf. [2,5]). Let the ideal I (H ) of integrals in H be defined by  I (H ) = h ∈ H | ha = hε(a) for all a ∈ H . Then an H -extension S of o is called a tame H -extension if (i) ranko (S) = ranko (H ); (ii) S is a faithful H -module; (iii) I (H )S = o. Further an H -extension S is called a Galois H -extension of o if a o-module homomorphism

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Y. Miyata / Journal of Algebra 276 (2004) 794–825

j : S ⊗o H → Endo (S) j (s ⊗ h)(t) = sh(t)

for s, t ∈ S, h ∈ H,

is an isomorphism. Now we summarize Childs’ results for n = 1 [4, Theorems 9.1, 12.2 and 14.1]. Theorem 7. Let K/k be a Kummer extension of degree p. Let α be a unit with valK (α − 1) = pe0 − c1 and b = [(valK (α − 1))/p]. Then (i), (ii) and (iii) hold. (i) There is a chain of Hopf Galois extensions Si = o[(α − 1)/π i ] of o for 0
i
b; S0 ⊂ S1 ⊂ · · · ⊂ Sb ⊆ O. Moreover Si is a Galois He0 −i -extension, where Hi is a Hopf order o[(σ − 1)/π i ]. (ii) If S is a Hopf Galois extension of o in O, then S = Si for some i. (iii) Let H be a Hopf order in kσ  with H = He0 (a maximal order). Then, if S is a tame H -extension, S is a Galois H -extension. Corollary 2. Let K/k be as in Theorem 7 and S be a tame H -extension with S ⊇ o[α]. Then S is a Galois H -extension. Proof. In case H = o[(σ − 1)/π e0 ], by Theorem 7, S is a Galois H -extension. For H = o[(σ − 1)/π e0 ], by [5, p. 132, (26.3) Proposition], o[α] = {β ∈ O | Hβ ⊆ O}, so o[α] ⊇ S and o[α] = S by the assumption. Thus S is a Galois H -extension by Theorem 7. 2 i

Lemma 9. Let S = o[{(α p − 1)/π ti }i ] as above. EndoG (S) ⊇ H (S). i

Proof. First we show generators (σ p − ai )/π ri belong to EndoG (S). For h = n − i − 1, by Lemma 3, we have



 

h   i σ p − ai /π ri α p − 1 /π th = 0.

For h = n − i − 1, by Lemma 3 again,



 

h 
n−1   h i σ p − ai /π ri α p − 1 /π th = ζ p − 1 α p /π ri +th

n−1 
h 
n−1  = ζ p − 1 α p − 1 + ζ p − 1 /π ri +th .

Y. Miyata / Journal of Algebra 276 (2004) 794–825

823

As ri = pn−1 e0 − tn−i−1 by the definition of ri , we have ri + th = pn−1 e0 by h = n − i − 1. h Therefore we see that any element of H (S) takes (α p − 1)/π th into S. As H (S) is a Hopf order, for an element γ of H (S) and x, y ∈ S, γ (xy) =



γ(1)(x)γ(2)(y) with γ(1), γ(2) ∈ H (S)




 cf. [5, p. 14] ,

from which it follows that H (S) ⊆ EndoG (S). The proof is completed. 2 We have the next theorem which is one of the main results of this paper. i

Theorem 8. Let S be S = o[{(α p − 1)/π ti }i ] as above. (i) S is a Galois H (S)-extension. Therefore H (S) = EndoG (S) and S = {β ∈ O | H (S)β ⊆ O}. (ii) Let S
be a subalgebra of O such that H (S
) = EndoG (S
) is a dual Larson order. Then S
is a tame H (S
)-extension. Moreover if S ⊆ S
and H (S
) ⊇ EndoG (O), S
= S and H (S
) = H (S). i

Proof. (i) Let Sh = o[{(α p − 1)/π ti }in−h ] for 0 < h < n and let S0 = o. We calculate the discriminant disc(S). As in the proof of Theorem 1, we have [S : Sn−1 [α]] = n−1 n−2 π p(p−1)t0p /2 and [Sn−1 : Sn−2 [α p ]] = π p(p−1)t1p /2 . Thus   

  n S : Sn−2 α p [α] = S : Sn−1 [α] Sn−1 [α] : Sn−2 α p [α] = π p ((p−1)/2)(t0+t1 ) . Repeating similar arguments, we have 

n S : o[α] = π p ((p−1)/2)(t0+t1 +···+tn−1 ) . We know easily disc(o[α]) = (normK/ k (pn α p −1 )) = opnp , and hence disc(S) = n n opnp π −p (p−1)(t0+t1 +···+tn−1 ) . For the ideal I (H ) of integrals, we have     
I H (S)∗ = o σ j /π (p−1)( 0
h
n

0
j

(cf. [9, p. 706], [7, p. 438, Proposition 4.1]). By [6, p. 52, Lemma 1.3], 

pn n n disc H (S)∗ = εI H (S)∗ = pnp π −p (p−1)( th )

and so disc(S) = disc(H (S)∗ ). By Lemma 9, we see S is an H (S)-module algebra and so a right H (S)∗ -comodule (cf. [2, p. 500], [5, p. 15]). Thus, by [6, p. 53, Lemma 1.5], we have S is a Galois H (S)-extension. (ii) By [2, p. 503, Corollary 3.4], we first know S
is a tame H (S
)-extension. Let H (S
) be the image of H (S
) under the canonical map G → G/σ p . Then by p p [2, p. 509, Lemma 4.8], S
σ  is a tame H (S
)-extension, so by Corollary 2, S
σ 

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Y. Miyata / Journal of Algebra 276 (2004) 794–825

is a Galois H (S
)-extension. Further, by the assumption S ⊆ S
, S σ  ⊆ S
σ  . Then, n−1 p
p − 1)/p] = tn−1 , we have S σ  = S σ  and by Theorem 7, noting b = [valK1 (α p H (S
) = H (S) = o[(σ − 1)/π r0 ]. As H (S
)∗ is a Larson order by the assumption, we
i
= have H (S
)∗ = o[{(σ p − 1)/π ti }i ] for some integers ti
satisfying (3). Immediately tn−1 n−1 n−1

∗ ∗ p e0 − r0 = p e0 − 1 − [c1/p]. By H (S ) ⊇ EndoG (O), H (S ) ⊆ A . Then by (26.1) and (26.2), we have that if rn−i−1 = sn−i−1 , ti

ti and if rn−i−1 = sn−i−1 + 1, ti

ti + 1. Suppose ti
= ti + 1, then by (25), p

ti
= ti + 1 = pn−1 e0 − rn−i−1 + 1 = pi e0 −



p

   c1 c1 i e − − 1 + 1 = p . 0 pn−i pn−i

By (3), we have
0
tn−1 − pn−i−1 ti
= pn−i−1

=p

n−i−1

 [ c1 ]  p pn−i−1



   c1 c1 − −1 n−i p p

 c1 − 1 < 0, − p 

a contradiction. Hence ti

ti , so H (S
)∗ ⊆ H (S)∗ and H (S
) ⊇ H (S). Since S is a Galois H (S)-extension, by [5, p. 132, (26.3) Proposition]  S = β ∈ O | H (S)β ⊆ O . Therefore 
 S ⊇ β ∈ O | H S
β ⊆ O ⊇ S
, hence S
= S and H (S
) = H (S). The proof is completed. 2 Corollary 3. Let S and i0 be as above. Then (i), (ii) and (iii) are equivalent. (i) i0 = n. (ii) H (S) = EndoG (O). (iii) S = O. Proof. (i) ⇒ (ii). As in the proof of Theorem 4, for 0
i < n, i < i0 and n − i − 1 < i0 . By (26.2), ri = si and hence H (S) = EndoG (o) by Theorems 3 and 6. (ii) ⇒ (iii). By Theorem 8(i), S = {β ∈ O | H (S)β ⊆ O}, so S ⊇ O by the assumption H (S) = EndoG (0). Thus S = O. (iii) ⇒ (i). by Theorem 8(i), O is a Galois H (S)-extension. Then, by [2, p. 514, Theorem 3], cn ≡ −1 (mod pn ) and so i0 = n. 2

Y. Miyata / Journal of Algebra 276 (2004) 794–825

825

Acknowledgment The author thanks the referee for his patience and advices.

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