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arXiv:math/0605011v2 [math.NT] 19 Nov 2006
A VALUATION CRITERION FOR NORMAL BASES IN
ELEMENTARY ABELIAN EXTENSIONS
NIGEL P. BYOTT AND G. GRIFFITH ELDER
Abstract. Let p be a prime number and let K be a finite extension of the
field Qp of p-adic numbers. Let N be a fully ramified, elementary abelian
extension of K. Under a mild hypothesis on the extension N/K, we show that
every element of N with valuation congruent mod [N : K] to the largest lower
ramification number of N/K generates a normal basis for N over K.
1. Introduction
The Normal Basis Theorem states that in a finite Galois extension N/K there
are elements α ∈ N whose conjugates {σα : σ ∈ Gal(N/K)} provide a vector space
basis for N over K. If K is a finite extension of the field Qp of p-adic numbers,
the valuation vN(α) of an element α of N is an important property. We therefore
ask whether anything can be said about the valuation of normal basis generators
in this case. We will prove
Theorem 1. Let K be a finite extension of the p-adic numbers, let N/K be a
fully ramified, elementary abelian p-extension, and let bmaxdenote the largest lower
ramification number. If the upper ramification numbers of N/K are relatively prime
to p, then every element α ∈ N with valuation vN(α) ≡ bmaxmod [N : K] generates
a normal field basis. Moreover, no other equivalence class has this property: given
any integer v with v ?≡ bmmod [N : K], there is an element ρv∈ N with vN(ρv) = v
which does not generate a normal basis.
This result arose out of work on the Galois module structure of ideals in exten-
sions of p-adic fields. For such extensions, it has been found that the usual ramifi-
cation invariants are, in general, insufficient to determine Galois module structure,
and thus that there is a need for a refined ramification filtration [BE02, BE05, BE].
This refined filtration is defined for elementary abelian p-extensions and requires
elements that generate normal field bases. Such elements are provided by Theo-
rem 1. Recent work [Eld] suggests that what is known for p-adic fields should also
hold in the analogous situation in characteristic p, where K is a finite extension of
Fp(X). Here Fpdenotes the finite field with p elements, and X is an indeterminate.
We therefore make the
Conjecture. Theorem 1 holds when K is a finite extension of Fp(X) as well.
Date: September 26, 2006.
1991 Mathematics Subject Classification. 11S15, 13B05.
Key words and phrases. Normal Basis Theorem, Ramification Theory.
Elder was partially supported by NSF grant DMS-0201080.
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2 NIGEL P. BYOTT AND G. GRIFFITH ELDER
2. Preliminary Results
Let K be a finite extension of the field Qp of p-adic numbers, and let N/K
be a fully ramified, elementary abelian p-extension with G = Gal(N/K)∼= Cn
Use subscripts to denote field of reference. So πN denotes a prime element in N,
vN denotes the valuation normalized so that vN(πN) = 1, and eK denotes the
absolute ramification index. Let TrN/Kdenote the trace from N down to K. For
each integer i ≥ −1, let Gi = {σ ∈ G : vN((σ − 1)πN) ≥ i + 1} be the ith
ramification group [Ser79, IV, §1]. Then G−1= G0= G1= G, and the integers
b such that Gb? Gb+1are the lower ramification break (or jump) numbers. The
collection of such numbers, b1< ··· < bm, is the set of lower breaks. They satisfy
b1 ≡ ··· ≡ bmmod p [Ser79, IV, §2, Prop. 11], where if bm ≡ 0 mod p then the
extension N/K is cyclic [Ser79, IV, §2, Ex. 3]. Let gi = |Gi|. Then the upper
ramification break numbers u1< ··· < umare given by u1= b1gb1/pn= b1 and
ui= (b1gb1+ (b2− b1)gb2+ ··· + (bi− bi−1)gbi)/pnfor i ≥ 2 [Ser79, IV, §3].
Now by the Normal Basis Theorem, the set
p.
NB =
?
ρ ∈ N :
?
σ∈G
K · σρ = N
?
of normal basis generators is nonempty. We desire integers v ∈ Z such that {ρ ∈
N : vN(ρ) = v} ⊂ NB. And so we are concerned by the following
Example 1. Suppose K contains a pth root of unity ζ, and let N = K(x) with
xp−πK= 0. Let σ generate Gal(N/K). Observe that (σ−1)xpi= 0 and TrN/Kxi=
0 for p ∤ i. So for each i ∈ Z, we have vN(xi) = i and xi?∈ NB. Here N/K has one
ramification break b = peK/(p − 1), which is divisible by p. [Ser79, IV, §2, Ex. 4].
Remark. Fortunately, these extensions provide the only obstacle. The restriction
in Theorem 1 to elementary abelian extensions with upper ramification numbers
relatively prime to p is a restriction to those extensions that do not contain a cyclic
subfield such as in Example 1 [Ser79, IV, §3 Prop. 14].
To prove Theorem 1 we need two results.
Lemma 2. Let N/K be as above with bm ?≡ 0 mod p, and let tG =
|Gbi\ Gbi+1|. If ρ ∈ N with vN(ρ) ≡ bmmod pn, then vN(TrN/Kρ) = vN(ρ) + tG.
Conversely, given α ∈ K there is a ρ ∈ N with vN(ρ) = vN(α) − tG≡ bmmod pn
such that TrN/K(ρ) = α.
?m
i=1bi·
Proof. Use induction. Consider n = 1 when Gal(N/K) = ?σ? is cyclic of degree
p. There is only one break b, which satisfies b < peK/(p − 1). Given ρ ∈ N with
vN(ρ) ≡ b mod p, we have TrN/Kρ ≡ (σ − 1)p−1ρ mod pρ. Since (p − 1)b < peK,
vN(TrN/Kρ) = vN(ρ) + (p − 1)b. And given α ∈ K, use [Ser79, V, §3, Lem. 4] to
find ρ ∈ N with vN(ρ) = vN(α) − (p − 1)b and TrN/Kρ = α.
Assume now that the result is true for n, and consider N/K to be a fully ramified
abelian extension of degree pn+1. Recall gi= |Gi|. Let H be a subgroup of G of
index p with Gb2⊆ H. Let L = NHand note that N/L satisfies our induction
hypothesis. Moreover the ramification filtration of H is given by Hi = Gi∩ H
[Ser79, IV, §1] . So |Hi| = gifor i > b1. Therefore tH= bm(gbm−1)+bm−1(gbm−1−
gbm) + ··· + b1(pn− gb2). Given ρ ∈ N with vN(ρ) ≡ bmmod pn+1, by induction
vN(TrN/Lρ) = vN(ρ)+tH. By the Hasse-Arf Theorem, pn+1| gbi(bi−bi−1) for 1 ≤
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A VALUATION CRITERION FOR NORMAL BASES3
i ≤ m. Thus tH≡ −bm+pnb1mod pn+1and vL(TrN/Lρ) ≡ b1mod p. Using [Ser79,
IV, §1, Prop. 3 Cor.], b1is the Hilbert break for the Cp-extension L/K. Applying
the case n = 1, we find vN(TrN/Kρ) = vN(ρ)+tH+pn(p−1)b1= vN(ρ)+tG. The
converse statement follows similarly, using tH+ pn(p − 1)b1= tG.
?
The following generalizes a technical relationship used in the proof of Lemma 2.
Lemma 3. Let N/K be a fully ramified, noncyclic, elementary abelian extension
with group G∼= Cn
is the largest lower break of N/K, b the only break of N/L, and ρ any element of
N with vN(ρ) ≡ bmmod pn, then vL(TrN/Lρ) ≡ b mod p.
p. Let H be a subgroup of G of index p, and let L = NH. If bm
Proof. In the proof of Lemma 2, H ⊇ Gb2so that Gb1H/H ? Gb1+1H/H following
[Ser79, IV, §1, Prop. 3, Cor.], and the break for G/H was b1.
no such luxury and we have to involve the upper numbers in our considerations,
although the argument is really no different. Note that there is a k such that
Guk+1H/H ? GukH/H. Thus ukis the upper ramification number of G/H. Since
there is only one break in the filtration of G/H, the lower and upper numbers for
G/H are the same, b = uk.
The ramification filtration for H is given by taking intersections: Hj= Gj∩ H.
Note that [Gbi: Gbi∩ H] = p for i ≤ k and Gbi⊆ H for i > k. Let hj = |Hj|.
Then hj = gj/p for j ≤ bk, and hj = gj for j > bk. Now let vN(ρ) = bm+ pnt.
Following the proof of Lemma 2 and using the Hasse-Arf Theorem,
Here we have
vN(TrN/Lρ) = bm+pnt+bm(hbm−1)+bm−1(hbm−1−hbm)+···+b1(hb1−hb2)
= pnt + (bm− bm−1)hbm+ (bm−1− bm−2)hbm−1+ ··· + (b2− b1)hb2+ b1hb1
≡ (bk− bk−1)hbk+ ··· + (b2− b1)hb2+ b1hb1≡ pnuk/p ≡ pn−1b mod pn
Therefore vL(TrN/Lρ) ≡ b mod p.
?
3. Main Result
Proof of Theorem 1. There are two statements to prove. We begin with the first:
We assume the upper breaks satisfy p ∤ ui, and prove that for ρ ∈ N
vN(ρ) ≡ bmmod pn=⇒ ρ ∈ NB.
The argument breaks up into two cases: the Kummer case where ζ ∈ K and the
non-Kummer case where ζ ?∈ K. Here ζ is a nontrivial pth root of unity.
We begin with the Kummer case, and start with n = 1. Let σ generate the
Galois group, and denote the one ramification number by b. Since in this case b is
also the upper number, p ∤ b. Therefore {vN((σ−1)iρ : 0 ≤ i < p} is a complete set
of residues modulo p. And since N/K is fully ramified, ρ generates a normal basis.
Now let n ≥ 2 and note that N = K(x1,x2,...,xn) with each xp
to show that K[G]ρ contains each element y = xj1
For y = 1 this is clear, since TrN/K(ρ) ∈ K. For any other y, let L = K(y) and let
b denote the ramification number of L/K. By Lemma 3, vL(TrN/L(ρ)) ≡ b mod p.
Since b is an upper number of the ramification filtration of G, p ∤ b. Now apply the
n = 1 argument, using TrN/L(ρ) in L/K. Thus y ∈ K[G]ρ.
We now turn to the non-Kummer case with ζ ?∈ K. Let E = K(ζ), let E/K have
ramification index eE/K, and let F = N(ζ). Then F/E is a fully ramified Kummer
extension of degree pn. Applying Herbrand’s Theorem [Ser79, IV, §3, Lem. 5] to
i∈ K. It suffices
n with 0 ≤ ji≤ p − 1.
1xj2
2···xjn
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4 NIGEL P. BYOTT AND G. GRIFFITH ELDER
the quotient G = Gal(N/K) of Gal(F/K), we find that the maximal ramification
break of F/E is eE/Kbm?≡ 0 mod p. The above discussion for the Kummer case
therefore applies to F/E. Suppose now for a contradiction that ρ ∈ N with vN(ρ) ≡
bmmod pn, and that K[G]ρ is a proper subspace of N. Then by extending scalars
(noticing that E and N are linearly disjoint as their degrees are coprime) we have
that E[G]ρ is a proper subspace of F. Moreover vF(ρ) ≡ eE/Kbmmod pn. This
contradicts the result already shown for the Kummer extension F/K, completing
the proof of the first statement of the theorem.
Consider the second statement: Given any integer v with v ?≡ bmmod pnthere
is a ρv∈ N with vN(ρv) = v such that TrN/Kρv= 0 and thus ρv?∈ NB.
To prove this statement note that given v ∈ Z, there is an 0 ≤ av< pnsuch that
v ≡ avbmmod pn, since p ∤ bm. If av?= 1 we will construct an element ρv∈ N with
vN(ρv) = v and TrN/Kρv= 0. To begin, observe that there is a integer k such that
0 ≤ k ≤ n − 1, av≡ 1 mod pkand av?≡ 1 mod pk+1. Recall gi= |Gi|. Since the
ramification groups are p-groups with gi+1≤ gi, there is a Hilbert break bs such
that gbs+1< pk+1≤ gbs. For i = k,k + 1 choose Hiwith |Hi| = piand Gbs+1⊂
Hk ⊂ Hk+1 ⊆ Gbs. Recall from Lemma 2 the expression for tG, and note that
tHk= bm(gbm−1)+bm−1(gbm−1−gbm)+···+bs(pk−gbs+1) ≡ −bm+bspkmod pn. Let
L = NHk. Since av?≡ 1 mod pk+1, av≡ 1 + rpkmod pk+1for some 1 ≤ r ≤ p − 1.
Using the fact that bs ≡ bmmod p, avbm+ tHk≡ (r + 1)bmpkmod pk+1. Since
pk| vN(α) for α ∈ L, we can choose α ∈ L with vN(α) = v + tHk− rpkbs.
So vL(α) ≡ bsmod p. Let σ ∈ G so that σHk generates Hk+1/Hk. Therefore
vN((σ − 1)rα) = v + tHk. Now using Lemma 2, we choose ρv ∈ N such that
vN(ρv) = v and TrN/Lρv= (σ − 1)rα. Since (1 + σ + ... + σp−1)TrN/Lρv= 0, we
have TrN/Kρv= 0.
?
Corollary 4. Let N/K be a fully ramified, elementary abelian extension of degree
pnwith n > 1 and one ramification break, at b. If ρ ∈ N with vN(ρ) ≡ b mod pn,
then ρ ∈ NB.
References
[BE]
[BE02] N. P. Byott and G. G. Elder, Biquadratic extensions with one break, Can. Math. Bull. 45
(2002), no. 2, 168–179.
[BE05] N. P. Byott and G. G. Elder, New ramification breaks and additive Galois structure, J.
Th´ eor. Nombres Bordeaux 17 (2005), no. 1, 87–107.
[Eld] G.G. Elder,
One-dimensionalelementary-abelian
arXiv.org:math/0511174 (2005-11-15).
[Ser79] J.-P. Serre, Local fields, Springer-Verlag, New York, 1979.
N. P. Byott and G. G. Elder, On the necessity of new ramification breaks, In preparation.
extensions oflocalfields,
Nigel P. Byott, School of Engineering, Computer Science and Mathematics, Univer-
sity of Exeter, Exeter EX4 4QE, United Kingdom
E-mail address: N.P.Byott@ex.ac.uk
G. Griffith Elder, Department of Mathematics, University of Nebraska at Omaha,
Omaha, NE 68182-0243 U.S.A.
E-mail address: elder@vt.edu
Current address: Department of Mathematics, Virginia Tech, Blacksburg VA 24061-0123
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