AMA CSC-AI
A. S. Classification
A.8. Final Comments on Regularity of Haar Measures
1. The Algebraic Closure Qp of Qp
1.7. Structure of Totally and Tamely Ramified Extensions
It is possible to improve the result (11.4.2) concerning the generation of totally ramified extensions.
Theorem. Let K C L C Q p be finite extensions of Qp. Assume that L/K is totally and tamely ramified of degree e. Then there exists a generator it of the maximal ideal P of R C K such that L is generated by an eth root of it in Q.
PROOF. By assumption e = [L : K] is prime to p. The proof will be accomplished in three steps.
(1) Consider arbitrary generators it of P C R C K and 7rL of PL C RL C L.
Since L/K is totally ramified of degree e, IJrL Ie = 17r l and 7rL/7r = u is a unit in RL. Since the residue degree of L/K is 1, the residue fields are the same, and there is a unit of R (one can take a root of unity in K) such that u (mod P)L.
Let us write
Ire = 7r - u,
u = +7rLV
(V E RL).Hence
7rL = in - ( +7rLV) = 7r +7r7rLV.
The element7r is also a generator of the ideal P of R. We are going to show that L is generatedby a root of the equation Xe - c7r. Let us replace the generator it by 7r' = 7r and simply denote it by it again. Thus we assume from now on that the generators 7rL and it are linked by a relation
7rL = 7r +7r7rLU (V E RL).
(2) The polynomial f = Xe - it is an Eisenstein polynomial (11.4.2) of R[X].
Hence itis irreducible over K[X]. We have
f(7rL) =7rL -7r = 7r7rLV, If(7rL)l = 17r7rLVI Letus factor f in Q° :
P
< 17r1.
f(X) = Xe -7r = fl (X -ai)
1 <i <e
19,
(where fi a, = ±7r). Since f is irreducible, the roots ai are conjugate and have the same absolute value in Qp, say jai 1 = c independent of i. Hence
Ce=fJ ICI, I=17r 1,
IaiI = C = Inlt/e = f rLl,
and
I7rL - ai 1 < max(lai 1, I7rLI) = l7TLl- If we come back to the polynomial f , then
Jfl (irL -ai) = If(lrL)I <
t<i <e
shows that at least one of the factors is smaller than IIrL I. Without loss of generality we may assume
I7rL-aII <InLI-
(3) The roots of f (X) = X e - rr = 0 are the ai = i a, where die = 1. Since e is prime to p, we have jai - 11 = I when l by Proposition 1 in (11.4.3). This proves
Iai-aI=Ice l=c=l7TL1 (i#1),
IlrL - aI < IULI = Ia - ai I (i 56 1)- By Krasner's lemma, we infer that
K(a) C K(JrL),
and since the element a has degree e, this inclusion is an equality.
Example. If we add a primitive pth root p of unity to Qp, we obtain a totally ramified extension K of degree p - 1. Hence K/Qp is tamely ramified and can be generated by a (p - 1)-th root of the generator -p of pZn.
For p = 3, we have seen in (11.4.6) that b = i works:
Q3(d J).
2. Definition of a Universal p-adic Field 2.1. More Results on Ultrametric Fields
Let us start with a couple of general results concerning (nondiscrete) ultrametric fields.
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2. Definition of a Universal p-adic Field 135
Proposition 1. Let K be an ultrametric field and K its completion. Then K is still an ultrametric field and
(a) I K I = IKI,
(b) K and K have the same residue field.
PROOF. Let A be the ring of Cauchy sequences in K. The ideal I of A consisting of Cauchy sequences a = (an) with a -+ 0 (also called null Cauchy sequences) is a maximal ideal: If an 0, then an 0 except for finitely many indices n and a is invertible in the quotient A/I. We can define K = A/I with a canonical injection K y K given by constant sequences. If a = (an) E A - I is a Cauchy sequence that is not null, the sequence (Ian I) is stationary (stationarity principle), and we define an absolute value on K by
Ial= limlanIEJK"ICR>ufora0 and101=0.
n-00
Obviously, the canonical injection K y K is an isometric embedding, and we view K as a subfield of K: The absolute value of K extends the absolute value of K. The residue field k of K parametrizes the open unit balls Bs1(a) (a = 0 or jai = 1) contained in the closed unit ball: k" parametrizes the open unit balls contained in the unit sphere S1 = {x E K : Ix I = 11. Any Cauchy sequence of the closed unit ball has all its final terms in an open unit ball; hence it corresponds to a fixed element in the residue field k.
An extension L of an ultrametric field K having same residue field kL = k and the same absolute values I L I = I K I is called an immediate extension of K. Hence the completion of K is an immediate extension of K.
Proposition 2. Let K be a nondiscrete ultrametric field and put A = {x E K : IxI < 1 } : maximal subring of K M = {x E K : Ix I < 1 } : maximal ideal of A.
Then, either M is principal, or M = M2 and the ring A is not Noetherian.
PROOF. By hypothesis r = I K" I { 11, and either r n (0, 1) has a maximal element 0 or it has a sequence tending to 1. In the first case we can choose it E M with lir I = 0, and M = it A is principal. In the second case, for each x E M, namely
IxI < 1, we can find an element y such that IxI < IyI < 1, so that x = y- (x/y)E M2.
Since y and x/y belong to M, this shows that x E M2, and we have proved M = M2.
In this last case, the subgroup F = I K" I is dense in R>0, and all the ideals
Ir=B<r=B<r(0;K)={xEK:lxl <r}
for r E r fl (0, 1) are distinct: The ring A is not Noetherian.
_-0
Proposition 3. With the same notation as before:
(a) If K is algebraically closed, so is the residue field k.
(b) If L is an algebraic extension of K, the residue field kL of L is also an algebraic extension of the residue field k of K.
PROOF. In any ultrametric field, I I > 1, Jai I 1 (i < n) implies
I
'I > (i < n),
IIn > YaiS'I,
i<n
and hence
1, i<n
+> ai
' 0.i<n
This proves that any root of a monic polynomial having coefficients Jai I < 1 belongs to the closed unit ball Ix I < 1.
(a) Let Xn +>i<n aiXi E k[X] be a monic polynomial of degree n > 1. Choose liftings ai E A of the coefficients, i.e., ai =ai (mod M), and consider the monic polynomial
Xn+>2aiX' E A[X].
i<n
Since the field K is algebraically closed, this polynomial has a root x E K. By the preliminary observation, x E A and x mod M is a root of the reduced polynomial Xn + Fi<n aiXi. This proves that k is algebraically closed.
(b) Let 0 E kL and choose a representative x E AL - ML of the coset
¢ 0: IxI = 1. By assumption, this element is algebraic over K, and hence x satisfies a nontrivial polynomial equation
Eaixn
= 0 (n > 1, ai E K).i<n
By the principle of competitivity, at least two monomials have maximal competing absolute values
Jai I = Jaix' I = Ja.tx' I = I aj I for some i
< j.
Dividing by ai, we obtain a polynomial equation with coefficients Jak I < 1. a' E A and at least two of them not in M. By reduction mod M we get a nontrivial
polynomial equation satisfied by . 0
'fl`fl moo,
fly
2. Definition of a Universal p-adic Field 137