J
OURNAL DE
T
HÉORIE DES
N
OMBRES DE
B
ORDEAUX
J. A. B
UCHMANN
H. W. L
ENSTRA
Approximatting rings of integers in number fields
Journal de Théorie des Nombres de Bordeaux, tome
6, n
o2 (1994),
p. 221-260
<http://www.numdam.org/item?id=JTNB_1994__6_2_221_0>
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Article numérisé dans le cadre du programme Numérisation de documents anciens mathématiques
Approximatting
rings
of
integers
in number fields.
par J. A. BUCHMANN AND H. W.
LENSTRA,
JR.RÉSUMÉ. 2014 Nous étudions dans cet article le problème algorithmique de la détermination de l’anneau des entiers d’un corps de nombres algébriques
donné. En pratique, ce problème est souvent considéré comme résolu mais
des résultats théoriques montrent que sa résolution ne peut être menée à
terme lorsque le corps étudié est défini par les équations dont les coefficients sont très gros. Or de tels corps apparaissent dans l’algorithme du crible
algébrique utilisé pour factoriser les entiers naturels.
En appliquant une variante d’un algorithme standard donnant l’anneau des entiers, on obtient un sous-anneau du corps de nombres qui peut être regardé comme le meilleur candidat possible pour l’anneau des entiers.
Ce meilleur candidat est probablement souvent le bon. Notre propos est
d’exposer ce qui peut être prouvé sur ce sous-anneau. Nous montrons que sa structure locale est transparente et rappelle celle des extensions modérément ramifiées de corps locaux. La plus grande partie de cet article
est eonsacrée à l’étude des anneaux qui sont "modérés" en un sens plus
général que celui habituel. Chemin faisant nous établissons des résultats de complexité qui prolongent un théorème de Chistov. L’article inclut
également une section qui discute des algorithmes en temps polynomial
pour les groupes abéliens de type fini.
ABSTRACT. - In this paper we study the algorithmic problem of finding the
ring of integers of a given algebraic number field. In practice, this problem is often considered to be well-solved, but theoretical results indicate that it
is intractable for number fields that are defined by equations with very large
coefficients. Such fields occur in the number field sieve algorithm for
facto-ring integers. Applying a variant of a standard algorithm for finding rings
of integers, one finds a subring of the number field that one may view as the
"best guess" one has for the ring of integers. This best guess is probably often correct. Our main concern is what can be proved about this subring.
We show that it has a particularly transparent local structure, which is
reminiscent of the structure of tamely ramified extensions of local fields. A
major portion of the paper is devoted to the study of rings that are "tame"
in our more general sense. As a byproduct, we prove complexity results that elaborate upon a result of Chistov. The paper also includes a section that
discusses polynomial time algorithms related to finitely generated abelian groups.
Key words: maximal order, tame extensions, algorithm.
Acknowledgements.
The first author wassupported by
the DeutscheForschungsgemeinschaft.
The second author wassupported by
theNa-tional Science Foundation under
grants
No. DMS 90-02939 and 92-24205. The second author isgrateful
to the Institute for AdvancedStudy
(Prince-ton),
wherepart
of the work on which this paper is based was done. Thefinal version of this paper was
prepared
while the second author was onap-pointment
as a Miller Research Professor in the Miller Institute for BasicResearch in Science.
1. Introduction
In this paper we are concerned with the
following problem
fromalgorith-mic
algebraic
numbertheory:
given
analgebraic
number fieldK,
determineits
ring
ofintegers
C7.Paradoxically,
thisproblem
is inpractice
considered well-solved(cf. [7,
Chapter
6]
and 7.2below),
whereas a result of Chistov[6]
(Theorem
1.3below)
suggests
that from a theoreticalperspective
theprob-lem is intractable. The
apparent
contradiction is easy to resolve.Namely,
allcomputational
experience
so far is limited to "small" number fieldsK,
such as number fields that aregiven
as K = whereQ
is thefield of rational numbers and
f
is an irreduciblepolynomial
of smallde-gree with small
integer
coefficients. Thealgorithms
that are used for smallfields will not
always
work whenthey
areapplied
to"large"
number fields.Large
number fields arealready making
their appearance inapplications
ofalgebraic
numbertheory
(see
[14]),
and the determination of theirrings
ofintegers
isgenerally
avoided(see
[5;
16, 9.4;
9]).
The results of thepresent
paper aremainly theoretically inspired,
butthey
may becomepractically
relevant if one wishes to do
computations
inlarge
number fields.In accordance with Chistov’s
result,
we shall see that there iscurrently
not much
hope
to find agood algorithm
for theproblem
ofconstructing
rings
ofintegers.
This is true if"good"
is taken to mean"running
inpolynomial
time" ,
and it isequally
trueif,
lessformally,
it is taken to mean"practically
usable,
also in hard cases". The sameapplies
to theproblem
ofrecognizing
rings
ofintegers,
i. e., theproblem
ofdeciding
whether agiven
subring
of agiven
algebraic
number field I~ isequal
to 0.To
appreciate
the centraldifficulty
it suffices to look atquadratic
fields. If m is aninteger
that is not a square, thendetermining
thering
ofintegers
of isequivalent
tofinding
thelargest
square divisor of m. Theof
recognizing
thering
ofintegers
of aquadratic
field isequivalent
to theproblem
ofrecognizing
squarefree
integers,
which is considered infeasible aswell.
In the
present
paper we obtain somepositive
results. We shall provethat,
eventhough
C~ may be hard todetermine,
one can at least constructa
subring
B of K that comes "close" to0,
that isperhaps
evenlikely
tobe
equal
to0,
that in any case has some of thegood
properties
of0,
andthat in
computational
applications
ofalgebraic
numbertheory
canprobably
play
the role of C7. Before we state our main result wegive
an informaloutline of our
approach.
Chistov
[6]
showed that theproblem
ofdetermining
thering
of inte-gers of agiven
number field ispolynomially equivalent
to theproblem
ofdetermining
thelargest squarefree
divisor of agiven positive integer
(see
Theorem 1.3
below).
For the latterproblem
nogood
algorithm
is known(see
Section7).
However,
there is a naiveapproach
thatoften
works. It isbased on the observation that
positive integers
with alarge repeated
prime
factor are scarce: for most numbers it is true that all
repeated
prime
fac-tors are small and therefore easy to find.
Thus,
dividing
agiven positive
integer
dby
all of itsrepeated
prime
factors that are less than a certainupper bound b one finds a number that may have a
good
chance ofbeing
the
largest squarefree
divisor ofd,
and that is often the best guess one has.The success
probability
of this methoddepends
on b and on the way inwhich d was obtained in the first
place.
Itis,
of course, easy to constructnumbers d that defeat the
algorithm.
One can
attempt
to determine thering
ofintegers
0 of agiven
numberfield K in a
similarly
naive manner. One starts from an order inK, i. e.,
asubring
A of C7 for which the index(C~ :
A)
of additive groups isfinite;
forexample,
one may take A =Z [a],
where a E K is analgebraic integer
withK =
Q (a) .
As we shall see, one can determine 0 if thelargest squarefree
divisor m of the discriminant
AA
of A is known. This resultsuggests
that one can determine a "bestguess"
for (?by working
with the best guess qthat one has for m instead of m itself.
If,
in the course of thecomputations,
the
hypothesis
that q issquarefree
is contradicted because aninteger a
> 1 is found for whicha2
divides q, then onesimply replaces q by
q/a
and onecontinues as if
nothing
hashappened.
This vague idea can be made
perfectly
precise.
Itgives
rise to apolyno-mial time
algorithm that,
given
an order A CK,
produces
an order B c Kcontaining
A as well as apositive integer
q. One knows that B = C~if q
issquarefree.
It will often be considered verylikely that q
is indeedsquare-free,
so that one is inclined to believe that B = 0. Our main concern iswhat one can prove about B without
relying
on anyunproved
assumptions
regarding
q. Inparticular,
we shall prove that Bequals
0 if andonty if q
is
squarefree,
and thatfinding
an order in K thatproperly
contains B isequivalent
tofinding
asquare a2 >
1dividing
q.Our results are derived from a local
property
of B that we refer to astameness at q.
Loosely speaking,
B is tame at q if B istrying
to resemblea full
ring
ofintegers
asclosely
as ispossible
in view of the factthat q
isnot known to be
squarefree.
Tameness is astrong
property,
whichprovides
us with substantial control over thering.
Before wegive
the definition weremind the reader of the local structure of full
rings
ofintegers.
Let 0 be the
ring
ofintegers
of analgebraic
numberfield,
let p be amaximal ideal of
C~,
and letOp
be thep-adic
completion
of C?(see
[1,
Chapter
10]).
Denoteby
p theunique prime
number thatbelongs
to p.If the ramification index
e(p/p)
of p over pequals
1,
then p is said to be unramified over p, and in that caseOp
is a local unramifiedalgebra
overthe
ring
Zp
ofp-adic integers
(see
Section3).
Local unramifiedZp-algebras
are easy to understand and to
classify,
andthey
have a verytransparent
structure
[25,
Section3-2];
forexample, they
are,just
likeZp
itself, principal
ideal domains that
have,
up tounits,
only
oneprime element,
namely
p.If,
moregenerally,
p does not dividee(p/p),
then p is said to be tame over p.In this case there is a local unramified
Zp-algebra
T and a unit v of T suchthat
Op ~
T[X]I(Xe(PIP) -
vp)T[X]
J =(see
[25,
Section3-4]).
Conversely,
let p be aprime number,
let T be a local unramifiedZp-algebra,
let v be a unit ofT,
and let e be apositive integer
that isnot divisible
by
p. Then there is analgebraic
number field whosering
ofintegers
0 has a maximal ideal pcontaining
p for which thering
T[(vp) 1I e]
is
isomorphic
toOp,
and thene(p/p)
= e. In summary, therings
which are
relatively simple
tounderstand, provide
a fulldescription
of thecompletions
of therings
ofintegers
of allalgebraic
number fields at all tamemaximal ideals.
In the wild case, in which p does divide
e(p/p),
the structure ofOp
is somewhat more
complicated,
but there isfortunately
no need for us toconsider it: it occurs
only if p
issmall,
and smallprimes
can be taken careof
directly.
Imitating
thedescription
above ofOp
we make thefollowing
definition.Let B be an order in a number
field,
andlet q
be apositive
integer.
We callB tame
at q
if for everyprime
number pdividing
q and every maximal ideal p of Bcontaining
p there exist a local unramifiedZp-algebra
T,
aninteger
ethat is not divisible
by
p, and a unit u of T such that thep-adic
completion
Bp
of Bis,
as aZp-algebra,
isomorphic
to(see
Section4).
If q
issquarefree
then p divides qonly
once; in that case uq = vp forsome unit v of
T,
and we are back at thering
considered above.However,
if p2
divides q thenuq)T[X]
occurs as aring
Op
asabove
only
in the trivial case that e = 1(cf. 3.5).
One of our main results now reads as follows.
THEOREM 1.1. There is a deterministic
polyreonzial
timealgorithrra
that,
given
a numberfield
K and an order A inK,
determines an order B in Kcontaining
A and apositive integer
q, such that B is tame at q and suchthat the
prime
numbersdividing
(0 :
B)
are the same as therepeated prime
divisors
of
q; here C~ denotes thering
of
integers
of
K.This theorem is
proved
in Section6,
along
with the other theorems stated in this introduction. Thealgorithms
referred to in our theorems will beexplicitly
exhibited.Clearly,
thering
B in Theorem 1.1equals
0 if andonly
if q issquarefree.
Generally
we shall see thatexhibiting
a squarea2
> 1dividing
qis,
underpolynomial
transformations,
equivalent
tofinding
anorder in K that
strictly
contains B(see
Theorem6.9).
Finding rings
ofintegers
iscustomarily
viewed as a localproblem,
in thesense that it suffices to do it
prime-by-prime. Algorithmically, however,
the bottleneck is of aglobal
nature: how tofind
theprime
numbers that oneneeds to look at? Once these are
known,
theproblem
admits a solution.This is
expressed
in our next result. If m is aninteger,
then an order A inK is said to be maximal at m if
gcd(m,
(0 : A))
= 1.THEOREM 1.2. There is a
polynomial
timealgorithm
that, given
analge-braic number
field
K,
an order A ircK,
and asquarefree
positive integer
m, determines an order B in K
containing
A that is maximal at m.From 1.2 we see in
particular
that if m isprime
one canfind,
inpolynomial
time,
an order in K that is maximal at m. If m is taken to be theproduct
of theprimes
p for whichp2
divides the discriminant ofA,
then the order B in Theorem 1.2equals
0.We next formulate a few
complexity
results ofpurely
theoretical interest.THEOREM 1.3. Under
polynomial transformations,
thefollowing
twoprob-lems are
equivalent:
(a)
given
analgebraic
numberfield
K,
find
thering
of algebraic
integers
of K;
(b)
given
apositive integer
d, find
thelargest squarefree
divisorof
d.Theorem 1.3
represents
aslight
improvement
over a theorem of Chistov(6),
asexplained
in 6.11. We shall prove that thecorresponding recognition
problems
are alsoequivalent
(Theorem 6.12).
Suppose
that an order A in analgebraic
number field isgiven.
In theproof
of Theorem 1.3 we shall seethat, given
thelargest squarefree
divisorof the discriminant
AA
ofA,
one can find thering
ofintegers
0 of K inpolynomial
time. In 6.13 we argue that it is hard to go in theopposite
direction: ifgiven
0 one caneasily
find thelargest squarefree
divisor ofAA,
thenproblem
1.3(b)
is easy as well. It ispossible, however,
tocompute
the
largest
square divisor ofAA
quickly
from 1;again
it is hard to go in theopposite
direction(see 6.14).
If the
ring
ofintegers
of a number field K isknown,
then the discriminant of K is easy tocompute.
One may wonderwhether,
conversely,
it is easyto
compute
thering
ofintegers
of K from the discriminant of K. In 6.10we shall see that this is
currently
not the case.However,
we do have thefollowing
result.THEOREM 1.4. There are
polynomial
timealgorithms
thatgiven
analge-braic number
field
K and oneof
(a), (b),
determine the other:(a)
thering
of algebraic
integers
of K;
(b)
thelargest squarefree
divisorof
the discriminantof
K.In the
body
of the paper we work with orders inproducts
of number fieldsrather than orders in number fields. This
presents
no additionaldifficulty.
One may
remark,
though,
that the case ofproducts
of number fields canin
polynomial
time be reduced to the case of asingle
numberfield, by
themain result of
[15].
Also,
several of our results are local in the sense thatthey
are directed not atconstructing 0,
but atconstructing
an order thatis maximal at a
given
integer
m, as in Theorem 1.2.We have refrained from
considering
moregeneral
baserings
than thering
Z of rationalintegers.
Over some baserings,
theproblem
offinding
maximal orders
is,
insubstance,
equivalent
to theproblem
ofresolving
sin-gularities
of curves(see [24]);
but in that context there is aquick
algorithm
for
problem
1.3(b),
so that the issues considered in this paper do not arise. It may beinteresting
to consider baserings
that arerings
ofintegers
ofnumber fields or, more
generally,
orders in number fields asproduced
by
ouralgorithms.
The contents of this paper are as follows. Sections
2,
3 and 4 contain the commutativealgebra
that we need. Noalgorithms
occur in these sections.In Section 2 we assemble some well-known results
concerning
orders.Sec-tions 3 and 4 are devoted to the notion of tameness,
locally
in Section 3 andglobally
in Section 4. Sections 5 and 6 deal withalgorithms.
In Section 5we recall a few basic
algorithms
for which a convenient reference islacking;
they mostly
concern linearalgebra
over therings
Z andZ/qZ,
where q
is apositive integer.
In Section 6 wepresent
thealgorithm
that underlies theproof
of Theorem 1.1. It is a variant of analgorithm
fordetermining
ma-ximal orders that is due to Zassenhaus
[26; 27].
Section 6 also contains theproofs
of the theorems stated above. In Section 7 we discuss thepractical
repercussions
of our results.For our conventions and notations on commutative
algebra
we refer toSection 2. For conventions
concerning
algorithms
we refer to Section 5 andto
[18].
2. Orders
In this section we establish the notation and
terminology
concerning
rings
and orders that we shall use, and we recall a few well-known facts.For
background
on commutativealgebra,
see[1].
2.1.
Rings
andalgebras.
Allrings
in this paper are assumed to be commutative with a unit element.Ring homomorphisms
are assumed topreserve the unit
element,
andsubrings
contain the same unit element.By
Z, Q,
Fp
we denote thering
ofintegers,
the field of rationalnumbers,
andthe field
of p
elements,
respectively,
where p is aprime
number. The groupof units of a
ring
R is denotedby
R* . Let R be aring.
An R-module Mis called free if it is
isomorphic
to the direct sum of a collection ofcopies
of
R;
ifR ~
0 then the number ofcopies
needed isuniquely
determinedby
M,
and it is called the rank ofM;
if R =0,
then the rank of M is definedto be 0. If an R-module M is free of finite rank n, then there is a basis of
M over
R, i. e.,
a collection of n elements wl, w2, ... , wn E M such thatfor each x E M there is a
unique
sequence of n elements ri, r2, ... , rn E R such that x = riwi-By
anR-algebra
we mean aring
Atogether
witha
ring
homomorphism
R - A. AnR-algebra
A is said to admit a finitebasis if A is free of finite rank when considered as an R-module. If this is
the case, then the rank of A as an R-module is called the
degree
of A overR,
notation:~A : R~
..2.2. Trace and discriminant. Let .R be a
ring
and let A be anR-algebra
admitting
a finite basis wi, - - - , wn . For each a EA,
the trace Tr a of aif awz = with rjj E
R,
then Tr a =3 = i
rjj. The trace Tr is an R-linear map A -> R. In case ofpossible ambiguity,
we may writeTr A
or Tr A/R instead of Tr . The discriminantAA
orAA/R
of A over Ris the determinant of the matrix
(Tr
The discriminant is- _ n
well-defined
only
up to squares of units of R. The R-idealgenerated by
AA
iswell-defined,
and we shall also denote itby
AA-
If R’ is anR-algebra,
then A’ =
AOtimes RR’
is anR’-algebra
that also admits a finite basis.The trace function As R’ is obtained from the trace function A - R
by
baseextension,
and the notationTr ,
Tr A, The A/R used for the latter willalso be used for the former. We have
AAI/Rl =
AA/RR’ as
ideals.2.3. Orders. Let R be a
principal
idealdomain,
and denoteby
F the fieldof fractions of R. An order over R is an
R-algebra
A that admits a finitebasis and that satisfies
AA 0
0. An order over Z issimply
called anorder;
equivalently,
an order can be defined as aring
without non-zeronilpotent
elements of which the additive group is free of finite rank as an abelian
group. Let A be an order over
R,
and writeAF
=AQ9RF.
ThenAF
is,
asan
F-algebra,
theproduct
offinitely
many finiteseparable
field extensionsof F.
By
a fractional A-ideal we mean afinitely generated
A-submoduleof
A~
that spansAF
as a vector space over F. If a and b are fractionalA-ideals,
then the index(a : b)
of b in a is defined to be the determinant of any F-linear mapAF --~ AF
that maps a ontob;
the index is an elementof F* that is well-defined
only
up to units of R. If b C a then the indexbelongs
to R -{0},
and if in addition R = Z then itis,
up tosign,
equal
tothe usual index. If a, b are fractional
A-ideals,
then we write a : b =Ix
EAF :
xb Ca};
this is also a fractional A-ideal. A fractional A-ideal a iscalled invertible if ab = A for some fractional A-ideal
b;
if this istrue,
then b = A : a, and a = A : b = A :(A :
a).
Anexample
of a fractional idealis the
complementary
moduleAt
=Ix
EAF :
Tr(xA)
CR}.
If(Wi)i::l
isa basis for A over
R,
then a basis forAt
over R isgiven
by
the dual basis!t %= 1,
which is characterizedby
Tr = 0 or 1according
as i0 j
ori =
j.
One has AC At
and(Afi :
A) =
AA.
By
an overorder of A we meana fractional A-ideal that is a
subring
ofAF.
If a is a fractionalA-ideal,
then a : a is an overorder of A. Each overorder B of A is an
order,
and itsatisfies A c B C
Bf
C At
andAA
=AB (B :
A)2.
Among
all overordersof A there is a
unique
one that is maximal underinclusion;
we shall denoteit
by
0. Thering
C~ isequal
to theintegral
closure of R inAF,
and it is theproduct
offinitely
many Dedekind domains. The discriminant of C~ is also called the discriminant ofAF.
We call A maximal if A =0;
this is thecase if and
only
if all fractional A-ideals are invertible. Ifm E R,
then the order A is said to be maximal at m ifgcd(m, (0 : A) )
= 1;
thishappens,
for
example,
if gcd(m,
AA)
=1,
because(0 :
A)2
dividesAA.
For the samereason, A itself is maximal if and
only
if it is maximal atAA-PROPOSITION 2.4.
Suppose
that R is aprincipal
idealdomain,
that m E Ris a non-zero
element,
and that A is an order over R. Then there areonly
finitely
manyprime
ideals pof
Acontaining
m, andthey
are all maximal.Moreover,
if
b denotes the intersectionof
theseprime ideals,
then we have:(a) b/mA
is the nilradicalof
thering
A/mA,
and there exists apositive
integers
t such that b D mA Dbt;
(b)
for
eachprime
ideal pof
Acontaining
m one has A :p ¢
A;
(c)
A is maximale atm if
andonly if
b : b = A.Proof.
Since A admits a finite basis over theprincipal
ideal domainR,
the R-moduleA/mA
is of finitelength.
Therefore thering
A/mA
is an Artinring.
From[1,
Chapter
8]
it follows that eachprime
ideal ofA/mA
ismax-imal,
that there areonly
finitely
many ofthem,
and that their intersectionis
nilpotent.
This proves the first two assertions of2.4,
as well as(a).
Toprove
(b),
we note that the annihilator of theprime
idealp/rraA
in the Artinring
A/rrtA
is non-zero, so mA : pproperly
contains mA. ThereforeA : p
properly
contains A. To prove(c),
first assume that A is maximalat m. From
(0 :
C A andgcd(m,
(0 : A))
= 1 it follows that for eachmaximal ideal
pR
of Rdividing
m the localizationsAPR
andOPR
areequal.
Hence the order
APR
overRp R
is aproduct
offinitely
many Dedekinddo-mains,
andbpR
is aproduct
of non-zero ideals in those Dedekind domains.Therefore
bpR : bpR
=ApR.
The sameequality
also holds for maximalideals
pR
of R that do not contain m, since in that casebp R
=APR.
Itfol-lows that b : b =
A,
asrequired.
For theconverse, assume that b : b = A.
The maximal ideals p of A
containing m
arepairwise coprime,
so theirintersection b is
equal
to theirproduct.
Hence b : b = Aimplies
that allthose p
satisfy
p : p = A. We claim that(A : p)p
= A for each p, so thateach p is invertible. If not, then from p C
(A : p)p
C A and themaximality
of p one derives that p =(A : p)p,
so A : p C p : p =A, contradicting
(b).
From the
invertibility
of all maximal idealscontaining
m one deducesby
induction that all A-ideals that contain a power
mk
of m, with k >0,
areinvertible,
and the same is then true for all fractional ideals H withApply
this to H =Ix
~ /:mix
~ A forsome
i > 0~.
This is aring,
so HH =H,
and theinvertibility
of Himplies
H = A. Therefore
(C~ : A)
iscoprime
to m. This proves 2.4..Remark.
Every
maximal ideal p of an order A over aprincipal
ideal domainR that is not a field contains a non-zero element of
R,
by
[1,
Corollary
5.8~.
PROPOSITION 2.5. Let A be an order over a
princiPal
idealdomain,
and let a be afractional
A-ideal. Then a is invertibleif
andonly if
the overorder(A :
a) :
(A : a)
of
Aequals A,
andif
andonly if
both A :(A :
a)
= a anda : a = A.
Proof.
If a isinvertible,
then as we saw in 2.3 we have A :(A : a)
= a, and a : a =(a : a)A
=(a : a)a(A :
a)
=a(A : a)
=A;
also,
b = A : a is theninvertible as
well,
so for the same reason we have(A : a) : (A : a)
= A.Next suppose that a is not invertible. Then the A-ideal
(A : a)a
is different fromA,
so there is a maximal ideal p of Acontaining
(A : a)a.
Wehave A :
p C A :
((A : a)a) =
(A :
a) :
(A :
a),
so from2.4(b)
we see that(A : a) : (A :
a) #
A. This proves that a is invertible if(A :
a) : (A :
a)
= A.Applying
this to b = A : a, we find that b is invertible if A : b = a and a : a =A,
and then its inverse a is invertible as well. This proves 2.5.2.6. Gorenstein
rings.
Let A be an order over aprincipal
ideal domain.We call A a Gorenstein
ring
if A :(A :
a)
= a for every ideal a of Athat contains a non-zero-divisor of A or,
equivalently,
for every fractionalA-ideal a. It is an easy consequence of
[3,
Theorem(6.3)]
that thisis,
fororders over
principal
idealdomains,
equivalent
to the traditional notion. Note that A is a Gorensteinring
if it is a maximal order. The converse isnot true
(cf. 2.8).
PROPOSITION 2.7. Let A be an order over a
principal
ideal domainR,
withcomPlementary
moduleAt.
Then thefollowing
properties
areequivalent:
(a)
A is a Gorensteinring;
(b)
for
anyfractional
A-ideal a, we have a : a = Aif
andonly if
a isinvertible;
(c)
At is
invertible.Proof.
From 2.5 and the definition of a Gorensteinring
it is clear that(a)
implies
(b).
To prove that(b)
implies
(c),
it suffices to prove thatAt : At
=A.
Generally,
put
at
={x
EAF :
Tr(xa)
cR}
for any fractional A-ideala, where
AF
is as in 2.3.Using
dual bases oneeasily
proves thatatt
= a,and from the definitions one sees that
at = At :
a.Applying
this to a= At
one obtainsAt : At
=A,
asrequired. Finally,
we prove that(c)
implies
(a).
Suppose
thatAt
isinvertible,
and let a be a fractional ideal.Applying
theequality
at = At :
a twice we find that a= att
=At :
(At :
a).
We need to prove that thisequals
A :(A : a).
If b = A :At
denotes the inverse ofAt,thenwehaveAt:a=(A:b):a=A:(ab)=(A:a):b=(A:a)At
andAt:(At:a)=At:((A:a)At)=(At:At):(A:a)=A:(A:a).
2.8.
Example.
Let R be aprincipal
ideal domain and letf
ER[X]
be amonic
polynomial
withnon-vanishing
discriminant. Then A =is an order over
R,
and if we write a =(X
modf )
E A thenAt
=f’(a)-lA
(cf.
[25,
Proposition
3-7-12]).
This shows thatAt
isinvertible,
so 2.7im-plies
that A is a Gorensteinring.
It is well-known that A is notnecessarily
maximal.
PROPOSITION 2.9. Let R be an Artin
ring,
let L be afree
R-rnodv,leof
finite
rank,
and let N c L be a subrraodule. Then N isfree
overR if
andonly if
L/N is
free
over R.Proof.
Since each Artinring
is aproduct
offinitely
many local Artinrings,
the
proof immediately
reduces to the case that R is local. It is convenientto use a few
properties
ofprojective
modules,
which can be found in[12,
Chapter
1,
Section1].
First suppose thatL/N
is free. Then the exactsequence 0 -> N -> L 2013~ 0
splits,
so N isprojective,
and thereforefree. For the converse, assume that N is free. Let m be the maximal
ideal of
R,
and let a E R a non-zero element annihilatedby
m. Thenis a
subspace
of theR/m-vector
spaceL/mL.
Supplementing
anR/m-basis of
N/mN
to one forL/mL
andapplying Nakayama’s
lemma onefinds a
surjection
N E9 Rn -->L,
where n = rank L - rank N.Comparing
the
lengths
of the two modules we see that it is anisomorphism.
HenceThis proves 2.9.
3. Tame
algebras
over thegradic integers
This section and the next one are devoted to a
study
of tameness, whichis one of the central notions of this paper.
We let in this section p be a
prime number,
and we denoteby
Zp
thering
ofp-adic integers.
We call aZp-algebra
T local if T is local as aring
with a residue class field of characteristic p. A local
Zp-algebra
T is said tobe unramified if T = for some monic
polynomial 9
EZp[Y]
for which
(g
modp)
EFp [Y]
is irreducible.Equivalently,
a local unramifiedZp-algebra
is theintegral
closure ofZp
in a finite unramified extension ofthe field
Qp
of p-adic
numbers(see
[25,
Section3-2)).
Throughout
thissection,
q denotes a non-zero element ofpZp.
Let Sbe a
Zp-algebra.
If S islocal,
then we call S tameat q
if there exist alocal unramified
Zp-algebra
T,
apositive integer
e that is not divisibleby
In
general,
we call S tame at q if S is theproduct
offinitely
many localZp-algebras
that are tame at q.If q is a
prime
element ofZp
then tamenessat q
isequivalent
to the traditionalnotion,
asexpressed by
thefollowing
result.PROPOSITION 3.1.
Suppose
that q EpZp,
q 0
p2Zp,
and let S be aZp-algebra.
Then S is local and tame atq if
andonly if
S isisorrtorphic
to theintegral
closureof Zp in
afinite tamely ramified field
extensionof
Qp.
Proof.
This follows from thedescription
oftamely
ramified extensionsgiven
in[25,
Sections3-2, 3-3,
3-4].
We now prove various
properties
ofZp-algebras
that are tame at q.PROPOSITION 3.2. Let T be a local
unramified
Zp -algebra, let
e be apositive
integer
that is not divisibleby
p, and let u E T* be a unit. Letfurther
S =
T[X]I(Xe - uq)T[X]
=T[7r],
where 7r =(X
mod Xe -uq).
ThenS is local and tame at q, and its rraaximal ideal is
generated by
p and 7r.Further,
the residue classfield
kof
S is the same as thatof
T,
and itsatisfies [k :
Fp] =
[T :
Zp] =
[S :
Zip]/6.
Proof.
It is easy to see that the S-idealpS
+ 7rS is maximal and that its residue class field k is the same as the residue class fieldT/pT
of T.Conversely,
letp C S
be a maximal ideal. Since S isintegral
overZp,
wehave
pflZp
=pZp (see
[1,
Corollary
5.8]),
so p E p.Also,
from ~e = uq E pit follows that 7r E p. This
implies
that p =pS
+7rS,
and that S is local.The fact that S is tame at q follows from the definition of tameness. The relations between the
degrees
follow from~T/pT : Zp/pZp] = [T : Zp]
and[S : T]
= e. This proves 3.2.PROPOSITION 3.3. Let
T,
e, u, S and 7r be as in3.2,
and let Tr be the tracefunction of S
overZp.
Then we have:(a)
thecomPlementary
module St
of S
overZp is
given
by
St
=7rq-’S,
and
Zp-module
isorraorphic
to the direct sumof
[k :
and
Sla is
as aZp/qZp-module
free of
rank~k :
Fp];
(d)
for
eachpositive integer
i theZp/qZp-module
+qS)(ai+1
+ +qS)2
isfree,
and its rankequals
0for
i=1=
e and[k :
Fp~
for
i = e.Proof.
Since T is unramified overZp,
we haveTt
= T.Combining
thiswith Tr = Tr T o Tr one finds that
St
is also thecomplementary
module of ,S over T. A T-basis of S is
given
by
1,
7r,7r 2
... ,7r e-1
Astraightforward
computation
shows that the dual basis isgiven
by
e- 1
(euq)-I7re-2,
... , 7 and this is a basis for7rq-IS.
Hence
St
= which is the first assertion of(a).
Another T-basis forSt
isgiven
by
1,
... ,q-’7r e-1
7 from which it follows thatst /s is,
as aT-module,
isomorphic
to the direct sum of e - 1copies
ofTlqT.
SinceTlqT
is free of rankII~ :
Fp]
overZp/qZp
thisimplies
the lastassertion of
(a) .
To prove
(b)
itsuffices, by
2.3,
tocompute
the determinantof a Zp-linear
map that mapsSt
onto S’(for
example, multiplication by
7re-I).
This is leftto the reader. For
(c)
we note that a =(est)
f1,S’ n S’=7r S,
so ae = =qS
andS’/a
=TlqT;
the lastisomorphism
follows from S =Finally,
from(c)
we obtain that =for any
positive integer i,
so -f-q~S’) 2
= 0 if i#
ewhich
implies
(d) .
This proves 3.3.3.4.Remark. Let S be a local
Zp-algebra
that is tame at q, and let k be its residue class field. We shall call[k :
Fp]
the residue class fielddegree
of S overZp.
From 3.2 it follows that T and e areuniquely
determinedby
S.Namely,
Tis,
as a local unramifiedZp-algebra,
determinedby
its residue classfield,
which is k.Using
Hensel’s lemma one can show that Tis even
uniquely
determined as asubring
of S(cf.
the construction of Tin the
proof
of3.7).
Next,
e is determinedby
e =[S :
Zp~ / ~l~ : Fp~ .
We shall call e the ramification index of S overZp .
If e >1,
then from3.3(a)
it follows that the idealqZp
is alsouniquely
determinedby
S. Hence alocal
Zp-algebra
that is not unramified cannot be tame at two values of q that are not divisibleby
the same power of p. From3.3(c)
one can deducethat,
for e >1,
notonly
the idealqZp
but also the set isuniquely
determinedby
S’.Conversely,
S isclearly
determinedby
T,
e andPROPOSITION 3.5. Let the notation be as in
3.2,
and let thepositive integer
g be such thatqZp =
PgZp.
Denote theintegral
closureof
S in S®zp Qp.
Then wehave
Ee- 1
and C S.Further,
9
isequal
to Sif
f
andonly if
e = 1 orq 0
p2Zp.
We haveP* P
and this
equals
(1)
if
andonly if
g is divisibleby e.
Proof.
We first prove theexpression
for S
under the addedassumption
that T contains aprimitive
eth root ofunity C.
In that case, there is aT-algebra automorphism a
of s with U7r = and agenerates
a group rof order e. The action of r on ,5’ extends to an action of r on S
0Zp
Qp
and on S. We consider the structure
of S
as a module over the groupring
T[r].
Frome #
0mod p
it follows that there is anisomorphism
ofT-algebras
T[r]
that sends a to ThereforeS
is,
as aT’[r]-module)
the direct sum of modulesSZ,
0 i e, whereSZ -
~x
ES’ :
ox =C’xl.
We have7ri
ES’2,
so a acts as theidentity
onand therefore is contained in the field T
Qp.
Because T is unramified overZp,
any T-submodule of that field is determinedby
theintegral
powersof p
that itcontains;
so it remains to see which powersof p
belong
toFor j
EZ,
wehave pi
E7r -’Si
if andonly
if7ripi
isintegral
over
~S’,
if andonly
if its eth power(uq)ipej
isintegral
overS’,
if andonly
if
e j > -gi,
if andonly
This shows that1T-iSi
=as
required.
Next we prove the
expression
for S
in thegeneral
case. Frome 0
0 mod pit follows that there exists a local unramified
Zp-algebra
T’containing
Tthat contains a
primitive
eth root ofunity.
Apply
the above to S’’ =SQ9TT’,
and use that S’
equals
the intersection of S‘Ozp
Qp
with theintegral
closureof ,5‘’ in
Cap.
This leads to the desired result.From
[gile] g
we see that = C ~S’. We have S’ _ ,S’ if andonly
if[gile] =
0 for0
i e -1,
if andonly
ifg (e - 1)
e, if andonly
if g
= 1 or e = 0. This proves the second statement of 3.5.The formula for the discriminant follows
by
an easycomputation
from3.3(b)
and the formulaS)2
from 2.3. The lastasser-tion is obvious. This proves 3.5.
Let now ,S’ be a
Zp-algebra
that is tame at q but that is notnecessarily
local. Then S’ is the
product
of the localizationsSp
of s at its maximal ideals p, of which there areonly
finitely
many, and each,S’p
is a localZp-algebra
that is tame at q. We shall denote the residue class fielddegree
and the ramification index of,S’p
overZp
by
f (p)
ande(p),
respectively.
PROPOSITION 3.6. Let S be a
Zp-algebra
that is tame at q, andput
a =Ix
E S’ : Tr CqZp},
where Tr is the traceof S
overZp.
the
integral
closureof S
ThenSla
isfree
as aZp/qZp -module,
(c)
if a
=qS,
then =(1)
and S =S;
(d)
if a 0 qS,
then q divides I we have CS,
equals
S
if
andonly
p2 Zp;
(e)
if e
denotes the least commonmultiple of
the numberse(p),
with pranging
over the maximal idealsof S,
then we have =(1)
if
and
only
if qZp is
the eth powerof
an idealof
Zp.
Proof.
The ideal a is theproduct
of thesimilarly
defined ideals ap of therings
Sp.
By
3.3(c),
each of theZp/qZp-modules SP/ap
isfree,
so the sameis true for
Sla.
To prove(a),
we may likewise assume that S islocal,
inwhich case it suffices to
apply
3.3(b),
3.2,
and3.3(c).
In the same way(b)
follows from
3.3(d).
If a =qS,
then we have[S: Zp] -
=0,
which
implies
the first statement of(c);
the second followsby
2.3. Next suppose thata ~ qS.
Then we have[S : Zp] - [Sla :
Zp/qZp]
>0,
which
implies
the first statement of(d).
Also,
for at least one p we have[Sp : Zp] -
Zp/qZp]
>0,
which means that ep > 1. Since S isintegrally
closed in S0Zp
Qp
if andonly
if eachSp
isintegrally
closed inSp
Qp,
it now follows from 3.5 that this is alsoequivalent
toq 0
p Zp .
This proves
(d).
Finally,
(e)
follows from the last statement of 3.5. This proves 3.6.The main result of this section enables us to
recognize
whether agiven
Zp-algebra
is tame at q,provided
that it hassufficiently
smalldegree
overZp.
THEOREM 3.7. Let p be a
prime
number,
and let q EpZp,
q ~
0,
whereZp
denotes thering
of p-adic
integers.
Letfurther
S be aZp-algebra
thatadmits a
,finite
basis,
with[S :
Zp]
p. Put a ={x
E S : Tr(xS)
CqZp},
where Tr is the trace
of
S overZp.
Then S is tame at qif
andonly if
a : a = S and both
a/qS
and(S : a)/S
arefree
asZp/qZp-modules.
Proof.
We first remark thatby
2.9,
applied
to R =ZplqZp,
L =S/qS,
andN =
a/qS,
theZp/qZp-module
a/qS
is free if andonly
ifS/a
is. Hence wemay
replace
a/qS
by
S/a
in the statement of Theorem 3.7.For the
proof
of the"only
if"part
we may assume that S is notonly
tame at q but alsolocal,
as in theproof
of 3.6. Thenby
3.3(c)
we have a =7rS’,
so a : a = S.Also,
S/a
is free overZp/qZp,
by
3.3(c),
and the sameapplies
to(S : a)/S
= =Sla.
This proves the"only
if"
part.
Next we prove the "if"
part.
Assume that a : a = S and that bothSla
and
(S :
a)/S
are free asZp/qZp-modules.
We first reduce to the case thatwith the
projective
limits of therings
0. From2.4(a)
we knowthat there is a
positive integer
t such thatIIp P J
pS D
Ilp
pt,
where pranges over the
prime
ideals of ,S’containing
p. It follows that thesystem
ofideals
(pnS)"O 1
is cofinal with thesystem
of ideals(IIp
so that S isalso the
projective
limit of therings
S/
(flp
pn) -
For each rt, the idealspn
are
pairwise coprime,
sopn ) ’--" rip
S/pn .
Hence if we letS’p
denotethe
projective
limit of therings
0,
then we have anisomorphism
S ^--’
Tjp
Sp
ofZp-algebras,
theproduct extending
over theprime
ideals p of S’containing
p. Inaddition,
eachSp
islocal,
and it isactually
the localization of ,S’ at p. As aZp-module,
eachSp
is a direct summand ofS,
so it isfree,
with[Sp : Zip] :5
[S :
Zp]
p.Also,
theassumptions
on a carry over to each,Sp .
Since S is tame if each of theSp
is,
we conclude that wemay assume that S is
local,
which we do for the remainder of theproof.
Denote
by
p the maximal ideal of S. Asabove,
we have p DpS D
p’
for some
positive
integer
t,
and ,5’ isp-adically complete.
We first prove that p =
p,S’
+ a. From[S :
Zp]
p it follows thatTr 1 =
[S :
Zp] -
1 ~ qZp,
so1 f/.:
a. Thisimplies
that a C p, sopS
+ a C p. To prove the otherinclusion,
we first note that the definition of agives
riseto an exact sequence
of
Zp/qZp-modules,
the third arrowmapping x
mod qS
to the mapsend-ing y
mod a to Tr(xy)
modqZp;
this arrow issurjective
becauseSla
andHom(S/a,
Zp/qZp)
are free of the same rank overZp/qZp
and hence havethe same
cardinality.
SinceS/a
isZp/qZp-free,
we have a naturalisomor-phism
Hence if we tensor the exact sequence above with
Fp
we obtain an exactsequence
of
Fp-vector
spaces. From this sequence we deduce thatp C pS
+ a, asfollows. Let x E p. We have
p’
Cps,
so forany y E S
themultiplication-by-xy
mapSips
isnilpotent
and has therefore trace 0 when considered as anFp-linear
map. Thisimplies
that x modpS’ belongs
to thekernel of the map
81pS Hom(S/(pS
+a), FP).
This kernelequals
theimage
ofa/(qS
+pa),
so that x EpS
+ a. Thiscompletes
theproof
of theequality
p =pS
+ a.The next
step
in theproof
is the construction of an unramifiedsubring
T of S that has the same residue class field as S. Let k =
SIP
be theresidue class field of
S,
and let T be theunique
unramified localZp-algebra
with residue class fieldTIPT k.
If T =Zp[Y]/gZp[Y],
thenby
Hensel’slemma g
has a zero in S(see
[1,
Exercise10.9~);
at thispoint
we use thatS is
p-adically complete.
Thisgives
aZp-algebra
homomorphism T - S,
which makes S into a
T-algebra.
Let e be the dimension of as a vectorspace over
T/pT
= k.By
Nakayama’s
lemma there is asurjective
T-linearmap S. We have e ~
[T :
Zp]
=e - ~k :
Fp] =
Fp] =
(S :
Zp],
socomparing
Zp-ranks
we see that the map Te --+ S must beinjective.
Thisimplies
inparticular
that the map T -> S isinjective.
Hence we may viewT as a
subring
ofS,
and S is free of rank e as a T-module.In the definition of a we may now
replace
Zp
by
T,
i. e., we have a ={x
ES : Tr
sIT(x8)
CqT},
where Tr SIT is the trace map for the extension T cS. This is an immediate consequence of the formula Tr = Tr o Tr SIT
and the fact that
qT
={x
E T : Tr CqZp 1;
the lastequality
holds because T is unramified over
ZP.
Any
T/qT-module
N that isfinitely generated
and free as aZp/qZp-module is also free as a
T/qT-module,
the rankbeing
[T :
Zp]
times assmall;
one proves thisby lifting
a k-basis ofN/pN
to aTlqT-basis
ofN,
in the same way as we
proved
above that S is free as a T-module. Hencethe
hypotheses
on a nowimply
thatS/a
and(S : a)/S
are free asT/qT-modules. The rank of
S/a
overT/qT
can becomputed
over the residueclass
field; using
thatpS
+ a = p we find that[Sla :
+a) :
T/pT~ _
[k : k]
=1,
so the natural mapS/a
is anisomorphism.
Next we prove that a is invertible. From a C p and
2.4(b)
we see thatS :
a # S,
so the module(S :
a)/S
is non-zero.Also,
it is free overT/qT
=Sla,
so the annihilator of(S :
a)IS
inS/a
is zero. This meansthat S :
(S : a)
= a. From ourhypothesis
a : a = S and 2.5 it now followsthat a is
invertible,
soa(S :
a)
= S.We deduce that a is
principal. Namely,
choose p E a withp(S :
a) 0
p.Then 1 E
p(S : a),
andmultiplying by
a we find a CpS.
Since we alsohave
pS
C a this proves that a =pS.
We claim that S =