C OMPOSITIO M ATHEMATICA
W OLMER V. V ASCONCELOS
The reduction number of an algebra
Compositio Mathematica, tome 104, n
o2 (1996), p. 189-197
<http://www.numdam.org/item?id=CM_1996__104_2_189_0>
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The reduction number of an algebra
WOLMER V. VASCONCELOS*
Department of Mathematics, Rutgers University, New Brunswick, New Jersey 08903, USA (Received 16 March 1995; accepted in final form 4 October 1995)
Abstract. A complexity of an algebra A over a field k is a measure of comparison to a polynomial ring over k. Here we bring to the fore the reduction number of a graded algebra A, and study its relationship to the arithmetic degree of A. The relationship between the reduction number and the Castelnuovo-Mumford regularity has been object of previous studies, but a presumed relationship
CM regularity and arithmetic degree breaks down.
Key words: Arithmetic degree, Castelnuovo-Mumford regularity, Cohen-Macaulay ring, Hilbert function, multiplicity, reduction number.
©
1996 Kluwer Academic Publishers. Printed in the Netherlands.1. Introduction
Let A be a
finitely generated, positively graded algebra
over a fieldk,
where
Ai
denotes the space ofhomogeneous
elements ofdegree
i. We furtherassume that A is generated by its 1- forms, A = k [A1 , in which case Ai
=Ali.
Suchalgebras
are said to be standard.Among
thecomplexities
of A are the Castelnuovo- Mumfordregularity reg(A)
of A and variousdegrees - multiplicity deg(A)
andarithmetic
degree arith-deg(A).
Another is the reduction numberr(A)
of A whichoccurs in the
study
of the distribution of thedegrees
of thegenerators
of A as a module over its Noether normalizations(see
below for alldefinitions).
Here weintroduce two
techniques,
one theoretical and the othercomputational,
to examinethe relative
strengths
of these numbers.There have been several
comparisons
between these numbers. Inaddition, [2]
is a
far-flung
survey ofreg(A)
andarith-deg(A)
in terms of thedegree
data of apresentation
A =k [x 1, ... , x,,] II. Moreover, [12]
carries out a detailed compar- ison between thedegrees
of an ideal and thedegrees
of some associated ideal of initial forms.From the
study of reg(A) (see [5], [9]
andparticularly [13]),
one hasHere we will show
(Theorem 7)
that for any standardalgebra
A* The author was partially supported by the NSF.
190
if k has characteristic zero.
Actually,
Theorem 7 deals withgeneral
affinealgebras
and the estimate above is one of its
applications
for standardalgebras.
Theapproach,
an
elementary
combination of linear andhomological algebras,
may be suited to obtain other estimates forr(A) .
The author is
grateful
toJürgen
&Maja Herzog
and Bemd Ulrich for severalenlightening
conversations.2. The
degrees
of a moduleWe recall the
settings
of threedefinitions,
two of which may not be in basic references. If & issufficiently large
and dim A =d,
there are forms XI, Xd EAI,
such thatis a Noether
normalization,
thatis, the xi
arealgebraically independent
over kand A is a finite R-module. Let
bl, b2,
...bs
be a minimal set ofhomogeneous generators
of A as an R-moduleWe will be
looking
at the distribution of the ri,particularly
at thefollowing integer.
DEFINITION 1. The reduction number
rR (A)
of A withrespect
to R is the supre-mum of all
deg(bi).
The(absolute)
reductionnumber r(A)
is the infimumof rR(A)
over all
possible
Noether normalizations of A.One of our aims is to make
predictions
about theseintegers,
but withoutavailing
ourselves of any Noether normalization. We
emphasize
thisby saying
that theNoether normalizations are invisible to us, and the information we may have about A comes from the
presentation
A =S/l.
The
integer r(A)
may be understood as a measure ofcomplexity
of thealgebra
A. Taken this way, it has been
compared
to another index ofcomplexity
of A:DEFINITION 2. Let S be a
polynomial ring,
let A =S/I
and letbe a minimal
graded
free resolution of A. TheCastelnuovo-Mumford regularity
ofA is the
integer
Another way to size A is
given
in[2]
and[8] through
the notion of the arithmeticdegree
of a module. Let A be a Noetherianring
and let M be afinitely generated
R-module. For each
prime
ideal pof A,
letPp (M)
denote the set of elements of the localizationMp
annihilatedby
some power of p. Note thatrp (M)
is an Artinianmodule over
Ap
that vanishes unless p is an associatedprime
of M. We denote thelength ofr?(M) by multM(p).
DEFINITION 3. The arithmetic
degree
of M is theinteger
The definition
(see
how these sums are calculated inProposition 5,
eventhough
the individual summands are not
always available) applies
toarbitrary modules,
not
just graded modules, although
its main use is forgraded
modules over astandard
algebra
A. If all the associatedprimes
of M have the samedimension,
then
arith-deg(M)
isjust
themultiplicity deg(M)
ofM,
which is obtained from its Hilbertpolynomial.
EXAMPLE 4. In the
following
families ofexamples,
weexplore possible
relationsbetween
arith-deg(A)
andreg(A).
(a)
In dimension1,
we havereg(A) arith-deg(A). Indeed,
let R =k[x]
be aNoether normalization of A and let
be the
decomposition
of A as the direct sum ofcyclic
R-modules.We have
while the minimal number of
generators
of A as an R-module isBecause A is
generated by
elements ofdegree 1,
there must be no gaps in the sequence ofdegrees
of its modulegenerators,
whichimplies bj v(A),
and nogaps either in the
degrees
of its torsion-freepart
sothat ai
e. This proves that for eachinteger Ê,
192
and therefore
(b)
In dimensiongreater
than1,
theinequality reg(A) arith-deg(A)
doesnot
always hold, according
to thefollowing example
of Bemd Ulrich. Let A =k [z ,
y, u,v]1 l,
with I =«x, y)2, xut
+yvt).
Ift >, 3,
(c) Despite
thepreceding example,
it issignificant
that theinequality reg(A) arith-deg(A)
has been established forlarge
classesof algebras (see [12]).
We now show how a program with the
capabilities of Macaulay ([3])
can be usedto
compute
the arithmeticdegree
of agraded
module M withoutavailing
itself ofany
primary decomposition.
Let ? =k[xi, ... 1 X,l
and suppose dim M =d
n.It suffices to construct
graded
modulesMi,
i = 1 ... n, such thatFor each
integer 1 # 0,
denoteLi
=Ext9 (M, S). By
localduality ([4]),
aprime idéal p
C S ofheight
i is associated to M if andonly
if(Li)p -# 0;
furthermoreR((Li).,) = multM(p).
This gives
what isrequired: Compute
for eachLi
itsdegree
e(Li)
and codi-mension ci . Then choose
Mi according
toIn other words:
PROPOSITION 5. For a
graded
S-module Mand for
eachinteger
i denoteby
cithe codimension
of Exts(M, S).
ThenEquivalently,
The
amusing
of this formula(one sets [ô,
=1)
is that itgives
a sum of sumsof terms some of which may not be available.
One consequence of this formulation of the arithmetic
degree
is the follow-ing.
Let T be a term order on S =k [x 1, - - - , zn]
and let I be ahomogeneous
ideal of S. Denote
by in(I)
the initial ideal of I for the term order. For eachinteger
r,Exts(S/I, S)
is a submodule of aquotient
ofExts(S/in(I), S).
Thisarises from the
spectral
sequence inducedby
the attached filtration(see [6], [11]).
Consequently,
we havegiving
Theorem 2.3 of[12].
3.
Integrality equations
From now on A is a standard
graded ring
and R =A;[z]
y A is a fixed Noether normalization. To determiner(A),
we look forequations
ofintegral dependence
ofthe elements of A with
respect
to R.A
simple approach
is to findgraded
R-modules on which A acts as endo-morphisms (e.g.
Aitself).
The most naivepath
to theequation
isthrough
theCayley-Hamilton
theoremworking
as follows. Let E be afinitely generated
R-module and let
f :
E -+ E be anendomorphism. Map
a freegraded
module overE and
lift f :
Let
be the characteristic
polynomial of c.p, n = rank(F).
It follows thatPw (f )
= 0. Thedrawback is that n, which is at least the minimal number of
generators
ofE,
may be toolarge.
One should do much betterusing
a trick of[1].
Liftf
to amapping
from a
projective
resolution of E into itself:Define
This rational function is
actually
apolynomial
inR[t] ([1]).
If E is agraded
module and
f
ishomogeneous,
thenPI (t)
is ahomogeneous polynomial, deg
E =deg PI (t).
194
PROPOSITION 6
(Cayley-Hamilton theorem). If
the rankof E
over R is e,Pf (t)
is a monic
polynomial of degree
e.Moreover, if E
istorsion-free
thenPf (f) .
E = 0.Furthermore, if
E isa faithful
A-module andf
ishomogeneous of degree 1,
thenProof.
Most of theseproperties
areproved
in[1]. Passing
over to the field of fractions ofR,
the characteristicpolynomial
of the vector spacemapping
is
precisely Pf (t) .
The existence of an
equation
and the fact that
HomR(E, E)
isgraded implies
that there is a similarequation where ci
E(z)B
Without the torsion-free
hypothesis
the assertions may fail. Forinstance,
if A =k [x, y] / (xy, y’), f is multiplication by y on A, then Pf (t)
=t, but Pf (f ) :,
0.In case the module E is A
itself,
we do not need the device ofProposition 6,
as we can argue
directly
as follows. For u eAI, R[u] ri R[t]1 l,
where I =f - J, height(J) >
2. But if A is torsion-free overR, R[u]
will have the sameproperty
and
necessarily
J =(1).
This means that the rank ofR[u],
which is thedegree
off, is
at moste(A).
A
question
ofindependent
interest is to find R-modules of smallmultiplicity
that afford
embeddings
For
example,
therelationship
between theirmultiplicities
may be aslarge
asminimal
primes
is at most3,
thendeg(
4. Arithmetic
degree
of a module and the reduction number of analgebra
Let A be an affine
algebra,
notnecessarily
standard. We now bound thedegrees
of the
equations
satisfiedby
the elements of A withrespect
to any of its’optimal’
Noether normalizations. These
simply
mean those normalizations that can be used to read thedegrees.
THEOREM 7. Let A be an
affine algebra
over aninfinite field k,
letk[z]
be anoptimal
Noether normalizationof A,
and let M be afinitely generated graded, faithful
A-module. Then every elementof
Asatisfies
a monicequation
overk [z] of
degree
at mostarith-deg(M).
Proof.
Letbe an
equidimensional decomposition
of the trivial submodule ofM,
derived froman
indecomposable primary decomposition by collecting together
thecomponents
of the same dimension. IfIi
= annihilator(MlLi),
then eachring 4/Ii
isunmixed, equidimensional
andSince k is
infinite,
there exists a Noether normalizationk[ZI,
...,Zd]
of A such thatfor each ideal
Ii,
a subset of theIZI, - - - , Zdj generates
a Noether normalization forAI li.
First,
we aregoing
to check thatarith-deg(M)
can be determinedby adding
thearithmetic
degrees
of the factors of the filtrationat the same time that we use the
Cayley-Hamilton
theorem.We write the arithmetic
degree
of M aswhere ei is the contribution of the
prime
ideals minimal overIi . (Waming:
Thisdoes not mean that ei =
arith-deg(M/Li ) .)
We first claim that(set Lo
=M)
Indeed,
there is anembedding
showing
thatFi
isequidimensional
of the same dimension asMi. If p
is anassociated
prime of Ii, localizing
weget (L n...
nLj )p
=(0)
which shows thatwhile the converse is clear. This shows that the
geometric degree
of themodule Fi
is
exactly
the contributionof e2
toarith-deg(M).
We are now
ready
to useProposition
6 on the modulesFi.
Letf
e A act oneach
Fi.
For eachinteger i,
we have apolynomial
with Consider the
polynomial
196
and evaluate it on
f
from left toright.
Asmeaning
thatPi ( f )
maps(Li
n...nLi-1)
into(Li
n... nLi),
asimple inspection
shows that
Pf (f )
=0,
since M is a faithful module.Observe that if A is a standard
algebra
andf
is an element ofAI,
thenPf(t)
gives
anequation
ofintegrality
off
relative to the idealgenerated by
z.COROLLARY 8. Let A be a standard
graded algebra
and denoteedeg(A)
=inff aàth-deg(M) ) 1 M faithful graded module}.
For any standard Noether normalization R
of A,
every elementof
Asatisfies
anequation of degree edeg(A)
over R.5. Reduction
equations
fromintegrality equations
If A is an standard
algebra,
for agiven
element u EAI,
atypical equation
ofreduction looks like
which is less restrictive than an
equation of integrality.
One should thereforeexpect
theseequations
to have lowerdegrees. Unfortunately
we do notyet
see how toapproach
it.The
following argument
shows how to pass fromintegrality equation
to somereduction
equations,
butunfortunately injects
the issue of characteristic into thefray.
PROPOSITION 9. Let A =
k[Al] be a
standardalgebra
over afield
kof
char-acteristic zero. Let R =
k[z]
y A be a Noethernormalization,
and suppose that every elementof Ai satisfies
a monicequation of degree
e overk[z].
Thenr(A)
e - 1.Proof.
Let u 1, ... , un be a set ofgenerators
ofA
1 overk,
and consider theintegrality equation
ofwhere the xi are elements of k.
By assumption,
we havewhere ai e
(z)’. Expanding
ue we obtainwhere a =
(cx1, ... , an )
is anexponent
of totaldegree
e, aa is the multinomial coefficient(â) ,
and ma is thecorresponding
’monomial’ in the xi . We must showfor each a.
To prove the
assertion,
it suffices to show that the span of the vectors(aama),
indexed
by
the set of all monomials ofdegree
e in nvariables,
has the dimension of the space of all such monomials.Indeed,
if these vectors lie in ahyperplane
we would have a
homogeneous polynomial
which vanishes on kl. This means that all the coefficients ca aa are zero, and therefore each Co; is zero since the a« do not vanish in characteristic zero.
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