Ann. Univ. Ferrara - Sez. VII - Sc. Mat.
Vol. IL, 19-30 (2003)
A Note on Symbolic and Ordinary Powers
of Homogeneous Ideals.
ALESSANDRO ARSIE (*) - JON EIVIND VATNE (**)
SUNTO - In questa nota siamo interessati ai moduli graduati Mk 4 I (k) /I k e Nk »4 I k /I k,
dove I è un ideale saturato nell’anello delle coordinate omogenee S »4 K[x0 , R , xn ]
di Pn, I (k) è la potenza simbolica e I k è la saturazione della potenza ordinaria. Poco è
noto su questi moduli e qui viene fornito un limite superiore ai loro diametri. Ne calcoliamo inoltre le funzioni di Hilbert e studiamo alcuni sottomoduli caratteristici nel
caso speciale di n 1 1 punti in posizione generale in Pn.
ABSTRACT - In this note we are interested in the graded modules Mk 4 I (k) /I k and Nk »4
»4 I k /I k , where I is a saturated ideal in the homogeneous coordinate ring S »4
»4 K[x0 , R , xn ] of P n , I (k) is the symbolic power and I k is the saturation of the ordinary power. Very little is known about these modules, and we provide a bound on
their diameters, we compute the Hilbert functions and we study some characteristic
submodules in the special case of n 1 1 general points in P n .
MSC(2002) Classification: 13C99, 14M10.
1. – Introduction and preliminaries.
Let K be any algebraically closed field of characteristic zero and let S »4
»4 K[x0 , R , xn ] be the ring of polynomials in n 1 1 indeterminates over K ,
graded in the usual way. Let X be a closed subscheme of the projective space
P n . To X , we can associate the homogeneous ideal IX % S , generated by all the
homogeneous polynomials that vanish on X , i.e. the saturated homogeneous
ideal of X; we have X 4 Proj (S/IX ). If we consider the (ordinary) power I n , we
can associate to it a new subscheme called the (n 2 1 )-th infinitesimal neigh-
(*) Indirizzo dell’autore: Dipartimento di Matematica, Università di Bologna,
Piazza di Porta S. Donato, 5, 40127 Bologna, Italy. E-mail: arsieHdm.unibo.it
(**) Indirizzo dell’autore: Deparment of Mathematics, University of Bergen, J.
Brunsgt. 12, N-5008 Bergen, Norway. E-mail: joneivind.vatneHmi.uib.no
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ALESSANDRO ARSIE
-
JON EIVIND VATNE
bourhood of X in P n , which encodes some properties of X , together with the
given embedding. In general, however, the ideal I n is not saturated. On the
other hand, if X is a reduced scheme, we can consider the symbolic power I (n) ,
the definition of which we recall:
DEFINITION 1.1. The kth symbolic power of the saturated ideal of a variety X % P n consists of all polynomials that vanish to order at least k at a general point of X .
From a more abstract and general point of view, we can give the following
definition of symbolic power. If R is any commutative ring with unity, S a multiplicative subset of R , and a % R any ideal, let us denote S(a) »4 f 21 (S 21 a),
where S 21 a is the localization of a with respect to the multiplicative system S
and f : R K S 2 1 R is the localization map. S(a) is called the saturation of a
with respect to S . Then we have:
DEFINITION 1.2. If p is a prime ideal of R , then the k-th symbolic power
of p is the ideal Sp (p n ), with the notation used above, where Sp »4 R 2 p and
p n is the ordinary power.
Obviously, we always have I n % I (n) and in general we have a strict inclusion. Our interest in the symbolic power stems from the fact that taking the
symbolic power of a homogeneous ideal gives the saturated ideal for a scheme
which is the scheme defined by the ordinary power, having subtracted the embedded components. Thus, in some special cases, via the symbolic power we
can obtain the saturation of the ordinary power. A meaningful case in this
sense is given by the following:
LEMMA 1.3. If X is a (reduced) zero-dimensional scheme in P n , with
saturated ideal I , then I (k) 4 I k, the saturation of I k .
PROOF. Obvious by the previous remark and the fact that a zero-dimensional scheme has no embedded components. r
We are naturally led to deal with the modules Mk »4 I (k) /I k , which measure, (in particular cases), how far I k is from being saturated. In literature,
very little is known about these objects and in this note we investigate some of
their properties (diameter, Hilbert function, submodules) in special but still
meaningful cases.
The modules Mk are also related to the problem of detecting complete intersections as shown by the following:
PROPOSITION 1.4. If X is a complete intersection in P n , with saturated
ideal I , then I (k) 4 I k for all k .
A NOTE ON SYMBOLIC AND ORDINARY POWERS ETC.
21
PROOF. Since the powers of a complete intersection ideal are arithmetically Cohen-Macaulay, they have no embedded components and there are no
embedded components in the vertex of the affine cone over it. Thus the ordinary powers and the symbolic powers coincide and both are saturated. r
Thus we may also view the modules Mk as a way of encoding «how far» a
variety is from being a complete intersection.
Our study of Mk is also related to an important invariant called Castelnuovo-Mumford regularity, which we now recall:
DEFINITION 1.5. For a coherent sheaf F on a projective variety X we define the Castelnuovo-Mumford regularity reg (F ) as the least integer m such
that H i (X , F (m2i) ) 4 0 ( iF1 , and in this case we say that F is m-regular.
For an embedded projective variety X % P n the Castelnuovo-Mumford regularity reg (X) is defined as reg (IX ), the regularity of the ideal sheaf of X as a
coherent sheaf on P n .
The geometric importance of this notion stems from the following well-known
result (see [7], lecture 14): if X is m-regular, then its saturated ideal IX is generated by homogeneous polynomials of degree at most m and hence thare are no
(m11)-secant lines to X. Furthemore, the Hilbert polynomial and the Hilbert
function of X have the same values for all integers tFm21. Indeed, the Castelnuovo-Mumford regularity is an important invariant which measures the complexity of the given ideal (or more generally module) I. If
0 K R K Fj K R K F 0 K I K 0 ,
is the minimal free resolution of I over S and fj is the maximum degree of the
generators of Fj , then we have:
reg (I) 4 max ] fj 2 jNj F 0 ( .
From this we immediately see that for any homogeneous ideal J , its saturation
degree s(J) is always less or equal to reg (J) (recall that the saturation degree
is the degree at which J starts to be equal to its saturation: (J)d 4 (J)d for any
d F s(J)).
2. – Some bounds on the diameter of Mk and Nk .
Let us start with the case of a reduced zero-dimensional scheme with the
following simple proposition:
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ALESSANDRO ARSIE
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JON EIVIND VATNE
PROPOSITION 2.1. Let X % P n be a reduced zero-dimensional subscheme, I
its defining saturated ideal and reg (I) the corresponding CastelnuovoMumford regularity. Then diam (Mk ) G k( reg (I) 2 1 ).
PROOF. As we have seen, if X is a reduced zero-dimensional subscheme,
then I (k) coincides with the saturated ideal I k of the ordinary power. Obviously
we have that reg (I k) G reg (I k ), but now, since dim (S/I) 4 1 , by [6] we have
that reg (I k ) G k reg (I). Via the characterization of Castelnuovo-Mumford
regularity in terms of minimal free resolutions, it turns out immediately that
I k is thus generated in degree less or equal to k reg (I). This implies that Mk is
zero in degree grater or equal to k reg (I) 1 1 . It is also clear that the module
Mk is zero in degree less or equal to k 2 1 , thus diam (Mk ) »4 (Mk )max 2
2 (Mk )min 4 k( reg (I) 2 1 ). r
Now we consider the case of a smooth projective variety X % P n . In this setting, the modules Mk do not measure anymore the lack of saturation, so we get
a bound on the diamater of the modules Nk »4 I k /I k in the following:
PROPOSITION 2.2. Let X % P n be a smooth variety of codimension e . Let
us assume that X is scheme-theoretically defined by hypersurfaces of degrees
d1 F d2 F R F ds . Then diam (Nk ) G kd1 1 d2 1 R 1 de 2 e 1 1 .
PROOF. In view of the results of [2], we have that H i (P n , Ik (t 2 i) ) 4 0 for
i F 1 provided that t 2 i F kd1 1 d2 1 R 1 de 2 n (this is proved for complex
smooth projective algebraic varieties, but using the so called Lefschetz’s principle we can extend this to any smooth projective algebraic variety defined
over any algebraically closed field of characteristic zero). This implies that t F
Fkd1 1 d2 1 R 1 de 2 n 1 i , but now the vanishing of the cohomology groups is
automatic for i D dim (X) 1 1 , so that we obtain reg (I k ) G kd1 1 d2 1 R 1 de 2
2e 1 1 , setting i 4 dim (X) 1 1 and taking into account the definition of reg (I k ).
Recalling again the characterization of the Castelnuovo-Mumford regularity
in terms of minimal free resolutions, we get that I k is generated in degree less
or equal to reg (I k ), so that Nk is zero in degree grater than kd1 1 d2 1 R 1
1de 2 e 1 1 . Thus as a rather rough estimate of diam (Nk ) we get diam (Nk ) G
G (Nk )max G kd1 1 d2 1 R 1 de 2 e 1 1 . r
Let us remark that if X % P n has degree d , then it can always be schemetheoretically defined by hypersurfaces of degree d (see [8]), so that by the previous proposition we have diam (Nk ) G d(k 1 e 2 1 ) 2 (e 2 1 ); this bound, even
if extremely rough, depends just on the degree of X and on its codimension.
The result of the previous proposition can be somewhat generalized, recalling the fundamental theorem proved in [4]:
A NOTE ON SYMBOLIC AND ORDINARY POWERS ETC.
23
THEOREM 2.3. Let I be an arbitrary homogeneous ideal and let d(I) denote the maximum degree of the homogeneous generators of I . Then there is a
constant f such that reg (I k ) G kd(I) 1 f.
Reasoning as before we immediately get that diam (Nk ) G kd(I) 1 f for any
homogeneous ideal I .
3. – Submodules of functions vanishing to higher order.
Of course we always have the inclusion I k % I (k) and the inclusion I (k 1 1 ) %
%I , but there are also polynomials vanishing to higher order at each point of
Proj (S/I) that are not in I k . That is, we can form a sequence of submodules
I (k 1 i) 1 I k
I (k 1 i 2 1 ) 1 I k
I (k)
R%
%
%
R
%
4 Mk
Ik
Ik
Ik
(k)
gfor notation M
[i]
k
»4
I (k 1 i) 1 I k
Ik
h.
This sequence terminates after a finite number of steps. A bound is given
by the following result of Ein, Lazarsfeld and Smith (see [5]) (which we state
in a special case):
PROPOSITION 3.1. If X % P n is a reduced subscheme, such that each component of X has codimension G e , with ideal sheaf ( . Then
((me) % (m (m N .
In the next section we will see that there is an example (i.e. n 1 1 general
points in P n) where this bound is very far from being sharp, that is the sequence of submodules inside Mk terminates much earlier.
4. – A special case: n 1 1 general points in P n.
We are interested in getting more information on the modules Mk 4 I (k) /I k
for a set of n 1 1 general (that is, not contained in a hyperplane) points in P n ,
with saturated ideal I . Our first theorem gives the Hilbert function of such a
module:
THEOREM 4.1.
The Hilbert function of Mk is given by
g h 2 (n 1 1 ) g
1(21 )
g hg
l h(l) 4
n1k21
n
n1l
n
n11
n11
n11
h1R1
h for
n 1 (n 1 1 ) k 2 nl 2 (n 1 1 )
n
l h(l) 4 0 otherwise.
n11
n
k G l G 2 k and
24
-
ALESSANDRO ARSIE
JON EIVIND VATNE
PROOF. This will follow from the two Propositions 4.3 and 4.4 and is a particular case of Theorem 4.5. r
Note that, given n 1 1 points in general position in P n , we can always
choose a set of coordinates x0 , R , xn such that the points are the coordinate
points
( 1 ; 0 ; R ; 0 ), ( 0 ; 1 ; 0 ; R ; 0 ), R , ( 0 ; R ; 0 ; 1 ).
The ideal of this reduced point set, X , is generated by all square free monomials of degree two; IX 4 (xi xj Ni c j) % S»4 k[x0 , R , xn ]. We will first of all
give a combinatorial description of the symbolic powers IX(k) of this ideal.
PROPOSITION 4.2. A monomial x I »4 x0i0 x1i1 R xnin is in (IX(k) )l , l 4
and only if ij G l 2 k (j .
!i
j
j
if
PROOF. The condition that x I IX(k) is that x I vanishes on each coordinate
point to order at least k . To vanish at ( 1 ; 0 ; R ; 0 ) to order at least k is the
same as saying that x1i1 R xnin has order at least k , that is i1 1 R 1 in F k `
`i0 G l 2 k , and similarly for the other ij’s. r
PROPOSITION 4.3.
The Hilbert function h of IX(k) is given by
h(l) 4 f(n 1 1 , l) 2 (n 1 1 ) f (n 1 1 , k 2 1 ) 1 R
R 1 (21 )n 1 1
for l F
n11
n
u v
n11
n11
f(n 1 1 , (n 1 1 ) k 2 nl 2 (n 1 1 ) )
k , h(l) 4 0 otherwise, where f (u , v) 4
g
u1v21
v
h
.
PROOF. In view of proposition 4.2, we see that the value h(l) of the Hilbert
function of IX(k) at the point l is just the cardinality of the set
Z »4 ](i0 , R , in ) N0n 1 1 N
! i 4 l , i G l 2 k(.
j
j
This combinatorial problem has generating function ( 1 1 x 1 R 1
1x l 2 k )n 1 1 (see any textbook on combinatorics, for instance [1]), and h(l) is just
the coefficient of x l in this polynomial. Now
( 1 1 x 1 R 1 x l 2 k )n 1 1 4
g
4 ( 1 2 (n 1 1 ) x (l 2 k 1 1 ) 1
1 2 x l2k11
12x
u v
n11
2
h
n11
4
x 2(l 2 k 1 1 ) 1 R 1 (21 )n 1 1 x (n 1 1 )(k 2 l 1 1 ) ) ˜
˜( 1 1 f(n 1 1 , 1 ) x 1 f(n 1 1 , 2 ) x 2 1 R) .
A NOTE ON SYMBOLIC AND ORDINARY POWERS ETC.
25
Since each power of x appears in the last factor of the product, we just need
to find the correct term in the first factor in order to extract the coefficient of
x l in the product. That is
f (n 1 1 , l) 2 (n 1 1 ) f(n 1 1 , k 2 1 ) 1
R 1 (21 )n 1 1
u v
u v
n11
2
n11
n11
f(n 1 1 , 2 k 2 l 2 2 ) 1 R
f(n 1 1 , (n 1 1 ) k 2 nl 2 (n 1 1 ) )
is the coefficient of x l . Of course, many of the terms here can be zero (by definition as soon as one of the entries inside f(i , j) is negative). Notice moreover
n11
k is just obtained by considering what is the
that the lower bound l F
n Z be empty. So we have computed the Hilbert
relation among k , l , n so that
function of IX(k) . r
Now we will compare the symbolic power with the ordinary power. Note
that in our case I k is generated in degree 2 k , and contains no polynomials of
lower degree.
PROPOSITION 4.4.
I2kk 4 I2(k)
k .
PROOF. Again there is a simple combinatorial argument which gives the
proposition: for a monomial, being in I2kk is the same as being the product of k
square free monomials of degree two. If we consider a monomial x I as corresponding to the sequence I 4 i0 , R , in N0n 1 1 then this condition is the same
as saying that we can subtract one from two different places in the sequence k
times, and end up with ( 0 , R , 0 ). The only way this can fail is if one of the ij’s
is larger than the sum of the rest. Thus we conclude by Proposition 4.2, using
l42k. r
For submodules Mk[i] »4
I (k 1 i) 1 I k
Ik
we have:
The Hilbert function of Mk[i] is given by
THEOREM 4.5.
h(l) 4 f(n 1 1 , l) 2 (n 1 1 ) f (n 1 1 , k 1 i 2 1 ) 1 R
R 1 (21 )n 1 1
u v
n11
n11
f(n 1 1 , (n 1 1 )(k 1 i) 2 nl 2 (n 1 1 ) ),
for
h(l) 4 0 otherwise.
n11
n
k1iGlG2k,
26
ALESSANDRO ARSIE
PROOF.
-
JON EIVIND VATNE
Let us consider the sequence
R%
I (k 1 i) 1 I k
Ik
%
I (k 1 i 2 1 ) 1 I k
Ik
%R%
I (k)
Ik
4 Mk
of submodules from (3) in the case of n 1 1 points in general position in P n .
First of all the fact that I (k 1 i) % I (k) together with Proposition 4.4 implies that
the Hilbert function of Mk[i] is zero in degrees F 2 k . Moreover, since I k does
not contain any polynomial of degree less than 2 k , the Hilbert function of Mk[i]
must coincide with the Hilbert function of I (k 1 i) up to degree 2 k 2 1 . This, together with the Hilbert function of I (k 1 i) , which is given in Proposition 4.3,
proves the Theorem. r
Let us remark that in view of Theorem 4.5, the chain of submodules
R Mk[i] % Mk[i 2 1 ] % R % Mk[ 0 ] 4 Mk ,
terminates much before the bound implicitly given in [5], as is immediate to
check. Indeed, according to the bound given in [5], we have for the ideal I of
n 1 1 general points in P n the relation: I (kn) % I k , so that Mk[i] 4 0 for any i F
Fk(n 2 1 ). On the other hand, by Theorem 4.5, we have that Mk[i] 4 0 as soon as
n11
n
g
k 1 i D 2 k , i.e. i D 2 2
n11
n
h k.
5. – Another example: coordinate p-planes in P n.
In this section we will present a description of the symbolic powers of the
ideal of the arrangement of coordinate p-planes in P n . These are the planes
given parametrically by (t0 , t1 , R , tp , 0, R , 0), (t0 , t1 , R , tp21 , 0, tp , 0, R , 0)
and so forth. The ideal of one such p-plane is given by (xi1 , xi2 , R , xin2p ) for
the corresponding subset ]i1 , R , in 2 p ( % ] 0 , R , n(. The ideal I of the collection of all such p-planes is thus given by all square free monomials of degree
p 1 2 in x0 , R , xn . Thus the powers of I are given as follows:
LEMMA 5.1. Let I be the ideal of the coordinate p-planes in P n . Then
I is generated in degree k(p 1 2 ) by all monomials which do not contain a
(k 1 1 )-th power.
k
PROOF. Obvious, following the approach developed in the previous case of
points. r
A NOTE ON SYMBOLIC AND ORDINARY POWERS ETC.
27
There is also a combinatorial description of the Hilbert function of the symbolic powers, as in the case of n 1 1 points in general position. Indeed, for a
monomial x I 4 x0i0 R xnin , being in Il(k) , l 4 ij is equivalent to vanishing to order at least k on each point of X . Vanishing to order k at a general point of the
p-plane given parametrically by (t0 , t1 , R , tp , 0 , R , 0 ) is the same as saying
!
p
that xp 1 1 , R , xn occurs in x I at least k times, or that
! i G l 2 k . Of course a
j40
j
similar statement is true for each of the other coordinate p-planes, and we are
led to consider the cardinality of the set
m
N ! i G l 2 k , ! i 4 ln ,
j4n
(i0 , R , in ) Nn0 1 1
jJ
j
j40
j
where J runs through all subsets of ] 0 , 1 , R , n( of cardinality p 1 1 . from
which, in principle, one can recover the dimension of Il(k) . Unfortunately, in
this case the knowledge of the Hilbert function of I (k) is not so useful to study
k
the modules Mk . Indeed, let us consider the difference between Ik(p
1 2 ) and
(k)
Ik(p 1 2 ) . The monomials in the latter are in the former if and only if, using the
combinatorial description, we can remove 1 from p 1 2 places at once, k times
to obtain ( 0 , 0 , R , 0 ). Of course this can not always be done. If, for instance,
x0 is in the monomial to such a high degree that we cannot get the first entry
to zero using the above procedure, then we can multiply the monomial with
any power of x0 and still have a monomial in I (k) 0 I k . Thus the two ideals are
different in all high degrees and this obviously implies that the modules Mk
have infinite diameter. This, of course, corresponds to the fact that the coordinate p-planes intersect, and that the ordinary powers of the ideal will have embedded components supported in the intersections of the coordinate p-planes.
On the other hand, if one considers special subsets X of arrangements of pplanes, such that these p-planes are not mutually intersecting, then the symbolic power of IX still agrees with the saturation of the ordinary power and the
modules Mk and Nk coincides too. In this setting, one should be able to get
some more information using again combinatorics. The simplest non trivial
example is the following:
EXAMPLE 5.2. Consider the coordinate lines in P 3 . According to the discussion above, the ideal I of these lines is generated by all square-free cubes
(xyz , xyw , xzw , yzw) % S 4 K[x , y , z , w]. Then
I 2 4 (x 2 y 2 z 2 , x 2 y 2 zw , x 2 yz 2 w , xy 2 z 2 w , x 2 y 2 w 2 , x 2 yzw 2 ,
xy 2 zw 2 , x 2 z 2 w 2 , xyz 2 w 2 , y 2 z 2 w 2 )
28
ALESSANDRO ARSIE
-
JON EIVIND VATNE
whereas the symbolic square also contains, for instance, x m yzw , where m is an
arbitrary natural number. In this case, one can find a completely combinatorial description of the Hilbert polynomial Pk for the modules Mk , as shown by
the following:
PROPOSITION 5.3. Let I be the defining homogeneous ideal for the coordinate lines in P 3 . Then the Hilbert polynomial Pk of Mk »4 I (k) /I k is given by
Pk 4
k(k 1 1 )( 2 k 1 1 )
Pk 4
6
2
k(k 1 1 )( 2 k 1 1 )
6
k
2
2
2
k
1
2
for odd k
for even k
2
PROOF. Let the above expression be pk ; we want to show that Pk 4 pk .
Note first that since we are taking into account the embedded components inside a one-dimensional scheme, the Hilbert polynomial is a constant. Also note
that we have the identity:
k(k 1 1 )( 2 k 1 1 )
6
i4k
4
!i .
2
i41
We will prove the proposition by induction, showing that if the formula is true
for k it is also true for k 1 2 (indeed, the formula is easily seen to be true for
k 4 1 , 2 , where P1 4 0 and P2 4 4 , corresponding to the four monomials
x m yzw , xy m zw , xyz m w , xyzw m of degree m 1 3 , which always belong to I ( 2 ) ,
but not to I 2 for large m). Note that the difference function is in any case (for k
odd or even):
pk 2 pk 2 2 4 4
uv
k
2
.
The induction will be in two parts: first we will show that the formula is true in
degree 3 k , and then show that this is the saturation degree of I k .
In degree 3 k we have to consider the cardinality of the set of monomials in
k
I3(k)
k 0 I3 k . The latter corresponds to monomials without a (k 1 1 )-th power, so
we consider the sets
Wi »4 ](a0 , a1 , a2 , a3 ) N40 N)i with ai F k 1 1 , ai 1 aj F k
for i c j , a0 1 a1 1 a2 1 a3 4 3 k( .
A NOTE ON SYMBOLIC AND ORDINARY POWERS ETC.
29
The conditions imply that exactly one of a0 , a1 , a2 , a3 is F k 1 1 . Thus we will
assume that it is a0 , and multiply our end result by four, to take into account
the fact that we have indeed four of these sets (exactly one for any ai such that
ai F k 1 1). So our goal is to prove that pk /4 4 pk 2 2 /4 1
g h (this is the differk
2
ence function divided by four and pk /4 represents the cardinality of each set
Wi) . To this aim note that
](a0 , R , a3 ) Na0 F k 1 1 R( 4
4 ](a0 , R , a3 ) Na0 F k 1 2 R( I2I](a0 , R , a3 ) Na0 4 k 1 1 R( .
Call these two sets RD and R4 . The set RD is in bijection with the set
](a08 , a18 , a28 , a38 )( Na08 F k 2 1 , ai8 1 aj8 F k 2 2 , a08 1 R 1 a38 4 3 k 2 6 (
by (a0 , a1 , a2 , a3 ) O (a0 2 3 , a1 2 1 , a2 2 1 , a3 2 1 ). This set has cardinality
pk 2 2 /4 by inductive hypothesis. The cardinality of the set R4 can be computed
by our previous results on points: since a0 4 k 1 1 we can subtract it from both
sides of a0 1 R 1 a3 4 3 k and get a1 1 a2 1 a3 4 2 k 2 1 , and the conditions
ai 1 aj F k still holds. By Theorem 4, with n 4 2 and l 4 2 k 2 1 , the number of
such triples is
u
212k21
2
v u
23
21k21
2
v
10
gh
(the later terms are all zero). But this is easily seen to be equal to k , and the
2
first part of the induction is complete.
(k)
k
0I3k
To see that the saturation degree is 3k, note that each monomial in I3k
k11
m
contains a (k11)-th power, say x
. Then x times this monomial is outside I k
for all m, and it is not in the saturation. Thus I k is saturated in degree 3k.
For degree l D 3 k , the monomials in Il(k) 0 Ilk are obtained from those in
k
I3(k)
k 0 I3 k just by multiplying with the variable occuring to degree at least k 1 1 .
This implies that the Hilbert polynomial of Mk is a constant for each k and that
its value is equal to four times the cardinality of any set Wi . r
REMARK 5.4. Note that the Hilbert polynomial Pk is not a polynomial in k.
Acknowledgements. We hearty thank A. Bigatti, A. Geramita, J. Migliore
and C. Peterson for having suggested the problem and for stimulating discussions. We also thank A.Holmsen for useful help with the combinatorics.
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ALESSANDRO ARSIE
-
JON EIVIND VATNE
REFERENCES
[1] I. ANDERSON, A first course in combinatorial mathematics, 2nd edition, Oxford
1989.
[2] A. BERTRAM - L. EIN - R. LAZARSFELD, Vanishing theorems, a theorem of Severi,
and the equations defining projective varieties, Journal of the AMS, 4, Number 3
(July 1991), pp. 587-602.
[3] K. A. CHANDLER, Regularity of powers of an ideal, Communications in Algebra, 25
(12) (1997), pp. 3773-3776.
[4] S. D.CUTKOSKY - J. HERZOG - N. V. TRUNG, Asymptotic behaviour of the Castelnuovo-Mumford regularity, to appear in Compositio Math.
[5] L. EIN - R. LAZARSFELD - K. E. SMITH, Uniform bounds and symbolic powers on
smooth varieties, math.AG/0005098.
[6] A. GERAMITA - A. GIMIGLIANO - Y. PITTELOUD, Graded Betti numbers of some embedded rational n-folds, Math. Ann., 301 (1995), pp. 363-380.
[7] D. MUMFORD, Lectures on curves on an algebraic surface, Annals of Math. Studies,
N. 59 (1966).
[8] D. MUMFORD, Varieties defined by quadratic equations, Corso CIME 1969. Questions on Algebraic Varieties, Rome, (1970), 30-100.
Pervenuto in Redazione il 7 dicembre 2001.