Maximal minors and linear powers

0.11 MB·

by  Winfried Bruns

MAXIMAL MINORS AND LINEAR POWERS
WINFRIEDBRUNS,ALDOCONCA,ANDMATTEOVARBARO
3
1
0 ABSTRACT. An ideal I in a polynomialring S has linear powersif all
2 thepowersIk ofI havealinearfreeresolution. We showthattheideal
of maximal minors of a sufficiently general matrix with linear entries
n
a has linearpowers. The requiredgenericityis expressedin termsof the
J heightsoftheidealsoflowerorderminors. Inparticularwe provethat

2 everyrationalnormalscrollhaslinearpowers.

]
C
A
h. 1. INTRODUCTION
t
a When I is a homogeneous ideal in a polynomial ring S, it is known
m
from work of Cutkosky, Herzog, Trung [CHT] and Kodiyalam [K] that the
[ Castelnuovo-Mumfordregularityof Ik isasymptoticallyalinearfunctionin

2 k. Many authors have studied the function reg(Ik) from various perspec-
v
tives,see forinstancetherecent papers ofEisenbudand Harris [EHa1] and
6
7 Chardin [Ch]. This function behaves in the simplest possible way when I
7
is generated by forms of a given degree, say d, and all its powers have a
1
. linear resolution, i.e. reg(Ik)=dk for all k. We term ideals with this prop-
3
0 erty ideals with linear powers. Similarly we say that a projective variety
2 haslinearpowerswhen itsdefiningideal has linearpowers.
1
The rationalnormal scrolls are importantprojectivevarieties. They have
:
v
both a toric and a determinantal presentation and play a prominent role
i
X in the Bertini-Del Pezzo classification theorem of irreducible varieties of
r minimaldegree,seethe“centennialaccount”ofEisenbudandHarris[EHa]
a
fordetails. A rational normal scroll of dimension n is uniquely determined
by a sequence of positive integers a ,...,a , see [Ha]. It is balanced if
1 n
|a −a |≤1foralli, j. In[CHV]Conca, HerzogandVallashowedthatthe
j i
Rees algebra of a balanced rational normal scroll is defined by a Gro¨bner
basis of quadrics and hence it is a Koszul algebra. It follows then from a
result of Blum, see [B] or Proposition 2.2, that balanced rational normal
scrollshavelinearpowers. Herzog,HibiandOhsugiaskin[HHO]whether
thesameistrueforeveryrational normalscroll.
2000MathematicsSubjectClassification. 13D02,13C40,14M12.
Key wordsandphrases. linearresolution,Castelnuovo-Mumfordregularity,syzygies,
rationalnormalscrolls.
1
2 WINFRIEDBRUNS,ALDOCONCA,ANDMATTEOVARBARO
One can ask the same question for irreducible varieties of minimal de-
gree. However,apartfromtherationalnormalscrolls,they areeitherquadric
hypersurfaces(forwhichthequestionistrivial)ortheVeroneseembedding
P2→P5. Thelattercanbetreatedbyadhocmethods(see[HHO]orPropo-
sition 3.12). So the question is really open only for the rational normal
scrolls. For them one could try to prove that the associated Rees algebra
is Koszul (for example by exhibiting a Gro¨bner basis of quadrics for their
defining ideals) but, despite to many efforts, there was no progress in this
direction.
The rational normal scrolls are determinantal; they are defined by the 2-
minors of a 2×n matrix with linearentries and expected height. The main
result of the paper asserts that the ideal maximal minors of a m×n matrix
X oflinearformshaslinearpowersprovidedtheidealsofminorssatisfythe
followinginequalities:
(1) heightI (X)≥n−m+1,
m
(2) heightI (X)≥min{(m+1− j)(n−m)+1,N}forevery j=2,...,
j
m−1,
where N = heightI (X), see Theorem 3.7. We prove also that, under the
1
above height conditions, the Rees algebra of I (X) is of fiber type and the
m
BettinumbersofI (X)k dependonlyonthesizeofthematrix,theexponent
m
kandN. Asacorollary,weanswerthequestionofHerzog,HibiandOhsugi
positively. In Section 2 we introduce themain notions and technical terms.
Wegiveacharacterizationofidealswithlinearpowersintermsofthe(1,0)-
regularityoftheirRees algbera, seeTheorem 2.5.
Thecomputationsthatledtothediscoveryofthetheoremsandexamples
presented in this paper have been performed with CoCoA[Co]. Our work
was partly supportedby the 2011-12Vigoni project “Commutativealgebra
andcombinatorics”.
2. IDEALS WITH LINEAR POWERS AND THEIR REES ALGEBRAS
Let S be the polynomial ring K[x ,...,x ] over an infinite field K with
1 n
maximalhomogeneousidealm =(x ,...,x ),andletI ⊂S beahomoge-
S 1 n
neousideal. TheCastelnuovo-MumfordregularityofagradedS-moduleM
isdenotedby reg(M).
Definition 2.1. We say that I has linear powers if all the powers of I have
a linear resolution. In other words, all the generators of I have the same
degree,say d, and reg(Ik)=kd forevery k∈N.
Examplesofidealswithlinearpowersare
(i) stronglystablemonomialidealsgenerated inonedegree,
(ii) productsofidealsoflinearforms,
MAXIMAL MINORSANDLINEARPOWERS 3
(iii) polymatroidalideals,
(iv) idealswithalinearresolutionand dimension ≤1,
see Conca and Herzog [CH]. Also, in [HHZ] Herzog, Hibi and Zheng
proved that monomial ideals generated in degree 2 have linear powers as
soon as they have a linear resolution. In general however ideals with a
linear resolution need not have linear powers; see Conca [C] for a list of
examples.
Propertiesofthepowersofanidealcanoftenbeexpressedin termsofthe
Rees algebra oftheidealitself. Thegoalof thissectionisthediscussionof
propertiesoftheReesalgebraofI thatarerelatedtoI havinglinearpowers.
SupposethatI is generated byelementsofdegree d. TheRees algebra
Rees(I)= Ij
Mj∈Z
hasaZ2-graded structurewith
Rees(I) =(Ij) .
(i,j) jd+i
It is standard graded in the sense that the K-algebra generators of Rees(I)
liveindegree(1,0)(theelementsofS )andindegree(0,1)(theelementsof
1
I ). Let m be theminimalnumberofgenerators of I and considerthepoly-
d
nomialringA=S[y ,...,y ]=K[x ,...,x ,y ,...,y ],bigradedbysetting
1 m 1 n 1 m
degx =(1,0)anddegy =(0,1)fori=1,...,nand j=1,...,m. Wehave
i j
aZ2-graded presentation
∼
A/P(I)=Rees(I).
Let Q(I) be the subideal of P(I) generated by elements of degree (∗,1).
By construction, Q(I) defines the symmetric algebra S(I) of I. The ideal I
is said to be of linear type if P(I)=Q(I), i.e. if Rees(I)=S(I). The fiber
coneF(I)ofI is, bydefinition,
F(I)=Rees(I)/m Rees(I)∼=K[I ]⊂S
S d
andcan bepresentedas
F(I)∼=K[y ,...,y ]/T(I)
1 m
where T(I)=P(I)∩K[y ,...,y ]. The ideal I is said to be of fiber type if
1 m
P(I)=Q(I)+T(I). Here we are identifying T(I) with its extension to A.
Inotherwords, I isoffibertypeiftheideal P(I)hasnominimalgenerators
ofdegree (a,b)witha>0 and b>1.
4 WINFRIEDBRUNS,ALDOCONCA,ANDMATTEOVARBARO
For a bigraded A-module M let b A (M) denote the Betti number of
i,(a,b)
M corresponding to homological position i and degree (a,b). The (1,0)-
regularity (also called x-regularity in [ACD, HHO, R]) of M is, by defini-
tion,
reg (M)=sup{a−i:b A (M)6=0forsomeb}.
(1,0) i,(a,b)
Furthermore P (M)denotesthemultigradedPoincare´ series of M, that is,
A
P (M)=(cid:229) b A (M)xizasb
A i,(a,b)
WehavethefollowingresultofBlum[B, Cor.3.6]:
Proposition2.2. If Rees(I)is Koszul,thenI has linearpowers.
The converse of Proposition 2.2 does not hold in general, see Examples
2.6,2.7and2.8below. Wewillmakeuseofthefollowingwell-knownfact,
seeforinstance[CH, Prop.1.2]:
Lemma 2.3. Let M be a finitely generated graded S-module and x ∈ S
1
such that 0 : x has finite length. Set a = sup{j : H0(M) 6= 0}. Then
M 0 m j
reg(M)=max{reg(M/xM),a }.
0
Let z ∈ S and set J = (I+(z))/(z)⊂ R =S/(z). We have a short exact
sequence:
0→W →Rees(I)/zRees(I)→Rees(J)→0 (1)
where
W = (Ik∩(z))/zIk= z(Ik :z)/zIk.
kM∈N kM∈N
Wehave:
Lemma 2.4. Assume that I has linear powers. Let z be a linear form in S
suchthat(Ik:z)/Ik hasfinitelengthforallk(agenerallinearformhasthis
property). Set J =(I+(z))/(z)⊂R=S/(z). Then:
(1) J haslinearpowersandtheBettinumbers(overR)ofthepowersof
J aredeterminedbythose(over S)ofthepowersof I.
(2) IfI isoffiber type, thenJ isof fibertype.
(3) The(1,0)-regularityofRees(I)is0.
Proof. (1) After a change of coordinates we may assume that z = x , and
n
R=K[x ,...,x ]. Denote the maximal homogeneous ideal of R by m .
1 n−1 R
By Lemma2.3we have
dk−1=reg(S/Ik)=max{reg(R/Jk),a (S/Ik)}.
0
ItfollowsthatJ haslinearpowers. Also a (S/Ik)=kd−1orH0 (S/Ik)=
0 m
S
0. Wealso haveIk :z=(V )+Ik where
k
V =(Ik :z) =(Ik :m ) =H0 (S/Ik)
k kd−1 S kd−1 m kd−1
S
MAXIMAL MINORSANDLINEARPOWERS 5
and m V ⊂ Ik. The Betti numbers of an ideal with linear resolution are
S k
determinedbyitsHilbertfunction. ComparingHilbertfunctionsandtaking
intoaccountthat dimV isb S (Ik)itfollowsthat
k n−1
n−1
b R(Jk)=b S(Ik)− b S (Ik)
i i (cid:18) i (cid:19) n−1
and hence the Betti numbers of the powers of J are determined by thoseof
thepowersof I.
(2) We have a presentation Rees(I)/x Rees(I)=R[y ,...,y ]/P where
n 1 m 1
P =(P(I)+(x ))/(x ) and we may represent the ideal W as (U+P )/P
1 n n 1 1
withU generated in degrees (0,∗) and (x ,...,x )U ⊂P . By construc-
1 n−1 1
tion,Rees(J)∼=R[y ,...,y ]/(U+P ). Notethatthismightnotbethemin-
1 m 1
imal presentation of Rees(J) because the numberof generators of J can be
smallerthan that of I. Indeed, m (I)−m (J)=dimU . We may choose the
0,1
y suchthaty ,...,y maptotheminimalgeneratorsof J andtheremaining
i 1 t
y ,...,y map to 0 (i.e. map to elements of x V in the presentation of
t+1 m n 1
Rees(I)). With these choices, P +U =P(J)+(y ,...,y ). By assump-
1 t+1 m
tion,P(I)has minimalgenerators only indegree (1,1)and (0,∗). Then the
same is true for P . Hence P(J)+(y ,...,y ) has minimal generators of
1 t+1 m
degree (1,1)and (0,∗). Thisis thedesired result.
Finally we prove statement (3) by induction on the dimension. Let A=
S[y ,...,y ]bethepolynomialringpresentingRees(I),andletB=R[y ,...,
1 m 1
y ]thecorrespondingpolynomialringforRees(J).
m
Since m annihilatesW, we can seeW as a K[y ,...,y ]-module. Since
R 1 m
K[y ,...,y ]isan algebraretract of B,wehave
1 m
P (W)=P (W)P (K[y]);
B K[y] B
see Herzog [H, Thm.1]orLevin [L, Thm.1.1]. Therefore the (1,0)-regula-
rity of W is 0. By induction on the dimension we may assume that the
(1,0)-regularity of Rees(J) is 0, and we conclude that Rees(I)/x Rees(I)
n
has (1,0)-regularity 0. This implies that the (1,0)-regularity of Rees(I) is
0. (cid:3)
So wehave:
Theorem2.5. TheidealI haslinearpowersifandonlyifreg Rees(I)=
(1,0)
0.
The implication ⇒ has been proved in Lemma 2.4. The converse has
beenprovedbyRo¨mer[R, Cor.5.5]. An alternativeproofisgivenin [HHZ,
Thm.1.1].
As the following example shows, ideals with linear powers need not to
beoffibertypenorhaveaKoszulRees algebra.
6 WINFRIEDBRUNS,ALDOCONCA,ANDMATTEOVARBARO
Example2.6. TheRees algebraRees(I)oftheideal
I =(a2b,a2c,abd,b2d)⊂K[a,b,c,d]
isdefined by theideal
P(I)=(−y b+y c,−y a+y d,−y a+y b,−y2c+y y d)
2 1 3 1 4 3 3 2 4
whose generators form a Gro¨bner basis. Hence the (1,0)-regularity of
Rees(I) is 0. It follows that I has linear powers, it is not of fiber type and
Rees(I)isnotKoszul.
The following two examples are ideals with linear powers and of fiber
type,butwithanon-quadratic(hencenon-Koszul)Rees algebra.
Example 2.7. A strongly stable ideal is a monomial ideal I satisfying I :
x ⊇I :x forevery i< j. Itisknownthattheregularityofastronglystable
i j
idealis equalto thelargest degreeofaminimalgenerator.
LetI beastronglystableidealgeneratedindegree d. ThepowersofI are
strongly stable as well, and hence I has linear powers. Moreover, strongly
stable ideals are of fiber type; see Herzog, Hibi and Vladoiu [HHV]. On
the other hand, Rees(I) need not be quadratic. For example, the smallest
strongly stable ideal of K[x ,x ,x ] containing the monomials x6,x2x2x2,
1 2 3 2 1 2 3
x3x3 hasanon-quadraticRees algebra.
1 3
Example 2.8. Let I be an ideal generated by monomials of degree 2 and
with a linear resolution. Then by [HHZ, Thm.2] I has linear powers. Fur-
thermore Villarreal proved in [V, Thm.8.2.1] that I is of fiber type if it is
square-free. On theotherhand Rees(I)need notbequadratic. Theideal
(x x ,x x ,x x ,x x ,x x ,x x ,x x ,x x ,x x ,x x )
3 6 1 3 5 6 4 6 2 3 1 5 3 4 1 6 1 2 4 5
hasalinearresolutionanditsRees algebraisnot quadratic.
3. MAXIMAL MINORS WITH LINEAR POWERS
Let X beam×n matrixwithentriesinaNoetherian ring R and letI (X)
j
denote the ideal of R generated by the size j minors of X. We will al-
ways assume that m ≤ n. The ideal I (X) of maximal minors is resolved
m
by the Eagon-Northcott complex provided grade(I (X)) ≥ n−m+1. In
m
[ABW, Thm.5.4] Akin, Buchsbaum and Weyman for every k∈N describe
a complex of free R-modules resolving the k-th powerof the ideal of max-
imal minorsof X undercertain genericity conditions. In theirnotations the
complex is denoted by Y(k,f ) where f is the R-linear map Rn⊗Rm → R
associated with X. Howeverwe prefer to use the notation Y(k,X) to stress
thedependenceon theinputmatrix X. They proved:
MAXIMAL MINORSANDLINEARPOWERS 7
Theorem 3.1. Suppose gradeI (X) ≥ (m+1− j)(n−m)+1 for every j.
j
Then I (X)k is resolved by the free complex Y(k,X). Furthermore the
m
lengthofY(k,X)ismin{k,m}(n−m).
Remark3.2. Thefact thatthelengthofY(k,X)ismin{k,m}(n−m)isnot
explicitly stated in [ABW]. Indeed in the proof of [ABW, Thm.5.4] it is
shown that the the length of Y(k,X) is ≤min{k,m}(n−m). To show that
equality holds one can assume right away that X is a matrix of variables
x and that R = K[x ] with K a field of characteristic 0. Then, accord-
ij ij
ing to [ABW, Section 5], TorR(I (X)k,K) is isomorphic to the kernel of a
i m
GL(V)×GL(W)-equivarianthomomorphismfrom
i
A= (V ⊗W)⊗(L V ⊗L W)
(mk) (mk)
^
to
i−1
B= (V ⊗W)⊗(L V ⊗L W),
(mk,1) (mk,1)
^
where V and W are K-vector spaces of dimension m and n and L de-
l
notes the Schur functor associated with the partition l . The decomposi-
tions in irreducible GL(V)×GL(W)-modules can be computed for all the
involved representations, using the (skew) Cauchy formula and the Little-
wood-Richardson rule. In particular, when i =min{k,m}(n−m), one can
check thefollowing:
(1) If k ≥ m, then L V ⊗L W is a direct summand of A
(mn+k−m) (nm,mk−m)
andnot ofB.
(2) If k < m, then L V ⊗L W is a direct summand of A and
(mk,kn−m) (nk)
notofB.
So, in both cases, Schur’s lemma implies that the kernel of the above map
is not zero. It follows that TorR(I (X)k,K)6=0 for i =min{k,m}(n−m).
i m
Hencetheprojectivedimensionof I (X)k is≥min{k,m}(n−m).
m
Forlaterapplicationswerecord thefollowinglemmaand its corollary:
Lemma3.3. SupposethatgradeI (X)≥n−m+1andgradeI (X)≥(m+
m j
1− j)(n−m)+2forevery j =1,...,m−1,then
Ass(R/I (X)k)=Ass(R/I (X))
m m
foreveryk>0.
Proof. We use induction on m. Let m = 1 and P be an associated prime
of I (X)k. After localization with respect to P we may assume that R is
m
local with maximal ideal P. The hypothesis implies that I =I (X) is now
1
generated by a regularsequence. Then the powersof I are perfect ideals of
8 WINFRIEDBRUNS,ALDOCONCA,ANDMATTEOVARBARO
graden,resolvedbytheEagon-Northcottcomplex. ThusdepthR =n,and
P
soP isan associated primeof I as well.
Now let m≥2, and let P be a prime ideal. Suppose first that P contains
I (X). We can apply Theorem 3.1. It shows that the projective dimension
1
ofR/I (X)k isatmostm(n−m)+1. TheinequalityforgradeI (X)implies
m 1
that depthR > projdim(R/I (X)k) . Thus P is not associated to I (X)k,
P m P m
andit isnotan associated primeof I (X)forthesamereason.
m
NowsupposethatPdoesnotcontainI (X). Thenwecan applythestan-
1
dard inversion argument to an entry of X, say x . This argument reduces
11
both m and n as well as the sizes of minors by 1, and since our bound on
grade depends only on differences of these numbers, they are preserved.
Since the inversion of an element outside P does not affect the property of
Pbeing an associatedprime,wecan apply theinductionhypothesis. (cid:3)
Remark3.4. Lemma3.3canalsobederivedfrom[BV,(9.27)(a)]thatgives
alowerboundand theasymptoticvalueforthedepth of R/I (X)k.
m
Corollary 3.5. Let X be a m×n matrix with entries in a Noetherian ring
R. Supposethatforsomenumber p with 1≤ p≤m, we have:
(i) gradeI (X)≥n−m+1,
m
(ii) gradeI (X)≥(m+1− j)(n−m)+2forall j= p+1,...,m−1.
j
Then
Ass(R/I (X)k)⊆Ass(R/I (X))∪{P:P⊇I (X)}
m m p
foreveryk>0.
Proof. Again we use induction on m. If a prime ideal P contains I (X),
p
thereisnothingtoprovefor P. Otherwiseitdoesnotcontaina p-minor. Its
inversionreduces all sizes by p. But then weare in thesituationofLemma
3.3. (cid:3)
From now one we assume that S is a polynomial ring overa field K and
X is a m×n matrix whose entries are linear forms in S. In this setting we
have:
Proposition3.6. Let X bea m×n matrixoflinear formswith m≤n. Sup-
poseheightI (X)≥(m+1− j)(n−m)+1for every j. Then
j
(1) I (X)haslinearpowers.
m
(2) TheRees algebraRees(I (X))isKoszul.
m
Proof. ThefirstassertionisaspecialcaseofTheorem3.1becausethecom-
plexesY(k,X)arelinearwhentheentriesof X arelinear. Thesecondasser-
tionfollowsfromtheresultofEisenbudandHuneke[EHu,Thm.3.5]since,
undertheassumptionofProposition3.6,theReesalgebraofI (X)isaquo-
m
tientofthegenericonebya regularsequence oflinearforms. (Thegeneric
MAXIMAL MINORSANDLINEARPOWERS 9
one is the Rees algebra of the ideal of maximal minors of a m×n matrix
of distinct variables over the base field). Hence it is enough to prove the
statementforamatrixofvariables. TheRees algebraoftheidealofminors
of a matrix of variables is a homogeneousASL [BV, Thm.9.14]and hence
itisdefined by aGro¨bnerbasisofquadrics. Therefore itis Koszul. (cid:3)
Ourmainresultis thefollowing:
Theorem 3.7. LetX beam×nmatrixwithm≤nwhoseentriesarelinear
forms in a polynomial ring S over a field K. Suppose that heightI (X)≥
m
n−m+1andheightI (X)≥min{(m+1− j)(n−m)+1,N}forevery j=
j
2,...,m−1where N =heightI (X). Then:
1
(1) I (X)haslinearpowers andit isoffiber type.
m
(2) TheBettinumbersof I (X)k depends onlyon m,n,k andN.
m
To proveTheorem3.7 weneed asortofdeformationargument.
Lemma 3.8. Let X be a m×n matrix whose entries are linear forms in a
polynomialringS over an infinitefield K. Let ybea newvariable. Then:
(1) ForeveryA∈M (K)and forevery j onehas
mn
heightI (X)≤heightI (X+yA)≤1+heightI (X).
j j j
(2) Thereexists A∈M (K)suchthat
mn
heightI (X+yA)=min{(m+1− j)(n+1− j),heightI (X)+1}
j j
forevery j.
Proof. (1) The inequality on the right follows from the inclusion I (X +
j
yA)⊆(y)+I (X). Fortheotherweconsidertheweightvector wthatgives
j
weight1tothevariablesofSandweight0toy. Onehasin (I (X+yA))⊇
w j
I (X). Sincetheheightdoesnotchangebytakingidealsoftheinitialforms,
j
weconcludethat heightI (X)≤heightI (X+yA).
j j
(2) The inequality ≤ follows from (1). Set c=dimS, h =heightI (X),
j j
g =(m+1−j)(n+1−j) andT ={j:h <g }. Byvirtueof(1)andsince
j j j
heightI (Y)≤g holds for every m×n matrixY, it is enough to show that
j j
thereexistsA∈M (K)suchthatheightI (X+yA)=h +1forevery j∈T.
mn j j
We choosea systemof coordinates x ,...,x for S such that forevery j the
1 c
set E ={x :k>h } is a system of parameters for the ring S/I (X). This
j k j j
can be done because the base field K is infinite. It follows that I (X) has
j
heighth alsomodulo(E ). WemaywriteX =(cid:229) c x A withA ∈M (K).
j j i=1 i i i mn
SinceX =(cid:229) hj xA mod(E ),wededucethatheightI ((cid:229) hj xA )=h ,that
i=1 i i j j i=1 i i j
is,theradical ofI ((cid:229) hj xA ) is(x ,...,x ),forevery j.
j i=1 i i 1 hj
Againbyvirtueof(1),itisenoughtoshowthatthereexists A∈M (K)
mn
such that heightI ((cid:229) hj xA +yA) = h +1 for every j ∈ T. We consider
j i=1 i i j
10 WINFRIEDBRUNS,ALDOCONCA,ANDMATTEOVARBARO
thesubvarietyV oftheprojectivespaceP(M (K))ofthematricesofrank
j mn
< j. Furthermore we consider the linear space L = hA : i = 1,...,h i ⊆
j i j
P(M (K)). By construction L ∩V =0/. Consider then the join L ∗V of
mn j j j j
L and V , that is L ∗V =∪BC where BC is the line joining B and C and
j j j j
(B,C) varies in L ×V . It is well know that L ∗V is a projective variety
j j j j
andthat
dimL ∗V =dimL +dimV +1=mn−1−(g −h ),
j j j j j j
see for instance [Ha, 11.37]. Therefore L ∗V is a proper subvariety of
j j
P(M (K)) if j ∈ T, and so is (L ∗V ). It follows that we can take
mn j∈T j j
A∈M (K)andA6∈L ∗V foreSvery j∈T. Weclaimthatwiththischoice
mn j j
of A one has heightI ((cid:229) hj x A +yA) =h +1 for every j ∈T as desired.
j i=1 i i j
Suppose, by contradiction, that for a j ∈ T one has heightI ((cid:229) hj xA +
j i=1 i i
yA)<h +1. Then there is a point (a ,...,a ,b) on the projective subva-
j 1 hj
riety of Phj defined by the ideal I ((cid:229) hj xA +yA) (where the last coordi-
j i=1 i i
nate correspond to the variable y). In other words (cid:229) hj a A +bA has rank
i=1 i i
< j. If b6=0 we obtain that A∈L ∗V , a contradiction. If b=0 we have
j j
(a ,...,a )∈L ∩V which isalsoa contradiction. (cid:3)
1 hj j j
ProofofTheorem 3.7. (1) It is harmless to assume that the base field is in-
finite. Set N = heightI (X). We may assume that N = dimS. We have
1
N ≤mn and we do decrising induction on N. If N =mn then we are in the
generic case and the assertion is true because of Proposition 3.6. Consider
a new variable y, and set R = S[y]. In view of Lemma 3.8 we may take
A ∈ M (K) such that the ideals of minors of the the matrix Y = X +yA
mn
satisfy
(3.1) heightI (Y)=min{(m+1− j)(n+1− j),heightI (X)+1}
j j
forevery j. Hence heightI (Y)=N+1 andsince, by assumption,wehave
1
heightI (X)≥min{(m+1− j)(n−m)+1,N}wemaydeducethat
j
(3.2)
heightI (Y)≥min{(m+1− j)(n+1− j),(m+1− j)(n−m)+2,N+1}
j
Inparticular,
(3.3) heightI (Y)≥min{(m+1− j)(n−m)+1,N+1}
j
holdsforevery j. Hence,byinduction,wemayassumethatI (Y)haslinear
m
powersandisoffibertype. SinceR/I (Y)⊗S=S/I (X)weconcludefrom
m m
Lemma 2.3 that I (X) has linear powers and is of fiber type provided we
m
showthat
(3.4) (I (Y)k :y)/I (Y)k
m m