On Landsberg's criterion for complete intersections

a rXiv:alg-ge o m/94806v124Aug1994On Landsberg’s criterion for complete intersections S.L’vovsky ∗August 24,1994Introduction In his preprint [1],ndsberg introduces an elementary characterization of complete intersections (Proposition 1.2in [1]).The proof of this proposition uses the method of moving frames.The aim of this note is to present an elementary proof of Landsberg’s criterion that is valid over any ground field.1Notation and statement of results Let k be an algebraically closed field and P N =Proj k [T 0,...,T N ]the N -dimensional projective space over k .If F is a homogeneous polynomial in T 0,...,T N ,we will denote by Z (F )⊂P N the hypersurface defined by F .If F is a homogeneous polynomial and x =(x 0:...:x N )∈P N ,put d x F =(∂F/∂T 0(z ),...,∂F/∂T N (z ))∈k N +1(actually d x F depends on the choice of homogeneous coordinates for x ;this abuse of notation should not lead to confu-sion).If x ∈X ,where X ⊂P N is a projective variety,then T x X ⊂P N denotes the embedded Zariski tangent space to X at x .If X ⊂P N is a projective variety,then its ideal sheaf will be denoted byI X ⊂O P N and its homogeneous ideal by I X ⊂k [T 0,...,T N ].We will say that a hypersurface Y =Z (F )trivially contains X iffF = G i F i ,where G i ’s and F i ’s are homogeneous polynomials,F i vanish on X for all i ,and deg F i <deg F for all i .If Y trivially contains X ,then Y ⊃X .We will say that a hypersurface W non-trivially contains X iffW contains X ,but not trivially.The following proposition is a slight reformulation of Landsberg’s criterion (cf.[1,Proposition 1.2]):Proposition 1.1For a projective variety X ⊂P N ,the following conditions are equivalent:(i)X is a complete intersection.(ii)There exists a smooth point x∈X having the following property:any hypersurface W⊂P N that non-trivially contains X must be smooth at x.(iii)For any smooth point x∈X and any hypersurface W that non-trivially contains X,W is smooth at x.(iv)For any smooth point x∈X and any hypersurface W that non-trivially contains X,T x W cannot contain an intersection i T x W i,where each W i is a hypersurface s.t.W i⊃X and deg W i<deg W(it is understood that the intersection of an empty family of tangent spaces is the entire P N).2ProofsFor the sequel we need two lemmas.Lemma2.1Let F1,...,F r be homogeneous polynomials over k in T0,...,T N. Assume that x=(x0:...:x N)∈P N is their common zero and that the vectors d x F1,...,d x F r are linearly dependent.Then one of the following alternatives holds:1.There is j∈[1;r]s.t.F j belongs to the ideal in k[T0,...,T N]generated byF i’s with i=j.2.There are homogeneous polynomials˜F0,...,˜F N s.t.the ideals(F0,...,F N)and(˜F0,...,˜F N)coincide,deg˜F i=deg F i for all i,and d x˜F j=0for some j.Proof.Let the shortest linear relation among d x F j’s have the formλ1d x F1+···+λs d x F s=0,whereλj=0for all j.Reordering F j’s if necessary,we may assume that deg F1≤deg F2≤···≤deg F s.Let t be such a number that deg F t=deg F s and deg F t−1<deg F s(if deg F1=deg F s,set t=1).If the polynomials F t,...,F s are linearly dependent,then it is clear that one of them lies in the ideal generated by the others and there is nothing more to prove.Assume from now on that F t,F t+1,...,F s are linearly independent. Then there exists an index j∈[t;s]and numbersµi,where i∈[t;s]s.t.µi F i+µj(λt F t+···+λs F s).(1)F j=i∈[t;s]\{j}For each i∈[1;t−1],choose a homogeneous polynomial G i s.t.deg G i= deg F s−deg F i and G i(x0,...,x N)=λi,and set˜F= i<t G i F i+ i≥tλi F i.(2)j2If˜F j=0,then F s∈(F1,...,F s−1)and thefirst alternative holds.Otherwise, deg˜F j=deg F j,d x˜F j=0by virtue of(2),and it follows from(1)and(2)thatµi F i+µj˜F j−µj i<t G i F i,F j=i∈[t;s]\{j}whence(F1,...,F j−1,˜F j,F j+1,...,F s)=(F1,...,F s).Hence in this case the second alternative holds,and we are done.The second lemma belongs to folklore.To state this lemma,let us introduce some notation.Denote by S the set of sequences of non-negative integersδ= (δ1,δ2,...)s.t.δM=0for all M≫0.Ifδ,η∈S,we will writeδ≻ηiffthere is an integer i s.t.δi>ηi andδj=ηj for all j>i.Lemma2.2Any sequenceδ1≻δ2≻···must terminate.Proof.For anyδ∈S,set n(δ)=max{j:δj=0},ℓ(δ)=δn(δ)>0.Ifδ≻ηand n(δ)=n(η),thenℓ(δ)≥ℓ(η).Let us prove the lemma by induction on n(δ1).If n(δ1)≤1,the result is evident.Assuming that the lemma is true whenever n(δ1)<m,suppose that there is an infinite sequenceδ1≻δ2≻...with n(δ1)= m.If n(δj)<n(δ1)for some j,we arrive at a contradiction by the induction hypothesis.Hence,n(δj)=n(δ1)=m for all j andℓ(δ1)≥ℓ(δ2)≥ 0Thus there exists an integer N s.tℓ(δj)is connstant for j≥N.For any j≥N, denote byδ′j∈S a sequence that is obtained fromδj by replacing its last positive term by zero.It is clear thatδ′N≻δ′N+1≻···,and this sequence is infinite by our assumption.This is again impossible by the induction hypothesis since n(δ′j)<n(δj)=m,whence the lemma.Proof of(ii)⇒(i).Put a=N−dim X.Let(F1,...,F r)be a system of(homogeneous)generators of I X.To any such system assign a sequence δ(F1,...,F r)∈S,whereδ(F1,...,F r)i=#{j∈[1;r]:deg F j=i}.I claim thatif r>a,then I X=(Φ1,...,Φs),whereΦi’s are homogeneous poly-nomials s.t.δ(F1,...,F r)≻δ(Φ1,...,Φs).To prove this claim,observe that d x F1,...,d x F r are linearly dependent since X is smooth at x and r>codim X.Now Lemma2.1implies that either one of the F j’s(say,F1)can be removed without affecting I X,or I X=(˜F1,...,˜F r), where deg˜F j=deg F j for all j and d x˜F j=0for some j.In thefirst case,the requiredΦ1,...,Φs can be obtained by merely removing F1;in the second case, hypothesis(ii)shows that˜F j= t i=1G iΨi,whereΨi∈I X and degΨi<deg˜F j for all j.Replacing˜F j byΨ1,...,Ψt in the sequence˜F1,...,˜F r and putting s=r+t−1,we obtain a sequenceΦ1,...,Φs s.t.I X=(Φ1,...,Φs)and δ(˜F1,...,˜F r)≻δ(Φ1,...,Φs).Since the degrees of˜F j’s and F j’s are the same, this means thatδ(F1,...,F r)≻δ(Φ1,...,Φs)as well,and the claim is proved.3Now we canfinish the proof as follows.If r=a,then X is the complete intersection of Z(F1),...,Z(F r)and there is nothing to prove.If r>a,then by virtue of our claim we can replace the system of generators F1,...,F r by Φ1,...,Φs.Let us iterate this process.By virtue of Lemma2.2this process must terminate and by virtue of our claim this is possible only when we have found a system of exactly a generators of the ideal I X.This means that X is a complete intersection,thus completing our proof.Proof of(iv)⇒(iii)⇒(ii).Trivial.Proof of(i)⇒(iv).Let X be a complete intersection of the hypersurfaces Z(F1),...,Z(F a).Assume that a hypersurface W=Z(F),with F irreducible, non-trivially contains X and that x=(x0:...:x N)∈P N is a smooth point of X;set m=deg F.Since Z(F)⊃X and X is a complete intersection of the Z(F i)’s,we see thatF= G i F i;(3) since W contains X non-trivially,at least some of the G j’s must be non-zero constants.Reordering F j’s if necessary,we may assume that G j is a constant (hence,deg F j=m)iff1≤j≤s.Taking d x of the both parts of(3),we seethatd x F=ai=1c id x F i,where c i=0for some i∈[1;s].(4)On the other hand,assume that W i=Z(B i)with irreducible B i’s.Then the hypothesis implies that d x F is a linear combination of d x B j’s,and the fact that X is a complete intersection of Z(F t)’s and Z(B j)⊃X implies that,for each j,there is a relationB j= t>s G jt F t(5) (it suffices to sum only over t>s since for t≤s we have deg F t=deg W> deg B j).If we take d x of both parts of(5),we see that,for each j,d x B j is a linear combination of d x F t’s with t>s.Hence d x F is also a linear combination of d x F t’s with t>s.Taking into account(4)we see that d x F i’s are linearly dependent.This is,however,impossible since x is a smooth point of the comlete intersection of Z(F j)’s.This contradiction completes the proof. References[1]Landsberg,J.M.Differential-Geometric Characterizations of Complete In-tersections.Preprint,alg-geom/9407002Email address:serge@schcorr.msk.sunskcsmoscow@4。