On the module of effective relations of a standard algebra
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On the mo dule of eective relations of a standard algebra By Francesc Planas-Vilanova Dept. Matematica Aplicada I. ETSEIB-UPC. Diagonal 647, E-08028 Barcelona . 1 Intro duction Let A be a commutative ring. We denote by a standard A -algebra a commutative graded A -algebra U = n 0 U n with U 0 = A and such that U is generated as an A -algebra by the elements of U 1 . Take x aset of (p ossibly innite) generators of the A -mo dule U 1 . Let V = A t ] b e the p olynomial ring with as manyvariables t (of degree one) as x has elements and let f : V ! U be the graded free presentation of U induced by the x . For n 2, we will call module of eective n - relations the A -mo dule E ( U ) n =ker f n =V 1 ker f n ; 1 .The minimum p ositiveinteger r 1such that the eective n -relations are zero for all n r +1 is known to be an invariant of U .It is called the relation typ e of U and is denoted by rt( U ). For an ideal I of A , we dene E ( I ) n = E ( R ( I )) n and rt( I ) = rt( R ( I )), where R ( I )= n 0 I n t n A t ] is the Rees algebra of I . In this pap er, wegivetwo descriptions of the A -mo dule of eective n -relations. In terms of Andre-Quillen homology wehave that E ( U ) n = H 1 ( A U A ) n (see 2.3). It turns out that this mo dule do es not dep end on the chosen x . In terms of Koszul homology we provethat E ( U ) n = H 1 ( x U ) n (see 2.4). Using these characterizations, weshow later some prop erties on the mo dule of eective n -relations and the relation typ e of a graded algebra. Meanwhile, our line of disquisition approaches us to several earlier works on the sub ject (see 2], 5 ], 6], 7], 9], 10 ], 13] and 14]). Section 2 is devoted to state the ab ove mentioned (co)homological characterizations of the A -mo dule of eective n -relations and compare them with some already known results. In section 3, we give some applications. The interest is sp ecially centered on the mo dule of n -relations of p owers of an ideal and the mo dule of n -relations of Veronese subrings. In particular, one concludes that rt( U ( p ) ) rt( U p + ) but, in general, rt( U ( p ) ) 6 =rt( U p + ), where U + = n> 0 U n is the irrelevant ideal of U and U ( p ) = n 0 U np is the p -th Veronese subring of U (see 3.12). Finally, in section 4 we characterize, in terms of a system of generators, which ideals have mo dule of eective n -relations zero. In particular, a new characterization of sequences of linear typ e is obtained. e-mail: [email protected]c.es 1
2 Homological description of eective relations Let U = n 0 U n be a standard A -algebra. Put U + = n> 0 U n its irrelevant ideal. If E = n 1 E n is a graded U -mo dule and r 1, we denote by F r ( E )the submo dule of E generated by the elements of degree at most r . Put (p ossibly innite) s( E ) = min f r 1 j E n = 0 for all n r +1 g : Since ( E=U + E ) n = E n =U 1 E n ; 1 , then the following three conditions are equivalent: F r ( E )= E ,s( E=U + E ) r , and, E n = U 1 E n ; 1 for all n r +1. Given h : W ! U , a surjective graded morphism of standard A -algebras, we are interested in the graded A -mo dule E ( h )=ker h=W + ker h . The following is an elementary, but useful lemma: Lemma 2.1 Let f : V ! U and g : W ! V be two surjective graded morphisms of standard A -algebras. Then, there exists a gradedexact sequenceof A -modules: E ( g ) ! E ( f g ) g ! E ( f ) ! 0 : (1) In particular, s( E ( f )) s( E ( f g )) max(s( E ( f )) s( E ( g ))) . Moreover, if V and W are two symmetric algebras, then E ( g ) n =0 and E ( f g ) n = E ( f ) n for al l n 2 . Proof . Exact sequence (1) follows from the snake lemma applied to the commutative diagram: W 1 ker g n ; 1 W 1 ker( f g ) n ; 1 W 1 ker f n ; 1 0 - - - - - - - 0 ker g n ker( f g ) n ker f n 0 ??? 1 g n ; 1 g n Moreover, if W = S ( W 1 ) and V = S ( V 1 ), then ker g = F 1 (ker g ). Denition 2.2 Let U be a standard A -algebra and let : S ( U 1 ) ! U be the graded morphism of standard A -algebras induced by the identityon U 1 .Given n 2, the module of eective n - relations of U is dened to be E ( U ) n = ker n =U 1 ker n ; 1 . Put E ( U ) = n 2 E ( U ) n =ker = S + ( U 1 ) ker . Then, the relation type of U is dened to be rt( U ) = s( E ( U )). Remark that if h : W ! U is any symmetric presentation of U ,that is, W is a symmetric algebra and h is a surjective graded morphism of standard A -algebras, then h can be factorized into h = f g , where g : S ( W 1 ) ! S ( U 1 ) is the induced morphism by h 1 : W 1 ! U 1 and f = . Thus, applying Lemma 2.1, E ( U ) n = E ( h ) n for all n 2 and s( E ( U )) = s( E ( h )). If I is an ideal of A , the module of eective n - relations of I is E ( I ) n = E ( R ( I )) n and the relation type of I is rt( I )=rt( R ( I )), where R ( I )= n 0 I n t n is the Rees algebra of I . An ideal with mo dule of eective 2-relations zero is called syzygetic . An ideal of relation typ e 1 is called of linear type (see, e.g., 8]). 2
Remark 2.3 In fact, sequence (1) is part of along exact sequence of Andre-Quillen homology. Indeed, the Jacobi-Zariski sequence asso ciated to the morphisms g : W ! V and f : V ! U , with resp ect to the U -mo dule A = U=U + , gives rise to ::: ! H 1 ( WVA ) ! H 1 ( WUA ) ! H 1 ( V U A ) ! H 0 ( WV A ) ! ::: : Using H 1 ( A A=I M ) = I=I 2 M and H 0 ( A A=I M ) = 0 for any ideal I of A and any A=I -mo dule M ,we get (1) (see 1]). On the other hand, the Jacobi-Zariski sequence asso ciated to the morphisms A ! S ( U 1 ) and : S ( U 1 ) ! U , with resp ect to the U -mo dule A ,is ::: ! H 1 ( A S ( U 1 ) A ) ! H 1 ( A U A ) ! H 1 ( S ( U 1 ) UA ) ! H 0 ( A S ( U 1 ) A ) ! ::: : Using H 1 ( A S ( U 1 ) A ) = 0 and H 0 ( A S ( U 1 ) A ) = H 0 ( A U A ), we get the graded isomorphism of A -mo dules H 1 ( A U A ) = H 1 ( S ( U 1 ) UA ) = ker = S + ( U 1 ) ker . Thus, H 1 ( A U A ) n = E ( U ) n is the mo dule of eective n -relations of U . In particular, rt( U ) = s( H 1 ( A U A )). There is also a description of the mo dule of eective n -relations in terms of Koszul (co)homology. Let f : V ! U be a surjective graded morphism of standard A -algebras. For each p 1, consider the map V p U ! U sending x y to f p ( x ) y and let K ( f p ) be the Koszul complex asso ciated to this U -linear form (see 1.6.1 of 3]). Since it is an homogeneous form of degree zero, K ( f p )is a complex of graded U -mo dules having dierentials homogeneous morphisms of degree zero. Concretely, K ( f p )= n 0 K ( f p ) n where K ( f p ) n is the following sub complex ( U n =0 for n< 0): ::: ;! A 2 ( V p ) A U n ; 2 p @ 2 ;! V p A U n ; p @ 1 ;! U n ;! 0 where @ q (( x 1 ^ ::: ^ x q ) y )= P q i =1 ( ; 1) i ; 1 x 1 ^ ::: ^ b x i ^ ::: ^ x q f p ( x i ) y , for all x i 2 V p and y 2 U n ; qp . In particular, for every q 0, H q ( K ( f p )) is a graded A -mo dule with H q ( K ( f p )) n = H q ( K ( f p ) n ). Theorem 2.4 Let f : V ! U and g : W ! V be two surjective graded morphisms of standard A -algebras. Let : S ( U 1 ) ! U be the canonical morphism and suppose W is a symmetric algebra. Given ( n 2 p =1) or ( n 2 p +1 p 2) , there are isomorphisms of A -modules H 1 ( K ( f p ) n )= ker ( f g ) n W p ker( f g ) n ; p = ker n S p ( U 1 ) ker n ; p : In particular, the module of eective n -relations of U is E ( U ) n = H 1 ( K ( f 1) n ) and the relation type of U is rt( U )=s( H 1 ( K ( f 1) )) . Proof .Put h = f g . Since n ; p p ,then W n ; p ker g p W p ker g n ; p W p ker h n ; p . Applying the snake lemma to the commutative diagram of exact rows 3
ker g p W n ; p W p ker h n ; p W p W n ; p V p U n ; p 0 - - - - - - - 0 ker h n W n U n 0 ? ? ? ? ? h n g p h n ; p we get the exact sequence of A -mo dules 0 ! ( g p h n ; p )( Z 1 (1 W p ) n ) !Z 1 ( f p ) n ! ker h n =W p ker h n ; p ! 0 where Z 1 (1 W p ) n , Z 1 ( f p ) n stand for the n -th comp onent of the 1-cycles mo dule of K (1 W p ), K ( f p ). If Z 1 (1 W p ) n = B 1 (1 W p ) n (the n -th comp onent of the 1-b oundaries mo dule of K (1 W p )), then ( g p h n ; p )( Z 1 (1 W p ) n )= B 1 ( f p ) n Thus, the rst isomorphism is demonstrated provided weprove H 1 ( K (1 W p )) n = 0 for a symmetric algebra W (see next lemma). In particular, if we take V = U and f =1 U ,then h = f g = g and one of the p ossible choices of h is the canonical morphism . Hence, applying twice the rst equalityto and to any h : W ! U arising from a symmetric algebra W ,wehave H 1 ( K (1 U p ) n )= ker n S p ( U 1 ) ker n ; p = ker h n W p ker h n ; p : Lemma 2.5 Let M be an A -module and W = S ( M ) the symmetric algebra of M . Then, for ( n 1 p =1) or ( n 2 p +1 p 2) , H 1 ( K (1 W p )) n =0 . Proof . Put T ( M ) the tensorial algebra of M and q = n ; p . Applying the snake lemma to the commutative diagram of exact rows T p ( M ) T q ( M ) T n ( M ) 0 - - - - - 0 ker ! W p W q W n 0 ? ? ? ? " ! we get the exact sequence 0 ! ker ! ker " ! ker ! ! 0. Thus, Z 1 (1 W p ) n = ker ! = (ker " )isthe A -mo dule generated by the elements ( x 1 x p ; 1 x p ) ( y 1 y 2 y q ) ; ( x 1 x p ; 1 y 1 ) ( x p y 2 y q ) where x i y j 2 M and x 1 x p stands for the pro duct in W = S ( M ). Clearly, if ( n 1, p = 1), then Z 1 (1 W p ) n = B 1 (1 W p ) n . Supp ose ( n 2 p +1, p 2), i.e., q > p . Then, H 1 ( K (1 W p ) n ) = 0 follows from the equality: ( x 1 x p ) ( y 1 y q ) ; ( x 1 x p ; 1 y 1 ) ( x p y 2 y q )= ( x 1 x p ) ( y 1 y q ) ; ( y 2 y p +1 ) ( x 1 x p y 1 y p +2 y q )+ ( y 2 y p +1 ) ( x 1 x p ; 1 y 1 x p y p +2 y q ) ; ( x 1 x p ; 1 y 1 ) ( x p y 2 y q ) : 4
Remark 2.6 Let f : S ( F ) ! S ( M ) b e the induced morphism on the symmetric algebras by an epimorphism : F ! M of A -mo dules. Then, the last three nonzero terms of K ( f p ) p + q , q p 1, dene the sequence: A 2 ( S p ( F )) A S q ; p ( M ) @ 2 ! S p ( F ) A S q ( M ) @ 1 ! S p + q ( M ) ! 0 (2) with @ 2 (( x 1 x p ) ^ ( y 1 y p ) z )=( y 1 y p ) f ( x 1 x p ) z ; ( x 1 x p ) f ( y 1 y p ) z and @ 1 (( x 1 x p ) t )= f ( x 1 x p ) t , x i y j 2 F , z 2 S q ; p ( M )and t 2 S q ( M ). On the other hand, Micali and Roby dened (in 10]) the sequence of A -mo dules T A p + q ( F ) ! S p ( F ) A S q ( M ) ! S p + q ( M ) ! 0 (3) with ( x 1 ::: x p + q )=( x 1 x p ) f ( x p +1 x p + q ) ; ( x 1 x p ; 1 x p +1 ) f ( x p x p +2 x p + q ) and = @ 1 . By a similar argument to that one of the end of Lemma 2.5, one can prove that Im @ 2 is always contained in Im andthatif q>p , then b oth mo dules are equal. Thus, the exactness of (2) (settled by Theorem 2.4 either for q p =1 or either for q>p 2) assures the exactness of (3). Nevertheless, if q = p 2, then (2) might not be exact (see pro of of Lemma 3.8) while (3) is always exact (see 10]). Corollary 2.7 Let U be a standard A -algebra and let x bea(possibly innite) set of forms of degree one generating U + .If H 1 ( x U ) denotes the rst Koszul homology group associated to x ,then E ( U ) n = H 1 ( x U ) n for al l n 2 . In particular, rt( U )=s( H 1 ( x U )) . Proof . Take in Theorem 2.4, f : S ( F ) ! U induced by a free presentation F ! U 1 asso ciated to x . Then, K ( f 1) = K ( x U )is the usual Koszul complex asso ciated to the elements x . Remark 2.8 Using duality between Koszul homology and cohomology (see 1.6.10 of 3]) we recover Schenzel's result rt( U )=s( H d ; 1 ( x U )) + d ,when x is nite of cardinal d (see 13]). Remark 2.9 Let I b e an ideal of A and R ( I )= n 0 I n t n its Rees algebra. Take f =1 R , the identityon R ( I ), in Theorem 2.4. Then, Z 1 ( f p ) n =ker ; I p I n ; p ! I n =Tor A 1 ( A=I p I n ; p ) whichisknown to b e isomorphic to Z 1 \ I n ; p F=I n ; p Z 1 , where 0 ! Z 1 ! F ! I p ! 0 is a presentation of I p with F free (see, e.g., 2.5 of 8]). Moreover, via the same isomorphism B 1 ( f p ) n =Im A 2 ( I p ) I n ; 2 p ! I p I n ; p = I n ; 2 p B 1 =I n ; p Z 1 : Thus, by Theorem 2.4, wehave H 1 ( f p ) n = ker n S p ( I ) ker n ; p = Z 1 \ I n ; p F I n ; 2 p B 1 which reproves an earlier result of K uhl (see 1.2 of 9]). 5
3 Some applications The purp ose of this section is to give some applications of Lemma 2.1 and Theorem 2.4. Example 3.1 Cyclic standard algebras Let U b e a cyclic standard A -algebra generated by a degree one form x 2 U 1 . Put f : A t ] ! U with f ( t )= x in Theorem 2.4. Then, E ( U ) n = H 1 ( K ( f 1) n ) = (0 : x ) \ U n ; 1 and rt( U ) = min f r 1 j (0 : x r +1 ) = (0 : x r ) g . Example 3.2 Change of base ring Let U be a standard A -algebra and let ' : A ! B be a homomorphism of rings. Take f : V ! U any surjective graded morphism of standard A -algebras in Theorem 2.4. It induces f 1 : V A B ! U A B . Since K ( f 1 p ) n = K ( f p ) n A B , one can deduce rt( U A B ) rt( U ). If ' is at, then H 1 ( K ( f 1 p ) n )= H 1 ( K ( f p ) n ) A B . In particular, rt( U ) = sup f rt( U p ) j p 2 Sp ec( A ) g . If ' is faithfully at, then rt( U A B ) = rt( U ). In particular, via the Nagata morphism A ! A t ] m t ] = B , m a maximal ideal of A , one can always supp ose, when calculating the relation typ e of U ,that A is a lo cal ring of maximal m and residual eld A= m = k innite. Let I be an ideal of A and G ( I ) = n 0 I n =I n +1 its asso ciated graded ring. Since G ( I )= R ( I ) A A=I , then (by3.2) rt( G ( I )) rt( R ( I )) = rt( I ). In 14 ], Valla showed that if rt( G ( I )) = 1, then rt( I )=1 too. Next prop osition is a generalization of that result. Prop osition 3.3 If I is an ideal, there exists E ( I ) n +1 ! E ( I ) n ! E ( G ( I )) n ! 0 , exact sequence of A -modules, for al l n 2 .In particular, if rt( I ) < 1 ,then rt( G ( I )) = rt( I ) . Proof .If 1 R , 1 G , denote the identity on R ( I ), G ( I ), resp ectively, then for each n 1, there is an exact sequence of complexes K (1 R 1) n +1 !K (1 R 1) n !K (1 G 1) n ! 0. Since the 0-th comp onent of the rst morphism is injectiveand H 0 ( K (1 R 1) n +1 ) = 0, we have enough to deduce the exact sequence E ( I ) n +1 ! E ( I ) n ! E ( G ( I )) n ! 0. In particular, if rt( I ) < 1 , one can pro ceed by decreasing induction. Remark 3.4 If rt( I )= 1 , then 3.3 might b e false as Example 4.4 of 11] shows. Note that, as a consequence of next prop osition, we will see that for the irrelevant ideal of a standard algebra hyp othesis rt( I ) < 1 can b e removed. Prop osition 3.5 Let U be a standard A -algebra and let U + = n> 0 U n denote its irrelevant ideal. Take f : W ! U a surjective graded morphism of standard A -algebras with W a symmetric algebra. Given ( n 2 p = 1) or ( n 3 p 2) , the module of eective n -relations of U p + is E ( U p + ) n = M q np ker f q W p ker f q ; p : In particular, E ( U p + ) n = 0 if, and only if, rt( U p + ) n ; 1 . For p = 1 , rt( U ) = rt( U + ) . Moreover, U + is a syzygetic ideal if, and only if, U is a symmetric algebra. 6
Proof .Let g : S U ( U p A U ) ! R ( U p + )be induced by the natural epimorphism of A - mo dules U p A U ! U p + .It is not hard to see K ( g 1) n = i 0 K (1 U p ) np + i . Moreover, if ( n 2 p =1), then np + i 2 and if ( n 3 p 2), then np + i 2 p +1. Therefore, by Theorem 2.4, E ( U p + ) n = H 1 ( K ( g 1) n )= M i 0 H 1 ( K (1 U p ) np + i )= M i 0 ker f np + i W p ker f ( n ; 1) p + i = M q np ker f q W p ker f q ; p : In particular, E ( U p + ) n E ( U p + ) n +1 .Thus, E ( U p + ) n = 0 is equivalentto rt( U p + ) n ; 1. For p = 1 and n 2, E ( U + ) n = i 0 ker f n + i =W 1 ker f n ; 1+ i = i 0 E ( U ) n + i = q n E ( U ) q . In particular, rt( U )=s( E ( U )) = s( E ( U + )) = rt( U + ). Moreover, E ( U + ) 2 = q 2 E ( U ) q = E ( U ). Thus, U + b e syzygetic is equivalentto U b e a symmetric algebra. Now, let us fo cus our attention into the relation typ e of Veronese subrings. Let U be a standard A -algebra. Recall that the p -th Veronese subring of U is dened to be the standard A -algebra U ( p ) = n 0 U np .Clearly, if f : V ! U is a (surjective) graded morphism of standard A -algebras, then it induces f ( p ) : V ( p ) ! U ( p ) another (surjective) graded morphism of standard A -algebras. Lemma 3.6 Let f : V ! U be a surjective graded morphism of standard A -algebras. Then, for al l p 1 , s( E ( f ( p ) )) 1 + (s( E ( f )) ; 1) =p ] ( a ] is the integer part of a ). Proof . Write s( E ( f )) ; 1= pa + b with 0 b<p . So (s( E ( f )) ; 1) =p ]= a . Take n 2+ a . Then ( n ; 1) p pa + p s( E ( f )). Thus, ker f np = V 1 ker f np ; 1 = ::: = V p ker f ( n ; 1) p and hence s( E ( f ( p ) )) 1+ a . Lemma 3.7 Let U be a standard A -algebra and let f : V ! U be a symmetric presentation of U .If ( n 2 p =1) or ( n 3 p 2) ,then the module of eective n -relations of U ( p ) is E ( U ( p ) ) n = ker f np V p ker f ( n ; 1) p : Proof . Take g : S ( V p ) ! U ( p ) induced by f p : V p ! U p in degree one. We have K ( g 1) n = K ( f p ) np . Moreover, if ( n 2 p =1), then np 2, and if ( n 3 p 2), then np 2 p +1. Thus, by Theorem 2.4, E ( U ( p ) ) n = H 1 ( K ( g 1) n )= H 1 ( K ( f p ) np )=(ker f np ) = ( V p ker f ( n ; 1) p ). Lemma 3.8 Let M bean A -module and S ( M ) its symmetric algebra. Then, for al l p 1 , rt( S ( M ) ( p ) ) 2 . Moreover, if p 2 and M is nitely generated, then rt( S ( M ) ( p ) )=1 if, and only if, M is local ly cyclic. Proof . By Lemma 3.7, E ( S ( M ) ( p ) ) n = 0 for all n 3. Thus, rt( S ( M ) ( p ) ) 2. Supp ose p 2 and ( A m k )is lo cal (see 3.2). If M is cyclic, then S ( M ) ( p ) = S ( S p ( M )) and rt( S ( M ) ( p ) ) = 1. Conversely, supp ose M nitely generated, but not cyclic. Take x y part of a basis of M k and x p y p x p ; 1 y in S p ( M ) k . Then, z = x p y p ; x p ; 1 y 7
xy p ; 1 2Z 1 ( f 1 k 1) 2 . Moreover, lo oking at the comp onents of an elementina k -basis of B 1 ( f 1 k 1) 2 , one sees that z = 2 B 1 ( f 1 k 1) 2 .Thus, H 1 ( K ( f 1 k 1) 2 ) 6 = 0, hence (by 3.2) H 1 ( K ( f 1) 2 ) 6 = 0 and rt( S ( M ) ( p ) )=2. Remark 3.9 Let I b e an ideal of linear typ e nitely generated, but not lo cally principal. Then, by Lemma 3.8, rt( I p ) = 2 for all p 2, which reproves 2.6 of 7 ]. Theorem 3.10 Let U be a standard A -algebra. Then, rt( U ( p ) ) max(1 + (rt ( U ) ; 1) =p ] 2) for al l p 1 . Moroever, if U is nitely generated and p 2 , then rt( U ( p ) )=1 if, and only if, U p is local ly generated by a d -sequenceoflength 1. Proof . Let : S ( U 1 ) ! U be the canonical morphism. Put g : S ( S p ( U 1 )) ! S ( U 1 ) ( p ) and f = ( p ) . Then, by Lemma 2.1, rt( U ( p ) ) max(s( E ( f )) s( E ( g ))) and, by Lemmas 3.6 and 3.8, we prove the inequality. Supp ose p 2and U nitely generated. By 3.2, one can supp ose that ( A m k )is a lo cal ring of innite residual eld k . If U p is generated by a d -sequence of length 1, then (by 3.1) rt( U ( p ) ) = 1. Conversely, supp ose rt( U ( p ) ) = 1. Take V = U k , so V ( p ) = U ( p ) k and rt( V ( p ) ) rt( U ( p ) ) = 1. Therefore, V ( p ) is a p olynomial ring of Krull dimension l = ( V p ) = dim V ( p ) =dim V (since V ( p ) V is an integral extension). Take W V a graded No ether normalization (it exists since k is innite, see 1.5.17 of 3]). Thus, dim W = dim V = l and so ; l + p ; 1 p = ( W p ) ( V p )= l , which forces l =1. Hence, ( U p )= ( V p )=1, U p = Ax is cyclic and, by 3.1 again, x is a d -sequence. Remark 3.11 The inequality of 3.10 was rstly proved byBackelin and Frob erg for nitely generated k -algebras (see 2]). Recently, Johnston and Katz showed a very similar statement to that of 3.10, but for U = R ( I ) the Rees algebra of an ideal I (see 7]). Since G ( I ) ( p ) = R ( I ) ( p ) A=I = R ( I p ) A=I , then (by 3.2) rt( G ( I ) ( p ) ) rt( I p ). In particular, for I = U + the irrelevant ideal of a standard algebra U , G ( I ) = U and rt( U ( p ) ) rt( U p + ). Thus, Johnston-Katz's result implies Backelin-Frob erg's result and the inequality of 3.10, when U is a No etherian ring. Nevertheless, the whole Theorem 3.10 can not b e deduced directly from earlier results since, in general, rt( U ( p ) ) 6 =rt( U p + ) as next example shows. Example 3.12 Put U = k x y z ] =J with J = ( x 3 y xy 3 z 4 x 2 y 2 z 3 ). Then, rt( U ) = 7, rt( U (2) )=2 and rt( U 2 + )=3 (remark that max(1 + (rt( U ) ; 1) = 2] 2) =4). Indeed, since E ( U ) n = ker n =U 1 ker n ; 1 , : S ( U 1 ) ! U the canonical morphism, then E ( U ) n = 0 for all n 2, n 6 = 4 7and E ( U ) 4 = k 3 and E ( U ) 7 = k .Thus, rt( U ) = s( E ( U )) = 7. Since ker 8 F 4 (ker ), then, by Lemma 3.7, E ( U (2) ) n =ker 2 n = S 2 ( U 1 ) ker 2( n ; 1) =0 for all n 3. Thus, rt( U (2) ) 2. Moreover, rt( U (2) )=2 since U 2 is not lo cally cyclic (see Theorem 3.10). Besides, using Prop osition 3.5, E ( U 2 + ) 4 = q 8 (ker q = S 2 ( U 1 )ker q ; 2 )=0, so rt( U 2 + ) 3. But, since ker 7 6 = S 2 ( U 1 ) ker 5 , E ( U 2 + ) 3 = q 6 (ker q = S 2 ( U 1 )ker q ; 2 ) 6 = 0. Hence, rt( U 2 + )=3. 8
4 Conditions on the generators In this section wecharacterize, in terms of a system of generators, which ideals have mo dule of eective n -relations zero. Our work here is inspired in previous results by Costa, see 5] and 6]. Concretely, in 6], it was dened a sequence of linear type as a sequence of elements x 1 ::: x d such that the ideals ( x 1 ::: x i ) are of linear typ e for i =1 ::: d . As a consequence of the main result of this section (see 4.7), we get a new characterization of sequences of linear typ e involving annihilator ideals (see 4.9). For an ideal I generated by d elements x 1 :::x d , we will denote by I i 1 :::i s the ideal generated by the x j , where j = 2f i 1 ::: i s g . For an A -mo dule M ,we will denote by A d ( M ) the set of alternating d d matrices with co ecients in M . Lemma 4.1 Let I be generatedby d elements x 1 :::x d and take n 2 . Then, E ( I ) n =0 if, and only if, for al l ( a 1 ::: a d ) 2 ( I n ; 1 ) d with a 1 x 1 + ::: + a d x d = 0 ,there exists ( b ij ) 2A d ( I n ; 2 ) such that 0 B B B @ a 1 a 2 . . . a d 1 C C C A = 0 B B B @ 0 b 1 2 ::: b 1 d ; b 1 2 0 ::: b 2 d . . . . . . . . . . . . ; b 1 d ; b 2 d ::: 0 1 C C C A 0 B B B @ x 1 x 2 . . . x d 1 C C C A : Proof . By Corollary 2.7, E ( I ) n = H 1 ( xt R ( I )) n , where K ( xt R ( I )) n is the n -th comp onent of the Koszul complex asso ciated to the elements x 1 t : : : x d t in R ( I ) = n 0 I n t n . That is, ! ( I n ; 2 ) ( d 2 ) @ 2 ;! ( I n ; 1 ) d @ 1 ;! I n ! 0, with @ 2 ( b 1 2 ::: b 1 d b 2 3 :::b d ; 1 d ) = ( a 1 :::a d ) dened by 0 B B B @ a 1 a 2 . . . a d 1 C C C A = 0 B B B @ 0 b 1 2 ::: b 1 d ; b 1 2 0 ::: b 2 d . . . . . . . . . . . . ; b 1 d ; b 2 d ::: 0 1 C C C A 0 B B B @ x 1 x 2 . . . x d 1 C C C A ,and @ 1 ( a 1 ::: a d ) = a 1 x 1 + + a d x d . Lemma 4.2 Let I be generated by d elements x 1 ::: x d and take n 2 .If E ( I ) n = 0 , then I 1 I n ; 1 : x n 1 = I 1 I n ; 2 : x n ; 1 1 . Proof .If a 2 I 1 I n ; 1 : x n 1 , then ax n 1 = a 2 x 2 + + a d x d , a i 2 I n ; 1 . In particular, (by 4.1) 0 B B B @ ax n ; 1 1 ; a 2 . . . ; a d 1 C C C A = 0 B B B @ 0 b 1 2 ::: b 1 d ; b 1 2 0 ::: b 2 d . . . . . . . . . . . . ; b 1 d ; b 2 d ::: 0 1 C C C A 0 B B B @ x 1 x 2 . . . x d 1 C C C A , b ij 2 I n ; 2 . Thus ax n ; 1 1 2 I 1 I n ; 2 . Remark 4.3 If d = 1, then the necessary condition of Lemma 4.2 becomes 0 : x n 1 =0: x n ; 1 1 , whichisknown to b e sucient to assure E ( I ) n = 0 (see Example 3.1). Lemma 4.4 Let I begenerated by d elements x 1 :::x d ( d 2) and n 2 . If E ( I ) n =0 , then (0 : x 1 ) \ I n ; 1 = 8 > < > : d X i =2 a i x i j a i 2 I n ; 2 x 1 0 B @ a 2 . . . a d 1 C A =( b ij ) 0 B @ x 2 . . . x d 1 C A for ( b ij ) 2A d ; 1 ( I n ; 2 1 ) 9 > = > : 9