aa r X i v : . [ m a t h . A C ] A p r A NOTE ON A SYSTEM OF PARAMETERS
MOHSEN ASGHARZADEHA
BSTRACT . Let u be in p ∈ Assh ( R ) . We present several situations for which ( u ) is (not) in an idealgenerated by a system of parameters. An application is given.
1. I
NTRODUCTION
Let ( R , m , k ) be a noetherian local ring of dimension d . We say a sequence x : = x , . . . , x d of elementsof m is a system of parameters if ℓ ( R / xR ) < ∞ . By I we mean the ideal generated by a system ofparameters. By Assh ( R ) we mean { p ∈ Ass ( R ) : dim R / p = d } . Let p ∈ Assh ( R ) and take u be in p . Question . (See [3, Question 6.4]) Can ( u ) ever be in I ?For the motivation, see [3, Introduction] by Fouli and Huneke. Their calculations strongly suggest theanswer is always ‘no’. For instance, over 1-dimensional rings. Also, over Gorenstein rings the answeris no, because of validity of the monomial conjecture, see Fact 2.3. We extend this by dropping theCohen-Macaulay assumption: Observation . Question 1.1 is not true over quasi-Gorenstein rings.It is easy to see that Question 1.1 is not true in each of the following three situations: i) k [[ X , Y , Z ]] J forsome unmixed ideal J , ii) Cohen-Macaulay rings of multiplicity two, and iii) k [[ X ,..., X n ]]( f , g ) where ( f , g ) is unmixed. Due to Observation 1.2, we pay a special attention to non quasi-Gorenstein rings withnontrivial zero-divisors. This enable us to check Question 1.1 in some new cases. Here, is a sample: Observation . Let P and Q be two prime ideals of S : = k [[ X , . . . , X d ]] generated by linear forms. ThenQuestion 1.1 is not true over R : = SPQ .For a related result concerning powers of a prime ideal, see Proposition 2.17. These observations havean application, see e.g. Corollary 2.16. It may be worth to note that Eisenbud and Herzog predicted thatproduct of ideals of height at least two in a regular ring is not Gorenstein. For an important progress,see [7]. In § n >
0, set R n : = k [ X , Y , Z , W ] m ( XY − ZW , W n , YW ) . In §
4, by mimicking from [9], we show Question 1.1 isnot true over R n if and only if n =
1. The ring R is two-dimensional, generically Gorenstein, Cohen-Macaulay, almost complete-intersection, of type two and of minimal multiplicity three. Also, in § p ∈ Assh ( R ) and u ∈ p such that ( u ) ⊂ I . In the final section we talk a little about a question by Strooker and St ¨uckrad: What is theset-theoretic union of all parameter ideals? Mathematics Subject Classification.
Primary 13C15.
Key words and phrases.
Associated prime ideals; limit closure; multiplicity; quasi-Gorenstein rings; system of parameters.
2. P
OSITIVE SIDE OF Q UESTION
Fact . (See [8, Thorem 14.1]) Let A be a local ring and let a : = a , . . . , a t be a part of system of param-eters. Then dim A / aA = dim A − t . Fact . (Fouli-Huneke) Let R be a 1-dimensional local ring and u be in p ∈ Assh ( R ) . Then ( u ) isnot in an ideal generated by a system of parameters. Proof.
Suppose by way of contradiction that there is a parameter ideal ( x ) such that ( u ) ⊂ ( x ) .Let r be such that ru = aux for some a . Thus, r − ax ∈ ( u ) ⊂ ( x ) . From this, r ∈ ( x ) . So, themap R / ( x ) u −→ R / ( ux ) is injective. In view of [3, Theorem 4.1] we see that ux is parameter. Since ux ∈ p ∈ Assh ( R ) , we get to a contradiction. (cid:3) Fact . (Fouli-Huneke and others) Let R be a Gorenstein local ring, u be in p ∈ Assh ( R ) . Then ( u ) is not in an ideal generated by a system of parameters. Proof.
Fouli and Huneke remarked that the desired claim follows from the validity of monomial conjec-ture. Recently, Andr´e [1] proved this. (cid:3) By µ ( − ) we mean the minimal number of elements that needs to generate ( − ) . Observation . Let ( S , n ) be a regular local ring and J ✁ S be unmixed. Adopt one of the followingsituations: i) dim S <
4, or ii) µ ( J ) <
3. Then Question 1.1 is not true over R : = S / J . Proof. i) In the light of Fact 2.2 we may assume that 1 < dim R <
4. If dim R =
3, then R is regular. Since R is a domain, the claim follows. It remains to assume that dim R = µ ( m ) =
3. It follows thatht ( J ) =
1. Over UFD, any height-one unmixed ideal is principal. Thus, R is hypersurface. It remains toapply Fact 2.3.ii) The case µ ( J ) = µ ( J ) =
2. Since J is unmixed and S is UFDwe deduce that ht ( J ) =
2. In particular, grade ( J , S ) = µ ( J ) =
2. This implies that J is generated by aregular sequence of length two. Thus, R is complete-intersection. Now, the desired claim follows fromFact 2.3. (cid:3) Corollary 2.5.
Question 1.1 is not true over a Cohen-Macaulay local ring R of multiplicity two.Proof.
Recall from Abhyankar’s inequality that µ ( m ) − dim R + ≤ e ( R ) . This implies that µ ( m ) ≤ dim R +
1. Thus, ˆ R is hypersurface, and so the claim follows. (cid:3) By H i a ( − ) we mean the i -th ˇCheck cohomology module of ( − ) with respect to a generating set of a .Also, in the sequel we will use the concept of limit closure. Let y : = y , . . . , y ℓ ⊂ m . Recall that R / ( y ) lim isthe image of R / ( y ) under the isomorphism H ℓ y ( R ) ∼ = lim −→ n R / ( y n , . . . , y n ℓ ) .Example . Let R : = k [[ x , y ]] . Then ( x , xy ) lim = R and ( x , y ) lim = ( x , y ) . In particular, limit closuredoes not preserve the inclusion. Proof.
Set J : = ( x , xy ) and u : = x xy = x y . In the light of Hartshorne-Litchenbaum vanishing,we see H J ( R ) =
0. By definition, J lim = R . One may see this more explicitly: 1 ∈ ( J [ ] : R u ) = (cid:0) ( x , x y ) : R x y (cid:1) . Since x , y is a regular sequence, we have ( x , y ) lim = ( x , y ) . (cid:3) However, by restriction over parameter ideals we have:Observation . Let J ⊂ J be two ideals generated by a part of system of parameters. Then J lim1 ⊂ J lim2 . Proof.
The claim is trivial if the ring is Cohen-Macaulay, and we are going to reduce to this case. Suppose J = ( y , . . . , y m ) . Set y : = y . · · · . y m . The sequence { ( y n + , . . . , y n + m ) : R y n | n ∈ N } is increasing and itsunion is J lim1 . There is an integer n such that J lim1 = ( y n + , . . . , y n + m ) : y n . By a theorem of Andr´e, thereis a big Cohen-Macaulay algebra A over R . In fact, B : = ˆ A is balanced. This yields that y , . . . , y m is aregular sequence over B . Thus, J lim1 = (( y n + , . . . , y n + m ) : B y n ) ∩ R = J B ∩ R . Since B is balanced, thesame argument implies that J lim2 = J B ∩ R . Since J ⊂ J , it follows that J lim1 ⊂ J lim2 . (cid:3) This observation suggests:
Definition 2.8.
For an ideal J of a local ring R , we set J BCM : = { x ∈ R | x ∈ JB for some balanced big Cohen-Macaulay R -algebra B } . Let E R ( k ) be the injective envelop of k as an R-module. A local ring R is called quasi-Gorenstein if H dim R m ( R ) ∼ = E R ( k ) . Here, we use a trick that we learned from [9] : Proposition 2.9.
Let ( R , m , k ) be quasi-Gorenstein, p ∈ Assh ( R ) and let u ∈ p . Then ( u ) * I lim . Inparticular, ( u ) is not in any ideal generated by a system of parameters.Proof. Set A : = R / uR and d : = dim R . Since u ∈ p ∈ Assh ( R ) , we deduce that d = dim A . Let m = p d ⊃ . . . ⊃ p = p be a strict chain of prime ideals of R . By going down property of flatness, thereis a chain m ˆ R = q d ⊃ . . . ⊃ q = : q of prime ideals of ˆ R such that q i lying over p i . In particular, there is a q ∈ Assh ( ˆ R ) lying over p , and so u ∈ q . Since ( ˆ R u ) = ( u ) ˆ R and I lim ˆ R = ( I ˆ R ) lim , without loss ofthe generality we may assume that R is complete, and A is as well.Let y : = y , . . . , y d be any system of parameter of R . Let x be the lift of y to A . Let µ Ax : A / x A → H dx ( A ) be the natural map. By the canonical element conjecture, which is now a theorem, we have µ Ax =
0. Since R is quasi-Gorenstein, H d m ( R ) = E R ( k ) . We set ( − ) v : = Hom R ( − , E R ( k )) . Denote themaximal ideal of A by n . Grothendieck’s vanishing theorem says that H > d n ( − ) =
0. We apply this alongwith the independence theorem of local cohomology modules to observe that H d n ( A ) = H d n ( R ) ⊗ R A = H d m ( R ) ⊗ R A . It turns out that µ Ax = µ Ry ⊗ R A .By definition, the map R / ( y ) lim → H d m ( R ) is injective, and its image is the submodule of H d m ( R ) which is annihilated by ( y ) lim . We have0 = ( µ Ax ) v = Hom R ( µ Ax , E R ( k )) = Hom R ( µ Ry ⊗ R A , E R ( k )) = Hom R ( A , ( µ Ry ) v ) = Hom R ( A , σ Ry ) ,where σ Ry : = (cid:16) R / ( y ) ։ R / ( y ) lim ∼ = −→ Ann E R ( k ) ( y ) lim ֒ → E R ( k ) (cid:17) v .The assignment E ′ Ann R E ′ induces a 1-1 correspondence from submodules E ′ of E R ( k ) to ideals of R . We have E R ( k ) v = R . The mentioned correspondence is given by ker ( E R ( k ) v ։ ( E ′ ) v ) too. Fromthese observations, for any ideal J we have ( Ann E R ( k ) J ) v ∼ = R / J . In particular, we can identify σ Ry , up toan isomorphism, with the composition of R ։ R / ( y ) lim and R / ( y ) lim ֒ → ( R / ( y )) v . It follows that thecomposition Hom R ( A , R ) −→ Hom R ( A , R / ( y ) lim ) −→ Hom R ( A , ( R / ( y )) v ) is nonzero. In particular, the map Hom R ( A , R ) −→ Hom R ( A , R / ( y ) lim ) is nonzero. It turns out that ( u ) * ( y ) lim . To see the particular case, it is enough to note that ( y ) ⊂ ( y ) lim . (cid:3) Remark . The quasi-Gorenstein assumption is needed: Let R be an equidimensional local ring withzero-divisors equipped with a parameter ideal I such that I lim = m . Such a thing exists, see [3, Example6.1]. Let p ∈ Assh ( R ) and let u ∈ p be nonzero. Clearly, ( u ) ⊂ m = I lim . In the next section we willshow that quasi-Gorenstein assumption is needed even in the particular case. Observation . Let R be of depth zero and p ∈ Assh ( R ) . There is a nonzero x ∈ p such that ( x ) * I . Proof.
Without loss of generality we may assume that dim R >
0. Thus, there is y ∈ m \ { p } . Sincedepth ( R ) =
0, there is an x such that m = ( x ) . Since p is prime and xy = x ∈ p . If ( x ) ⊂ I , then we should have m ⊂ I . This implies that R is regular, a contradiction. (cid:3) An R-module K R is called canonical if K R ⊗ R ˆ R ∼ = H dim R m ( R ) v . In the case the ring is Cohen-Macaulay, wedenote it by ω R .Fact . A local ring is quasi-Gorenstein if and only if K R (exists and) becomes free and of rank one. Proposition 2.13.
Let P be a nonzero prime ideal of a local ring ( A , n ) . Then Question 1.1 is not true overR : = AP n . Also, R is quasi-Gorenstein if and only if n is principal.Proof. Let p : = PR (resp. m : = n R ). We have Assh ( R ) = { p } . Let u ∈ p . We may assume that u = pm = u =
0, it follows that m = ( u ) . If ( u ) were be a subset of an ideal I , generatedby a parameter sequence, then we should have m ⊂ I . It turns out that R is regular. But, R is not even areduced ring. This contradiction yields a negative answer to Question 1.1.Suppose n is principal. Then d : = dim R ≤
1. First, assume that d =
0. Since P =
0, we have P = n . If d =
1, since n is principal, it follows that R is a discrete valuation domain (see [8, Theorem 11.7]). Again,since P = P = n . In each cases, P = n . Thus, socle of R is m . Since µ ( m ) = R isquasi-Gorenstein. Now assume that n is not principal. We have two possibilities: i) P = n , or ii) P = n .In the first case, Soc ( R ) = m . Since µ ( m ) = µ ( n ) > R is not quasi-Gorenstein. Then, without loss ofthe generality we may and do assume that P = n . As ˆ R is complete, and in view of [2, (1.6)], K ˆ R exists.Recall that ˆ R ∼ = ˆ A ˆ P ˆ n . We apply this along with Ass ( K ˆ R ) = Assh ( ˆ R ) (see [2, (1.7)]) to deduce thatAss ( K ˆ R ) = Assh ( ˆ R ) ⊂ min ( ˆ R ) = min ( p ˆ R ) = { min ( p ˆ R ) , m ˆ R } ⊂ Ass ( ˆ R ) .Thus, K ˆ R is not free. By Fact 2.12, ˆ R is not quasi-Gorenstein. Recall that a local ring is quasi-Gorensteinif and only if its completion is as well. So, R is not quasi-Gorenstein. (cid:3) Proposition 2.14.
Let P and Q be two prime ideals of S : = k [[ X , . . . , X d ]] generated by linear forms. LetR : = SPQ . Then Question 1.1 is not true over R.Proof.
We may assume that neither P nor Q is zero. Let p (resp. q ) be the image of P (resp. Q ) in R .Then Assh ( R ) ⊂ { p , q } . Let G ( P ) (resp. G ( Q ) ) be the minimal monomial generating set of P (resp. Q ). Let I be a parameter ideal. Suppose first that G ( P ) ∩ G ( Q ) = ∅ . After rearrangement, we mayassume that X ∈ G ( P ) ∩ G ( Q ) . Also, without loss of generality, we set G ( Q ) : = { X , . . . , X i } and G ( P ) : = { X , . . . } . By symmetry, we may and do assume that p ∈ Assh ( R ) . Let u ∈ p and suppose onthe contradiction that ( u ) ⊂ I . Since pq = q ⊂ ( u ) ⊂ I . Recall that { x , . . . , x i } ⊂ m modulo m is k -linearly independent. Since I m ⊂ m , we deduce that { x , . . . , x i } ⊂ I modulo I m is k -linearly independent. In particular, { x , . . . , x i } is part of a minimal generating set of I . From this, x is a parameter element. Thus, x / ∈ S r ∈ Assh ( R ) r ⊂ p ∪ q . This is a contradiction. Then, withoutloss of generality we may assume that G ( P ) ∩ G ( Q ) = ∅ . After rearrangement, we can assume that G ( P ) : = { X , . . . , X m } and G ( Q ) : = { X m + , . . . , X m + n } . We take ℓ : = d − ( m + n ) . Also, by symmetry,we may assume that m ≤ n . We have two possibilities: i) m < n , or ii) m = n .i) Since m < n , we have Assh ( R ) = { p } . Let u ∈ p and suppose on the contradiction that ( u ) ⊂ I .We have q ⊂ ( u ) ⊂ I . It turns out that { x m + , . . . , x m + n } is part of a minimal generating set of I , andso part of a system of parameters. In view of Fact 2.1 we see: ℓ + m = dim S / Q = dim R / q = dim R / ( x m + , . . . , x m + n ) = dim R − n = ( ℓ + n ) − n = ℓ .This implies m =
0, and consequently P =
0. This is a contradiction.ii) The condition m = n implies that Assh ( R ) = { p , q } . The same argument as i) yields the desiredclaim. (cid:3) The behavior of quasi-Gorenstein property under certain flat ring extensions is subject of [2] . Lemma 2.15.
Let A be a complete local ring. Then A is quasi-Gorenstein if and only if A [[ X ]] is quasi-Gorenstein.Proof. By Cohen’s structure theorem, A is quotient of a Gorenstein local ring G . Also, A [[ X ]] is quotientof a Gorenstein local ring G [[ X ]] . In particular, K A and K A [[ X ]] exist, see [2, (1.6)]. Set r : = dim G − dim A = dim ( G [[ X ]]) − dim ( A [[ X ]]) . In view of [2, (1 .6 )] we have K A [[ X ]] ∼ = Ext rG [[ X ]] ( A [[ X ]] , G [[ X ]]) ∼ = Ext rG ( A , G ) ⊗ A A [[ X ]] ∼ = K A ⊗ A A [[ X ]] ( ∗ ) Suppose A [[ X ]] is quasi-Gorenstein. By applying − ⊗ A [[ X ]] A [[ X ]]( X ) along with A [[ X ]] ∼ = K A [[ X ]] ( ∗ ) ∼ = K A ⊗ A A [[ X ]] we deduce that A ∼ = K A . By Fact 2.12 A is quasi-Gorenstein. The converse part follows by ( ∗ ) . (cid:3) The following result inspired from [7] . Corollary 2.16.
Let P and Q be two nonzero prime ideals of S : = k [[ X , . . . , X n ]] generated by linear forms suchthat G ( P ) ∩ G ( Q ) = ∅ and let R : = SPQ . The following are equivalent:i) R is hypersurface,ii) R is complete-intersection,iii) R is Gorenstein,iv) R is quasi-Gorenstein,v) R is Cohen-Macaulay.Proof.
First, we prove that the first four items are equivalent. Among them, the only nontrivial impli-cation is iv ) ⇒ i ) : We assume that R is quasi-Gorenstein. Suppose on the way of contradiction thatone of Q and P is not principal. By symmetry, we may and do assume that P is not principal. Since G ( P ) ∩ G ( Q ) = ∅ , both of PR and QR are minimal prime ideals of R . Recall that quasi-Gorensteinrings are equidimensional. It turns out that ht ( P ) = ht ( Q ) . Let G ( P ) (resp. G ( Q ) ) be the minimalmonomial generating set of P (resp. Q ). Without loss of generality, we set G ( Q ) : = { X , . . . , X ℓ } and G ( P ) : = { X ℓ + , . . . , X ℓ } . Since P is not principal, ℓ ≥
2. The extension k [[ X ,..., X ℓ ]]( X ,..., X ℓ )( X ℓ + ,..., X ℓ ) −→ R is ei-ther the identity map or is the power series extension. Then, in view of Lemma 2.15, we may and do as-sume that 2 ℓ = n . For each i ≤ ℓ , we set a i : = x i + x ℓ + i and we denote the ideal generated by them with I . Since x i a i = x i and x ℓ + i a i = x ℓ + i we deduce that { a i } is a system of parameters. We set ζ : = ∏ ℓ i = a i .Then m ζ ⊂ ( a i ) ℓ i = (here, we need ℓ ≥ I lim = S n ≥ (( a n , . . . , a n ℓ ) : ζ n − ) to deduce that I lim = m . In the light of Proposition 2.9 we see that R is not quasi-Gorenstein. This is acontradiction that we searched for it. Here, we show iv ) ⇒ v ) : It is enough to use iv ) ⇔ i ) .Finally, we show v ) ⇒ i ) : As R is equidimensional and by using the above argument, we deducethat R : = k [[ X ,..., X ℓ ]]( X ,..., X ℓ )( X ℓ + ,..., X ℓ ) → R is either the identity map or is the power series extension. It followsthat R is Cohen-Macaulay. We claim that depth of R is one. The element x + x ℓ + is regular, becausezd ( R ) = S p ∈ Ass ( R ) p = ( x , . . . , x ℓ ) ∪ ( x ℓ + , . . . , x ℓ ) . We need to show depth ( R ) ≤
1. In view of[5, Corollary 3.9], a way to see this, is that its punctured spectrum is disconnected. The closed subsetsV ( x , . . . , x ℓ ) \ { m } and V ( x ℓ + , . . . , x ℓ ) \ { m } are disjoint, non-empty and their union is Spec ◦ ( R ) . Thissays that Spec ◦ ( R ) is disconnected. So, depth ( R ) =
1. Since R is Cohen-Macaulay, ℓ = dim R = depth ( R ) =
1. From this, R is hypersurface. (cid:3) Proposition 2.17.
Let P be a prime ideal of S : = k [ X , . . . , X d ] generated by linear forms and R : = S / P n forsome n > . Let u ∈ q ∈ Assh ( R ) . Then ( u ) is not in an ideal I generated by a homogeneous system ofparameters.Proof. Without loss of generality, we assume that P = n >
1, because the claim is clear overintegral domains. After rearrangement, X ∈ P . Let p : = PR . Then Assh ( R ) = { p } , i.e., q = p . Supposeon the contradiction that ( u ) ⊂ I . Since p n = x n − ⊂ ( u ) ⊂ I . Let i be the smallestinteger such that x i ∈ I . Then i ≤ n −
1. First, we deal with the case x i ∈ m I . There are a j ∈ m and b j ∈ I such that x i = ∑ j a j b j . By looking at this equation in S we get X i − ∑ j A j B j ∈ P n . Since S isUFD and by a degree-consideration, there is an ℓ > X ℓ = A j and B j = X i − ℓ for some j .Since b j ∈ I , we see that x i − ℓ ∈ I . This is impossible, because of the minimality of i . This implies that x i / ∈ m I . In particular, x i is a parameter element, because it is part of a minimal generating set of I . So, x i / ∈ S q ∈ Assh ( R ) q = p , a contradiction. The proof is now complete. (cid:3) Remark . Adopt one of the following situations:i) Let S : = k [[ X , . . . , X d ]] and p be a prime ideal generated by linear forms.ii) Let S be a 4-dimensional unramified complete regular local ring and p be any prime ideal.Let R : = S / p n for some n >
1. Then R is quasi-Gorenstein if and only if p is principal. Proof.
The if part is clear. Now, suppose p is not principal.i) After rearrangement, there is an 0 < ℓ < d such that p = ( X i ) di = ℓ . Set A : = k [[ X ℓ ,..., X d ]]( X i | ℓ ≤ i ≤ d ) n . Its socle isnot principal. Thus, A is not Gorenstein. Recall that R = A [[ X , . . . , X ℓ − ]] . In view of Lemma 2.15 wededuce that R is not quasi-Gorenstein.ii) Suppose on the way of contradiction that R is quasi-Gorenstein, i.e., R = K R . Recall that K R satisfies Serre’s condition S ( ) . Since p is not principal, ht ( p ) ≥
2, i.e., dim R ≤
2. From these, R isCohen-Macaulay. It follows that R is Gorenstein. By [7] this is impossible. (cid:3) In the same vein we have:Example . Let S : = k [ X , . . . , X m ] be a polynomial ring, n be its irrelevant ideal, and let R : = SP n forsome homogeneous prime ideal P of S containing a linear form. Let q ∈ Assh ( R ) and take u be in q .Then ( u ) is not in an ideal I generated by a homogeneous system of parameters.3. M ORES ON NON QUASI -G ORENSTEIN RINGS
The following yields another proof of Corollary 2.16, because PQ = P ∩ Q. Observation . Let ( A , n ) be Cohen-Macaulay. Let I and J be two unmixed ideals of A of same heightand ht ( I + J ) ≥ ht ( I ) +
2. Then R : = AI ∩ J is not quasi-Gorenstein. In fact H dim R m ( R ) decomposable. Proof.
Since A is Cohen-Macaulay, I and J are unmixed and of same height we deduce that d : = dim R = dim A / I = dim A / J . Similarly, dim ( A / I + J ) ≤ d −
2, because ht ( I + J ) ≥ ht ( I ) +
2. Weuse Grothendiek’s vanishing theorem along with a long exact sequence of local cohomology mod-ules induced by 0 → R → A / I ⊕ A / J → A / ( I + J ) → d m ( R ) ∼ = H d m ( A / I ) ⊕ H d m ( A / J ) . By Grothendieck’s non-vanishing theorem, the decomposition is nontrivial. Dueto flat base change theorem, we know completion behaves well with local cohomology modules. We ap-ply Matlis’ functor over ˆ R to see that K ˆ R equipped with a nontrivial decomposition. In particular, K ˆ R isnot of rank one. Thus, ˆ R is not quasi-Gorenstein. So, R is not quasi-Gorenstein. (cid:3) We left to the reader to deduce the third proof of Corollary 2.16 from the following result that its proof is moretechnical than Observation 3.1:Fact . (Hochster-Huneke) Let R be d -dimensional local, complete and equidimensional. Then H d m ( R ) is indecomposable if for any p , q ∈ min ( R ) there are minimal prime ideals p : = p , . . . , p n : = q such thatht ( p i + p i + ) ≤ i . As complete rings are catenary, the following may be considered as a slight generalization of Fact 3.2.
Corollary 3.3.
Let A be catenary and equidimensional. Let p and q be prime ideals of A of same height and ht ( p + q ) ≥ ht ( p ) + . Set R : = A p ∩ q . Then H dim R m ( R ) decomposable. In particular, R is not quasi-Gorenstein.Proof. The assumptions guarantee that d : = dim R = dim A / p = dim A / q = dim A − ht ( p ) , see [8, § ( A / p + q ) ≤ dim A − ht ( p + q ) ≤ d −
2. By the proof of Observation 3.1we get the claim. (cid:3)
Proposition 3.4.
Let ( R , m ) be catenary and equidimensional. Let p , . . . , p n be prime ideals of R of same height,n > and ht ( p i + p i + ) ≥ ht ( p i ) + for all ≤ i < n. Set d : = dim R p ... p n . Then H d m ( R p ... p n ) decomposesinto n nonzero submodules. In particular, R p ... p n is not quasi-Gorenstein.Proof. We argue by induction on n ≥
2. First, we deal with the case n : = p : = p and q = p . The assumption ht ( p + q ) ≥ ht ( p ) + p = q . Since p and q are of sameheight, q " p . Thus, q R p = R p . Also, flat extensions behave well with respect to the intersection ofideals. Consequently, ( p ∩ qpq ) p =
0. This yields that dim ( p ∩ qpq ) < dim ( R pq ) . By Grothendiek’s vanishingtheorem, H > d − m ( p ∩ qpq ) =
0. We look at the short exact sequence 0 → p ∩ qpq → R pq → R p ∩ q →
0. Thisinduces the following exact sequence0 = H d m ( p ∩ qpq ) −→ H d m ( R pq ) −→ H d m ( R p ∩ q ) −→ H d + m ( p ∩ qpq ) = d m ( R pq ) ∼ = H d m ( R p ∩ q ) ∼ = H d m ( R p ) ⊕ H d m ( R q ) .This completes the proof when n =
2. Now suppose, inductively, that n ≥
3, and the result has beenproved for n −
1. By repeating the above argument, we see H d m ( R p ... p n ) ∼ = H d m ( R p ... p n − ) ⊕ H d m ( R p n ) . Bythe inductive step, H d m ( R p ... p n ) ∼ = L ni = H d m ( R p i ) . In the light of Grothendieck’s non-vanishing theorem,H d m ( R p i ) =
0. This completes the proof. (cid:3)
Corollary 3.5.
Let R be any noetherian local ring. Let p , . . . , p n be prime ideals of R of same codimension, n > and dim ( R p i + p i + ) ≤ dim R p i − for all ≤ i < n. Then R p ... p n is not quasi-Gorenstein. Example . The bound ht ( p + p ) ≥ ht ( p ) + p : = ( x ) and p : = ( y ) in R : = k [[ x , y ]] . Corollary 3.7.
Let R be any noetherian local ring. Let p , q be prime ideals of R of same codimension such that dim ( R p + q ) ≤ dim R p − . Then R pq is not Cohen-Macaulay.Proof. On the way of contradiction we assume that R pq is Cohen-Macaulay. Let d : = dim R pq . By Corollary3.5, H d m ( R pq ) equipped with a nontrivial decomposition. The same thing holds for H d m ( c R pq ) . By Matlisduality, ω c R pq decomposes into nontrivial submodules. This is a contradiction. (cid:3) Corollary 3.8.
Adopt the notation of Corollary 3.5. Then R p ... p n is not Cohen-Macaulay. Corollary 3.9.
Adopt the notation of Observation 3.1. Then R : = AI ∩ J is not Cohen-Macaulay.Proof. Combine Observation 3.1 along with the argument of Corollary 3.7. (cid:3)
4. N
EGATIVE SIDE OF Q UESTION
Over a Cohen-Macaulay local ring ( R , m ) we have µ ( m ) − dim R + ≤ e ( R ) . If the equality holds we say Ris of minimal multiplicity. Here, we show the multiplicity two (resp. quasi-Gorenstein) assumption of Corollary2.5 (resp. Proposition 2.9) is important. Also, both assumptions dim S < and µ ( J ) < (resp. 1-dimensionalassumption) of Observation 2.4 (resp. Fact 2.2) are really needed.Example . For each n >
0, set R n : = Q [ X , Y , Z , W ] m ( XY − ZW , W n , YW ) .i) Question 1.1 has negative answer over R n if and only if n = R n is Cohen-Macaulay if and only if n < R is two-dimensional, generically Gorenstein, Cohen-Macaulay, almost complete-intersection,of type two and of minimal multiplicity 3. Proof.
The ring R is hypersurface. By Fact 2.3, we get the claim for n =
1. Then we may assumethat n >
1. We set u : = x and p : = ( x , w ) . Recall that xy ∈ ( z ) , and so xy ∈ rad ( x + y , z ) . Since x = x ( x + y ) − xy = x ( x + y ) − zw we see x ∈ rad ( x + y , z ) . Similarly, y ∈ rad ( x + y , z ) . Also, w ∈ rad ( x + y , z ) , because it is nil. Hence, rad ( x + y , z ) = m . Since Ass ( R ) = { p , ( w , y ) } , dim ( R n ) = { x + y , z } is a parameter sequence.Set P : = Q [ X , Y , Z , W ] m . The free resolution of R over P is 0 → P A −→ P → P → R →
0, where A : = − W X − W − Z Y .Since p. dim P ( R ) =
2, and in view of Auslander-Buchsbaum formula, we deduce that R is Cohen-Macaulay. Suppose n >
2. The primary decomposition of J : = ( XY − ZW , W n , YW ) is given by ( W , X ) ∩ ( W , YW , Y , XY − ZW ) ∩ ( Z , Y , W n ) .Then, Ass ( R n ) = { p , ( w , y ) , ( w , y , z ) } . Since R n has an embedded prime ideal, it is not Cohen-Macaulay.Recall that m / ∈ Ass ( R n ) we deduce that depth ( R n ) = From xy = xy − yzw = y ( xy − zw ) =
0, we conclude that y ∈ ( u ) . We claim that ( y ) = ( u ) .If n = ( ) = ( y ) ∩ ( x , w ) . Now, let n >
2. The onlyprimary components of J that contains X is ( W , X ) . Now we compute the intersection of reminder: I : = ( W , YW , Y , XY − ZW ) ∩ ( Z , Y , W n ) = ( W n , YW , Y , XY − ZW ) .Since w n = yw = xy − zw =
0, the image of I in R n is ( y ) . From this ( x ) = ( y ) . We conclude from y = y ( x + y ) − yx = y ( x + y ) − wz that ( u ) is in an ideal generated by a system of parameters.Since ω R = ( y , w ) R we know that R is generically Gorenstein and of type two. Due to the equality m = ( x + y , z ) m we remark that ( x + y , z ) R is a reduction of m . It turns out that e ( R ) = e ( x + y , z ; R ) .The chain ( x + y , z ) ⊂ ( x + y , z , y ) ⊂ ( x + y , z , y , w ) ⊂ R shows that ℓ ( R / ( x + y , z )) =
3. Since R is Cohen-Macaulay, e ( R ) = e ( x + y , z , v + w ; R ) = ℓ ( R / ( x + y , z , v + w )) =
3. In particular, R is ofminimal multiplicity. (cid:3) Here, we present an example of multiplicity two. In particular, the Cohen-Macaulay assumption of Corollary2.5 is important.Example . Let R : = Q [ X , Y , Z , W , V ] m ( XY − ZW , WV , YW ) . Then Question 1.1 has positive answer over R for certain p and u . Also, e ( R ) = ( R ) = < = dim R . Proof.
Recall that xy ∈ ( z ) , and so xy ∈ rad ( x + y , z ) . Since x = x ( x + y ) − xy we see x ∈ rad ( x + y , z ) ,and y ∈ rad ( x + y , z ) . In view of v = v ( v + w ) we see v ∈ rad ( x + y , z , v + w ) . In the same vein, w ∈ ( x + y , z , v + w ) . In sum, rad ( x + y , z , v + w ) = m . In order to show { x + y , z , v + w } is a parametersequence, we remark that dim R =
3. To see this, we recall Ass ( R ) = { ( v , z , y ) , ( w , x ) , ( w , y ) , ( v , w , y ) } .In fact, the primary decomposition of J : = ( XY − ZW , WV , YW ) is given by ( V , Z , Y ) ∩ ( W , X ) ∩ ( W , Y ) ∩ ( V , W , YW , Y , XY − ZW ) .Since R has an embedded associated prime ideal, it is not Cohen-Macaulay. Set P : = Q [ X , Y , Z , W , V ] m .The projective resolution of R , as a P -module, is given by 0 → P B −→ P A −→ P → P → R →
0, where A : = − YW − WV − V XY − ZW XVY − ZW and B : = ZW − VY .Since p. dim P ( R ) = ( R ) =
2. We set u : = x and p : = ( x , w ) . From xy = xy − yzw = y ( xy − zw ) =
0, we conclude that y ∈ ( u ) . From xyv = xyv − zvw = v ( xy − zw ) = yv ∈ ( u ) . Thus, ( y , yv ) ⊂ ( u ) . We are going to show the reverse inclusion. Theonly primary components of J that contains X is ( W , X ) . Now we compute the intersection of reminder: I : = ( V , Z , Y ) ∩ ( W , Y ) ∩ ( V , W , YW , Y , XY − ZW ) = ( WV , YV , YW , Y , XY − ZW ) .Since wv = yw = xy − zw =
0, the image of I in R n is ( y , yv ) . From this ( x ) = ( y , yv ) .We conclude from y = y ( x + y ) − wz that y ∈ ( x + y , z ) ⊂ ( x + y , z , v + w ) . Since yw = yv = y ( v + w ) − yw = y ( v + w ) ∈ ( v + w ) ⊂ ( x + y , z , v + w ) .These observations yield that ( u ) ⊂ ( x + y , z , v + w ) . The later is generated by a system of parameters.It is easy to see that m = ( x + y , z , v + w ) m . By definition, ( x + y , z , v + w ) is a reduction of m . Recallthat e ( R ) = e ( x + y , z , v + w ; R ) . The following chain ( x + y , z , v + w ) ⊂ ( x + y , z , v + w , y ) ⊂ ( x + y , z , v + w , y , w ) ⊂ R shows that ℓ ( R / ( x + y , z , v + w )) =
3. Since R is not Cohen-Macaulay, e ( x + y , z , v + w ; R ) < ℓ ( R / ( x + y , z , v + w )) . Note that Assh ( R ) is not singleton. We put this along with the associativity formula forHilbert-Samuel multiplicity to deduce that e ( R ) =
1. In view of2 ≤ e ( R ) = e ( x + y , z , v + w ; R ) < ℓ ( R / ( x + y , z , v + w )) = ( R ) = (cid:3) The above ring is not reduced: ( zw ) = .
5. A
REMARK ON THE UNION OF PARAMETER IDEALS
We denote the family of all ideals generated by a system of parameters by Σ . We are interested in S I ∈ Σ I.Parameter ideals may have nontrivial nilpotent elements. This may happen even over Cohen-Macaulay rings. Forinstance over R : = k [[ X , Y ]]( X ) the nilpotent element xy is in the parameter ideal ( y ) . More generally:Remark . Let R be a local ring of positive depth. If S I ∈ Σ I has no nontrivial nilpotent elements, then R is reduced. Indeed, let 0 = y ∈ nil ( R ) , and let x be a regular element. By extending x to a systemof parameters, we see x ∈ S I ∈ Σ I . So, xy ∈ S I ∈ Σ I . Since x is regular, xy = ( S I ∈ Σ I ) ∩ nil ( R ) =
0, we get to a contradiction.
The above depth condition is important:Example . Let R : = k [[ X , Y ]] X ( X , Y ) . Then nil ( R ) = ( S I ∈ Σ I ) ∩ nil ( R ) = Proof.
Clearly, nil ( R ) = ( x ) =
0. Let f ∈ I be nilpotent for some I ∈ Σ . We have f = a x + a y + a y + · · · . We set a : = a + a y + a y + · · · . Then, f = a x + ya , since x = xy =
0. From these f = y a . There is an n ∈ N such that 0 = f n = y n a n . As Ann ( y n ) = ( x ) we deduce that a n ∈ ( x ) . As ( x ) is prime, a ∈ ( x ) . Consequently, f = a x + ya = a x . It is enough to show a =
0. On the way ofcontradiction we assume that a ∈ k ∗ . We conclude that x ∈ I . Let g ∈ m \ ( x ) be such that I = ( g ) . Let c and d be such that g = cx + dy . Take r be such that x = rg . Since g / ∈ ( x ) , we have r ∈ ( x ) . Let s besuch that r = sx . Therefore x = rg = sxg = sx ( cx + dy ) =
0, a contradiction. (cid:3)
Proposition 5.3.
Let I ∈ Σ and let R be one of the following three classes of local rings: i) quasi-Gorenstein, ii) aCohen-Macaulay ring of dimension one, or iii) a Cohen-Macaulay ring of multiplicity two. Then R is an integraldomain if and only if Q ⊂ I for some Q ∈ Spec ( R ) .Proof. The only if part is trivial. Conversely, assume that Q ⊂ I for some Q ∈ Spec ( R ) . Let Q ⊂ Q be a minimal prime ideal. Since R is equidimensional, Q ∈ Assh ( R ) . By definition, there is an x ∈ R such that ( x ) = Q . Suppose on the way of contradiction that Q =
0. Since xQ = x ∈ zd ( R ) = ∪ q ∈ Ass ( R ) q = ∪ q ∈ Assh ( R ) q . There is q ∈ Assh ( R ) such that x ∈ q . Recall that ( x ) = Q ⊂ Q ⊂ I . In the case i), Proposition 2.9 lead us to a contradiction. In the case ii) (resp. iii) it is enough toapply Fact 2.3 (resp. Corollary 2.5). (cid:3) Corollary 5.4.
Let ( R , m ) be as Proposition 5.3 and assume in addition that R has a prime element (e.g., R ishypersurface). Then R is an integral domain if and only if m = S I ∈ Σ I.Proof.
The only if part is trivial. Conversely, assume that m = S I ∈ Σ I . Let p ∈ m be a prime element.Then p ∈ I for some I ∈ Σ . Since the ideal ( p ) is prime, the desired claim is in Proposition 5.3. (cid:3) We denote the set of all unit elements of ( − ) by U ( − ) . The following example presents a connection from S I ∈ Σ I to U ( − ) :Example . Let R : = k [[ X , Y ]] / ( X ) . Then m \ S I ∈ Σ I = U ( R ) x ≃ set U ( R ) . Proof.
Clearly, U ( R ) x ⊂ m \ S I ∈ Σ I . For the reverse inclusion, let f ∈ m \ S I ∈ Σ I . There are a ij ∈ k suchthat f = a x + a y + a xy + a y + · · · . We set a : = a + a x + a y + · · · . Then, f = a x + ya .We use f / ∈ S I ∈ Σ I and the fact that ya is in the parameter ideal ( y ) to conclude a =
0. Recall that f is not a parameter element. We apply this to see f ∈ S p ∈ Assh ( R ) p = ( x ) . We plug this in f = a x + ya to observe that ya ∈ ( x ) . Since y / ∈ ( x ) , we have a ∈ ( x ) . Let r ∈ R be such that a = rx . Then f = a x + ya = a x + yrx = x ( a + ry ) ∈ x U ( R ) , because a ∈ U ( k ) . (cid:3) Acknowledgement .
We used Macaulay2 several times.R
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