aa r X i v : . [ m a t h . A C ] D ec A NOTE ON THE HOMOLOGY OF DIFFERENTIALS
MOHSEN ASGHARZADEHA
BSTRACT . We deal with the complete-intersection property of maximally differential ideals. Also, we con-nect Gorenstein homology of derivations to the Gorenstein property of the base rings. These equipped withsome applications.
1. I
NTRODUCTION
Let R be a commutative noetherian ring. In the case of prime characteristic, each iteration of theFrobenius map defines a new R -module structure on R , and this R -module is denoted by n R . By Peskine-Szpiro functor we mean F n ( M ) = M ⊗ n R . Also, let B be a maximally differential ideal. By G -resolutionof B we mean the Gorenstein resolution of B , i.e., an acyclic complex of totally reflexive modules thatapproximates B . For more details, and for the unexplained definitions see §2. We start with: Question . (See [13, Question 4.9]) If R is Cohen-Macaulay and T • is a minimal G -resolution of B such that F n ( T • ) is acyclic for all n ≥
1, then must B be a complete intersection?In §3 we give negative answers to Question 1.1 from different point of views. Our list of examplescontains zero and prime characteristic cases, zero and higher dimension cases, and the case of integraldomains. Our next goal is to understand the following amazing conjecture: Conjecture . (See [13, Conjecture 3.12]) Suppose that R is a local k -algebra for which the R -moduleDer k ( R ) is finitely generated.a) If Gdim ( Der k ( R )) < ∞ , then R is Gorenstein;b) If CI-dim ( Der k ( R )) < ∞ , then R is a complete intersection.Recall that Maloo [10] presented a nontrivial ring R such that Gdim ( Der k ( R )) = CI-dim ( Der k ( R )) = < ∞ . Toward supporting his conjectures, Miranda-Neto asked: Question . (See [13, Page 11]) Is the Maloo’s ring Gorenstein?In §4 we show Question 1.3 has positive answer by proving that the proposed ring is complete-intersection. Despite this, we show Conjecture 1.2 is not true in the above setting. Then, we try to rescueConjecture 1.2. In this regard, we present a number of cases for which the mentioned conjecture remainstrue. We do this by using different computational methods. We close §4 by presenting a tiny connectionto the rigidity conjecture from deformation theory. Mathematics Subject Classification.
Primary 13D05, 13N15.
Key words and phrases.
Complete-intersection; derivastion; differential ideals; Gorenstein rings; homological dimensions; to-tally reflexive modules.
2. P
RELIMINARY DEFINITIONS
Definition 2.1.
We say that the ideal a is complete intersection if either a = a can be generated byan R -sequence.Here, ( − ) ∗ means Hom R ( − , R ) . Definition 2.2. An R -module M is called totally reflexive provided that:i) the natural map M → M ∗∗ is an isomorphism,ii) Ext iR ( M , R ) = Ext iR ( M ∗ , R ) = i ≥ Gorenstein dimension of M , denoted Gdim R ( M ) , is defined to be the infimum of all nonnegativeintegers n , such that there exists an exact sequence 0 → G n → · · · → G → M →
0, in which each G i isa totally reflexive R -module. We call such a resolution, by a minimal G -resolution.Every finitely generated module over a Gorenstein ring has finite Gorenstein dimension. Moreover,if R is local and Gdim R ( M ) < ∞ , then it follows that Gdim R ( M ) = depth R − depth R ( M ) , and we callit Auslander-Bridger formula. Also, if ( R , m , k ) is local and that Gdim R ( k ) < ∞ , then it follows that R is Gorenstein, and we call it the local-global principal. To see more information we recommend themonograph [2]. Definition 2.3. i) A derivation is an additive map D : R → R such that satisfies D ( ab ) = aD ( b ) + bD ( a ) .ii) If R is a k -algebra, the set of all derivations vanishes at k is denoted by Der k ( R ) .iii) Let ∅ = D be a set of derivations. An ideal I is said to be D -differential if D ( I ) ⊂ I for all D ∈ D .By noetherian condition, there is a maximal D -differential proper ideal with respect to the inclusion.If the ring is local, there is exactly one such ideal for the given D we denote it by B : = B D . Definition 2.4.
An ideal B is called maximally differential ideal if there is a set of derivations ∅ = D such that B = B D .By Ω R / k we mean the module of Kähler differential. It is well-known that Der k ( R ) = Ω ∗ R / k . Let I bean ideal of R and set R : = R / I . We use the conormal sequence0 −→ I ( ) I −→ I / I −→ Ω R / k ⊗ R R −→ Ω R / k −→ I ( ) is the second symbolic power. From now on we restrict the k -algebra R be such that Der k ( R ) is finitely generated as an R -module. To see more informations on the module of differentials we rec-ommend the survey article [5], and the related chapters of the books [4] and [12]. Finally, we would liketo cite the excellent work [17] for prime characteristic methods in local algebra.3. C OMPLETE - INTERSECTION AND DIFFERENTIAL IDEALS
Here, we give negative answer to Question 1.1 by presenting several examples.
Example . Let R : = F [ X , Y ]( X , Y ) . There is a derivation D such that the following holds:i) ( x , y ) = B : = B D ,ii) B is totally reflexive, i.e., T • : = → B id −→ B → G -resolution of B ,iii) F n ( T • ) is acyclic for all n ≥ B is not a complete intersection. Also, there is a derivation D ′ such that B D ′ = ( x ) . Proof. i) We define D : = X ∂ / ∂ X + Y ∂ / ∂ Y . We first check that for D is zero over any of the relations:1) D ( X ) = XD ( X ) = D ( Y ) = YD ( Y ) = D induces a well defined derivation on R . We use the small letter for the residue class of elements in R . Note that D ( x ) = x and D ( y ) = y . If we set m : = ( x , y ) then D ( m ) ⊂ m . Since m is maximal, and inview of the definition, we see that m is the maximally differential ideal with respect to D . So, B = ( x , y ) .ii) The ring R is Gorenstein (in fact, complete-intersection) and is of dimension zero. It follows thatany module is totally reflexive, as claimed.iii) Any linear functor sends an isomorphism to an isomorphism. From this F n ( T • ) is acyclic for all n ≥ m = B is not a complete intersection.If we set D ′ : = x ∂ / ∂ x + ∂ / ∂ y then B D ′ = ( x ) . (cid:3) Remark . The ring is differentially simple with respect to all of derivations, i.e., B Der ( R ) = Example . Let R : = F p [ X ..., X n ]( X p ..., X pn ) . There is a derivation D such that:i) ( x , . . . , x n ) = B : = B D ,ii) B is totally reflexive, i.e., T • : = → B id −→ B → G -resolution of B ,iii) F n ( T • ) is acyclic for all n ≥ B is not a complete intersection.Here, the ring is 1-dimensional and the example is of any characteristic: Example . Let k be any field and let R : = k [ X , Y ]( XY ) . There is a derivation D such that:i) ( x , y ) = B : = B D ,ii) B is totally reflexive, i.e., T • : = → B id −→ B → G -resolution of B ,iii) F n ( T • ) is acyclic for all n ≥ B is not a complete intersection. Proof. i) We define D : = X ∂ / ∂ X − Y ∂ / ∂ Y . We first check that for D is zero over any of the relations: D ( XY ) = X ∂ / ∂ X ( XY ) − Y ∂ / ∂ Y ( XY ) = XY − YX = D induces a well defined derivation on R . Note that D ( x ) = x and D ( y ) = y . If we set m : = ( x , y ) then D ( m ) ⊂ m . Since m is maximal, B = ( x , y ) .ii) The ring R is Gorenstein (in fact, complete-intersection) and is of dimension one. It follows thatany module is of G -dimension at most one. In view of 0 → B → R → R / B → B is of G -dimension zero. In other words, B is totally reflexive.iii) Any linear functor sends an isomorphism to an isomorphism. From this F n ( T • ) is acyclic for all n ≥ m = m is generated by a regular sequence. By definition, B is not a completeintersection. (cid:3) Proposition 3.5.
Let R be any zero-dimensional Gorenstein ring of prime characteristic p. There is a derivationD such that the following holds:i) B is totally reflexive, i.e., T • : = → B id −→ B → is a minimal G-resolution of B ,ii) F n ( T • ) is acyclic for all n ≥ ,iii) B is not a complete intersection.Proof. If R is of the form k [ X ..., X n ]( X p ..., X pn ) then we get the claim, see Example 3.3. Then we may assume R is notof that form. In particular, R is not differentially simple with respect to all of derivations. From this, B is not zero with respect to any subset of derivations. By adopting the proof of Example 3.1 we see all ofclaims are true. (cid:3) Here, we present an example which is an integral domain:
Example . Let R : = F [ T , T , T ] which is 1-dimensional. There is a derivation D such that thefollowing holds:i) ( T , T , T ) = B : = B D ,ii) B is totally reflexive, i.e., T • : = → B id −→ B → G -resolution of B ,iii) F n ( T • ) is acyclic for all n ≥ B is not a complete intersection.Also, there is another derivation D ′ such thata) ( T , T ) = B : = B D ′ ,b) B is totally reflexive, i.e., T • : = → B id −→ B → G -resolution of B ,c) F n ( T • ) is acyclic for all n ≥ B is not a complete intersection. Proof.
Note that R = F [ X , Y , Z ]( XZ − Y , X − Z ) . For the first part, we define D : = Y ∂ / ∂ Y . We need to check that D is zero over any of the relations:1) D ( XZ − Y ) = YD ( Y ) = Y = D ( X − Z ) = D induces a well defined derivation on R . Note that D ( X ) = D ( Z ) = D ( Y ) = Y . We use thesmall letter for the residue class of elements in R . If we set m : = ( x , y , z ) then D ( m ) ⊂ m . Since m ismaximal, B = ( x , y , z ) .ii) The ring R is Gorenstein (in fact, complete-intersection) and is of dimension one. It follows thatany module is of G -dimension at most one. In view of 0 → B → R → R / B → B is of G -dimension zero. In other words, B is totally reflexive.iii) Any linear functor sends an isomorphism to an isomorphism. From this F n ( T • ) is acyclic for all n ≥ m = m is generated by a regular sequence. By definition, B is not a completeintersection.For the second part, we set D ′ : = ∂ / ∂ y . Since D ′ is zero over any of the relations, i.e., it induces awell defined derivation on R . Note that D ′ ( X ) = D ′ ( Z ) = D ′ ( Y ) = D ′ ( Y ) =
1. We combine thiswith the fact that ℓ ( R / ( x , y , z )) = B = ( x , y , z ) . Since xz − y = B = ( x , z ) , and we see the desired claims. (cid:3)
4. D
ERIVATIONS AND THE G ORENSTEIN PROPERTY By µ ( − ) we mean the minimal number of elements that needs to generate ( − ) . Recall that the em-bedding dimension of a local ring ( R , m ) is embdim R : = µ ( m ) . Now, we recall the following: Fact . (See [12, Exercise 21.2]) Let A be a local ring with embdim A = dim A +
1. If A is Cohen-Macaulay, then it is complete-intersection.Here, we affirmatively answer Question 1.3. Proposition 4.2.
Maloo’s ring is complete-intersection, and so Gorenstein.Proof.
Recall that Maloo’s ring is a 1-dimensional noetherian local integral domain which is not regular.In particular, it is Cohen-Macaulay. By looking at the example, we see its maximal ideal generated bytwo elements. Since ( R , m ) is not regular, µ ( m ) =
2. So, embdim R = µ ( m ) = = dim R +
1. In view ofFact 4.1, R is complete-intersection. (cid:3) Here, we show both parts of Conjecture 1.2 are not true:
Example . There is a local k -algebra R for which the R -module Der k ( R ) is finitely generated such thata) Gdim ( Der k ( R )) < ∞ ,b) CI-dim ( Der k ( R )) < ∞ ,c) R is not Gorenstein, and so it is not complete-intersection. Proof.
Let R : = F [ X , Y ]( X , X Y , Y ) . For simplicity, we set k : = F . We set f : = X , f : = X Y and f : = Y .For any f ∈ K [ X , Y ] we set d f : = ∂ ( f ) / ∂ Xdx + ∂ ( f ) / ∂ Ydy . From the conormal sequence, the module ofKähler differential is of the from Ω R / k ∼ = Rdx ⊕ Rdy / ( d f , d f , d f ) ,where Rdx ⊕ Rdy is a free R -module with base { dx , dy } . Also, we have1) d f = ∂ ( X ) / ∂ Xdx + ∂ ( X ) / ∂ Ydy = X dx = d f = ∂ ( X Y ) / ∂ Xdx + ∂ ( X Y ) / ∂ Ydy = Y Xdx + YX dy = d f = ∂ ( Y ) / ∂ Xdx + ∂ ( Y ) / ∂ Ydy = Y dy = Ω R / k is a free R -module of rank 2. Note thatDer k ( R ) = Ω ∗ R / k ∼ = Hom R ( Ω R / k , R ) ∼ = R .From this, Gdim ( Der k ( R )) = CI-dim ( Der k ( R )) = p. dim ( Der k ( R )) = < ∞ ,i.e., a ) and b ) are valid. In order to show c), we compute the socle of R and it is enough to show that itis not one-dimensional. Indeed, Soc ( R ) = ( R m ) ⊇ ( x y , y x ) . (cid:3) We need the following result:
Fact . (Jacobian criterion, see [4, Corollary 16.22]) Let R be an equi-dimensional affine ring (notnecessarily reduced) over a perfect field k . The following assertions are true:a) Suppose that Ω R / k is locally free over R and R is reduced or char k =
0, then Ω R / k has rank d : = dim R . b) The module Ω R / k is locally free over R and of rank d iff R p is a regular local ring for each primeideal p of R .There is a natural way to rescue Conjecture 1.2 by putting some restrictions, e.g., k is of zero charac-teristic and may be R is of finite type over k and etcetera. In what follows we present series of cases forthe validity of Conjecture 1.2. Proposition 4.5.
Let ( S , n , k ) be a regular local ring containing k, of zero characteristic and of dimension d ≥ .Let n > be an integer. Let R : = S n n . The following are equivalent:a) Gdim ( Der k ( R )) < ∞ ,b) n = ,c) R is Gorenstein.Proof. a ) ⇒ b ) : It is easy to see that Der k ( R ) is finitely generated. Then, by Auslander-Bridger formulaGdim ( Der k ( R )) =
0. In particular, Der k ( R ) is (totally) reflexive. By a result of Ramras [15], over thering R any reflexive module is free. Thus, Der k ( R ) is free. In general, freeness of a dual module does notimply the freeness of a module. But, if a ring is of depth zero, this happens, see [15, Lemma 2.6]. Sinceour ring is of depth zero, we have Ω R / k is free. Since char ( R ) = R is regular. This implies that n = b ) ⇒ c ) : This is trivial. c ) ⇒ a ) : Over Gorenstein local rings any finitely generated module is of finite G -dimension. (cid:3) In the same vein, one can prove that:
Proposition 4.6.
Let ( R , m ) be a local algebra of finite type over a field k of zero characteristic such that m = , µ ( m ) > and that m = ( m ) . Then Gdim ( Der k ( R )) < ∞ iff R is Gorenstein. The particular case of the next result follows from Proposition 4.6. Here, we present a direct methodto reprove it.
Proposition 4.7.
Let R be a local algebra of finite type over a field k of zero characteristic such that m = . Then Der k ( R ) syzygetically is equivalent to some nontrivial copies of k, i.e., Syz i ( Der k ( R )) ∼ = Syz j ( k n ) . In particular,the following are equivalent:a) Gdim ( Der k ( R )) < ∞ ,b) R is Gorenstein.Proof. Recall that Syz i ( − ) stands for the i -th syzygy module of ( − ) . First, assume that Syz ( Ω R / k ) = Ω R / k is free. Since char ( R ) = R is regular, and so Syz ( k ) =
0. Then we may assume that Syz ( Ω R / k ) =
0. Since we work with theminimal free resolution, Syz ( Ω R / k ) ⊂ m R β ( Ω R / k ) . We apply Hom ( − , R ) to0 −→ Syz ( Ω R / k ) −→ R β ( Ω R / k ) −→ Ω R / k −→ −→ Ω ∗ R / k −→ R β ( Ω R / k ) d −→ Syz ( Ω R / k ) ∗ .Let S : = im ( d ) ⊂ Syz ( Ω R / k ) ∗ .Claim i) Let L be an R -module such that m L =
0. Then m Hom R ( L , R ) =
0. Indeed, let f ∈ Hom R ( L , R ) and r ∈ m . Also, let ℓ ∈ L . By definition, r f ( ℓ ) = f ( r ℓ ) = f ( ) = Recall that m Syz ( Ω R / k ) ⊂ m ( m R β ( Ω R / k ) ) =
0. In view of Claim i) we deduce that m S ⊂ m Syz ( Ω R / k ) ∗ =
0. If S = Ω ∗ R / k ∼ = R β ( Ω R / k ) is free. Since our ring is of depth zero and in view of [15, Lemma 2.6],we have Ω R / k is free. In view of Jacobian criterion R is regular. Then, without loss of the generality wemay and do assume that S =
0. In particular, S has a R / m -module structure and since the R / m is afield, S = ⊕ n k for some n >
0. We have the following short exact sequence0 −→ Der k ( R ) −→ R β ( Ω R / k ) −→ ⊕ n k −→ ( ∗ ) So, Der k ( R ) ∼ = Syz ( k n ) . Now we prove the particular case: a ) ⇒ b ) : It follows from Syz i ( Der k ( R )) ∼ = Syz j ( k n ) that Gdim ( k ) = Gdim ( ⊕ k ) = ( i − j ) + Gdim ( Der k ( R )) < ∞ . By the local-global principal, R is Gorenstein. So, b ) follows. b ) ⇒ a ) : This implication always holds. (cid:3) By id ( − ) we mean the injective dimension. The previous observation leads us to: Proposition 4.8.
Let R be any artinian algebra of finite type over a field k of zero characteristic. The followingare equivalent:a) id ( Der k ( R )) < ∞ ,b) p. dim ( Der k ( R )) < ∞ ,c) R is regular.Proof. All of things behave well with respect to localization. Then, we may assume the ring is local. Inview of Fact 4.4 we may assume that Ω k ( R ) =
0. Since R is of depth zero its follows that Ω k ( R ) ∗ = a ) ⇒ b ) : Any dual module is a submodule of a free module. Then, there is an exact sequence0 −→ Der k ( R ) f −→ R n −→ coker ( f ) −→ ( ∗ ) Since depth ( R ) = ( Der k ( R )) = depth ( R ) =
0. This implies that ( ∗ ) splits, i.e., Der k ( R ) ⊕ coker ( f ) = R n . By the Krull–Remak–Schmidt theorem, Der k ( R ) is free. So, id ( R ) =
0. In particular, R is Gorenstein. Over Gorenstein rings, finiteness of injective dimension is same as finiteness of projectivedimension (see [8, Theorem 2.2]). So, p. dim ( Der k ( R )) < ∞ . b ) ⇒ c ) : By Auslander-Buchsbaum formula, p. dim ( Der k ( R )) =
0. In view of [15, Lemma 2.6], wesee Ω R / k is free. We apply Jacobian criterion to deduce that R is regular. c ) ⇒ a ) : This is trivial. (cid:3) Question . Let ( R , m ) be an algebra of finite type over a field k of zero characteristic. Find situationsfor which the following are equivalent:a) id ( Der k ( R )) < ∞ ,b) p. dim ( Der k ( R )) < ∞ ,c) R is regular. Remark . i) The existence of finitely generated module of finite injective dimension implies that thering is Cohen-Macaulay.ii) Let R be the Maloo’s ring and recall that R is a non-regular ring of zero characteristic such thatDer k ( R ) is free. In view of Proposition 4.2 we see R is Gorenstein. So, any free module of finite rank isof finite injective dimension. In fact, id ( Der k ( R )) = < ∞ . But, R is not regular.iii) Recall that Maloo’s ring is not of finite type over k . Then part ii) suggests that in Question 4.9 weneed to assume R is of finite type over k . iv) Question 4.9 is not true over rings of prime characteristic. Indeed, we look at R : = F [ X ] / ( X ) .Then Der k ( R ) is free, and since R is Gorenstein, id ( Der k ( R )) = < ∞ . But, R is not regular.v) Let k be an algebraically closed field of zero characteristic, G be a group acts on k -vector space kX ⊕ kY without pseudo-reflection. Let R : = k [[ X , Y ]] G . Then id ( Der k ( R ) ∗ ) < ∞ iff R is regular. Indeed,suppose id ( Der k ( R ) ∗ ) < ∞ . The main idea is to use the fundamental sequence0 −→ ω R −→ A −→ R −→ k −→ ( ∗ ) ,where A is a finitely generated module. Under the stated assumptions, it is shown in [11, Page 180] that A ∼ = Der k ( R ) ∗ . In view of ( ∗ ) we have 0 → ω R −→ Der k ( R ) ∗ → m →
0. Since id ( ω R ) < ∞ we deducethat id ( m ) < ∞ . We left to the reader to check that R is regular (see [8, Page 318]).v) Adopt the notation of Question 4.9 and suppose the ring is reduced and of finite type over k . Theconjectural implication b ) ⇒ c ) was asked independently by Herzog and Vasconcelos. It may be worthto note that this implication is a generalization of the Zariski-Lipman conjecture, see [9].A ring R is called almost complete intersection if there is a regular ring S with an ideal I such that µ ( I ) ≤ ht ( I ) + R = S / I . This is slightly different from [6]. Observation . Let R be a 1-dimensional almost complete intersection local domain which is essen-tially of finite type over a field k of zero characteristic. Suppose Der k ( R ) is nonzero. The following areequivalent:a) id ( Der k ( R )) < ∞ ,b) p. dim ( Der k ( R )) < ∞ ,c) R is regular. Proof.
Since R is homomorphic image of a regular ring, it has a canonical module. Also, R is Cohen-Macaulay. Then the canonical module is denoted by ω R . Since Der k ( R ) = Hom R ( Ω R / k , R ) is a dualmodule, its depth is one. By definition, Der k ( R ) is maximal Cohen-Macaulay. a ) ⇒ b ) : Maximal Cohen-Macaulay modules of finite injective dimension are of the form ω ⊕ nR forsome n >
0. A ring is called quasi-reduced if it is ( S ) and generically Gorenstein. For example, reducedrings are quasi-reduced. Over quasi-reduced rings any dual module is reflexive, see e.g. [1, Fact 5.13]and references therein. Since, R is quasi-reduced and Der k ( R ) is a dual module, we deduce that Der k ( R ) is reflexive. Direct summand of reflexive modules is again reflexive. Since Der k ( R ) ∼ = ω ⊕ nR , it followsthat ω R is reflexive. Now, we recall the following result of Kunz [6, Theorem 1]:Fact: Suppose A is almost complete-intersection. The following are equivalenti) K A is reflexiveii) For all p ∈ Spec ( A ) with ht ( p ) = A p is a complete intersection.We apply this fact to see R is complete-intersection. In particular, R is Gorenstein. Over Gorenstein rings,finiteness of injective dimension is same as finiteness of projective dimension. So, p. dim ( Der k ( R )) < ∞ . b ) ⇒ c ) : Recall that depth ( Der k ( R )) =
1. By Auslander-Buchsbaum formula, Der k ( R ) is free. In thelight of Lipman’s result, R is regular, see [9, Theorem 1]. Since R is 1-dimensional, R is regular. c ) ⇒ a ) : This is trivial. (cid:3) Example . Let R : = k [ T , T , T ] where k is a field of zero characteristic. Then id ( Der k ( R )) = ∞ = p. dim ( Der k ( R )) . Proof.
It is easy to see that R ∼ = k [ X , Y , Z ]( X Y − Z , XZ − Y , YZ − X ) . By definition, R is almost complete-intersection.Since R is not regular, id ( Der k ( R )) = ∞ = p. dim ( Der k ( R )) . (cid:3) A ring R is called purely almost complete intersection if there is a regular ring S with an ideal I suchthat µ ( I ) = ht ( I ) + R = S / I . Corollary 4.13.
Let R be a -dimensional purely almost complete intersection local domain which is essen-tially of finite type over a field k of zero characteristic. Suppose Der k ( R ) is nonzero. Then id ( Der k ( R )) = p. dim ( Der k ( R )) = ∞ .Proof. By a result of Kunz [7], purely almost complete intersection rings are not Gorenstein. Now, thedesired claim is in Observation 4.11. (cid:3)
The next result is an extension of Proposition 4.7:
Proposition 4.14.
Let R be a graded algebra over a field k of zero characteristic. Assume R is Cohen-Macaulayand of minimal multiplicity. The following are equivalent:a)
Gdim ( Der k ( R )) < ∞ ,b) R is Gorenstein.Proof. Assume Gdim ( Der k ( R )) < ∞ . Suppose on the way contradiction that R is not Gorenstein. Applythis along with the Cohen-Macaulay and minimal multiplicity assumptions to deduce that any totallyreflexive module is free. For more details see e.g., [1, Observation 4.17]. Thus, p. dim ( Der k ( R )) < ∞ . Bya result of Platte [14] (also see [5, Theorem 4.5]), R is quasi-Gorenstein. Since R is Cohen-Macaulay, wededuce that R is Gorenstein. This completes the proof of a ) ⇒ b ) . The reverse implication is trivial. (cid:3) Example . Let R : = k [ T , T , T ] where k is a field of zero characteristic. Then Gdim ( Der k ( R )) = ∞ . Proof.
It is easy to see that R / T R ∼ = k [ X , Y ] m . Thus, R is Cohen-Macaulay and of minimal multiplicity.Since R is not Gorenstein, Gdim ( Der k ( R )) = ∞ . (cid:3) Let us compute the previous observations in a prime characteristic case:
Example . Let S : = F [ X , Y ] and set R : = S ( X , Y ) . Then Gdim ( Der F ( R )) = ∞ = id ( Der F ( R )) . Proof.
For simplicity, we set k : = F . We set f : = X , f : = Y and f : = XY . Recall that Ω R / k ∼ = Rdx ⊕ Rdy / ( d f , d f , d f ) where1) d f = ∂ ( X ) / ∂ Xdx + ∂ ( X ) / ∂ Ydy = Xdx = d f = ∂ ( Y ) / ∂ Xdx + ∂ ( Y ) / ∂ Ydy = Ydx = d f = ∂ ( XY ) / ∂ Xdx + ∂ ( XY ) / ∂ Ydy = Ydx + Xdy .Since ( d f ) R = { ( ry , rx ) : r ∈ R } ∼ = R / ( R ( y , x )) = R / m , there is an exact sequence0 −→ R / m −→ R −→ Ω R / k −→ R ( R / m , R ) = { r ∈ R : r m = } = m . Dualizing this yields that0 −→ Der k ( R ) −→ R −→ m −→ Ext R ( Ω R / k , R ) −→ (+) Now, we compute the projective resolution of Ω R / k :. . . −→ R ( y , x ) −→ R ( yx ) −→ R −→ Ω R / k −→ Apply Hom R ( − , R ) to it, yields thatExt R ( Ω R / k , R ) = H (cid:18) R ( y , x ) −→ R ( yx ) −→ R (cid:19) = (+) , we have the following exact sequence:0 −→ Der k ( R ) −→ R −→ m −→ ( ∗ ) Suppose on the way of contradiction Gdim ( Der k ( R )) < ∞ . From ( ∗ ) we deduce that Gdim ( m ) < ∞ .In view of 0 → m → R → R / m → ( k ) < ∞ . It follows from the local-globalprincipal that R is Gorenstein. So, its socle is one-dimensional. On the other hand, Soc ( R ) = ( R m ) = m = ( x , y ) is 2-dimensional. This contradiction shows that Gdim ( Der k ( R )) = ∞ .Suppose on the way of contradiction that id ( Der k ( R )) < ∞ . Then it is injective. In view of ( ∗ ) wehave Der k ( R ) ⊕ m ∼ = R . The left hand side has at least three generators and the right hand side needstwo elements, a contradiction. (cid:3) Here, we work with a non-Cohen-Macaulay ring:
Example . Let R : = F [ X , Y ]( X , XY ) . Then Gdim ( Der F ( R )) = id ( Der F ( R )) = ∞ . Proof.
For simplicity, we set k : = F . It is easy to see that Ω R / k ∼ = Rdx ⊕ Rdy / ( Y dx ) . Note that ( Y dx ) R = { ( ry , 0 ) : r ∈ R } ∼ = R / ( R ( y )) = R / xR . There is an exact sequence0 −→ R / xR −→ R −→ Ω R / k −→ ( † ) Note that . . . −→ R ( x ) −→ R ( y ) −→ R −→ Ω R / k −→ Ω R / k . ApplyHom R ( − , R ) to it, yields the following complex0 −→ Der k ( R ) −→ R ( y ) −→ R x −→ R −→ . . . (+) Note that im (( y )) = y R = R / ( R y ) = R / xR and ker ( x ) = ( x , y ) . So,Ext R ( Ω R / k , R ) = H (cid:18) R ( y ,0 ) −→ R ( x ) −→ R (cid:19) = ( x , y )( y ) = ( ∗ ) Since (+) is left-exact, we have the following exact sequence0 −→ Der k ( R ) −→ R −→ im (( y )) = R / xR −→ ( ∗ , ∗ ) Suppose on the way of contradiction Gdim ( Der k ( R )) < ∞ . Note that depth ( R ) =
0. By Auslander-Bridger formula, Der k ( R ) is totally reflexive. From ( ∗ , ∗ ) and Auslander-Bridger formula, we deducethat Gdim ( R / xR ) =
0. In view of ( † ) , and Auslander-Bridger formula, we deduce that Ω R / k is totallyreflexive. Thus, Ext R ( Ω R / k , R ) =
0. This is in contradiction with ( ∗ ) . We proved that Gdim ( Der k ( R )) = ∞ .In view of ( ∗ , ∗ ) we deduce that id ( Der k ( R )) = ∞ . (cid:3) A 1-dimensional local integral domain is called rigid if it admits no infinitesimal deformations. Inother words, T R =
0. Our interests comes from T R ∼ = Ext R ( Ω R / k , R ) . The rigidity conjecture says thatrigid rings of zero characteristic are regular. For more details see [5]. Proposition 4.18.
Let R be a -dimensional local irreducible k-algebra over a perfect field k of any characteristic.Assume R is of minimal multiplicity. The following are equivalent: a) Ext iR ( Ω R / k , R ) = for all ≤ i ≤ ,b) R is regular.Proof. Since R is reduced, it satisfies Serre’s condition ( S ) . In particular, R is Cohen-Macaulay. a ) ⇒ b ) : First, we show that R is Gorenstein. Suppose on the way of contradiction that R is notGorenstein, and search for a contradiction. We are in the situation of [1, Observation 4.17]. It saysthat over a Cohen-Macaulay ring of minimal multiplicity which is not Gorenstein, the vanishing ofExt iR ( Ω R / k , R ) = ≤ i ≤ Ω R / k is free. One may use Jacobian criterion to see R isregular (or use the Lipman’s result). But, regular rings are Gorenstein. This contradiction allow us toassume R is Gorenstein. In particular, the rigidity conjecture is equivalent with the Berger’s conjecture,see [3] for the statement. We are going to prove the latter one. Since R is of minimal multiplicity, there isa parameter element r such that r m = m . Set R : = R / rR . Then ( R , n ) is a zero-dimensional Gorensteinring with n =
0. Without loss of the generality we may and do assume that n =
0. Then Soc ( R ) = n .Since the socle is one-dimensional, there is some s ∈ m such that n = ( s ) , where s : = s + rR . Since n = m rR we conclude that m = ( s , r ) . In view of Fact 4.1 R is complete-intersection. But, Berger’s conjecture isvalid in this case, and so R is regular, as claimed. a ) ⇒ b ) : This is trivial because Ω R / k is free. (cid:3) We cite to [16] for an additional reference. R
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