THE HARTSHORNE-LICHTENBAUM VANISHING
THEOREM
GABRIEL CHIRIACESCU
1. Cohomological dimension
The folowing definition was introduced by Hartshorne in his famous
paper [6].
Throughout this chapter R will denote a non-trivial commutative
Noetherian ring.
Definition 1.1. Let R be a Noetherian ring and a a proper ideal of it.
We shall define the following number
cd(a, R) := sup{i ∈ N|(∃)R − module M such that Hai (M ) 6= 0}
and we shall call it cohomological dimension of R relatively to the ideal
a.
Remarks. 1) Since R is a Noetherian ring we know that for all i >
ara(a) we have Hai (−) = 0. Thus cd(a, R) < ∞, and so cd(a, R) is
indeed a natural number.
2) The definition tell us that cd(a, R) ≤ ara(a).
3) It is easily to see that we can define cohomological dimension in the
following way
cd(a, R) = min{n ∈ N|Hai (−) = 0, f or all i > n}
4) If R is local with maximal ideal m then cd(m, R) = dim(R), since we
allready know, by Grothendieck’s non-vanishing theorem, that Hmd (R) 6=
i
0 where d = dim(R),
√ and Hm (−) = 0 for all i > dim(R).
5) cd(a, R) = cd( a, R).
6) Since every module is a direct limit of finitely generated sub-modules
we can define cohomological dimension as
sup{i ∈ N|(∃) a f initely generated module M such that Hai (M ) 6= 0}
Proposition 1.2. cd(a, R) = 0 if and only if there exists k ∈ N such
that ak = 0.
Proof: Suppose that there exists k ∈ N such that ak = 0. Then we
have Hai (M ) ∼
ExtiR (R/an , M ) = 0 since the first argument of Exts
= lim
−→
n≥k
is a free module.
For the converse let us observe that ak = 0, for a natural number k,
is equivalent with Γa (R) = R. Indeed, if ak = 0 then 0 : ak = R hence
1
2
GABRIEL CHIRIACESCU
Γa (R) = R. Conversely, if Γa (R) = R then Supp(Γa (R)) = Spec(R).
But we know that Supp(Γa (R)) ⊆ Var(a). It results that Var(a) =
Spec(R) which means that a is a nilpotent ideal.
Thus, under the hypothesis that cd(a, R) = 0, we have to show that
Γa (R) = R. Suppose that this equality is not true. If R = R/ Γa (R) 6=
0 then Ha0 (R) = 0. On the other hand, we have the following isomorphisms Hai (R) ∼
= Hai (R) for all i ≥ 1, therefore Hai (R) = 0 for all i ≥ 0
and so grade(a, R) = ∞. We infer that aR = R which is equivalent
to say that a + Γa (R) = R. Since Ass(Γa (R)) = Var(a) ∩ Ass(R) we
deduce that Var(Γa (R)) ⊆ Var(a). Finally we get that
∅ = Var(a + Γa (R)) = Var(a) ∩ Var(Γa (R)) = Var(Γa (R))
and so Γa (R) = R, which is a contradiction.
Proposition 1.3. Let R be a Noetherian ring and s ∈ N. Then the
following statements are equivalent :
(i) Hai (M ) = 0 for all i > s and all finitely generated modules M .
(ii) Hai (R) = 0 for all i > s.
Proof: (i) ⇒ (ii) is obvious.
(i) ⇒ (ii). Let n = ara(a). We know that Hai (−) = 0, for all i ≥ n + 1.
By using descending induction on i, starting with i = n + 1 we may
assume that Hai+1 (M ) = 0, for all finitely generated modules M . We
have to show that Hai (M ) = 0, where i > s.
Let 0 −→ K −→ Rt −→ M −→ 0 be a finite presentation of M .
From the long exact sequence of local cohomology we get the exact
sequence
Hai (Rt ) −→ Hai (M ) −→ Hai+1 (K)
Since the two outside terms vanish it results that Hai (M ) = 0.
Corollary 1.4. cd(a, R) = sup{i ∈ N|Hai (R) 6= 0}.
Proposition 1.5. (i) Let A be a Noetherian ring, a an ideal of it and
B a Noetherian A-algebra. Then cd(aB, B) ≤ cd(a, A).
(ii) We have the following equalities:
cd(a, A) = cd(aAred , Ared ) = max{cd(a(A/p), A/p)| p ∈ Min(A)}
Proof. (i) If we denote cd(a, A) = n then Hai (M ) = 0 for all i > n and
for all modules M , in particular for the A-module B we have Hai (B) =
i
0. But 0 = Hai (B) ∼
(B) for all i > n so we get cd(aB, B) ≤ n.
= HaB
(ii) Since we have the following surjections A → Ared → A/p for all
p ∈ Spec(A), we can deduce, from (i), that
cd(a, A) ≥ cd(aAred , Ared ) ≥ cd(a(A/p), A/p)
Let n = max{cd(a(A/p), A/p)| p ∈ Min(A)}. It is enough to show
that n = cd(a, A). From (i) we know that n ≤ cd(a, A). Now, A as an
THE HARTSHORNE-LICHTENBAUM VANISHING THEOREM
3
A-modul has a filtration:
0 = I0 ⊆ I1 ⊆ . . . ⊆ It = A
such that Ij /Ij−1 ∼
= A/pj where pj ∈ Spec(A). If qj is a minimal prime
ideal contained in pj then cd(A/pj ) ≤ cd(A/pj ) ≤ n since we have the
natural homomorphism A/pj → A/qj . We deduce that Hai (Ij /Ij−1 ) =
0, for all j = 1, . . . , t and all i > n. Since I1 ∼
= A/p1 it results that
Hai (I1 ) = 0 for all i > n. Applying induction on j to the following
exact sequences :
0 −→ Ij−1 −→ Ij −→ Ij /Ij−1 −→ 0
and since It = A, we deduce, from the long exact sequence of local
cohomology which arises from the above exact sequence, that Hai (A) =
0 for all i > n.
2. Preparatory lemmas
Before proving the first lemma we shall recall a standard fact about
injective modules. Let p be a prime ideal of R and E(R/p) the injective
envelope of R/p. If a is an ideal of R then the functor Γa (−) acts on
E(R/p) as follows:
E(R/p) if I ⊆ p
Γa (E(R/p)) =
0
if I 6⊆ p
Indeed, if we denote by E the injective envelope E(R/p), then, in
the case
p, S
we have 0 : pn ⊆ 0 : an for all n ∈ N. Therefore
S a ⊆
E = (0 : pn ) ⊆ (0 : an ) = Γa (E) and so Γa (E) = E.
n
n
If a 6⊆ p then there is x ∈ a − p so that multiplication by xn induces
an automorphism of E for all n ∈ N. Let e ∈ Γa (E) then there exists
n ∈ N such that an · e = 0 and so xn e = 0. Therefore e = 0 and
Γa (E) = 0.
Now,
L let J be an injective module over the Noetherian ring R. Let
J=
E(R/pα ) be the decomposition of J into indecomposable injecα
tive modules. By the above
L remark and from the aditivity of Γa (−)
it results that Γa (J) =
E(R/pα ) which is an injective module. In
a⊆pα
L
particular, for an element f ∈ R we have Γ(f ) (J) =
E(R/pα ).
f ∈pα L
On the other hand, for an element f ∈ R, one has Jf =
E(R/pα ),
f 6∈pα
since
E(R/p)S =
ERS (RS /pS ) if p ∩ S = ∅
0
if p ∩ S 6= ∅
and for a prime ideal p such that p ∩ S = ∅ we have ER (R/p) ∼
=
ERS (RS /pS ) and this isomorphism is over R and also over RS .
4
GABRIEL CHIRIACESCU
Lemma 2.1. Let f ∈ R, and let J be an injective R-module. Then the
following sequence of R-modules is split exact:
0 −→ Γ(f ) (J) −→ J −→ Jf −→ 0
Application of the aditive functor Ga (−) to the above exact sequence
yields a further split exact sequence:
0 −→ Γa+(f ) (J) −→ Γa (J) −→ Γa (Jf ) −→ 0
Proof: Using the remark above it is clear that the first sequence is split
exact. For the second sequence, we recall that Γa1 (Γa2 (−)) = Γa1 +a2 (−)
for two ideals a1 , a2 of the ring R. Therefore Γa (Γ(f ) (−)) = Γa+(f ) (−)
and the second sequence is split exact.
Lemma 2.2. Let f ∈ R and let M be an R-module. Then, there is a
long exact sequence of R-modules
0
0
0
0 −→ Ha+(f
) (M ) −→ Ha (M ) −→ Ha (Mf )
−→
···
i
i
i
−→ Ha+(f
) (M ) −→ Ha (M ) −→ Ha (Mf )
i+1
−→ Ha+(f
···
) (M ) −→
such that, whenever ϕ : M −→ N is a homomorphism of R-modules,
the following diagram
i
Ha+(f
) (M )
- H i (M )
a
(ϕ)
Hi
a+(f )
?
i
Ha+(f
) (N )
i (ϕ)
Ha
?
- H i (N )
a
-H i (Mf )
-H i+1 (M )
a+(f )
i (ϕ )
Ha
f
H i+1 (ϕ)
a+(f )
?
-H i (Nf )
?
- H i+1 (N )
a+(f )
a
a
is commutative for all i ∈ N.
Proof: Let
α
d1
di
I • : 0 −→ M −→ I 0 −→ I 1 −→ · · · −→ I i −→ I i+1 −→ · · ·
be an injective resolution of M where α is the augmentation homomorphism. Similarly, we have an exact sequence for N
β
δ1
δi
J • : 0 −→ N −→ J 0 −→ J 1 −→ · · · −→ J i −→ J i+1 −→ · · ·
By standard facts of homological algebra we know that there is a chain
map from the first exact sequence in the second one. It is obvious from
the first lemma of this section that there exists a commutative diagram
0
- Γa+(f ) (I • )
- Γa (I • )
- Γa ((I • )f )
-0
0
?
- Γa+(f ) (J • )
?
- Γa (J • )
?
- Γa ((J • )f )
-0
THE HARTSHORNE-LICHTENBAUM VANISHING THEOREM
5
of complexes of R-modules and chain maps of such complexes such
that, for each i ∈ N, the sequence
0 −→ Γa+(f ) (I i ) −→ Γa (I i ) −→ Γa ((I i )f ) −→ 0
is exact, and a similar property holds for the lower row. Thus the rows
of the above diagram give rise to two long exact sequences of local cohomology modules and a chain map of the first long exact sequence into
the second. Since, by the first lemma of this section, (I • )f provides an
injective resolution of Mf as R-module and (J • )f provides an injective
resolution for Nf , all the claims follows.
Lemma 2.3. Let R be a local domain with dim(R) = n. We suppose
that the following set
B = {b | b ideal of R, Han (R) 6= 0 and dim(R/b) > 0}
is non-empty. Let p a maximal member of B.
Then p is a prime ideal and dim(R/p) = 1.
Proof: Suppose that the statement “p a prime ideal and dim(R/p) = 1”
is false. Then, there is a prime ideal q such that p ⊂ q and dim(R/q) =
1. Let f ∈ q − p. From the previous lemma we have the following exact
sequence
n
n
n
Hp+(f
) (R) −→ Hp (R) −→ Hp (Rf ) −→ 0
By the Independence of Local Cohomology Theorem we have Hpn (Rf ) ∼
=
n
HpRf (Rf ). Since f is contained in the maximal ideal of R one has that
dim(Rf ) < n. By Grothendieck’s Vanishing Theorem it follows that
n
HpR
(Rf ) = 0 and so Hpn (Rf ) = 0. Therefore we have the exact sef
quence
n
n
Hp+(f
) (R) −→ Hp (R) −→ 0
n
Since p ∈ B we have Hpn (R) 6= 0 so that Hp+(f
) (R) 6= 0 and, in addition,
dim(R/p + (f )) > 0. We deduce that p + (f ) ∈ B and, since p + (f )
properly contains p, we have obtained a contradiction to the maximality
of p.
Lemma 2.4. Suppose that (R, m) is local ring, and (R0 , m0 ) is a local
subring such that R is a finitely generated as an R0 -module. Let p
a prime ideal of R be such that dim(R/p) = 1. Then, there is an
0
element
p f ∈ R such that, if B denotes the subring R [f ] of R, such
that (p ∩ B)R = p.
Proof: Of course, R is an integral extension of R0 since it is a finite
extension of R0 . By the Incomparability Theorem for integral extension, p must be a minimal prime over (p ∩ R0 )R. Let p2 , p3 , . . . , pt the
other minimal prime ideals over (p ∩ R0 )R (there may be none). By the
t
S
Prime Avoidance Theorem, there is f ∈ p −
pi . If B = R0 [f ] is the
i=2
6
GABRIEL CHIRIACESCU
R0 algebra generated by f and, since f is integral over R0 , we deduce
that B is finite over R0 and so integral over R0 . Will be enough to show
that p is the one and only minimal prime of (p ∩ B)R. It certainly
is one, as we can see by applying again the Incomparability Theorem.
Let us suppose that q is another minimal prime ideal of (p ∩ B)R,
q 6= p. Obviously q 6= m since otherwise q will be not minimal. Since
R0 ⊂ R is an integral extension it follows that q ∩ R0 6= m0 . Since
p ∩ R0 ⊆ p ∩ B ⊆ q it follows p ∩ R0 ⊆ q ∩ R0 ⊂ m0 . Since R/p is an
integral extension of R0 /p ∩ R0 it follows that dim(R0 /p ∩ R0 ) = 1 and
so p ∩ R0 = q ∩ R0 . Another use of Incomparability Theorem shows us
that q must be a minimal ideal of (p ∩ R0 )R and so q ∈ {p2 , . . . , pt }.
But f ∈ p ∩ B ⊆ (p ∩ B)R ⊆ q and we get a contradiction because
t
S
f ∈p−
pi .
i=2
The following lemma represents a well known result so we shall omit
its proof.
Lemma 2.5. Let R be a subring of the integral domain S, and suppose
that R is integrally closed and S = R[u] where u ∈ S is integral over
R. Let K be the field of fractions of R. Then u is algebraic over
K and its minimal polinomial f over K belongs to R[X]. Also, if
φ : R[X] −→ R[u] is the surjective homomorphism sending X to u,
then φ induces an isomorphism of R-algebras
∼
=
φ : R[X]/f R[X] −→ R[u] = S
Lemma 2.6. Let (R, m, k) be a Gorenstein local ring with residual field
at m equal to k. Then depth(M ) > 0 if and only if ExtnR (M, R) = 0
Proof: Since R is Gorenstein and dim(R) = n it follows that idR (R) =
n.
(⇒) There is an element r ∈ m such that r is a non-zerodivisor on
r
M . The exact sequence 0 −→ M −→ M −→ M = M/rM −→ 0
induces an exact sequence
r
ExtnR (M , R) −→ ExtnR (M, R) −→ ExtnR (M, R) −→ 0
n
n
since Extn+1
R (M , R) = 0. We deduce that r ExtR (M, R) = ExtR (M, R)
n
and by Nakayama’s lemma we get ExtR (M, R) = 0.
(⇐) Since R is Gorenstein it follows that ExtnR (k, R) ∼
= k 6= 0. We
claim that m 6∈ Ass(M ) which is equivalent with depth(M ) > 0. Indeed
if m ∈ Ass(M ) then we have an exact sequence
0 −→ k −→ M −→ C −→ 0
of R-modules which induces an exact sequence
ExtnR (M, R) −→ ExtnR (k, R) −→ Extn+1
R (C, R) = 0
which provides a contradiction.
The following result is due to C.Chevalley.
THE HARTSHORNE-LICHTENBAUM VANISHING THEOREM
7
Lemma 2.7. Let (R, m) be a local ring, m-adically
complete and (an )n∈N
T
a descending sequence of ideals such that
an = 0. Then, for each
n∈N
n ∈ N, there exists t(n) ∈ N such that at(n) ⊆ mn .
Proof: Suppose that the conclusion is false. Thus there exists an r ∈ N
such that am 6⊂ mr , for all m ∈ N. Hence am 6⊂ mn for all n ≥
r. For each n the ring R/mn is Artinian. Therefore the sequence of
am + mn
ideals am (R/mn ) =
eventually stabilizes. Therefore there is
mn
a natural number t(n) such that at(n) + mn = al + mn , for all l ≥ t(n).
It is clear that we can choose t(n) < t(n + 1) for all n ≥ r and so
at(n) ⊆ at(n) + mn = at(n+1) + mn
for all n ≥ r. Starting with an element xr ∈ at(r) − mr one can find
an element xr+1 ∈ at(r+1) such that xr − xr+1 ∈ mr . Therefore there
exists a sequence (xn )n∈N with the property that xn − xn+1 ∈ mn and
xn ∈ at(n) for all n ≥ r. It follows that (xn ) is a Cauchy sequence and,
since R is a complete ring, there exists lim xn := x∗ ∈ R. For each
n→∞
n ≥ r we have that xn , xn+1 , . . . ∈ at(n) . Since at(n) is a T
closed ideal
one has x∗ ∈ at(n) for all n ≥ r, so we deduce that x∗ ∈
at(n) = 0.
n≥r
On the other hand, since xn − xr ∈ mr for all n ≥ r, it follows that
xr ∈ mr , which is a contradiction.
Lemma 2.8. Let (R, m) be a complete local domain, and let p be a
prime ideal such that dim(R/p) = 1. Then:
(i) For each n ∈ N, there exists a positive integer t which depends
on n such that p(t) ⊆ mn .
(ii) The family (p(n) )n∈N is a system of ideals of R over N and the
negative strongly connected sequence of covariant functors
!
Hpi (−) i∈N and
lim ExtiR (R/p(j) , −)
−→
j
i∈N
are isomorphic.
Proof: (i) It is enough to prove thatTall conditions of Chevalley’s theorem are fulfilled. We know that
p(n) = ker(R → Rp ) = 0 and
n∈N
(n)
(p )n∈N is a descending sequence of ideals.
(ii) Since (p(n) )n is an inverse family of ideals will be enough to show
that for each n there exists a nonnegative integer t(n) such that p(t(n)) ⊆
pn ⊆ p(n) . The inclusion pn ⊆ p(n) is clear. For the second inclusion
we have to observe that p is the unique minimal prime ideal of pn ,
and p(n) is the uniquely determined p-primary component in any minimal primary decomposition of pn . Since dim(R/p) = 1 it follows that
Ass(R/pn ) ⊆ {p, m} It results that either pn = p(n) and then we have
8
GABRIEL CHIRIACESCU
nothing to prove, or pn = p(n) ∩ q, where q is a m-primary ideal and so
there exists h ∈ N such that mh ⊆ q. It follows that p(n) ∩ m ⊆ pn . By
using Chevalley’s theorem we deduce that there exists t = t(h) such
that p(t) ⊆ mh . Therefore we get p(n) ∩ p(t) = p(max{t,n}) ⊆ pn ⊆ pn ⊆(n) .
By using a result proved in a previous chapter we know that the following negative strongly conected sequence of functors
!
!
∼
Hpi (−) i∈N ∼
lim ExtiR (R/pj , −)
lim ExtiR (R/p(j) , −)
= −→
= −→
j
i∈N
j
i∈N
are isomorphic.
3. The proof of Hartshorne-Lichtenbaum vanishing
theorem
The following statement is known as the local version of the HartshorneLichtenbaum Vanishing Theorem
Theorem 3.1. Let (R, m) be a local ring with dimR = n and a a proper
ideal of it. The following statements are equivalent:
(i) Han (R) = 0 , and so cd(a, R) < n.
b with dim R/P
b
(ii) For each prime ideal P ∈ Spec(R)
= n, we have
b R
b + P) > 0
dim(R/a
.
Proof: (i)⇒(ii) Assume that Han (R) = 0 and that there exists a prime
b such that dim(R/a
b R
b + P) = 0. Since R
b is a flat R-algebra
ideal P of R
b∼
b and so
we have, by the flat base change, that Han (R) ⊗ R
= HanRb (R)
b = 0. The ideal mR
b is the maximal ideal of R
b and the ring
H n (R)
b
aR
b
b
b + P)/P
R/P
is local with maximal ideal mR/P
and the ideal (aR
b
is mR/P-primary
ideal. By the non-vanishing theorem it follows that
n
b
b
H
(R/P)
6= 0 because dim(R/P)
= n. By the Independence
b
(aR+P)/P
n
∼
b
b
Theorem we get HaRb (R/P)
(R/P)
6= 0. By using the
= H(a
b
R+P)/P
b 6= 0 which
properties of cohomological dimension it follows that HanRb (R)
is a contradiction.
b R
b + P) > 0, for all P ∈ Spec R
b with
(ii) ⇒ (i) Suppose that dim(R/a
n
b
b is faithfully flat algebra over
dim(R/P)
= n and Ha (R) 6= 0. Since R
n
b∼
b 6= 0.
R it follows that Ha (R) 6= 0 if and only if Han (R) ⊗ R
= HanRb (R)
So we may suppose that R is a complete local ring. By using the
properties of cohomological dimension proved in the first section we
have that
n = cd(a, R) = max{cd((a(R/p), R/p) |p ∈ MinR}
THE HARTSHORNE-LICHTENBAUM VANISHING THEOREM
Let p ∈ Min(R) such that cd((a(R/p), R/p) = n. Since
9
R
=
a+p
R/p
we may suppose that R is an integral domain. By a previa + p/p
ous lemma we know that there exists p ∈ Spec R with Hpn (R) 6= 0 and
dim(R/p) = 1. By the structure theorem of complete local rings there
exists a complete regular local ring (R0 , m0 ) ⊆ (R, m) such that R is a
finitelly generated R-module. Therefore p
there is an element f ∈ R such
that, if we denote by B = R0 [f ], then (p ∩ B)R = p. Since R is a
finite extension of B and B is a finite extension of R0 we deduce that B
is a complete local ring with maximal ideal m∩B and dim B = n. Since
R0 is a normal domain it follows that there exists a monic polynomial
h ∈ R0 [X] such that
R0 [X]/hR0 [X] ∼
= R0 [f ] = B
Since R0 [X] is regular and h is a non-zerodivisor
in R0 [X] it follows
p
that B is a Gorenstein local ring. Since (p ∩ B)R = p we deduce
n
that H(p∩B)R
(R) 6= 0 and so, by Independence Theorem, it follows
n
that Hp∩B (R) 6= 0. Applying again the properties of cohomological
n
(B) 6= 0. Let us denote by q = p ∩ B.
dimension we deduce that Hp∩B
We have dim(B/q) = dim(R/p) = 1 since R is a finite extension of B.
On the other hand since B is a Gorenstein, local and complete ring
with dim(B/q) = 1 one has the following isomorphism
Hqn (B) ∼
ExtnB (B/q(j) , B)
= lim
−→
j
Since q(j) is q-primary ideal it follows that depth(B/q(j) ) > 0. We get
that ExtnB (B/q(j) , B) = 0 for all j ∈ N. Thus Hqn (B) = 0 which is a
contradiction.
4. A second proof of HLVT
In this section we shall present a second proof of Hartshorne-Lichtenbaum
Vanishing Theorem which esentially uses the Local Duality Theorem.
The next lemma is elementary but is very useful for our purpose.
We need some notations and remarks.
T
Let R be a Noetherian ring and let (0) =
qj be a irredundat
j∈ΛR
√
primary decomposition of (0) where ΛR is a finite set and qj = pj ∈
Ass(R). For two ideals a and b of R we define the following set
p
def
ΛR (b, a) = {j ∈ ΛR | Var(pj +a)∩D(b) 6= φ} = {j ∈ ΛR | b 6⊂ pj + a}
where D(b) := Spec(R) − Var(b). In the case b = R one has
def
ΛR (R, a) = { j ∈ ΛR | Var(pj +a) 6= φ} = { j ∈ ΛR | pj +a 6= R} = ΛR (a)
We have the following straightforward properties:
10
GABRIEL CHIRIACESCU
a)ΛR (b, a) = ΛR (b + a, a).
b) ΛR (b, a) ∩ ΛR (c,
= ΛR (b + c, a).
√a) √
c) ΛR (b, a) = ΛR ( b, a)
d) For an element f ∈ R, if we denote by σ : Spec(Rf ) → Spec(R)
the canonical map given by contraction, then σ(ΛRf (af )) = ΛR ((f ), a)
where σ is induced by σ in an obvious way.
b is the a-adic
If S is the multiplicative set {1 + x | x ∈ a} and R
completion of R, then it is well known that Ker(R → RS ) = Ker(R →
b = T an . Since there is a bijection of sets beetwen Spec(RS ) and the
R)
n
set {p ∈ Spec(R)
| p + a 6= R}, it results that we have the equality:
T
T
n
a =
qj .
n
j∈ΛR (a)
Let b := (f1 , f2 , . . . , ft ) be an ideal of R and let α and β be the
β
α
cf that can be defollowing homomorphisms R → ⊕ti=1 Rfi → ⊕ti=1 R
i
βi
αi
cf
duced from the canonical homomorphisms R → Rfi and Rfi → R
i
cf means the af -adic completion of Rf . Let us denote by
where R
cf .
ϕ = β ◦ α : R → ⊕t R
i=1
i
Lemma 4.1. With the above notations we have
T
√
(
qj if b 6⊂ a
Ker(ϕ) = j∈ΛR (b,a)
√
R
if b ⊆ a
Ker(ϕ) neither depends on the choice of the irredundant primary
decomposition of (0) nor on the choice of the system of generators of
b.
In particular,√if (R, m) is a local ring and a + b is m-primary ideal
such that b 6⊂ a, then Ker(ϕ) is the intersection of those p-primary
ideals, p ∈ Ass(R), with the property that dim(R/p + a) > 0.
Proof: By induction on t we may reduce our considerations to the case
√
cf = 0 and Ker(ϕ) = R. If f 6∈ √a, then
where t = 1. If f ∈ a then R
\
\
\
Ker(ϕ) = α−1 (Ker(β)) =
anf ∩ R =
(qj )f ∩ R =
qj
n
j∈ΛRf (af )
Now, if we denote by a : b∞ =
S
j∈ΛR ((f ),a)
(a : bs ), then
s
a : b∞ =
t
\
(a : fi∞ ) =
i=1
t
\
(afi ∩ R)
i=1
T
T
On the other hand, Ker(β) = ( anf1 , . . . , anft ), therefore we get
n
n
t
\
\
\\
\
n
n
Ker(ϕ) = α ( af1 , . . . , aft ) =
(anfi ∩ R) = (an : b∞ )
−1
n
n
n i=1
n
THE HARTSHORNE-LICHTENBAUM VANISHING THEOREM
11
The last part of the lemma can be deduced from the fact that
ΛR (b, a) = ΛR (b + a, a) = ΛR (m, a) = {j ∈ ΛR | dim(R/pj + a) > 0}
Let (R, m, k) be a local ring and M an R-module. If E := ER (k)
is the injective envelope of k we shall denote by M ∨ the Matlis dual
of M . With these preliminaries we can prove the main result of this
section.
Proposition 4.2. Let (R, m, k) be a Gorenstein local and complete ring
with dim(R/a) = t and let f = f1 , . . . , ft be a system of parameters for
R/a.
(i) We have the following exact sequence:
0 −→ Had (R)∨ −→ R −→
t
M
cf
R
i
i=1
cf is the af -adic completion of Rf .
where R
i
i
T i
∨
d
qj
(ii) Ha (R) =
(iii)
j∈Λ(m,a)
d
Ass(Ha (R)∨ ) = Ass(R)
− {pj | j ∈ ΛR (m, a)}
Proof: (i) Let b be the ideal (f1 , . . . , ft ). Then, for any n ∈ N we have
an exact sequence, as a part of the Čech complex associated to the
R-module R/an and to the elements f1 , . . . , ft ,
0 −→
Hb0 (R/an )
n
−→ R/a −→
t
M
(R/an )fi
i=1
Taking the inverse limit and taking into account that lim is a left exact
←−
functor we get:
0 −→
lim Hb0 (R/an )
←−
n
−→ R −→
t
M
cf
R
i
i=1
By the Ring Independence Theorem we have
H 0 (R/an ) ∼
= H 0 n (R/an ) ∼
= H 0 (R/an )
b
(b+a )
m
Applying Local Duality Theorem we get
lim Hb0 (R/an ) ∼
H 0 (R/an )
= lim
←−
←− m
n
n
∼
HomR (ExtdR (R/an , R), E)
= lim
←−
n
∼
ExtdR (R/an , R), E)
= HomR (lim
−→
n
∼
= HomR (Had (R), E)
∼
= Had (R)∨
cf ). Now, one can
(ii) By (i) we know that Had (R) = Ker(R → ⊕ti=1 R
i
aplly Lemma 4.1.
12
GABRIEL CHIRIACESCU
(iii) Taking into account that Ass(R) has no embedded primes, the
equality follows from (ii) and, for example, from [1. Chapitre IV, 1,
n0 2, Prop. 6]
Before proving the Harsthorne-Lichtenbaum theorem we shall present
a result due to Watts, see [9, (33.3)].
Proposition 4.3. Let CR be the category of R − modules and F :
CR −→ CR be a right exact functor which commutes with direct sums.
Then
F(M ) ∼
= M ⊗R F(R)
Proof: For an element m ∈ M we define the homomorphism λm :
R −→ M , λm (a) := am. Then, we have the homomorphism F(λm ) :
F(R) −→ F(M ). There exists an R-bilinear mapping
M × F(R) −→ F(M )
(m, x) −→ F(λm )(x)
We deduce that there exists a homomorphism τM : M ⊗ F(R) −→
F(M ). If M is a free module then τM is clear an isomorphism. Now,
take a presentation G0 −→ G −→ M −→ 0, where G0 , G are free
modules. We have the following commutative diagram
G0 ⊗R F(R)
?
F(G0 )
- G ⊗R F(R)
-M ⊗R F(R)
?
- F(G)
?
- F(M )
- 0
-
0
where the first two vertical homomorphisms are isomorphisms. It follows that the third one is an isomorphism.
Corollary 4.4. Let R be a Noetherian ring, M an R-module and a an
ideal of R. Suppose that there exists m ∈ N such that Hai (−) = 0 for
all i > m. Then
Ham (M ) = M ⊗R Ham (R)
Proof: Since Ham (−) is a right exact functor and commutes with direct
sums we can apply the above proposition.
Proof of HLVT. Since the implication (i) ⇒ (ii) is the trivial part of
the proof we shall only prove (ii) ⇒ (i). By the faithfully flatness of
b we may assume that R is a complete local ring. As in [8] we can
R
find a complete Gorenstein ring (A, n, k) with dim(A) = dim(R) and a
surjective homomorphism ψ : A → R. Set c = Ker(ψ) and b := ψ −1 (a).
Let us observe that (ii) is equivalent to the following inclusion of sets
THE HARTSHORNE-LICHTENBAUM VANISHING THEOREM
13
{p ∈ Ass(R) | dim(R/p) = d} ⊆ {pj | j ∈ ΛR (m, a)}
By the ring independence of local cohomology and from the fact that
Hbd (−) is a right exact functor we deduce that
H d (R) ∼
= H d (R) ∼
= H d (A) ⊗A R
a
b
b
By using this isomorphism, the adjoint property and the fact that
ER (k) ∼
= HomA (R, EA (k)) we shall estimate the Matlis dual of Had (R):
Had (R)∨ ∼
= HomA (Hbd (R), EA (k)) ∼
= HomA (R ⊗A Hbd (A), EA (k))
∼
= HomA (A/c, Hbd (A)∨ )
Considering the Ass and applying Proposition 4.2 we get:
AssA (Had (R)∨ ) = Var(c) ∩ AssA (Hbd (A)∨ )
= Var(c) ∩ Ass(A) − Var(c) ∩ {pj | j ∈ ΛA (n, b)}
where pj ∈ Ass(A).
It is clear that there exists a bijection between the last set and the
set
{p ∈ Ass(R) | dim(R/p) = d} − {pj | j ∈ ΛR (m, a)}.
But, by hypothesis, this set is empty. Therefore Had (R)∨ = 0 and so
Had (R) = 0.
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New York, 1994.
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