INJECTIVE MODULES AND TORSION FUNCTORS

arXiv:1508.07407v2 [] 12 Oct 2016

PHAM HUNG QUY AND FRED ROHRER
Abstract. A commutative ring is said to have ITI with respect to an ideal a if the
a-torsion functor preserves injectivity of modules. Classes of rings with ITI or without
ITI with respect to certain sets of ideals are identified. Behaviour of ITI under formation
of rings of fractions, tensor products and idealisation is studied. Applications to local
cohomology over non-noetherian rings are given.

Introduction
Let R be a ring1. It is an interesting phenomenon that the behaviour of injective Rmodules is related to noetherianness of R. For example, by results of Bass, Matlis and
Papp ([14, 3.46; 3.48]) the following statements are both equivalent to R being noetherian:
(i) Direct sums of injective R-modules are injective; (ii) Every injective R-module is a
direct sum of indecomposable injective R-modules.
In this article, we investigate a further property of the class of injective R-modules,
dependent on an ideal a ⊆ R, that is shared by all noetherian rings without characterising
them: We say that R has ITI with respect to a if the a-torsion submodule of any injective
R-module is again injective2. It is well-known that ITI with respect to a implies that every
a-torsion R-module has an injective resolution whose components are a-torsion modules.
We show below (1.1) that these two properties are in fact equivalent.
Our interest in ITI properties stems from the study of the theory of local cohomology
(i.e., the right derived cohomological functor of the a-torsion functor). Usually, local cohomology is developed over a noetherian ring (e.g., [4]), although its creators were obviously
interested in a more general theory – see [7]. While some deeper theorems about local
cohomology might indeed rely on noetherianness, setting up its basic machinery does not
need such a hypothesis at all. The ITI property with respect to the supporting ideal serves
as a convenient substitute for noetherianness, and thus identifying non-noetherian rings
2010 Mathematics Subject Classification. Primary 13C11; Secondary 13D45.
Key words and phrases. ITI, non-noetherian ring, injective module, torsion functor, local cohomology,
weakly proregular ideal.
PHQ was partially supported by NAFOSTED (Vietnam), grant no. 101.04-2014.25. An earlier
version of this article was cited as Bad behaviour of injective modules.
1Throughout the following rings are understood to be commutative.
2ITI stands for “injective torsion of injectives”.
1

INJECTIVE MODULES AND TORSION FUNCTORS

2

with ITI is a first step in extending applicability of local cohomology beyond noetherian
rings, in accordance with natural demands from modern algebraic geometry and homological algebra.3 Besides the approach via ITI, there is also the extension of local cohomology
to non-noetherian rings via the notion of weak proregularity. This originates in [7], was
studied by several authors ([1], [17], [20]), and is quite successful from the point of view
of applications. However – and in contrast to ITI – it applies only to supporting ideals of
finite type, and when working in a non-noetherian setting, it seems artificial to restrict
ones attention a priori to ideals of finite type.
The first section contains mostly positive results. Besides giving a new proof of the fact
that noetherian rings have ITI with respect to every ideal (1.16), we prove that absolutely
flat rings have ITI with respect to every ideal (1.12) and that 1-dimensional local domains
(e.g. valuation rings of height 1) have ITI with respect to ideals of finite type (1.19). We
also show that ITI properties are preserved by formation of rings of fractions (1.8) and
discuss further necessary or sufficient conditions for ITI.
Unfortunately, there are a lot of nice rings without ITI, even with respect to some
principal ideals. In the second section, we provide several such examples (2.2, 2.3, 2.9)
and also show that ITI may be lost by natural constructions such as idealisation (2.10)
or base change (2.5), even when performed on noetherian rings.
The goal of the third section is twofold. First, we show that ITI with respect to an
ideal a of finite type is strictly stronger than weak proregularity of a (3.1, 3.2). Second,
to round off, we sketch how to swap ITI for noetherianness in some basic results on local
cohomology.
Our study of rings with or without ITI left us with at least as much questions as we
found answers; some of them are pointed out in the following. We consider this work as
a starting point in the search for non-noetherian rings with ITI and hope this naturally
arising and interesting class of rings will be further studied and better understood.
General notation and terminology follows Bourbaki’s Éléments de mathématique; concerning local cohomology we follow Brodmann and Sharp ([4]).
1. Rings with ITI
Let R be a ring and let a ⊆ R be an ideal. We choose for every R-module M an injective
hull eR (M) : M ֌ ER (M) with eR (M) = IdM if M is injective; for basics on injective hulls
we refer the reader to [2, X.1.9]. The a-torsion functor Γa is defined as the subfunctor of
S
the identity functor on the category of R-modules with Γa (M) = n∈N (0 :M an ) for each
R-module M. For basics on torsion functors we refer the reader to [4], but not without
3Concerning

“theorems that might indeed rely on noetherianness”, let us mention that in 3.7 we exhibit
rings with ITI with respect to every ideal but without Grothendieck vanishing.

INJECTIVE MODULES AND TORSION FUNCTORS

3

a word of warning that over non-noetherian rings these functors may behave differently
than over noetherian ones – see [18].4
(1.1) Proposition The following statements are equivalent:
(i) Γa (M) is injective for any injective R-module M;
(ii) Every a-torsion R-module has an injective resolution whose components are a-torsion
modules;
(iii) ER (Γa (ER (M))) = Γa (ER (M)) for every R-module M;
(iv) Γa (ER (Γa (M))) = ER (Γa (M)) for every R-module M.
Proof. The equivalences “(i)⇔(iii)” and “(ii)⇔(iv)” are immediate.
“(iii)⇒(iv)”: Let M be an R-module. Then,
Γa (M) ⊆ Γa (ER (Γa (M))) ⊆ ER (Γa (M)).
Taking injective hulls yields
ER (Γa (M)) ⊆ ER (Γa (ER (Γa (M)))) ⊆ ER (Γa (M)).
As the R-module in the middle equals Γa (ER (Γa (M))) by (iii) we get the desired equality.
“(iv)⇒(iii)”: Let M be an R-module. Then,
Γa (ER (M)) ⊆ ER (Γa (ER (M))) ⊆ ER (M).
Taking a-torsion submodules yields
Γa (ER (M)) ⊆ Γa (ER (Γa (ER (M)))) ⊆ Γa (ER (M)).
As the R-module in the middle equals ER (Γa (ER (M))) by (iv) we get the desired equality.

We say that R has ITI with respect to a if the statements (i)–(iv) in 1.1 hold. It is
well-known that noetherian rings have ITI with respect to every ideal ([4, 2.1.4]); we give
a new proof of this fact below (1.16). But let us begin by exhibiting first examples – albeit
silly ones – of possibly non-noetherian rings with ITI properties.
(1.2) Examples A) Every ring has ITI with respect to nilpotent ideals.
B) If R is a 0-dimensional local ring, then proper ideals of finite type are nilpotent, and
hence R has ITI with respect to ideals of finite type.

N

C) If K is a field, then K[(Xi )i∈N ]/hXi Xj | i, j ∈ iK[(Xi)i∈N ] is a non-noetherian 0dimensional local ring whose proper ideals are all nilpotent, hence it has ITI with respect
to every ideal.
•
4Although

some of the following can be expressed in the language of torsion theories we avoid this
mainly because Γa is not necessarily a radical if a is not of finite type – see 2.12.

INJECTIVE MODULES AND TORSION FUNCTORS

4

Next we look at a necessary condition for R to have ITI with respect to a which proves
to be very useful in producing rings without ITI later on. Moreover, the question about its
sufficiency is related to the question whether ITI properties are preserved along surjective
epimorphisms.
(1.3) Proposition If R has ITI with respect to a, then ER (R/a) is an a-torsion module.
Proof. Immediately from 1.1 (iv), since R/a is an a-torsion module.



(1.4) Lemma If a, b ⊆ R are ideals with b ⊆ a and M is an R-module with bM = 0
such that ER (M) is an a-torsion module, then ER/b (M) is an a/b-torsion module.
Proof. Straightforward on use of the canonical isomorphism
HomR (R/b, ER (M)) ∼
= ER/b (M) ([4, 10.1.16]).

of R/b-modules


(1.5) Question It would be useful to know whether the converse of 1.3 holds, i.e.:
(A) Suppose ER (R/a) is an a-torsion module. Does then R have ITI with respect to a?
For example, if b ⊆ R is an ideal with b ⊆ a and ER (R/a) is an a-torsion module, then
ER/b ((R/b)/(a/b)) is an a/b-torsion module by 1.4. So, if Question (A) could be positively
answered and R has ITI with respect to a, then R/b would have ITI with respect to a/b.
In particular, ITI properties would be preserved along surjective epimorphisms.
•
While we do not know whether or not ITI properties are preserved along surjective
epimorphisms (let alone arbitrary epimorphisms), we show now that they are preserved
along flat epimorphisms, hence in particular by formation of rings of fractions.5
(1.6) Proposition Let h : R → S be a flat epimorphism of rings. If R has ITI with
respect to a, then S has ITI with respect to aS.
Proof. Let I be an injective S-module, let b ⊆ S be an ideal, and let f : b → ΓaS (I) be a
morphism of S-modules. Then, f induces a morphism of R-modules
b ∩ R → Γa (I ↾R ) = ΓaS (I) ↾R . The R-module I ↾R is injective by [14, 3.6A], hence
so is Γa (I ↾R ) by our hypothesis on R. So, there exists y ∈ Γa (I) such that if x ∈ b ∩ R,
then f (h(x)) = xy. Let x ∈ b. Since h is a flat epimorphism, b = (b ∩ R)S by [15, IV.2.1],
hence x is an S-linear combination of elements of b ∩ R, and thus f (x) = xy. Now, ΓaS (I)
is injective by Baer’s criterion, and the claim is proven.

(1.7) Corollary Let h : R → S be a flat epimorphism of rings. If R has ITI with respect
to every ideal (of finite type), then S has ITI with respect to every ideal (of finite type).
5This

result could raise hope to obtain non-noetherian rings with ITI by considering flat epimorphisms with noetherian source. Alas, Lazard showed that flat epimorphisms with noetherian source have
noetherian target ([15, IV.2.3]).

INJECTIVE MODULES AND TORSION FUNCTORS

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Proof. Immediately by [15, IV.2.1] and 1.6.



(1.8) Corollary Let T ⊆ R be a subset. If R has ITI with respect to a, then T −1 R has
ITI with respect to T −1 a; if R has ITI with respect to every ideal (of finite type), then
T −1 R has ITI with respect to every ideal (of finite type).
Proof. Immediately by 1.6 and 1.7.



(1.9) Questions Since ITI properties are inherited by rings of fractions, it is natural to
ask for a converse, i.e., whether they can be delocalised in one of the following ways:
(B) If Rp has ITI with respect to ap for every p ∈ Spec(R), does then R have ITI with
respect to a?
(C) Let (fi )i∈I be a generating family of the ideal R. If Rfi has ITI with respect to afi for
every i ∈ I, does then R have ITI with respect to a?
The following special case of Question (B) deserves to be mentioned separately.
(D) Do pointwise
ideal? 6

noetherian

rings

have

ITI

with

respect

to

every
•

Now we turn to a first class of rings with ITI properties, namely, the absolutely flat
ones. Recall that R is absolutely flat if it fulfills the following equivalent conditions
([14, 3.71; 4.21]): (i) Every R-module is flat; (ii) Every monogeneous R-module is flat;
(iii) Rm is a field for every maximal ideal m ⊆ R; (iv) R is reduced and dim(R) ≤ 0;
(v) R is von Neumann regular. Since products of infinitely many fields are absolutely
flat and non-noetherian, this provides a class of interesting non-noetherian rings with
ITI. (In view of Question (D) in 1.9 we may note that absolutely flat rings are pointwise
noetherian.)
(1.10) Lemma If the R-module R/a is flat, then a is idempotent.
Proof. Applying the exact functor • ⊗R R/a to the exact sequence 0 → a ֒→ R → R/a
h
yields an exact sequence 0 → a/a2 → R/a → R/a with h = IdR/a . So, we get a = a2 and
thus the claim.

(1.11) Proposition If the R-module R/a is flat, then R has ITI with respect to a.
Proof. Let M be an a-torsion R-module. The ideal a is idempotent by 1.10, hence aM = 0,
and thus we can consider M canonically as an R/a-module. Now, we consider the injective
R/a-module ER/a (M) and its scalar restriction ER/a (M) ↾R , which is an a-torsion module.
Moreover, the canonical morphism R → R/a being flat, ER/a (M) ↾R is injective by [14,
6A

ring R is pointwise noetherian if Rp is noetherian for every p ∈ Spec(R). For facts on pointwise
noetherian rings, including examples of non-noetherian but pointwise noetherian rings, we suggest [9].

INJECTIVE MODULES AND TORSION FUNCTORS

6

3.6A]. Scalar restriction to R of eR/a (M) : M → ER/a (M) being a monomorphism of
R-modules M → ER/a (M) ↾R we thus get a monomorphism of R-modules ER (M) →
ER/a (M) ↾R . Therefore, ER (M) is an a-torsion module, and the claim is proven.

(1.12) Corollary Absolutely flat rings have ITI with respect to every ideal.
Proof. Immediately by 1.11.



(1.13) Corollary Let m ⊆ R be a maximal ideal such that the R-module R/m is injective
or that Rm is a field. Then, R has ITI with respect to m.
Proof. Both conditions are equivalent by [14, 3.72], and they are equivalent to R/m being
flat by [10, Lemma 1]. The claim follows then from 1.11.

Next we study the relation between the property of an R-module M of being an atorsion module and the assassin of M, i.e., the set AssR (M) of primes associated with
M. If R is noetherian, it is well-known that M is an a-torsion module if and only if its
assassin is contained in Var(a) (i.e., the set of primes containing a). Since assassins are
not well-behaved over non-noetherian rings, we also consider the so-called weak assassin
AssfR (M). Recall that a prime ideal p ⊆ R is weakly associated with M if it is a minimal
prime of the annihilator of an element of M. For basics about weak assassins we refer the
reader to [3, IV.1 Exercise 17], [12] and [21].
These considerations lead to a new proof of the fact that noetherian rings have ITI with
respect to every ideal, and to the result that 1-dimensional local domains (e.g., valuation
rings of height 1) have ITI with respect to ideals of finite type.
(1.14) Proposition Let M be an R-module. We consider the following statements: (1)
M is an a-torsion module; (2) AssfR (M) ⊆ Var(a); (3) AssR (M) ⊆ Var(a).
We have (1)⇒(2)⇒(3); if a is of finite type, we have (1)⇔(2)⇒(3); if R is noetherian,
we have (1)⇔(2)⇔(3).
Proof. Recall that AssR (M) ⊆ AssfR (M), with equality if R is noetherian ([3, IV.1 Exercice
17]). Hence, (2) implies (3), and the converse holds if R is noetherian.
If M is an a-torsion module and p ∈ AssfR (M), there exists x ∈ M such that p is
with an x = 0, thus we get
a minimal prime of (0 :R x), hence there exists n ∈
an ⊆ (0 :R x) ⊆ p, and therefore a ⊆ p. This shows (1)⇒(2).
Suppose
now that a is of finite type. If AssfR (M) ⊆ Var(a), then for x ∈ M we have
p

a ⊆ (0 :R x), and thus there exists n ∈ with an x = 0. This shows (2)⇒(1).

N

N

(1.15) Proposition If every R-module M with AssR (M) ⊆ Var(a) is an a-torsion
module, then R has ITI with respect to a.

INJECTIVE MODULES AND TORSION FUNCTORS

7

Proof. Every R-module has the same assassin as its injective hull ([14, 3.57]). So, if M is
an a-torsion R-module, then AssR (ER (M)) = AssR (M) ⊆ Var(a) by 1.14, hence ER (M)
is an a-torsion module, and thus 1.1 yields the claim.

(1.16) Corollary Noetherian rings have ITI with respect to every ideal.
Proof. Immediately from 1.14 and 1.15.



(1.17) Proposition If AssfR (M) = AssfR (ER (M)) for every R-module M, then R has
ITI with respect to ideals of finite type.
Proof. Replacing AssR by AssfR in the proof of 1.15 yields the claim.



(1.18) Proposition Local rings whose non-maximal prime ideals are of finite type have
ITI with respect to ideals of finite type.
Proof. Let R be such a ring and let m denote its maximal ideal. If m is of finite type then
R is noetherian by Cohen’s theorem ([8, 0.6.4.7]) and the claim follows from 1.16. So,
suppose m is not of finite type.
Let M be an R-module. By 1.17, it suffices to show AssfR (M) = AssfR (ER (M)). If
M = 0, this is fulfilled. We suppose M 6= 0 and assume AssfR (M) $ AssfR (ER (M)). By
[22, 1.8], a prime ideal of R of finite type is associated with M if and only if it is weakly
associated with M. Therefore, our assumptions imply m ∈ AssfR (ER (M)) \ AssfR (M). So,
there exists x ∈ ER (M) such that m is a minimal prime of (0 :R x), and as M ֒→ ER (M)
is essential there exists r ∈ R with rx ∈ M \ 0. The ideal (0 :R rx) ⊆ R has a minimal
prime p which is weakly associated with M, so that p 6= m, but it contains (0 :R x),
yielding the contradiction p = m.

(1.19) Corollary 1-dimensional local domains have ITI with respect to ideals of finite
type.
Proof. Immediately from 1.18.



(1.20) Questions The successful application of 1.17 in the proof of 1.18 lets us ask for
more:
(E) For which rings R do we have AssfR (M) = AssfR (ER (M)) for every R-module M?
By [14, 3.57], one class of such rings, containing the noetherian ones, are those over which
assassins and weak assassins coincide:
(F) For which rings R do we have AssR (M) = AssfR (M) for every R-module M?
One may note that assassins and weak assassins do not necessarily coincide over a 1and valuation ν,
dimensional local domain: If R is a valuation ring with value group

Q

INJECTIVE MODULES AND TORSION FUNCTORS

8

and a := {x ∈ R | ν(x)2 > 2}, then AssR (R/a) is empty and AssfR (R/a) contains the
maximal ideal of R.7
•
We end this section with two examples, showing that the hypotheses in 1.14 cannot be
omitted.

(1.21) Examples A) Let K be a field. We consider the absolutely flat ring R := K N , the
ideal b ⊆ R whose elements are precisely the elements of R of finite support, the R-module
M := R/b, and f = (fn )n∈N ∈ R with f0 = 0 and fn = 1 for every n > 0. Then, f ∈ R \ 0
is idempotent, and hence the principal ideal a := hf iR is not nilpotent. Yassemi showed
in [21, Section 1, Example] that AssR (M) = ∅ and in particular AssR (M) ⊆ Var(a). If M
is an a-torsion module and x = (xn )n∈N ∈ R, then setting x′0 = 0 and x′n = xn for n > 0
it follows x′ := (x′n )n∈N ∈ b, and so there exists n0 ∈ such that if n ≥ n0 , then xn = 0
– a contradiction. This show that M is not an a-torsion module.
In particular, the implication (3)⇒(1) in 1.14 does not necessarily hold, even if a is
principal.

N

B) Let R be a 0-dimensional local ring whose maximal ideal m is not nilpotent, e.g.
the ring constructed below in 2.2 c). Then, R is not an m-torsion module, but AssfR (R) ⊆
Spec(R) = Var(m).
In particular, the implication (2)⇒(1) in 1.14 does not necessarily hold.
•
2. Rings without ITI
In this section, we exhibit several examples of rings without the ITI property. The first
bunch of examples relies on the following observation about local rings with idempotent
maximal ideal.
(2.1) Proposition If R is a local ring whose maximal ideal m is idempotent, then the
following statements are equivalent: (i) R is a field; (ii) R is noetherian; (iii) R has ITI
with respect to m; (iv) ER (R/m) is an m-torsion module.
Proof. By 1.3 and 1.16 it suffices to show that (iv) implies (i). Suppose ER (R/m) is an
m-torsion module and let x ∈ ER (R/m) \ 0. By idempotency of m we have mx = 0, hence
hxiR ∼
= R/m is simple. The canonical injection R/m ֒→ ER (R/m) being essential we have
(R/m) ∩ hxiR 6= 0, hence, by simplicity, x ∈ R/m. Thus, R/m = ER (R/m) is injective,
and therefore R is a field by [14, 3.72].

Recall that a valuation ring is a local domain whose set of (principal) ideals is totally
ordered with respect to inclusion, and that a Bezout ring is a domain whose ideals of finite
type are principal. A quotient of a valuation ring or of a Bezout ring has the property that
7This

example was suggested by Neil Epstein via MathOverflow.net.

INJECTIVE MODULES AND TORSION FUNCTORS

9

the set of its ideals is totally ordered with respect to inclusion or that its ideals of finite
type are principal, respectively. Moreover, valuation rings are Bezout rings, and Bezout
rings are coherent rings whose local rings are valuation rings ([3, VI.1.2; VII.1 Exercice
20; VII.2 Exercices 12, 14, 17]).
(2.2) Proposition Let K be a field, let Q denote the additive monoid of positive rational
numbers, let R := K[Q] denote the algebra of Q over K, let {eα | α ∈ Q} denote its
canonical basis, and let m := heα | α > 0iR and a := heα | α > 1iR . Furthermore, set
S := Rm , n := mm , b := am , T := S/b and p := n/b.
a) R is a 1-dimensional Bezout ring that does not have ITI with respect to its maximal
ideal m.8
b) S is a 1-dimensional valuation ring that has ITI with respect to non-maximal ideals,
but does not have ITI with respect to its maximal ideal n.
c) T is a 0-dimensional non-coherent local ring that has ITI with respect to non-maximal
ideals, but does not have ITI with respect to its maximal ideal p.
Proof. First, we note that m is an idempotent maximal ideal of R and the only prime
ideal of R containing a. Hence, S and T are local rings with idempotent maximal ideals
n and p, respectively, and T ∼
= R/a is 0-dimensional. So, there is a surjective
= (R/a)m/a ∼
morphism of rings R → T , mapping m onto p. The ring R is the union of the increasing
family (K[e 1 ])n∈N of principal subrings, hence a Bezout ring, and it is integral over
n!
its 1-dimensional subring K[e1 ] by [5, 12.4], hence 1-dimensional (cf. [11, Example 31]).
Therefore, S is a 1-dimensional valuation ring, and in particular not a field. Moreover,
p = (0 :T e11 + b) is not of finite type, hence T is not coherent ([14, 4.60]), and in particular
not noetherian. Thus, the negative claims about ITI follow from 1.4 and 2.1.
For an ideal c ⊆ S we set α(c) := inf{α ∈ Q | eα ∈ c}. We have α(n) = 0, and if c, d ⊆ S
are ideals with α(c) > α(d) then c ⊆ d. Since ideals of S of finite type are of the form heα iS
with α ∈ Q, it follows that an ideal c ⊆ S is of finite type if and only if eα(c) ∈ c. Now,
let c ⊆ S be an ideal and let n ∈ . If α ∈ Q with eα ∈ cn , then there are α1 , . . . , αn ∈ Q
with eα1 , . . . , eαn ∈ c and eα = eα1 +···+αn , implying α = α1 + · · · + αn ≥ nα(c). Therefore,
α(cn ) ≥ nα(c). Conversely, if β ∈ Q with β > nα(c), then nβ > α(c), implying e β ∈ c and
n
thus eβ ∈ cn . Therefore, α(cn ) ≤ nα(c). It follows α(cn ) = nα(c).
Next, let c, d ⊆ S be non-maximal, non-zero ideals and let n ∈ , so that α(cn ) 6= 0 and
α(dn ) 6= 0. There exist k, m ∈ with α(cnk ) = kα(cn ) ≥ α(dn ) and α(dnm ) = mα(dn ) ≥
α(cn ), implying cnk ⊆ dn and dnm ⊆ cn , and therefore Γc = Γd (cf. [18, 2.2]). It follows
that S has ITI with respect to non-maximal ideals if and only if it has ITI with respect
to some non-maximal ideal, and by 1.19 this is fulfilled.

N

N

8In

N

[2, X.1 Exercice 27 f)] the reader gets the cruel task to show that R is a valuation domain – but
its element 1 + e1 is neither invertible nor contained in m.

INJECTIVE MODULES AND TORSION FUNCTORS

10

N

Finally, if c ⊆ S is a non-maximal ideal then α(c) > 0, hence there exists n ∈ with
α(cn ) = nα(c) > 1 and therefore cn ⊆ b. Thus, non-maximal ideals of T are nilpotent, so
that T has ITI with respect to non-maximal ideals by 1.2 A).

We draw some conclusions from these examples. While a 1-dimensional local domain
has ITI with respect to ideals of finite type (1.19), it need not have so with respect to
every ideal (2.2 b)). While a reduced 0-dimensional ring (i.e., an absolutely flat ring) has
ITI with respect to every ideal (1.12), an arbitrary 0-dimensional ring need not have so
(2.2 c)). John Greenlees conjectured (personal communication, 2010) that a ring with
bounded a-torsion9 for some ideal a has ITI with respect to a; in particular, domains
would have ITI with respect to every ideal. Example 2.2 b) disproves this conjecture.
(2.3) Corollary The polynomial algebra K[(Xi )i∈I ] in infinitely many indeterminates
(Xi )i∈I over a field K does not have ITI with respect to hXi | i ∈ IiK[(Xi)i∈I ] .
Proof. Let U := K[(Xi )i∈I ] and let q := hXi | i ∈ IiU . Assume that U has ITI with
respect to q. Then, Uq has ITI with respect to qq by 1.8. Furthermore, in the notations of
2.2, there is a surjective morphism of rings U → K[Q] mapping q onto m. This morphism
induces a surjective morphism of rings Uq → S mapping qq onto n. Now, 1.3 and 1.4 imply
that ES (S/n) is an n-torsion module, and thus S has ITI with respect to n by 2.1. This
contradicts 2.2 b).

One may note that R = K[(Xi )i∈N ] is not only a coherent domain, but in fact a filtering
inductive limit of noetherian rings with flat monomorphisms (cf. [6, p. 48]), and that its
hXi | i ∈ iR -adic topology is separated. (In 2.11 we will give a direct proof of the fact
that ER (R/hXi | i ∈ iR ) is not a hXi | i ∈ iR -torsion module.)

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(2.4) Question Inspired by 2.3 we ask the following question:
(G) Do polynomial algebras in (countably) infinitely many indeterminates over fields
have ITI with respect to (monomial) ideals of finite type?
•
As an application of 2.1, we show that ITI is not necessarily preserved by tensor products. In particular, it is not stable under base change.
(2.5) Proposition Let p be a prime number, let K be a non-perfect field of characteristic
p, and let L be its perfect closure. Then, L ⊗K L has ITI with respect to ideals of finite
type, but does not have ITI with respect to its maximal ideal Nil(L ⊗K L).10
9A

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ring R is of bounded a-torsion for some ideal a if there exists n ∈ with Γa (R) = (0 :R an ).
is also the standard example showing that noetherianness is not necessarily preserved by tensor
products (cf. [8, I.3.2.5]).
10This

INJECTIVE MODULES AND TORSION FUNCTORS

11

Proof. It is well-known that L ⊗K L is a non-noetherian 0-dimensional local ring with
maximal ideal Nil(L ⊗K L) ([8, I.3.2.5]). So, the first claim is clear by 1.2 B), and by 2.1,
it suffices to show that Nil(L ⊗K L) is idempotent. Let f ∈ Nil(L ⊗K L) \ 0. There exist
Pm
m
m ∈ and families (xi )m
i=0 and (yi )i=0 in L \ 0 with f =
i=0 xi ⊗ yi . Since L is perfect,
p
there exist for every i ∈ [0, m] elements vi , wi ∈ L with vi = xi and wip = yi . It follows
P
Pm
p
p
p
p
f= m

i=0 vi ⊗ wi = (
i=0 vi ⊗ wi ) and therefore f ∈ Nil(L ⊗K L) as desired.

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Up to now, we saw only rings without ITI with respect to ideals not of finite type. The
second bunch of examples relies on an observation about valuation rings with maximal
ideal of finite type (2.8).
(2.6) Lemma Let a, b ⊆ R be ideals and suppose the canonical injection b ֒→ R is
essential. Then, ER (b) is an a-torsion module if and only if a is nilpotent.
Proof. As b ֒→ R is essential, R ⊆ ER (R) = ER (b). If ER (b) is a a-torsion module, then
so is R, hence a is nilpotent. The converse is clear.

(2.7) Corollary Let R be a quotient of a valuation ring, let x ∈ R \ 0, and let a ⊆ R
be an ideal. Then, ER (hxiR ) is an a-torsion module if and only if a is nilpotent.
Proof. The set of principal ideals of R is totally ordered with respect to inclusion, hence
the canonical injection hxiR ֒→ R is essential, and thus 2.6 yields the claim.

(2.8) Proposition If R is a valuation ring whose maximal ideal m is of finite type, then
the following statements are equivalent: (i) R is noetherian; (ii) R has ITI with respect
to m; (iii) ER (R/m) is an m-torsion module.
Proof. By 1.3 and 1.16, it suffices to show that (iii) implies (i). Suppose (iii) holds and
assume R is non-noetherian. By [3, VI.3.6 Proposition 9], a valuation ring with maximal
ideal m of finite type is noetherian if and only if its m-adic topology is separated. Hence,
T
there exists x ∈ ( n∈N mn ) \ 0. We consider the local ring R := R/xm, its maximal ideal
m := m/xm, and x := x + xm ∈ R. If there exists n ∈ with mn = 0, then mn ⊆ xm ⊆
mn+1 , and Nakayama’s Lemma yields the contradiction mn = 0. As (0 :R x) = m and
therefore R/m ∼

= hxiR , we get a contradiction from 1.4 and 2.7.

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(2.9) Proposition Let p be a prime number, let R := [(Xi )i∈N ] be the polynomial
algebra in indeterminates (Xi )i∈N over , let a := hpj−i Xj − Xi | i, j ∈ , i < jiR , let
m := hpiR + hXi | i ∈ iR , let S := R/a, and let n := m/a. Furthermore, denote by Yi
the canonical image of Xi in S for i ∈ , let T := Sn /hpY0iSn , and let p := nn /hpY0 iSn .
a) Sn is a 2-dimensional valuation ring that does not have ITI with respect to its principal maximal ideal nn .

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b) T is a 1-dimensional local ring that does not have ITI with respect to its principal
maximal ideal p.

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Proof. We claim first that if i, m, n, r, s ∈ , then pm Yir and pn Yis are comparable with
respect to divisibility in S. Indeed, without loss of generality, we can suppose r < s, so
if m ≤ n, then the claim is clear. Otherwise, pn Yis = pm Yir Yis−1−r Yi+m−n and thus the
and f ∈ S, then
claim holds. Keeping in mind the definition of a, we see that if b ∈
P
r
with i ≥ b and a finite family (ak )k=0 in with f = rk=0 ak Yik . In
there exist i ∈
particular, if f, g ∈ S, then there exist i ∈
and finite families (ak )rk=0 and (bk )sk=0 in
Pr
Ps
P
k
k
with f = k=0 ak Yi and g = k=0 bk Yi . Moreover, for f = rk=0 ak pnk Yik ∈ S \ 0
with i ∈
and finite families (nk )rk=0 in
and (ak )rk=0 in \ hpi there exists k0 :=
min{k ∈ [0, r] | ak 6= 0}, and pnk0 Yik0 divides pnk Yik for every k ∈ [0, r] with ak 6= 0. In
particular, for f ∈ Sn \ 0 there exist a unit u of Sn and i, k, n ∈ with f = upn Yik , and
we have f ∈ nn if and only if (n, k) 6= (0, 0). It follows that any two elements of Sn are
comparable with respect to divisibility.
/ a. Assume aX0n ∈ a. There exists
Next, we show that if a ∈ \0 and n ∈ , then aX0n ∈
m ∈ with aX0n ∈ a′ := hpj−i Xj − Xi | i, j ∈ [0, m], i < jiR′ , where R′ := [(Xi )m
i=0 ]. If
j−i
m−i
j−i m−j
i, j ∈ [0, m] with i < j, then p Xj − Xi = (p Xm − Xi ) − p (p
Xm − Xj ), hence
m−i
′
m−i
a = hp Xm −Xi | i ∈ [0, m−1]iR′ . Factoring out hp Xm −Xi | i ∈ [1, m−1]iR′ we get
aU n = f (pm V −U) in the polynomial algebra [U, V ] with f ∈ [U, V ]\0, a contradiction,
and thus our claim. From this we get pX0 ∈
/ a and thus c := hXi | i ∈ iR /a 6= 0.
Moreover, it also follows that S is a domain. Indeed, if f, g ∈ S \ 0 with f g = 0, then
there exist i, k, l, m, n ∈ , a, b ∈ \ hpi, and f ′ , g ′ ∈ c with f = pm Yik (a + f ′ ) and
g = pn Yil (b + g ′). Furnishing R with its canonical -graduation and S with the induced
-graduation, we get abpm+n Yik+l = 0, hence abpr X0s ∈ a for some r, s ∈ , and finally
the contradiction ab = 0. So, S is a domain, and hence Sn is a valuation ring.
then Yi = pYi+1 , hence p divides Yi in S, and therefore nn =
If i ∈
hpiSn + hYi | i ∈ iSn = hpiSn is principal. Thus, p = hpiT is principal, too. Now, let
q ∈ Spec(Sn ) \ {0, nn }, so that p ∈
/ q. Then, there exists an i ∈ with Yi ∈ q, since elen k
ments of nn are of the form up Yi with a unit u of Sn , i, k, n ∈ and (n, k) 6= (0, 0). This
implies pn Yi+n = Yi ∈ q and thus Yi+n ∈ q for every n ∈ , and also Yi−k = pk Yi ∈ q for
every k ∈ [0, i]. It follows q = cn , hence Spec(Sn ) = {0, cn , nn }, and therefore dim(Sn ) = 2.
then Yi = pn Yi+n , hence pn divides Yi in S, and therefore
Furthermore, if i, n ∈
T
0 6= pY0 ∈ cn ⊆ n∈N nnn . Thus, T is 1-dimensional and Sn is non-noetherian, so a) follows
from 2.8.
If n ∈ ∗ , then pn divides Y0 = pn Yn , so if pY0 divides pn , then Yn is a unit of Sn ,
which is a contradiction. Therefore, hpY0 iSn contains no power of p, and hence p is not
nilpotent. Moreover, Y0 ∈
/ hpY0 iSn , so denoting by Z the canonical image of Y0 in T it

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follows p = (0 :T Z), hence T /p ∼
= hZiT , and thus 1.3 and 2.7 imply that T does not have
ITI with respect to p.

Consider again the ring T from 2.9. By 1.17 there exists a T -module M with AssfT (M) 6=
AssfT (ET (M)), and we can explicitly describe such a module. Indeed, T has precisely
one non-maximal prime and ET (T /p) is not a p-torsion module, so 1.14 implies {p} =
AssfT (T /p) $ AssfT (ET (T /p)) = Spec(T ).
We draw from the examples in 2.2 and 2.9 the conclusion that while a 1-dimensional
local domain has ITI with respect to ideals of finite type (1.19), an arbitrary 1-dimensional
local ring need not have so. Further examples of non-noetherian valuation rings whose
maximal ideal is of finite type can be found in [16, p. 79, Remark] and [11, Example 32].
Based on 2.9 we show now that the idealisation of a module over a noetherian ring
need not have ITI. (Recall that for a ring R and an R-module M, the idealisation of M
is the R-algebra with underlying R-module R ⊕ M and with multiplication defined by
(r, x) · (s, y) = (rs, ry + sx).)
(2.10) Proposition We use the notations from 2.9 and denote by U the idealisation of
the hpi -module obtained from hYi | i ∈ iSn /hpY0 iSn by scalar restriction. Then, U is a
local ring that does not have ITI with respect to its principal maximal ideal.

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Proof. For i ∈ we denote by Zi the canonical image of Yi in M := hYi | i ∈ iSn /hpY0 iSn .
If i, j ∈ , then Yi Yj = Y0 Yi+j = pY0 Yi+j+1 in Sn . Therefore, an element m of M \ 0 has
P
the form m = kj=1 vj Zij with a strictly increasing sequence (ij )kj=1 in
and a family
ik −ij
k
Zik for j ∈ [1, k − 1], it follows that m can be
(vj )j=1 in hpi \ {0}. Since Zij = p
written in the form m = vZi with i ∈
and v ∈ hpi \ {0}. Suppose that i is chosen
minimally with this property, and assume v ∈ hpiZhpi . Then, v = v ′ pn with v ′ ∈ hpi \ hpi
and n > 0. If n > i, we get the contradiction vZi = v ′ pn Zi = v ′ pn−i−1 pZ0 = 0. If n ≤ i,
we get vZi = v ′ pn Zi = v ′ Zi−n , contradicting the minimality of i. Therefore, v ∈ hpi \ hpi.
and
The above shows that an element of U \ 0 has the form (upn , vZi ) with i, n ∈
u, v ∈ ( hpi \ hpi) ∪ {0} such that (u, v) 6= (0, 0). Using this, it is readily checked that
U is a local ring with maximal ideal q = h(p, 0)iU , that the morphism of U-modules
U → U with 1 7→ (0, Z0) induces an isomorphism of U-modules U/q ∼
= h(0, Z0)iU , and
T
T
T
n
n
n
that n∈N q = n∈N h(p , 0)iU = n∈N (hp i ⊕ M) = 0 ⊕ M = h(0, Zi ) | i ∈ iU 6= 0.
and
In particular, q is not nilpotent. Now, we consider (upn , vZi ) ∈ U \ 0 with i, n ∈
n
−1
u, v ∈ ( hpi \ hpi) ∪ {0}. If u 6= 0, then (up , vZi )(0, u Zn ) = (0, Z0 ), and if u = 0, then
v 6= 0, and therefore (upn , vZi )(pi , 0) = (0, Z0 ). From this it follows that the canonical
injection h(0, Z0)iU ֒→ U is essential, thus 1.3 and 2.6 yield the claim.


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We end this section by considering again a 0-dimensional local ring R with maximal
ideal m. If m is nilpotent, then R has ITI with respect to every ideal (1.2 A)). If m is

INJECTIVE MODULES AND TORSION FUNCTORS

14

idempotent, then R has ITI with respect to every ideal if and only if it is a field (2.1). All
examples of 0-dimensional local rings without ITI constructed so far had an idempotent
maximal ideal (2.2, 2.5). We present now a further 0-dimensional local ring R; its maximal
ideal m is neither nilpotent nor idempotent, but T-nilpotent11, and its m-adic topology is
separated. But still R does not have ITI with respect to m.
(2.11) Lemma Let K be a field, let A := K[(Xi )i∈N ] be the polynomial algebra in
countably infinitely many indeterminates (Xi )i∈N , let n := hXi | i ∈ iA and let a :=
h{Xi Xj | i, j ∈ , i 6= j} ∪ {Xii+1 | i ∈ }iA . Then, there exists f ∈ EA (A/n) \
Γn (EA (A/n)) with af = 0.

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denote the set of monomials in A. We furnish E := HomK (A, K) with
Proof. Let
its canonical structure of A-module. Its elements are families in K indexed by . For
we have sg = (αst )t∈M . The morphism of A-modules A → E
g = (αt )t∈M ∈ E and s ∈
mapping 1 to (αt )t∈M with α1 = 1 and αt = 0 for t 6= 1 has kernel n and hence induces
a monomorphism of A-modules A/n ֌ E by means of which we consider A/n as a
sub-A-module of E.
/ {Xii | i ∈ }.
Let f = (αt )t∈M with αt = 1 for t ∈ {Xii | i ∈ } and αt = 0 for t ∈
with i 6= j and t ∈ , then Xi Xj f (t) = f (Xi Xj t) = 0. If i ∈
and t ∈ ,
If i, j ∈
i+1
i+1
n
then Xi f (t) = f (Xi t) = 0. This implies af = 0. If n ∈ , then Xn ∈ nn and
Xnn f (1) = f (Xnn ) = 1. Thus, if n ∈ , then nn f 6= 0. Furthermore, X1 f (1) = f (X11 ) = 1,
and for t ∈
\ {1} we have X1 f (t) = f (X1 t) = 0, so that X1 f is a non-zero element of
A/n. Therefore, the canonical injection A/n ֒→ A/n + hf iA is essential, and so it follows
f ∈ EA (A/n) as desired.


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(2.12) Proposition Let K be a field, let A, n and a be as in 2.11, let R := A/a, denote
by Yi the canonical image of Xi in R, and let m := hYi | i ∈ iR . Then, R is
for i ∈
a 0-dimensional local ring; its maximal ideal m is neither nilpotent nor idempotent, but
T -nilpotent; its m-adic topology is separated; the m-torsion functor Γm is not a radical 12;
R does not have ITI with respect to m.

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Proof. We set P := {Xi Xj | i, j ∈ , i 6= j} and Q := {Xii+1 | i ∈ }. The ideal m is
it follows m ⊆ Nil(R) ⊆
maximal since R/m ∼
= K. As Yi is nilpotent for every i ∈
m, hence m = Nil(R). Thus, R is a 0-dimensional local ring with maximal ideal m. If
n∈
and mn = 0, then Ynn = 0, hence Xnn ∈ a – a contradiction; therefore, m is not
nilpotent. If m is idempotent, then Y1 ∈ m2 = hYi2 | i > 1iR , hence X1 is a polynomial in
P ∪ Q ∪ {Xi2 | i > 1}, thus a polynomial in P ∪ {X0 } ∪ {Xi2 | i > 0} – a contradiction;
Q
therefore, m is not idempotent. Assume now there is a family (fi )i∈N in m with ni=0 fi 6= 0

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ideal a of R is T-nilpotent if for every family (xi )i∈N in a there exists n ∈ with ni=0 xi = 0.
12A radical is a subfunctor F of Id
Mod(R) with F (M/F (M )) = 0 for every R-module M .

11An

INJECTIVE MODULES AND TORSION FUNCTORS

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for every n ∈ . Without loss of generality, we can suppose all the fi are monomials in
with i 6= j, there exists k ∈
such that all the
{Yj | j ∈ }. As Yi Yj = 0 for i, j ∈
Qk+1
fi are monomials in Yk , yielding the contradiction i=0 fi = 0. Thus, m is T-nilpotent.
If n ∈ , then mn = hYin | i ≥ niR , so in elements of mn there occurs no Yi with i < n.
T
T
It follows that in elements of n∈N mn there occurs no Yi at all, hence n∈N mn = 0 and
thus R is m-adically separated. It is readily seen that 1 + Γm (R) is a non-zero element
of Γm (R/Γm (R)), hence Γm is not a radical. Finally, the A-module (0 :EA (A/n) a) is not
an n-torsion module by 2.11, hence by base ring independence of torsion functors and
[4, 10.1.16] we see that ER (R/m) ∼
= (0 :EA (A/n) a) is not an m-torsion module. Thus, 1.3
implies that R does not have ITI with respect to m.


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If, in 2.11 and 2.12, we take K instead of a field to be a selfinjective ring, then the
conclusions still hold, except that R need not be a 0-dimensional local ring and that m
need not be a maximal ideal of R.
3. Weak proregularity, and applications to local cohomology
In this section we clarify the relation between ITI and weak proregularity, and then
sketch some basic results on local cohomology for rings with ITI, but omit proofs. Results
under ITI hypotheses on the closely related higher ideal transformation functors can be
found in [19, 2.3].
Let R be a ring and let a ⊆ R be an ideal. The right derived cohomological functor of the a-torsion functor Γa is denoted by (Hai )i∈Z , and Hai is called the i-th local cohomology functor with respect to a. There is a canonical isomorphism of functors
HomR (R/an , •) that can be canonically extended to an isomorphism of
Γa (•) ∼
= lim
−→n∈N
ExtiR (R/an , •))i∈Z ([4, 1.3.8]).
δ-functors (Hai (•))i∈Z ∼
= (lim
−→n∈N
Suppose now that a is of finite type and let a = (ai )ni=1 be a generating family of a.
Čech cohomology with respect to a yields an exact δ-functor, denoted by (Ȟ i (a, •))i∈Z .
There is a canonical isomorphism Γa (•) ∼
= Ȟ 0 (a, •) that can be canonically extended to a
morphism of δ-functors γa : (Hai (•))i∈Z → (Ȟ i (a, •))i∈Z ([4, 5.1]). The sequence a is called
weakly proregular if γa is an isomorphism. (By [20, 3.2] this is equivalent to the usual –
but more technical – definition of weak proregularity ([1, Corrections], [17, 4.21], cf. [7,
Exposé II, Lemme 9]).) The ideal a is called weakly proregular if it has a weakly proregular
generating family. By [17, 6.3], this is the case if and only if every finite generating family
of a is weakly proregular. Ideals in noetherian rings are weakly proregular ([4, 5.1.20]),
but the converse need not hold – see below for examples. Local cohomology with respect
to a weakly proregular ideal a of finite type behaves quite well, and so we may ask about
the relation between this notion and ITI with respect to a. The next two results clarify
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INJECTIVE MODULES AND TORSION FUNCTORS

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(3.1) Proposition There exist a ring R and a weakly proregular ideal a such that R
does not have ITI with respect to a.
Proof. The 2-dimensional domain Sn from 2.9 does not have ITI with respect to its principal maximal ideal nn = hpiSn . But p is regular, and hence nn is weakly proregular by
[17, 4.22].

The proof of the next result makes use of Koszul homology and cohomology. We briefly
recall our notations and refer the reader to [2, X.9] and [4, 5.2] for details. Let a = (ai )ni=1
be a finite sequence in R. We denote by K• (a) the Koszul complex with respect to a
and by K • (a) := HomR (K• (a), R) the Koszul cocomplex with respect to a. We define
functors K• (a, ) := K• (a) ⊗R and K • (a, ) := HomR (K• (a), ), and for i ∈ we set
we set au = (aui )ni=1 .
Hi (a, ) := Hi (K• (a, )) and H i (a, ) := H i (K • (a, )). For u ∈
For u, v ∈
with u ≤ v there is a morphism of functors K• (au , ) → K• (av , ), and
these morphisms give rise to an inductive system of functors (K• (au , ))u∈N . We denote
its inductive limit by K• (a∞ , ), and we set Hi (a∞ , ) := Hi (K• (a∞ , )).

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(3.2) Proposition Let R be a ring and let a ⊆ R be an ideal of finite type. If R has
ITI with respect to a, then a is weakly proregular.
Proof. Let a = (ai )ni=1 be a finite generating family of a. Let i ∈
5.2.5] that there is a canonical isomorphism of functors

Z. It follows from [4,

Ȟ i (a, •) ∼
= Hn−i (a∞ , •).

(1)

Since inductive limits are exact, there is a canonical isomorphism of functors
(H (au , •)).
Hi (a∞ , •) ∼
= lim
−→ i

(2)

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If u ∈ , then the components of K• (au ) and K • (au ) are free of finite type, hence there are
∼
HomR (K • (au ), R) of complexes and
canonical isomorphisms K• (au )
=
HomR (K • (au ), R) ⊗R ∼
= HomR (K • (au ), ) of functors ([2, II.2.7 Proposition 13; II.4.2
Proposition 2]). Therefore, there is a canonical isomorphism of functors
lim (H (au , )) ∼
H (HomR (K • (au ), )).
= lim
−→ i
−→ i

(3)

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u∈

If I is an injective R-module, then HomR (•, I) is exact and thus commutes with formation
of homology, so that there is a canonical isomorphism of R-modules
(4)

lim H (HomR (K • (au ), I)) ∼
HomR (H i (au , R), I).
= lim
−→
−→ i

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If u ∈ , then H i (au , R) is an a-torsion module ([2, X.9.1 Proposition 1 Corollaire 2]).
Hence, there is a canonical isomorphism of functors
(5)

HomR (H i (au , R), Γa (•)).
lim HomR (H i (au , R), •) ∼
= lim
−→
−→

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u∈

Now, if I is an injective R-module, then so is Γa (I) by our hypothesis, and thus assembling the above, we get canonical isomorphisms of R-modules
Ȟ i (a, I)

(1)−(4)

∼
=

(5)

lim HomR (H n−i (au ), I) ∼
=
−→

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u∈

lim HomR (H n−i (au ), Γa (I))
−→

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(1)−(4)

∼
=

Ȟ i (a, Γa (I)).

u∈

Since the components of nonzero degree of the Čech cocomplex with respect to a of an
a-torsion module are zero, it follows that Ȟ i (a, •) is effaceable for i ∈ ∗ , and thus a is
weakly proregular.


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The results from Section 1 together with 3.2 imply, for example, that ideals of finite
type in absolutely flat rings or in 1-dimensional local domains are weakly proregular. The
statement about absolutely flat rings can be proven directly by first noting that idempotent elements are proregular and thus generate weakly proregular ideals ([20, 2.7]), and
then using the fact that an ideal of finite type in an absolutely flat rings is generated by an
idempotent ([13, 4.23]). Interestingly, the same observation shows that the principal ideal
a in Example 1.21 A) is weakly proregular. Finally, let us point out that 3.2 together with
1.16 yields a new proof of the fact that ideals in noetherian rings are weakly proregular.
Now we turn to basic results on local cohomology for rings with ITI. The interplay
between Γa and local cohomology functors is described by the following result, proven
analogously to [4, 2.1.7] on use of 1.1.
(3.3) Proposition Suppose R has ITI with respect to a, let i > 0 and let M be an
R-module. Then, Hai (Γa (M)) = 0, and the canonical morphism Hai (M) → Hai (M/Γa (M))
is an isomorphism.
In case a is of finite type, this follows immediately from 3.2 and the definitions of weak
proregularity and of Čech cohomology.
Using the notion of triad sequence we get the important Comparison Sequence.

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(3.4) Proposition Let n ∈ , let b ∈ R and b := hbiR , and suppose R has ITI with
respect to a and with respect to b. Then, there is an exact sequence of functors
n
0 −→ Hb1 ◦ Han−1 −→ Ha+b
−→ Γb ◦ Han −→ 0.

INJECTIVE MODULES AND TORSION FUNCTORS

18

In [20, 3.5], Schenzel gets the same conclusion under the hypothesis that a is of finite
type and that a, b and a + b are weakly proregular. The Comparison Sequence allows for
an inductive proof of an ITI variant of Hartshorne’s Vanishing Theorem ([4, 3.3.3]), but
this also follows immediately from 3.2.
The next two results concern the behaviour of local cohomology under change of rings:
the Base Ring Independence Theorem and the Flat Base Change Theorem. Both can be
proven analogously to [4, 4.2.1; 4.3.2], relying on a general variant of the vanishing result
[4, 4.1.3]. The latter can be obtained in the noetherian case on use of the Mayer-Vietoris
Sequence, and in general on use of the Comparison Sequence.
(3.5) Proposition Let R → S be a morphism of rings. Suppose a is of finite type and
R has ITI with respect to ideals of finite type. Then, there is a canonical isomorphism of
δ-functors
i
(HaS
(•) ↾R )i∈Z ∼
= (Hai (• ↾R ))i∈Z .

(3.6) Proposition Let R → S be a flat morphism of rings. Suppose a is coherent and
S has ITI with respect to extensions of coherent ideals of R.13 Then, there is a canonical
isomorphism of δ-functors
i
(• ⊗R S))i∈Z .
(Hai (•) ⊗R S)i∈Z ∼
= (HaS

Under the hypothesis that a and aS are weakly proregular, the conclusion of the Base
Ring Independence Theorem is shown to hold in [17, 6.5], while the conclusion of the Flat
Base Change Theorem follows immediately from the definitions of weak proregularity and
Čech cohomology.
Up to now, we saw that several basic results on local cohomology can be generalised
by replacing noetherianness with ITI properties (and maybe some coherence hypotheses).
But alas!, this does not work for Grothendieck’s Vanishing Theorem ([4, 6.1.2]). We end
this article with a positive and a negative result in this direction for absolutely flat rings.
Keep in mind that absolutely flat rings have ITI with respect to every ideal by 1.12
(hence their ideals of finite type are weakly proregular by 3.2) and that they are moreover
coherent.
(3.7) Proposition Let R be an absolutely flat ring.
a) If a ⊆ R is an ideal of finite type, then Hai = 0 for every i > 0.
b) If R is non-noetherian, then there exist an ideal a ⊆ R not of finite type, an R-module
M and i > dim(M) such that Hai (M) 6= 0.
13The

hypothesis on S is fulfilled if it has ITI with respect to ideals of finite type.

INJECTIVE MODULES AND TORSION FUNCTORS

19

Proof. An ideal a ⊆ R is idempotent, hence Γa (•) ∼
= HomR (R/a, •), and thus Hai = 0 if
and only if the R-module R/a is projective. In case a is of finite type, the R-module R/a
is flat and of finite presentation, hence projective. This shows a).
Assume now R is non-noetherian and Hai = 0 for every i > 0. By the above, this
implies projectivity of every monogeneous R-module, thus by [2, VIII.8.2 Proposition 4]
the contradiction that R is semisimple (and in particular noetherian). This shows b). 
Acknowledgement: For their various help during the writing of this article we thank
Markus Brodmann, Paul-Jean Cahen, Neil Epstein, Shiro Goto, John Greenlees, Thomas
Preu, Rodney Sharp, and Kazuma Shimomoto. We are grateful to the referee for his
careful reading.
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Department of Mathematics, FPT University, 8 Ton That Thuyet, 10307 Hanoi, Vietnam
E-mail address: quyph@fpt.edu.vn
Grosse Grof 9, 9470 Buchs, Switzerland
E-mail address: fredrohrer@math.ch

