NONZERO INJECTIVE COVERS OF MODULES

Richard Belshoff and Jinzhong Xu

Abstract. We show that if R is a ring such that every nonzero left R–module has a nonzero injective cover, then R is left Artinian. The converse is not true. If R is commutative, then the properties are equivalent.

1. Background. A projective cover of a left R–module M (first introduced by H. Bass [2] in the early 1960s) can be characterized as a pair (P, f), where P is a projective left R–module and f: P → M is a linear map satisfying the following two properties:

(i) if P′ is projective and g: P′ → M is a linear map, then there exists a linear map h: P′ → P such that f ∘ h = g.

(ii) any linear map h: P → P such that f ∘ h = f is an automorphism of P.

The dual notion of an injective envelope of an R–module M consists of an injective left R–module E and a linear map f: M → E such that

(i) if E′ is injective and g: M → E′ is a linear map, then there exists a linear map h: E → E′ such that h ∘ f = g, and

(ii) any linear map h: E → E such that h ∘ f = f is an automorphism of E.

Using these categorical definitions of projective cover and injective envelope as motivation, other types of covers and envelopes can be defined. This was initiated by E. Enochs in [4]. An injective cover of a left R–module M consists of an injective module E and a linear map f: E → M satisfying the following two properties:

(i) if E′ is injective and g: E′ → M is a linear map, then there exists a linear map h: E′ → E such that f ∘ h = g.

(ii) any linear map h: E → E such that f ∘ h = f is an automorphism of E.

It was shown in [4] that a ring R is left Noetherian if and only if every left R–module has an injective cover. However, we find that injective covers of many nonzero modules are zero. For example, the injective cover of ℤ, or any finitely generated ℤ–module, is zero. Therefore, we are interested in studying a nonzero map ϕ: E → M, where M ≠ 0 is a left R–module, and E is an injective module. Using the result cited above, it is easy to see that for a left Noetherian ring R, if every nonzero module M has the property that there is a nonzero map E → M where E is injective, then M has a nonzero injective cover.

In this paper, R will denote an associative ring with identity, and all modules will be unital left R–module. We denote by l(x) = {r ∈ R | rx = 0} the left annihilator of x ∈ R, and if R is commutative we use the notation Ann(x).

2. Perfect Rings and Injective Modules. Perfect rings were originally defined and studied using projective and flat modules. If R is any ring, then by Theorem P of [2], R is left perfect if and only if R satisfies any one of the following equivalent conditions:

• every module has a projective cover.

• the Jacobson radical J = J(R) is T –nilpotent and R/J is Artinian

• every flat module is projective.

The results in this section provide a new way of viewing perfect rings by using injective modules.

\underline{Theorem 1}. Let J = J(R) be the Jacobson radical of a ring R and suppose that for every a ∈ R, the set of left annihilator ideals {l(a^{n}) | n ≥ 1} satisfies the ascending chain condition (ACC). If for every nonzero left R–module M there is a nonzero map E → M where E is injective, then J is a nil–ideal.

\underline{Proof}. By Zorn’s Lemma, there is a maximal nil–ideal N contained in J. We claim that N = J. If J/N ≠ 0, then there is a nonzero map ϕ: E → J/N where E is injective. Define Tr_{J/N}(E) to be the sum of all the images h(E), where h: E → J/N is a linear map. That is, Tr_{J/N}(E) = {∑ h(E) | h ∈ Hom_{R}(E, J/N)}. Certainly Tr_{J/N}(E) is a submodule of J/N. Thus, Tr_{J/N}(E) = N_{1}/N for some ideal N_{1} containing N, and N_{1}/N ≠ \overline{0}.

For any a ∈ N_{1}, there is an integer n such that l(a^{n}) = l(a^{2n}). Let \overline{c} = \overline{a}^{n} = a^{n} + N. Then \overline{c} ∈ Tr_{J/N}(E) and so there is an integer t so that \overline{c} = h_{1}(e_{1})+⋯+h_{t}(e_{t}), for some linear maps h_{i}: E → J/N, and some elements e_{i} ∈ E, for 1 ≤ i ≤ t. If we let e^{∗} = e_{1} + ⋯ + e_{t}, let E^{∗} = E ⊕ ⋯ ⊕ E denote t copies of E, and let h = ⊕_{i=1}^{t}h_{i}, then h: E^{∗} → J/N, e^{∗} ∈ E^{∗} and h(e^{∗}) = \overline{c}. Define f: Rc^{2} → E^{∗} by f(rc^{2}) = rce^{∗}. Then we have the diagram and since E^{∗} is injective, there is a map g which extends f. Let e^{∗} = g(1) ∈ E^{∗}. Then

ce^{∗} = f(c^{2}) = g(c^{2}) = c^{2}g(1) = c^{2}e_{1}^{∗}.

Now h(ce^{∗}) = h(c^{2}e_{1}^{∗}) implies that \overline{c}^{2}(1 − h(e_{1}^{∗})) = \overline{0} ∈ J/N ⊂ R/N. It follows that \overline{c}^{2} = \overline{0} ∈ R/N, so a^{2n} = c^{2} ∈ N which means that a is nilpotent in R. Therefore, N_{1} is a nil–ideal of R, and N ⊂ N_{1}.

\underline{Corollary 1}. If R is left self–injective and for every a ∈ R the set of left annihilator ideals {l(a^{n}) | n ≥ 1} has ACC, then J(R) is nil and R/J(R) is regular.

\underline{Proof}. If R is left self–injective, then every nonzero cyclic left R–module is a homomorphic image of an injective module, hence by Theorem 1, J(R) is nil and by Anderson–Fuller, [1] R/J(R) is regular.

Recall that a ring R is said to be left perfect if and only if every left R–module has a projective cover. This is the case if and only if J(R) is left T–nilpotent and R/J(R) is Artinian semi–simple. See [2].

\underline{Theorem 2}. If R satisfies the following conditions, then R is a left perfect ring.

(a) For every nonzero R–module M there is a nonzero map E → M where E is injective.

(b) For any sequence {a_{n} | n ≥ 1} in R, the set of left annihilator ideals {l(a_{1} ⋯ a_{n}) | n ≥ 1} has ACC.

(c) The union of any ascending chain of left T–nilpotent ideals is left T–nilpotent.

\underline{Proof}. We will first show that the Jacobson radical J is left T–nilpotent. By (c) and Zorn’s Lemma, there is a maximal left T–nilpotent ideal N ⊂ J. If J ≠ N, i.e. if J/N ≠ 0, then by hypothesis (a) we have a nonzero map ϕ: E → J/N where E is injective. Set Tr_{J/N}(E) = L/N, where L is an ideal of R containing N. By (b), for any sequence {a_{n} ∈ L | n ≥ 1}, there is an integer n_{1} such that l(a_{1} ⋯ a_{n_{1}−1}) = l(a_{1} ⋯ a_{n_{1}}). Now \overline{a_{n_{1}}} = a_{n_{1}} + N ∈ L/N. By definition of Tr_{J/N}(E), there is some integer t so that \overline{a_{n_{1}}} = h_{1}(e_{1}) + ⋯ + h_{t}(e_{t}), where h_{i}: E → J/N is a linear map, and e_{i} ∈ E for i = 1, …, t. Let E^{∗} be the direct sum of t copies of E, let h = ⊕_{i=1}^{t}h_{i}, and let e^{∗} = e_{1} ⊕ ⋯ ⊕ e_{t}. Then h: E^{∗} → J/N and h(e^{∗}) = \overline{a_{n_{1}}}. Now define f: Ra_{1} ⋯ a_{n_{1}−1}a_{n_{1}} → E^{∗} by f(ra_{1} ⋯ a_{n_{1}−1}a_{n_{1}}) = ra_{1} ⋯ a_{n_{1}−1}e^{∗}, and let g be a map which extends f. So we have the commutative diagram

If we let e_{1}^{∗} = g(1) ∈ E^{∗}, then

a_{1} ⋯ a_{n_{1}−1}e^{∗} = f(a_{1} ⋯ a_{n_{1}}) = a_{1} ⋯ a_{n_{1}}e_{1}^{∗},
h(a_{1} ⋯ a_{n_{1}−1}e^{∗}) = h(a_{1} ⋯ a_{n_{1}}e_{1}^{∗}),
a_{1} ⋯ a_{n−1}\overline{a_{n_{1}}} = a_{1} ⋯ a_{n_{1}}h(e_{1}^{∗}),
\overline{a_{1} ⋯ a_{n_{1}}} = \overline{a_{1} ⋯ a_{n_{1}}}h(e_{1}^{∗}).

So \overline{a_{1} ⋯ a_{n_{1}}}(\overline{1} − h(e_{1}^{∗})) = \overline{0} implies that a_{1} ⋯ a_{n_{1}} ∈ N. Set b_{1} = a_{1} ⋯ a_{n_{1}} ∈ N.

Similarly, for the sequence {a_{k} | k ≥ n_{1}+1}, we can find an integer n_{2} such that a_{n_{1}+1} ⋯ a_{n_{2}} ∈ N. Set b_{2} = a_{n_{1}+1} ⋯ a_{n_{2}}. Consequently we get a sequence {b_{t} | t ≥ 1} ⊂ N, where b_{t} = a_{n_{t}+1} ⋯ a_{n_{t+1}} with n_{0} = 0. Since N is left T–nilpotent, for some integer s we have b_{1}b_{2} ⋯ b_{s} = 0, and then a_{1} ⋯ a_{n_{s+1}} = 0. Hence L is left T–nilpotent. It follows that L = N is left T–nilpotent, and Tr_{J/N}(E) = L/N is zero. Therefore, J = N is left T–nilpotent.

Next we will prove that R/J is Artinian semi–simple. First of all, any set of orthogonal idempotent elements of R/J can be lifted to a set of orthogonal idempotent elements of R. The set of left annihilator ideals {l(a_{1} ⋯ a_{n}) | n ≥ 1} has ACC by assumption, and R/J has an indecomposable decomposition as a left R–module. Let \overline{R} = R/J = N_{1} ⊕ ⋯ ⊕ N_{t}, where each N_{i} is an indecomposable submodule of R/J. Now we will show that N_{i} is a simple R/J–module.

Let H be a nonzero submodule of N_{i}. Since N_{i} is indecomposable, in order to prove H = N_{i} we only need to show that H contains a nonzero direct summand of N_{i}. By the correspondence between idempotents and direct summands, we only need to construct a special idempotent element w in R/J such that it creates a direct summand of N_{i}.

By (a), there is an injective module E and a map ϕ: E → H with ϕ ≠ 0. Set Tr_{H}(E) = I/J ⊂ H, and I/J ≠ 0. For any a ∈ I, by (b) there is an integer n such that l(a^{n}) = l(a^{n+2}). Now \overline{a} = a + J ∈ Tr_{H}(E), so there is an integer t such that \overline{a} = h_{1}(e_{1})⊕⋯⊕h_{t}(e_{t}) for some linear maps h_{i}: E → H and for some e_{i} ∈ E. If we let E^{∗} be the direct sum of t copies of E, let e^{∗} = e_{1} ⊕ ⋯ ⊕ e_{t} and let h = ⊕_{i=1}^{t}h_{i}, then h: E^{∗} → H and h(e^{∗}) = \overline{a}. Now define f: Ra^{n+2} → E^{∗} by f(ra^{n+2}) = ra^{n}e^{∗}, and let g be a map which extends f. We have a commutative diagram

and if we set g(1) = e_{1}^{∗} ∈ E^{∗}, then

a^{n}e^{∗} = f(a^{n+2}) = g(a^{n+2}) = a^{n+2}e_{1}^{∗}.

Since h(E^{∗}) is contained in Tr_{H}(E) = I/J, if we set \overline{b} = h(e_{1}^{∗}) ∈ I/J, then since h(e^{∗}) = \overline{a},

\overline{a}^{n+1} = \overline{a}^{n+2}\overline{b}.

We can also find an integer m and an element \overline{c} ∈ I/J such that \overline{b}^{m} = \overline{b}^{m+1}\overline{c}. Consequently,

\overline{a}^{(n+1)+m+(n+1)+1}\overline{b}^{m+(n+1)+1} = \overline{a}^{n+1},
\overline{a}^{2(m+(n+1)+1)}\overline{b}^{m+(n+1)+1} = \overline{a}^{(n+1)+m+1},
\overline{b}^{m+(m+n+1)+1}\overline{c}^{m+n+1+1} = \overline{b}^{m},
\overline{b}^{2(m+(n+1)+1)}\overline{c}^{m+(n+1)+1} = \overline{b}^{m+(n+1)+1}.

Set x = \overline{a}^{n+1+m+1}, y = \overline{b}^{m+n+1+1} and z = \overline{c}^{m+n+1+1}. Then x^{2}y = x and y^{2}z = y. And because z − x ∈ I/J, there is an integer k and an element d ∈ I/J such that (z − x)^{k+1}d = (z − x)^{k}. From these three relations

x^{2}y = x,
y^{2}z = y,
(z − x)^{k+1}d = (z − x)^{k},

we can prove that x = xyx. (See the Appendix.)

Since x ≠ 0, if we set w = yx, then w ∈ I/J ⊂ H and w^{2} = yxyx = yx = w. Hence, \overline{R} = \overline{R}w ⊕ K, where \overline{R}w ⊂ N_{i}, and N_{i} = \overline{R}w ⊕ (K ∩ N_{i}). This implies that N_{i} = H, since N_{i} is indecomposable, and then R/J is Artinian semi–simple.

\underline{Corollary 2}. If every nonzero left R–module has a nonzero injective cover, then R is left Artinian.

\underline{Proof}. If every left R–module has an injective cover, then R is left Noetherian, and it follows that the hypotheses (a), (b) and (c) in Theorem 2 are all satisfied. Hence, R is left Artinian.

The following example shows that the converses to Theorem 2 and Corollary 2 are not true. We will give an example of a left Artinian ring R for which hypothesis (a) of Theorem 2 is not satisfied.

Let F be a field, let

e_{11} = ( \begin{matrix}1 & 0 \\ 0 & 0 \end{matrix} ) , e_{12} = ( \begin{matrix}0 & 1 \\ 0 & 0 \end{matrix} ) , e_{22} = ( \begin{matrix}0 & 0 \\ 0 & 1 \end{matrix} ) ,

and let

R = F e_{11} + F e_{12} + F e_{22} = \{ ( \begin{matrix}a & b \\ 0 & c \end{matrix} ) | a, b, c ∈ F \}.

We know that R is left Artinian. Now we will show that the hypothesis (a) is not satisfied by R.

Since R = Re_{11} ⊕ (Re_{12} + Re_{22}), Re_{11} is a projective simple module of R. If there is a nonzero map E → Re_{11} with E injective, then Re_{11} is injective.

However, the homomorphism f: Re_{12} → Re_{11} given by f(re_{12}) = re_{11}, can not be extended. For if we consider the diagram

where g | Re_{12} = f, if

g(1) = ( \begin{matrix}a & 0 \\ 0 & 0 \end{matrix} ) ∈ Re_{11},

then f(e_{12}) ≠ 0 but

g(e_{12}) = e_{12}g(1) = ( \begin{matrix}0 & 1 \\ 0 & 0 \end{matrix} ) ( \begin{matrix}a & 0 \\ 0 & 0 \end{matrix} ) = ( \begin{matrix}0 & 0 \\ 0 & 0 \end{matrix} ) .

\underline{Theorem 3}. If R is commutative, then the following statements are equivalent.

1. R is perfect.

2. (a) For every nonzero R-module M there is a nonzero linear map E → M where E is injective.

(b) For any sequence a_{1}, a_{2}, … in R, the set of annihilator ideals {Ann(a_{1} … a_{n}) | n ≥ 1} has ACC.

\underline{Proof}. (1) ⟹ (2). Because R has DCC on principal ideals [2], there is an integer n such that Ra_{1} ⋯ a_{n} = Ra_{1} ⋯ a_{n+k}, for all k ≥ 1. This implies that Ann(a_{1} ⋯ a_{n}) = Ann(a_{1} ⋯ a_{n+k}) for k ≥ 1. Hence, (b) is satisfied. For (a), consider the decomposition R = R_{1} ⊕ ⋯ ⊕ R_{t}, where each R_{i} is a local ring with maximal ideal m_{i}. For any nonzero R_{i}–module M_{i}, the simple module S_{i} = R_{i}/m_{i} can be embedded in M_{i}. Let E_{i}(M_{i}) be the injective envelope of M_{i} as an R_{i}– module. Then we have a nonzero map E_{i}(M_{i}) → E_{i}(M_{i})/m_{i}E_{i}(M_{i}) → S_{i} ↪ M_{i}.

Since R is commutative, E_{i}(M_{i}) is R–injective. Therefore, for any nonzero R– module M, we have M = R_{1}M ⊕ ⋯ ⊕ R_{t}M, where each M_{i} = R_{i}M is an R_{i}– module. Now it is easy to see that there is a nonzero linear map E → M where E is injective.

(2) ⟹ (1). By Theorem 1, J is nil. For any sequence {a_{n} | n ≥ 1} in J, there is an integer n such that Ann(a_{1} ⋯ a_{n+1}) = Ann(a_{1} ⋯ a_{n}). Since a_{n+1} is nilpotent, a_{n+1}^{k} = 0 and a_{n+1}^{k−1}a_{1} ⋯ a_{n}a_{n+1} = 0. Then a_{n+1}^{k−1} ∈ Ann(a_{1} ⋯ a_{n}), and so a_{n+1}^{k−1}a_{1} ⋯ a_{n} = 0. Continuing in this way we eventually get that a_{n+1}^{2}a_{1} ⋯ a_{n} = 0, which implies that a_{n+1} ∈ Ann(a_{1} ⋯ a_{n}). This shows that J is T–nilpotent. Then by the proof of Theorem 2, R is perfect.

\underline{Corollary 3}. If R is commutative, then R is Artinian if and only if every nonzero R–module has a nonzero injective cover.

3. Appendix. The purpose of this appendix is to prove the following proposition. The main idea of the proof is contained in the article by M. F. Dischinger [3].

\underline{Proposition 1}. If x, y, z, d are elements of R, and if

(i) x^{2}y = x, (ii) y^{2}z = y, (iii) (z − x)^{k+1}d = (z − x)^{k} for some positive integer k,

then xyx = x.

\underline{Proof}. Using (i) and (ii) it is easy to see that

xz = x^{2},  (1)
xyz = x.  (2)

The first equation follows because xz = x^{2}yz = x(x^{2}y)yz = xx^{2}y = xx. The second equation follows by a similar calculation: xyz = x^{2}yyz = x^{2}y = x.

Using equation (1), we have (z − x)^{2} = z^{2} − zx − xz + x^{2} = z^{2} − zx and so

(z − x)^{2} = z(z − x).

Then an induction argument shows that for any integer n ≥ 1,

(z − x)^{n} = z^{n−1}(z − x).  (3)

Now use this and equations (1) and (2) to see that

xy(z − x)^{2} = xyz(z − x) = x(z − x) = xz − x^{2} = 0.  (4)

Also, by hypothesis (iii), there is some positive integer k so that

(z − x)^{k} = (z − x)^{k+1}d = (z − x)^{k}(z − x)d = z^{k−1}(z − x)^{2}d.

Using hypothesis (ii) as the basis for another induction argument, it follows that y^{n}z^{n−1} = y for every positive integer n. Hence by (3),

y^{n}(z − x)^{n} = y^{n}z^{n−1}(z − x) = y(z − x)  (5)

for every positive integer n. Now using equation (3), hypothesis (iii) and the equation above, we have

y(z − x)^{2}d = y^{k}z^{k−1}(z − x)^{2}d = y^{k}(z − x)^{k+1}d = y^{k}(z − x)^{k} = y(z − x) (6)

for some positive integer k. By equation (5) and this equation,

y(z − x)d = y^{2}(z − x)^{2}d = y^{2}(z − x).

By equations (4) and (6)

xy(z − x) = xy(z − x)^{2}d = 0.

Therefore, xyz = xyx. Now since x = xyz, we conclude that xyx = x.

\underline{References}

1. F. Anderson and K. Fuller, “Rings and Categories of Modules,” Springer– Verlag, Berlin and New York, 1974.

2. H. Bass, “Finitistic Dimension and a Homological Generalization of Semi– primary Rings,” Transactions of the American Mathematical Society, 15 (1960), 466–488.

3. M. F. Dischinger, “Sur les Anneaux Fortement π–Réguliers,” C. R. Acad. Sc. Paris, 283 (1976), A571–A573.

4. E. Enochs, “Injective and Flat Covers, Envelopes and Resolvents,” Israel Journal of Mathematics, 39 (1981), 189–209.

Richard Belshoff Department of Mathematics Southwest Missouri State University Springfield, MO 65804 email: belshoff@math.smsu.edu

Jinzhong Xu Department of Mathematics University of Kentucky Lexington, KY 40506 email: abc@ms.uky.edu