A note on simple modules over quasi-local rings

Matlis showed that the injective hull of a simple module over a commutative Noetherian ring is Artinian. Many non-commutative Noetherian rings whose injective hulls of simple modules are locally Artinian have been extensively studied recently. This property had been denoted by property $(\diamond)$. In this paper we investigate, which non-Noetherian semiprimary commutative quasi-local rings $(R, m)$ satisfy property $(\diamond)$. For quasi-local rings $(R,m)$ with $m^3=0$, we prove a characterisation of this property in terms of the dual space of $Soc(R)$. Furthermore, we show that $(R,m)$ satisfies $(\diamond)$ if and only if its associated graded ring $gr(R)$ does. Given a field $F$ and vector spaces $V$ and $W$ and a symmetric bilinear map $\beta:V\times V\rightarrow W$ we consider commutative quasi-local rings of the form $F\times V \times W$, whose product is given by $(\lambda_1, v_1,w_1)(\lambda_2,v_2,w_2) = (\lambda_1\lambda_2, \lambda_1v_2+\lambda_2v_1, \lambda_1w_2+\lambda_2w_1+\beta(v_1,v_2))$ in order to build new examples and to illustrate our theory. In particular we prove that any quasi-local commutative ring with radical cube-zero does not satisfy $(\diamond)$ if and only if it has a factor, whose associated graded ring is of the form $F\times V \times F$ with $V$ infinite dimensional and $\beta$ non-degenerated.


Introduction
The structure and in particular finiteness conditions of injective hulls of simple modules have been widely studied. Rosenberg and Zelinsky's work [16] is one of the earliest studies of finiteness conditions on the injective hull of a simple module. Matlis showed in his seminal paper [13] that any injective hull of a simple module over a commutative Noetherian module is Artinian. Jans in [11] has termed a ring R to be left co-noetherian if every simple left R-module has an Artinian injective hull. Vamos showed in [21] that a commutative ring R is co-noetherian if and only if R m is Noetherian for any maximal ideal m ∈ Max(R)generalizing in this way Matlis' result. In connection with the Jacobson Conjecture for non-commutative Noetherian rings, Jategaonkar showed in [12] (see also [5,18]) that the injective hulls of simple modules are locally Artinian, i.e. any finitely generated submodule is Artinian, provided the ring R is fully bounded Noetherian. We say that a ring R satisfies condition (⋄) if (⋄) Injective hulls of simple left R-modules are locally Artinian.
In this paper we study (⋄) for, not necessarily Noetherian, quasi-local commutative rings R with maximal ideal m such that m 3 = 0. A description of such rings is given in terms of the dual space of Soc(R) seen as a vector space over R/m (Theorem 12). Furthermore, we relate property (⋄) of (R, m) with its associated graded ring gr(R) = R/m ⊕ m/m 2 ⊕ m 2 in Corollary 14. Given a field F and vector spaces V and W and a symmetric bilinear map β : V × V → W we consider commutative quasi-local rings of the form F × V × W , whose product is given by to build new examples and to illustrate our theory. In particular we prove in Proposition 17 that a quasi-local commutative ring with radical cube-zero does not satisfy (⋄) if and only if it has a factor whose associated graded ring is of the form F × V × F with V of infinite dimension and β non-degenerate.

Preliminaries
The following Lemma shows that conditon (⋄) is intrinsically linked to Krull's intersection Theorem: Lemma 1. Let R be a (not necessarily commutative) ring with Jacobson radical J, such that finitely generated Artinian modules have finite length. If R has property (⋄), then for any left ideal I of R one has ∞ n=0 (I + J n ) = I Proof. Let I be any left ideal of R. Then R/I embeds into a product of cyclic modules R/K i with essential simple socle by Birkhoff's theorem, where I ⊆ K i and the intersection K i = I. By hypothesis each of these modules R/K i is Artinian and hence has finite length. Thus there exists a number n i ≥ 1 such that J n i R/K i = 0 ⇔ J n i ⊆ K i . Hence I + J n i ⊆ K i for all i and as the intersection of the K i 's is I, we have

Remark 2.
(1) Assuming the hypotheses of Lemma 1, one can easily adapt the above proof to show that ∞ n=0 J n M = 0 for any finitely generated left R-module M. Furthermore, if M is a finitely generated essential extension of a simple left Rmodule, then there exists n > 0 such that J n M = 0. Recall that a ring R with Jacobson radical J is called semilocal if R/J is semisimple. A semilocal ring with nilpotent Jacobson radical is called semiprimary. The second socle of a module M is the submodule Soc  (c) ⇒ (a) is clear since if I is a left ideal such that M = R/I is a cyclic essential extension of a simple left R-module, then (I : J)/I is cyclic and by assumption (I : J 2 )/I is finitely generated. Hence (I : J 2 )/I has finite length. Applying our hypothesis to I ′ = (I : J), we can conclude that (I : J 3 )/I has finite length. Continuing we have also that R/I = (I : J n )/I has finite length.
A sufficient condition for a ring to satisfy (⋄) is given by the following Lemma. Proof. Suppose I ⊂ K ⊆ R are ideals such that K/I is a simple R-module and essential in R/I. If Soc(R) ⊆ I, then R/I is a factor of R/Soc(R) and hence Artinian. If Soc(R) I, then (Soc(R) + I)/I is a semisimple submodule of R/I and hence must equal K/I, i.e. Soc(R) + I = K. As a quotient of R/Soc(R), the module R/K = R/(Soc(R) + I) is a Artinian and so is R/I. Clearly it is not necessary for a ring R with (⋄) to satisfy R/Soc( R R) being Artinian. Moreover, Example 16 shows that there are commutative rings R such that R/Soc(R) satisfies (⋄), but R does not.

Local-Global Argument
Jans in [11] defined a ring R to be left co-noetherian if for every simple left R-module its injective hull is Artinian. Vamos has shown in [21] that a commutative ring R is co-noetherian if and only if R m is Noetherian for all m ∈ Max(R).
The following Lemma shows the relation between co-Noetherianess and condition (⋄) for commutative quasi-local rings. The proof follows the ideas of [20, Theorem 1.8]. It follows that m = B and hence m is finitely generated. Suppose that R is not Noetherian. Let Q be maximal among the ideals C of R such that C is not finitely generated. Then Q is a prime ideal of R by a standard argument (see [6,Theorem 2]). Clearly Q = m. Let p ∈ m with p / ∈ Q. By the choice of Q the ideal Q + Rp is finitely generated, say

Lemma 5. The following statements are equivalent for a commutative quasi-local ring
It follows that Q = D and hence Q is finitely generated, a contradiction. Thus R is Noetherian.
The diamond condition is equivalent to the condition that any injective hull of a simple R-module is locally Artinian, i.e. finitely generated submodules are Artinian. As seen in Remark 2(2) a commutative ring satisfies (⋄) if and only if any injective hull E of a simple R-module is locally of finite length, i.e. any finitely generated submodule of E has finite length.
The following Proposition is well-known and can be found for example in [19, Proposition 5.6] Proposition 6. Let R be commutative ring, m a maximal ideal of R and denote by R m the localisation of R by m. Then the injective hull It would be good to have a kind of local-global argument for condition (⋄) in comparison to Vamos' result on co-Noetherian rings. The following Proposition intends to find this kind of argument.
Proposition 7. Let R be a commutative ring, m ∈ Max(R) and E = E(R/m) the injective hull of R/m. The following statements are equivalent: (a) E is locally of finite length as R m -module; (b) E is locally of finite length as R-module and the R-submodule generated by an element x ∈ E and the R m -submodule generated by x coincide.
Hence the R-submodule generated by any set of elements of E coincides with the R m -submodule generated by that set. In particular any finitely generated R-submodule of E is also a finitely generated R m -submodule of E, which has finite length.
The condition that R-submodules and R m -submodules generated by a set coincides means that the lattice of submodules of R E and Rm E are identical. Hence E is locally of finite length as R-module implies also that it is locally of finite length as R m -module. Question: Does there exist a commmutative ring R that satisfies (⋄), but R m does not satisfy (⋄), for some m ∈ Max(R) ?

Commutative semiprimary quasi-local rings
A quasi-local commutative ring is co-noetherian if and only if it is Noetherian. Recall, that an ideal I of a ring R is called subdirectly irreducible if R/I has an essential simple socle. Clearly a ring R satisfies (⋄) if and only if R/I is Artinian, for all subdirectly irreducible ideals I of R.

4.1.
Commutative quasi-local rings with square-zero maximal ideal. Given any vector space V over a field F , the trivial extension (or idealization) is defined on the vector space R = F × V with multiplication given by (a, v)(b, w) = (ab, aw + vb), for all a, b ∈ F and v, w ∈ V . Any such trivial extension R is a commutative quasi-local ring that satisfies (⋄). However R is Noetherian if and only if V is finite dimensional.
Proof. Let (R, m) be a commutative quasi-local ring with m 2 = 0. Since m is a vector space over R/m it is semisimple. Let K be any subdirectly irreducible ideal of R. If K = m, then R/m is simple. So assume K ⊂ m. Then there exists a complement L such that m = L ⊕ K and Soc(R is a short exact sequence. Hence R/K has length 2.

4.2.
Commutative quasi-local rings with cube-zero maximal ideal. In this section we will characterise commutative quasi-local rings (R, m) with m 3 = 0 satisfying (⋄). Recall that Soc(R) = Ann(m) = {r ∈ R : rm = 0}. Hence m 2 ⊆ Soc(R). We start with a simple observation. The last Lemma raises the question, whether the converse of (1) holds? That is, whether m/Soc(R) needs to be finitely generated for a commutative quasi-local ring R with m 3 = 0 and satisfying property (⋄)? As we will see in Example 18, this need not be the case. To show that V f = Soc(R) + I we use the essentiality of Soc(R/I) = (Soc(R) + I)/I in R/I: For any a ∈ V f \ I, there exists r ∈ R such that ra + I is a non-zero element of Soc(R/I) = (Soc(R) + I)/I. Note that r ∈ m since otherwise f (ra) = 0 and hence ra ∈ Ker(f ) ⊆ I. Therefore r is invertible and a + I = r −1 ra + I ∈ Soc(R/I), i.e. V f = Soc(R) + I.
On the contrary, let f be any non-zero element f ∈ Hom F (Soc(R), F ). Then there exists a non-zero element x ∈ Soc(R) with f (x) = 1 and hence Soc(R) = F x ⊕ Ker(f ). Let I be an ideal of R that contains Ker(f ) and that is maximal with respect to x ∈ I. Thus I is subdirectly irreducible and Soc(R/I) = Soc(R) + I = (F x ⊕ I)/I is simple and essential in R/I. By construction Ker(f ) = I ∩ Soc(R).
Note that m(Soc(R) + I) = mI ⊆ m 2 ∩ I ⊆ Soc(R) ∩ I = Ker(f ), i.e. Soc(R) + I ⊆ V f . To show the converse, let a ∈ V f \ I, then by essentiality there exists r ∈ R with ra = x + y ∈ Soc(R) + I with x ∈ Soc(R) and y ∈ I and ra ∈ I. If r ∈ m, then ra ∈ Ker(f ) ⊆ I, contradicting essentiality. Hence r ∈ m and a ∈ Soc(R) + I, i.e. V f = Soc(R) + I. Suppose m/V f is finite dimensional for any f ∈ Hom(Soc(R), F ). Let I be a subdirectly irreducible ideal of R. If Soc(R) ⊆ I, then R/I is an R/m 2 -module. Since R/m 2 is a quasi-local ring with square-zero radical, we have by Lemma 9, that R/m 2 satisfies (⋄). Hence R/I must be Artinian. If Soc(R) ⊆ I, then by Lemma 11, there exists a non-zero map f : Soc(R) → F such that Soc(R) + I = V f . By hypothesis m/V f is finite dimensional and is therefore Artinian as R-module. As R/m and V f /I are simple modules, also R/I is Artinian, proving that R/I is Artinian for any subdirectly irreducible ideal I of R, i.e. R satisfies (⋄).
Let (R, m) be any commutative quasi-local ring. The associated graded ring of R with respect to the m-filtration is the commutative ring gr(R) = n≥0 m n /m n+1 with multiplication given by For any ideal I of R, the associated graded ideal is gr(I) = n≥0 (I ∩ m n + m n+1 )/m n+1 .
In particular gr(m) = n≥1 m n /m n+1 is the unique maximal ideal of gr(R). Hence (gr(R), gr(m)) is a commutative quasi-local ring with residue field F = R/m. Furthermore, gr(Soc(R)) is contained in Soc(gr(R)).
Lemma 13. Let (R, m) be a commutative quasi-local ring with residue field F and m 3 = 0. For any f ∈ Hom F (gr(Soc(R)), F ) there exists g ∈ Hom F (Soc(R), F ) such that where V g is the ideal of R defined by Lemma 11. Proof. Let f : gr(Soc(R)) → F and denote by π : Soc(R) → m 2 the projection onto m 2 , since m 2 is a direct summand of Soc(R). Define g : Soc(R) → F by g(a) = f (0, a, π(a)), for all a ∈ Soc(R). Then (0, x, y) ∈ V f if and only if f (0, 0, tx) = 0, for all t ∈ m. Since tx = π(tx) ∈ m 2 , the later is equivalent to g(tx) = 0 for all t ∈ m, i.e. x ∈ V g (in the ring R). Proof. If R satisfies (⋄) and f : Soc(gr(R)) → F is a non-zero map, then by Lemma 13 there exists g : Soc(R) → F such that V f = gr(V g ). Since V g contains Soc(R) and hence m 2 , we have V f = gr(V g ) = 0 × V g /m 2 × m 2 . Thus, gr(m)/V f ≃ m/V g . By Theorem 12, m/V g is finite dimensional as R satisfies (⋄). Hence gr(m)/V f is finite dimensional for all f ∈ Soc(gr(R)) * . Again by Theorem 12, gr(R) satisfies (⋄).
Let f : Soc(R) → F be any non-zero linear map and let V f = {a ∈ m | f (ma) = 0}. If m 2 ⊆ Ker(f ), then V f = m. If m 2 ⊆ Ker(f ), then there exists x ∈ m 2 with f (x) = 1. Note that V 2 f ⊆ Ker(f ), hence I = 0 × V f /m 2 × ker(f ) is a subdirectly irreducible ideal of gr(R). To see this note that E = 0×0×F x is a simple submodule of gr(R)/I ≃ F ×m/V f ×F x. We will show that E is essential in gr(R)/I. Let (0, a, b) ∈ gr(R)/I. If a ∈ V f , there exists c ∈ m such that f (ac) = 0 and ac − f (ac)x ∈ Ker(f ) . Hence (0, a, b) a = 0, and b = 0, then (0, 0, b) is a non-zero element of E. Hence E is an essential simple submodule of gr(R)/I and if gr(R) satisfies (⋄), the quotient gr(R)/I and therefore also m/V f must be Artinian, thus finite dimensional. By Theorem 12, R satisfies (⋄).

Examples
Let (R, m) be a commutative quasi-local ring with m 3 = 0. The associated graded ring gr(R) is of the form gr(R) = F ⊕ V ⊕ W where F = R/m and V = m/m 2 and W = m 2 are spaces over F . Moreover, the multiplication of R induces a symmetric bilinear map β : V × V → W . Hence gr(R) is uniquely determined by (F, V, W, β) and its multiplication can be identified with the multiplication of a generalized matrix ring. Writing the elements of S = F × V × W as 3-tuples (λ, v, w) we have that the multiplication is given by The units are precisely the elements (λ, v, w) with λ = 0 and the unique maximal ideal of S is given by Jac . Recall, that β is called non-degenerate or non-singular if V ⊥ β = 0. In general β need not be non-degenerate: Example 15. Let F be any field and V be any vector space over F with countably infinite basis {v 0 , v 1 , v 2 , . . .}. Define a symmetric bilinear form as β : . The bilinear form of the last example was not non-degenerate. Since 0 ×V ⊥ β ×0 is always an ideal, we can pass to F × V /V ⊥ β × W where the bilinear form β is now non-degenerate. The following is a natural example of such a ring with non-degenerate bilinear form: Then , : V × V → R is a non-degenerate symmetric bilinear form on V . Hence, by Lemma 10, R = R × V × R is a commutative quasi-local ring with cube-zero radical, that does not satisfy (⋄), because its socle Soc(R) = 0 × 0 × R is one-dimensional, hence finitely generated, but Jac(R)/Soc(R) is infinite dimensional, hence not finitely generated as R-module.
These kind of rings must occur as the associated graded ring of a quotient of a commutative quasi-local ring (R, m) with m 3 = 0 that does not satisfy (⋄). On the other hand if R has a factor R/I whose associated graded ring gr(R/I) is of the form F × V × F with infinite dimensional vector space V and non-degenerate bilinear form β : V × V → F , then Soc(gr(R/I)) = 0 × 0 × F is a simple submodule of gr(R/I), which is essential since β is non-degenerate. As V is infinite dimensional, the semisimple gr(R/I)-module (0 × V × F )/(0 × 0 × F ) is not artinian and hence gr(R/I) does not satisfy (⋄). By Corollary 14, R/I does not satisfy (⋄) and therefore also R does not.
Let V = A be any unital commutative F -algebra. Consider the multiplication of A as a symmetric non-degenerate bilinear map µ : A × A → A and form the ring S = F × A × A as before. In order to apply Theorem 12 recall that m = 0 × A × A and Soc(S) = 0 × 0 × A, as the multiplication of A is non-degenerate. Hence elements of Soc(S) * can be identified with elements of A * . For any f ∈ A * we defined where I(f ) is the largest ideal of A that is contained in Ker(f ). Theorem 12 says that S satisfies (⋄) if and only if m/V f ≃ A/I(f ) is finite dimensional for any f ∈ A * . From the theory of coalgebras, we borrow the notion of the finite dual A • of an algebra, which is the subspace of A * consisting of the elements f ∈ A * that contain an ideal of finite codimension in their kernel. Hence S satisfies (⋄) if and only if A • = A * .
Example 18. The trivial extension A = F × V of a vector space V (see 4.1) is an example of an algebra A satisfying A • = A * . To see this, note that for any linear subspace U of A, U ∩ V is an ideal of A. Thus, if f ∈ A * , then Ker(f ) ∩ V is an ideal of codimension less or equal to 2 and f ∈ A • . In particular for such A, S = F × A × A satisfies (⋄). However if V is infinite dimensional, then m/Soc(S) ≃ V is not finitely generated as S-module, which shows that the converse of Lemma 10(1) does not hold.
However, it might happen that the kernel of an element f ∈ A * does not contain an ideal of finite codimension as the following example shows. Thus v = (λ 0 , λ 1 , . . . , λ n ) is in the kernel of the linear map given by the matrix: In particular det(B n ) = 0. Hence if the sequence (f (b n )) n∈N produces a sequence of matrices (B n ) that have all non-zero determinant, then for each 0 = a = n i=0 λ i b i ∈ A, there exists 0 ≤ m ≤ n such that f (b m a) = 0, i.e. β = f • β is non-degenerate.
Matrices of the form of B n are called Toeplitz or Hankel matrices. A particular example of such a matrix is the Hilbert matrix, which is the matrix , where c n = n−1 i=1 i n−i (see [9]). Hence if we define f (b n ) = 1 n+1 for any n ≥ 0 , then the kernel of f does not contain any non-zero ideal and the bilinear form β = f •µ is non-degenerate. An algebra with a multiplicative basis as above is for example the polynomial algebra A = F [x]. Thus S = F × F [x] × F [x] does not satisfy (⋄).