A note on zero-divisors of commutative rings

In this paper we show that if a ring R has finite Goldie dimension, then every finitely generated ideal of R consisting of zero-divisors has non-zero annihilator. We also construct an example of a ring of infinite Goldie dimension such that above condition does not hold.

Bartosiewicz asked (private communication), whether the converse holds, i.e. (P1) If the set of zero-divisors of a commutative ring R is an ideal of the ring R then R satisfies the condition ( ). The answer to the above question is negative, because there are examples of rings with ideals consisting entirely of zero-divisors and such that the ideals contain finite subsets with zero annihilator. General examples of such rings are presented in [7,16]. In Sect. 3 we give one more, as a simpler and better illustration of the situation considered in this paper.
It is well-known (cf. [1]) that the set of non-units of a commutative ring is an ideal if and only if the ring is local (i.e. has a unique maximal ideal). It seems that obtaining of similar characterization for zero-divisors is difficult. In this paper, we reduce that problem to the following question: for which classes of rings the condition ( ) is satisfied? The most natural class of rings satisfying this condition is the class of Noetherian rings. It is well-known that if I is an ideal in a Noetherian ring and if I consists of zero-divisors, then the annihilator of I is non-zero. Therefore, zero-divisors form an ideal in the class of Noetherian rings. In Sect. 3, we generalize that fact to the class of rings with finite Goldie dimension. Precisely, we show that the condition ( ) holds for each proper ring with finite Goldie dimension (Theorem 3.4). For different classes of rings, condition ( ) was considered by other authors, e.g. [8,11,15]. In our paper the obtained result concerns the class of rings that was never examined before. Moreover, in many cases, it is not difficult to verify whether a ring has a finite Goldie dimension. Therefore, Theorem 3.4 is a useful tool for studying the condition ( ). For example, this theorem can be applied to the ring of distributions K mentioned above for which the Goldie dimension is finite.
The related topics to the one examined in this paper one can find in [4,9,[12][13][14].

Preliminaries
All rings in this paper are associative and commutative but we do not assume that each ring has an identity element. To denote that I is an ideal of a ring R we write I R. For undefined terms and used facts we refer the reader to [1,10]. Note that if the set of zero-divisors of a ring R forms an ideal I , then I is a ring without identity consisting of zero-divisors. Thus it is quite natural to consider rings without identity.
It is easy to check that the (P1) is equivalent to the following question.
(P2) Is it true that the annihilator of every finite set is non zero for every commutative ring consisting of zero-divisors?
Now we recall some notions and a result, which will be used later. An ideal I of a ring R is called essential if for every non-zero ideal J of R, I ∩ J = 0. A non-zero ideal I of a ring R is called uniform if every nonzero ideal of R contained in I is essential in I . A ring R is said to have finite Goldie dimension if it does not contain infinite direct sums of non-zero ideals.
It is well-known (cf. [5,6]) that a ring R has a finite Goldie dimension if and only if it contains a direct sum I = I 1 ⊕ · · · ⊕ I n of uniform ideals I i and I is an essential ideal of R.
It is clear that Noetherian rings have a finite Goldie dimension.

Results
In this section we will construct a ring with infinite Goldie dimension, giving a negative answer to (P2). We also show that (P2) has a positive answer for rings with finite Goldie dimension. Suppose that R is a ring and M is a left and right R-module. M is called R-bimodule if for arbitrary a, b ∈ R and m ∈ M, a(mb) = (am)b.
If R is a ring and V is an R-bimodule, then the set is a ring with respect to canonical matrix addition and multiplication.
Example 3.1 Let P = F[x, y] be the polynomial ring in two commutative variables x, y over a field F and let A = x P + y P. Clearly, A is a commutative ring (without identity) and for every w ∈ A, w P P and P/w P has a natural structure of P-bimodule. Let N be the P-bimodule 0 =w∈A P/w P. Obviously N is also A-bimodule so we can form the ring R = a n 0 a | a ∈ A, n ∈ N .
Note that for arbitrary w ∈ A and n ∈ N , w n 0 w 0 e w 0 0 = 0 0 0 0 , where e w is the element of N , whose wth component is equal to 1 and all other components are equal to 0. Thus R consists of zero-divisors. Now we will show that ann R x 0 0 x , y 0 0 y = 0. Indeed, take a n 0 a ∈ ann R x 0 0 x , y 0 0 y . Then ax = ay = 0, so a = 0. Moreover xn = yn = 0. This means that for every 0 = w ∈ A, xn w = yn w = 0, where n w denotes the w-component of N . However each n w is of the form p w + w P for some p w ∈ P. Since xn w = yn w = 0 we get that x p w ∈ w P and yp w ∈ w P. Let x p w = wp 1 and yp w = wp 2 for some p 1 , p 2 ∈ P. Then x yp w = yx p 1 = xwp 2 , so yp 1 = x p 2 . This implies that p 1 = x p 1 and p 2 = yp 2 for some p 1 , p 1 ∈ P. Consequently, x p w = wx p 1 , so p w = wp 1 . This however means that p w + w P = w P, so each component of n is equal to 0. Therefore n = 0 which shows that a n 0 a = 0 and we are done.
One easily sees that the above constructed ring R contains the infinite direct sum 0 =w∈A 0 P/w P 0 0 of ideals, i.e., the Goldie dimension of R is infinite.
It is not hard to find an example of non-Notherian ring of finite Goldie dimension. Let F be a field and let X ∪ {θ } be a set of symbols {x α }, where α is an element of the set of positive real number with zero. Multiplication in X is defined as Under this operation the set X ∪ {θ } is a commutative semigroup. Then, let P = F 0 [X ] be the contracted semigroup algebra of X over F. It is clear that P has a finite Goldie dimension. In order to prove that P is non-Noetherian, it is enough to take the elements x α for α > 0 and consider the ideals I α = x α P. Now we will show that the condition ( ) holds for each proper ring with finite Goldie dimension. We will need the following two lemmas which are in fact known. The latter one is just a very classical result, which one can find in [1], the former is not so classical but also known. We include their simple proofs for completeness.
Recall that an ideal I of a ring R is called prime if for arbitrary elements x, y ∈ R \ I , x y ∈ I . Suppose that x, y ∈ J and x y ∈Ī J . Then there exists 0 = z ∈ I such that zx y = 0. If zx = 0 then x ∈Ī J . If zx = 0, then, since zx ∈ I , y ∈Ī J . These show that the idealĪ is prime.

Lemma 3.3
If I 1 , . . . , I n are prime ideals of a ring R and R = I 1 ∪ · · · ∪ I n , then R = I i for some 1 ≤ i ≤ n.

Theorem 3.4 If a proper ring R has a finite Goldie dimension, then every finitely generated ideal of R consisting of zero-divisors has non-zero annihilator.
Proof Since R is of finite Goldie dimension it contains an essential ideal I = I 1 ⊕ · · · ⊕ I n , where I 1 , . . . , I n are uniform ideals of R. Let J =< a 1 , a 2 , . . . , a k > be an ideal generated by a 1 , a 2 , . . . , a k ∈ R consisting of zero-divisors. Since J consists of zero-divisors, for every x ∈ J , ann R (x) = 0. Obviously ann R (x) is an ideal of R, so essentiality of I implies that I ∩ ann R (x) = 0. Thus there are x i ∈ I i , not all equal 0, such that (x 1 + · · · + x n )x = 0. However x i x ∈ I i , so x i x = 0 for all 1 ≤ i ≤ n. This shows that if x i = 0, then x ∈Ī Ji . Consequently J =Ī J 1 ∪ · · · ∪Ī J n . Applying Lemmas 3.2 and 3.3 we get that J =Ī Ji for some 1 ≤ i ≤ n. In particular a 1 , . . . , a k ∈Ī Ji , so ann R (a j ) ∩ I i = 0 for 1 ≤ j ≤ k. However I i is a uniform ideal of R, so T = ann R (a 1 ) ∩ · · · ∩ ann R (a k ) ∩ I i = 0. Clearly T ⊆ ann R ({a 1 , . . . , a k }) and we are done.