ANNIHILATORS OF NILPOTENT ELEMENTS

Let x be a nilpotent element of an infinite ring R (not necessarily with 1 ). We prove that A ( x ) —the two-sided annihilator of x —has a large intersection with any infinite ideal I of R in the sense that card ( A ( x ) ∩ I ) = card I . In particular, card A ( x ) = card R ; and this is applied to prove that if N is the set of nilpotent elements of R and R ≠ N , then card ( R \ N ) ≥ card N .

For an element x of a ring R, let A (x), A r (x), and A(x) denote, respectively, the left, right and two-sided annihilator of x in R. For a set X, we denote cardX by |X|; and say that a subset Y of X is large in X if |Y | = |X|. We prove that if x is any nilpotent element and I is any infinite ideal of R, then A(x) ∩ I is large in I, and in particular |A (x)| = |A r (x)| = |A(x)| = |R|. The last result is applied to obtain a generalization of a result of Putcha and Yaqub [2] which shows that an infinite nonnil ring has infinitely many nonnilpotent elements. A short proof of their result is given in [1]. We prove a much stronger result showing that the set of nonnilpotent elements of a nonnil ring is at least as large as is its set of nilpotent elements. The following lemma is simple but crucial.
Proof. Consider the map y → yx from (S,+) onto (Sx,+). The kernel is A (x) ∩ S, so |S| = |Sx||A (x) ∩ S| and the result follows since S is infinite.
A subset of a ring R is said to be root closed if whenever it contains a power of an element, it also contains the element itself.
Theorem 2. Let R be an infinite ring and α an infinite cardinal. Then, the following hold.
Proof. (i) Let |A r (x n ) ∩ I| = α for some n ≥ 2 and consider x n−1 (A r (x n ) ∩ I). By Lemma 1, Applying the previous theorem for I = R, we obtain the following corollary. The previous corollary will be applied in the proof of the above-mentioned generalization of a result of Putcha and Yaqub [2]. We also need the following result.

Lemma 4. Let b be a nonnilpotent element of an infinite ring R. If R\N is infinite, then
In a ring with 1, the map x → 1 + x from N into R\N is 1 − 1, so |R\N | ≥ |N |. The next theorem shows that the same result holds in any nonnil ring. In particular, we get the result of Putcha and Yaqub [2] stating that R is finite when R\N is finite and not empty.
Proof. We start with R infinite. Suppose |R\N | < |N |, then |N | = |R| and |R\N | < |R|. By the previous lemma, if b ∈ R\N, |A (b)| ≤ |R\N |, so |A (b)| < |R| and by Lemma 1, |Rb| = |R|. Now |R| = |Rb| ≤ |Nb| + |(R\N )b| and |(R\N )b| ≤ |R\N | < |R|, so |Nb| = |R|. Therefore, |{b + xb|x ∈ N}| = |R|, so since |R\N | < |R|, there exists x ∈ N such that b + xb ∈ R\N, namely b + xb ∈ N. Since x ∈ N, 1 + x is formally invertible, so A r (b + xb) = A r (b). By Corollary 3, |A r (b + xb)| = |R| and by Lemma 4, |A r (b)| ≤ |R\N | < |R|, a contradiction. Now let R be finite and let J be its radical. Since J is nilpotent, if a ∈ R, a + J is nilpotent in R/J if and only if a is nilpotent, and if a ∈ N, (a + J) ∩ N = ∅. Since R/J is a finite semisimple ring, it has 1, so at least half of its elements are nonnilpotent, hence at least half of the distinct cosets a + J, a ∈ R, do not intersect N, and therefore at least half of the elements of R are not nilpotent, so |R\N | ≥ |N |.