On a Property of Nilpotent Matrices over an Algebraically Closed Field

Suppose 𝐹 is an algebraically closed field. We prove that the ring ∏︀ ∞ 𝑛 =1 M 𝑛 ( 𝐹 ) has a special property which is, somewhat, in sharp parallel with (and slightly better than) a property established by ˇSter (LAA, 2018) for the rings ∏︀ ∞ 𝑛 =1 M 𝑛 ( Z 2 ) and ∏︀ ∞ 𝑛 =1 M 𝑛 ( Z 4 ) , where Z 2 is the finite simple field of two elements and Z 4 is the finite indecomposable ring of four elements.

All rings are assumed here to be associative, containing the identity element 1 which differs from the zero element 0 of . Recall that a ring is nil-clean provided that each its element is a sum of a nilpotent and an idempotent, is -regular provided that for every element ∈ there is ∈ N such that ∈ , and is strongly -regular provided that ∈ +1 . In his seminal paper [4],Šter showed that the ring ∏︀ ∞ =1 M (Z 2 ) is nil-clean but not strongly -regular, whereas the ring ∏︀ ∞ =1 M (Z 4 ) is nil-clean but not -regular. He utilizes an innovation of the method used in [1]. Specifically, for any ∈ N, it was proved there that, for every × matrix over the finite field Z 2 , there exists an idempotent matrix such that ( − ) 4 = 0, while the index of nilpotence over the finite ring Z 4 is precisely 8. As usual, the symbol will stand in the sequel the standard matrix identity. Thereby, = + for some 4 = 0 and hence ( − ) = ( − ) , but it is not clear at all whether [( − ) ] 4 = 0 will hold eventually.
On the other side, in [2] we have examined rings having the property that, for each ∈ , there is an idempotent ∈ such that (1 − ) is nilpotent. We shall be here even rather more precise by considering an existing idempotent ∈ with [(1 − ) ] 2 = 0. It is well known that finite fields are, surely, not algebraically closed. So, the purpose of this very short note is to show that some (although little) improvement is possible by a strengthening of the technique utilized in [2] in the case of algebraically closed fields.
Before proceed by proving our chief result, we need the next two technical statements. Lemma 1. Let be a unital ring, ≥ 2, and = ∑︀ −1
Proof. First, suppose that = 2. Then and hence, taking = 0 ∈ M ( ) , we have (( − ) ) 2 = 2 = 0. Let us therefore assume that ≥ 3, and let Then ∈ M ( ) , is clearly an idempotent, and Proof. For each let denote the projection of onto the component M ( ) in . Since is algebraically closed, for each we can find an invertible matrix ∈ M ( ) such that = −1 is in Jordan canonical form. By Lemma 2, for each we can find an idempotent matrix ∈ M ( ) such that (( − ) ) 2 = 0. Now, for each let = −1 , and let = ( 1 , 2 , . . . ) ∈ . Since each is idempotent, the same holds for each , and hence also for . Also, since ∈ M ( ) and is invertible, we have for each that and hence ∈ . Finally, since (( − ) ) 2 = 0, for each we have from which it follows that (( − ) ) 2 = 0, as required. 2 We end our work with the following challenging query: Problem 1. Extend the considered above property for any field which is not necessarily algebraically closed.
An intuitive idea could be the following one: It is enough to establish the claim for a given M ( ) with the index of the nilpotent (1 − ) bounded independent of . Since every matrix is the direct sum of a unit and a nilpotent (we do not need the field to be algebraically closed for this), it is enough to do the assertion for units and for nilpotents. For a unit , we take = 1. Now suppose is nilpotent. It is enough to do the statement for the Weyr canonical form of -for more details the interested reader can see [3]. Thus assume has Weyr structure ( 1 , 2 , ..., ). The idea is to get an idempotent in that is diagonal, has 0's in the first 1 places and the last , and such that (1 − ) has zero blocks (relative to the partition 1 , .., ) except in the (1, 2) block. Then index of the nilpotent (1 − ) is exactly 2.
We will illustrate in the case of a homogeneous structure (3, 3, 3, 3) but the argument in the nonhomogeneous case is similar although a little trickier. Thus, in terms of 3 × 3 blocks and = 3 , we will have that