If $[A,A]A[\lambda,A] = 0$ then $\lambda \in Z(A).$

Question

Suppose that $A$ is a unital ring and $([A,A]) = A.$ If $[A,A]A[\lambda,A] = 0$ prove that $\lambda \in Z(A).$

Comments: This is part of an exercise I'm doing, I'm posting this part because I am unable to see and thereby end up exercise.

Answer 1

Well, if we assume that

$[A, A] = \{ [x, y] \mid x, y \in A \}, \tag{1}$

$[\lambda, A ] =\{ [\lambda, x ] \mid x \in A \}, \tag{2}$

and in general, for $C, D \subset A$,

$CD = \{ cd \mid c \in C, d \in D \}, \tag{3}$

then I think we can proceed as follows: first of all, we describe $(C)$ for $C \subset A$; we have:

$(C) = $ $S = \{ \sum_1^n r_i c_i s_i \mid r_i, s_i \in A, c_i \in C, n \in \Bbb N \}, \tag{4}$

where of course $\Bbb N$ is the set of natural numbers. (4) asserts that $(C)$ is precisely the set of all finite sums of elements of $A$ of the form $rcs$, where $r,s \in A$ and $c \in C$; indeed, it is easy to see that this must be the case: the set described on the right-hand side of (4) is manifestly closed under ring addition, manifestly contains the zero element ($0c0$ where $c \in C$), manifestly contains additive inverses of all its elements (since $-rcs + rcs = 0$), and is manifestly closed under both left and right multiplication by elements of $A$; furthermore, every ideal containing the set $C$ must also contain every sum of the form $\sum_1^n r_i c_i s_i$, $r_i, s_i \in A$, $c_i \in C$; thus in fact the set $S$ is the same as the ideal $(C)$.

Now suppose that for some $C \subset A$ we have $(C) = A$; then since $A$ is unital there exist $n \in \Bbb N$, $r_i, s_i \in A$, and $c_i \in C$, with $1 \le i \le n$, with

$\sum_i^n r_i c_i s_i = 1_A, \tag{5}$

where $1_A$ is the unit of $A$. Now if in fact we also have

$CA[\lambda, A] = 0, \tag{6}$

then for every $c \in C$ and $s \in A$ we also have

$cs[\lambda, A] = 0; \tag{7}$

in particular for those $c_i$, $s_i$ occurring in (5) we find

$c_i s_i [\lambda, A] = 0, 1 \le i \le n; \tag{8}$

thus, for any $r_i$

$r_i c_i s_i [\lambda, A] = 0, \tag{9}$

yielding

$\sum_1^n r_i c_i s_i [\lambda, A] = 0; \tag{10}$

in particular, choosing the $r_i$ also as in (5),

$[\lambda, A] = 1_A [\lambda, A] = (\sum_0^n r_i c_i s_i) [\lambda,A] = 0, \tag{11}$

by (10); and since $[\lambda, A] = 0$, according to (2) we have

$\lambda x - x \lambda = [\lambda, x ] = 0, \tag{12}$

that is,

$\lambda x = x \lambda \tag{13}$

for all $x \in A$; that is, $\lambda \in Z(A)$.

Nota Bene: I suppose it is worth pointing out that the above argument actually proves something more general than the title assertion, since $C$ may be any subset of $A$ with $(C) = A$; taking $C = [A, A]$ returns to the specific question at hand.

The essential proposition submits to even further generalization. If we assume that $CAD = 0$ with $(C) = A$, then an argument which is basically the same as the above yields $D = 0$. Again, taking $D = [\lambda, A]$ brings us back to this specific problem. End of Note.