Page 244 - Leech, Jonathan E. 2020. Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond. Koper: University of Primorska Press
P. 244
Jonathan E. Leech │ Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond
Theorem 6.4.15 In any ring R, K = KQ(R) is a nil ideal and Q(R/K) = Q(R)/K. Thus R is
idempotent-closed if and only if R/K is abelian. (Note that RKR need not be {0} here.)
Proof. Indeed, for any ideal I in R, KI is a nil ideal of R. Thus for K = KQ(R), each idempotent in
R/K has the form e + K for some idempotent e in R, so that Q(R/K) = Q(R)/K. Suppose R/K is
abelian. Then Q(R)/K = Q(R/K) is also, in which case Q(R) is idempotent-closed by Theorem
6.4.10, and so is R. Conversely, if R is idempotent-closed, then Q(R)/K is abelian, and since
Q(R) contains all the idempotents of R, Lemma 6.4.14 above assures that all idempotents in R/K
commute, making it abelian. £
What can one say about R/Q(R) in general? In particular, is E(R/Q(R)) = {0}? This would
be an extreme case of idempotent-closure. The answer is affirmative when R is idempotent-
closed. We begin with a special case:
Lemma 6.4.16 If a ring R is abelian, then E(R/Q(R)) = {0}.
Proof. Letting Q = Q(R), suppose that x + Q is idempotent in R/Q. Thus x2 − x ∈Q. It follows
that e in E(R) exists such that x2 − x = e(x2 − x) = (ex)2 − ex. Moreover, x decomposes as ex + y
where y = x − ex so that ey = 0. Then
x2 − x = (ex + y)2 − (ex + y) = (ex)2 + y2 − (ex + y) = [(ex)2 − ex] + [y2 − y].
But since x2 − x = (ex)2 – ex, y2 − y = 0 in R, that is, y is in E(R), so that x ∈Q and x + Q is the
zero element in R/Q. £
Theorem 6.4.17 If R is idempotent-closed, then E(R/Q(R)) = {0} = Q(R/Q(R)).
Proof. Again, Q = Q(R) is an ideal and K = KQ is a nil ideal of R. From R/Q ≅ (R/K)/(Q/K) and
the preceding two results, E(R/Q(R)) = {0}, and hence Q(R/Q(R)) = {0} follows. £
Thus squeezing Q(R) to a point eliminates any nonzero idempotents in R/Q(R) provided
R is idempotent-closed. In general, this need not be so. In any case, every idempotent-closed
ring R is a ring extension of an idempotent-dominated subring Q by a ring T = R/Q for which
E(T) = {0}. From the standpoint of the idempotents, Q and T are polar opposites: the
idempotents are maximally engaged in Q, while they are minimally engaged in T. In turn Q is an
extension of a nilpotent ring K for which E(K) = {0} by an idempotent-covered abelian ring A.
Indeed, in many cases Q is a “subdirect sum” of K and an internal copy of A. We consider other
decompositions in the following section.
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Theorem 6.4.15 In any ring R, K = KQ(R) is a nil ideal and Q(R/K) = Q(R)/K. Thus R is
idempotent-closed if and only if R/K is abelian. (Note that RKR need not be {0} here.)
Proof. Indeed, for any ideal I in R, KI is a nil ideal of R. Thus for K = KQ(R), each idempotent in
R/K has the form e + K for some idempotent e in R, so that Q(R/K) = Q(R)/K. Suppose R/K is
abelian. Then Q(R)/K = Q(R/K) is also, in which case Q(R) is idempotent-closed by Theorem
6.4.10, and so is R. Conversely, if R is idempotent-closed, then Q(R)/K is abelian, and since
Q(R) contains all the idempotents of R, Lemma 6.4.14 above assures that all idempotents in R/K
commute, making it abelian. £
What can one say about R/Q(R) in general? In particular, is E(R/Q(R)) = {0}? This would
be an extreme case of idempotent-closure. The answer is affirmative when R is idempotent-
closed. We begin with a special case:
Lemma 6.4.16 If a ring R is abelian, then E(R/Q(R)) = {0}.
Proof. Letting Q = Q(R), suppose that x + Q is idempotent in R/Q. Thus x2 − x ∈Q. It follows
that e in E(R) exists such that x2 − x = e(x2 − x) = (ex)2 − ex. Moreover, x decomposes as ex + y
where y = x − ex so that ey = 0. Then
x2 − x = (ex + y)2 − (ex + y) = (ex)2 + y2 − (ex + y) = [(ex)2 − ex] + [y2 − y].
But since x2 − x = (ex)2 – ex, y2 − y = 0 in R, that is, y is in E(R), so that x ∈Q and x + Q is the
zero element in R/Q. £
Theorem 6.4.17 If R is idempotent-closed, then E(R/Q(R)) = {0} = Q(R/Q(R)).
Proof. Again, Q = Q(R) is an ideal and K = KQ is a nil ideal of R. From R/Q ≅ (R/K)/(Q/K) and
the preceding two results, E(R/Q(R)) = {0}, and hence Q(R/Q(R)) = {0} follows. £
Thus squeezing Q(R) to a point eliminates any nonzero idempotents in R/Q(R) provided
R is idempotent-closed. In general, this need not be so. In any case, every idempotent-closed
ring R is a ring extension of an idempotent-dominated subring Q by a ring T = R/Q for which
E(T) = {0}. From the standpoint of the idempotents, Q and T are polar opposites: the
idempotents are maximally engaged in Q, while they are minimally engaged in T. In turn Q is an
extension of a nilpotent ring K for which E(K) = {0} by an idempotent-covered abelian ring A.
Indeed, in many cases Q is a “subdirect sum” of K and an internal copy of A. We consider other
decompositions in the following section.
242
