A ring of matrices over a Rickart ring need not be a Rickart ring. Under these conditions the rings of endomorphisms prove to be Baer rings see Regular ring in the sense of von Neumann. A ring of polynomials over a commutative Rickart ring is a Rickart ring. The analogous property for right annihilators is automatically fulfilled in this case.

## Differential Extensions of Weakly Principally Quasi-Baer Rings

This is a complete lattice if and only if the annihilator of any set is generated by projections. The term "Rickart ring" was introduced in honour of C. Rickart, who studied the corresponding property in rings of operators see [1]. Log in. Namespaces Page Discussion. Historical note. Von Neumann showed, conversely, that if. Kaplansky Examples: any division ring with involution ; any integral domain in particular, any field with the identity involution.

A ring with these properties is called a Baer ring. See [53, Excr. The supremum of a nonempty set of projections in A may be calculated coordinatewise. See also [g 18, Exer. A nontrivial example is the algebra of bounded sequences that are 'real at infinity', defined as follows.

## Differential Extensions of Weakly Principally Quasi-Baer Rings

The ring A of all 2 x 2 matrices over the field of three elements. VP be the set of all such cp conceivably,. Let 2' be a Hilbert space and let. The following conditions on d are equivalent: a. The following conditions are equivalcnt: a. Y is any set of positive elements of d that is increasingly directed i. CP as calculated in the partially ordered set of self-adjoint elements of Y 2 is also in d. The answer, in principle, is to modify the arguments or attempt to adjoin a unity element, and in practice it is Chapter 1.

First, let us review the unit case : Proposition 1. I Condition ii alone is of no help Exercise 1. The appropriate extension to the unitless case is as follows: Definition 1. The notion of annihilating lcji projection ALP is defined dually. Immediate from Proposition 1 and Definition 1. I Proposition 5. Same as , Prop.

Explicitly, if x is in the center oj A, then RP x is a centralprojeclion. Corollary 2. Let A be a a-ring. We define a unitification A , of A, provided there exists an auxiliary ring K , called the ring of Lsculur. The proofs of the first two are straightforward. Lemma 1. Notation as in Definition 3. Lemma 2.

Notation ah in Definition 3. Tlze projections of A , arc flze projections e, 1 -e, where e is a projection in A. Lemma 3. Tlzen e is an ARP of a in A if and on1. I It is in the following key lemma that the torsion-free hypothesis condition 3 of Definition 3 is used: Lemma 4. I Theorem 1. Assirme A O. Thus A, is the set of all 'ultimately constant' sequences.

Then i I is infinite; ii if A has a unitification A, in the sense of Definition 3, then the family e, ,,, has supremum 1 in A,. Central cover works best when the lattice of central projections is complete, but a few remarks can be made in a more general setting: Proposition 1. Assuming lz is a central projection such that C e, 5 6. Clearly h x and h f have the same left-annihilator, therefore LP 1z. We give here an alternate proof that exhibits the desired formulas.

I The condition on A in Corollary 2 is an import recurring theme: Definition 3. F is a Hilbert space, then 9 X is a factor. Fix a unit vector x in ,it. By hypothesis,. Atreduces every operator in Y. F o r example, let d be 4 6. Central Cover 37 a commutative von Neumann algebra on a Hilbert space.

X is I by Example 4 above. Tlze following conditions on f are equivalent: a J' is a central projection in eAe; b. If A is a juctor, then so is eAe. Obviously b implies c , and c implies a. Citing Proposition 3, we have -. It follows that if O and 1 are the only central projections in A, then O and e are the only central projections in PAP. Proof: i is obvious. Corollary 3. I The final result of the section is for application later on cf.

Disjointify the C e, , that is, let k,, Proposition 2 is a special case. Central Cover 39 4A. Thc central projections of hA are the central projections k in A such that k 5 h; in other words, a projection in hA is central in hA if and only if it is central in A. Exercise 8 , then every direct summand has this property. Every projection e has at least I -e as a complement.

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If e is a minimal projection , Exer. Let e be a projection in A. In particular, the central idempotcnts of A and B may be identified. This is a reason for considering central covcr for clcments other than projections.

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## Rickart ring - Encyclopedia of Mathematics

The following answer is the main rcsult of the section: Theorem 1. For clarity, we separate the proof of sufficiency Proposition 1 and necessity Proposition 2. The first lemma is valid in any topological space: Lemma. Then the fumily 6 has U as supremunz in tlze cluss of ull clopen sets ordered by inclusion. On the one hand, P, c P for all I. I cc Proposition 1. Proof: Since T is Stonian, it is clear from the lemma that the lattice of clopen sets is complcte, in othcr words, the projection lattice of A is complete.

I We approach the converse through a pair of lemmas. I ] ; what remains to be vcrificd is the explicit formula for supremum indicated in the above lemma. I a, In essence, thc next lemma is a result about thc Stone representation space of a Boolean algebra: Lemma 2. I Proposition 2.

Let U be an open set in T; it is to be shown that U is open. As remarked following Proposition 1, the clopen sets form a complete lattice; let P be the supremum of the family P,. Assume to the contrary that the opcn set P - U is nonempty.

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Form example: Proposition 3.